Mechanical BE (Engineering mechnics) by IMTS INSTITUTE

I ns t i t ut eofManage me nt & Te c hni c alSt udi e s ENGI NEERI NGMECHNI CS

500

BACHELORI NMECHANI CALENGI NEERI NG www. i mt s i n s t i t u t e . c o m

IMTS (ISO 9001-2008 Internationally Certified) ENGINEERING MECHANICS

ENGINEERING MECHANICS

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ENGINEERING MECHANICS CONTENTS: UNIT – 1

01-68

Introduction,Objectives,Force,Forces

acting

point

parallelogram

law

forces,Triangle law of forces,Lami’s Theorem,Resultant of a system of explainer forces acting at a point,Parallel forces,Moment of a force,Varignon’s Theorem,Couple,coplanar forces acting on a rigid body

UNIT – 2

69-109

Introduction,Objectives,Friction,Laws

friction,Coefficient

friction,Angle

friction,Common Catenary,Important Formulae,Suspension Bridge

UNIT – 3

110-173

Introduction,Objectives,Displacement,Velocity,Parallelogram law of velocities,Motion in a straight line under constant acceleration,Simple harmonic motion,Impulsive force,Direct

impact

two

smooth

spheres,Oblique

impact

two

smooth

spheres,Impact of a sphere on a smooth fixed plane

UNIT-4

174-219

Introduction,Objectives,Projectile,Equation of the trajectory

,Time of flight,Range on

an inclined plane

UNIT – 5 Introduction,Objectives,Moment

220-271 of

Inertia,Tjeore,

parallel

axes,Theorem

perpendicular axes ,Moment of Inertia of Standard bodies. ,Motion of a rigid

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body,Conservation of angular momentum,Compound pendulum ,Centre of Suspension and centre of oscillation ,Motion of a circular disc down an inclined plane

UNIT QUESTIONS :

272-329

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UNIT – 1 1.0 Introduction 1.1 Objectives 1.2 Force 1.3 Forces acting at a point parallelogram law of forces 1.4 Triangle law of forces 1.5 Lami’s Theorem 1.6 Resultant of a system of explainer forces acting at a point 1.7 Parallel forces 1.8 Moment of a force 1.9 Varignon’s Theorem 1.10 Couple 1.11 coplanar forces acting on a rigid body 1.12 Answers to self assessment problems

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1.0 INTRODUCTION Mechanies is that part of mathematics which deals with forces and their actions on particles and rigid bodies. Statics is that part of mechanics discussing about the effect of forces on bodies at rest. In this unit first we will discuss about the forces acting on a particle. Next we will see parallel forces. After that the resultant of a system of co-planar forces acting at a point is calculated. The moment of a force about a point is defined. The conditions of equilibrium of a system of coplaner forces are derived many theorems are given with proof. Illustrated examples are given in detail based on those theorems. 1.1 OBJECTIVES After going through this unit, you will be able to understand 

A force



Parallelogram law of forces



Triangular law of forces



Lami’s theorem



Polygon of forces



Parallel forces



Moment of a force



Couple



The resultant of a system of coplanar forces acting at a point



The conditions of equilibrium of a system of coplanar forces acting on a rigid body

STATICS Mechanics is the branch of science which deals with motion or change in position of physical objects. It is broadly divided into two parts. Statics is that part of mechanis which deals with the action of forces on bodies at rest. Dynamics is other part of mechanics dealing with the action of forces on bodies in motion. 1.2 Definition

Force

Force is that which either changes or tends to change thestate of rest or the uniform motion of a body in a straight line. It is a vector quantity. It is represented by a directed like segment.

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1.3 Forces acting on a particle 1.3.1 The Parallelogram law of forces Statement: If two forces acting at a point 0 be represented in magnitude and direction by two sides of a parallelogram, their resultant is represented in magnitude and direction by the diagonal of the parallelogram through that point. Proof:

Let

P and Q be two forces acting at a point o. Let  be the angle between them. Let OA and

OB represent P and Q in magnitude and direction. Complete the parallelogram OABC. Then the diagonal

OC represents the resultant force R in

magnitude and direction.

R  PQ





 R.R  P  Q . P  Q



R  P.P  P.Q  Q.P  Q.Q 2

 P  P.Q  P.Q  Q 2

 P  2P.Q  Q 2

 R 2  P2  2PQcos   Q2

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ENGINEERING MECHANICS  R  P2  Q2  2PQcos 

4  1

Next, Let

AOC  

In COD

AD Q

tan  

CD OD

cos  

tan  

CD OA  AD

AD  Qcos 

Qsin    2 P  Q COS 

sin  

 tan  

CD , Q

CD  Qsin 

Magnitude of the resultant force

R  P2  Q2  2PQcos  Its direction is given by

tan  

Qsin  P  Q cos 

Note 1: If

P and Q are perpendicular,   90

sin   sin 90  1

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cos   cos90  0

 R  P2  Q2  2PQcos  R  P2  Q2  2PQ.0

R  P 2  Q2

tan  

Qsin  P  Q cos 

tan  

Q.1 P  Q.0

tan  

Q P

 1

  2

Note: 2 If P and Q are equal in magnitude P= Q

R  P2  Q2  2PQcos  R  P2  P2  2PP cos 

R  2P2  2P2 cos 

R  2P2 1  cos  

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R  2P 2 .2cos 2 R  2P cos

 2

 1

tan  

Qsin  P  Q cos 

tan  

P sin  P  P cos 

tan  

P sin  P 1  cos  

tan  

  .cos 2 2  2 cos 2 2

2sin

 2 tan    cos 2 sin

tan   tan 

 2

1.3.2 Problems 1. Two forces of magnitudes 6p and 8p act at a point along two perpendicular directions. Find the resultant. Two forces 6p and 8p act at 0 along two

 r directions.

  90

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Let R be the resultant force. Then

R 2  P 2  Q2 R 2   6P   8P  2

R 2  100P2 R  10P Let R make an angle Then

tan  

 with 6p

Q 8P 4   P 6P 3

4   tan 1   3 2. Two equal forces P and P act at a point at an angle  . If the resultant is of magnitude p, find

.

Resultant force

R  2P.cos

 2

Given R = P

 P  2P.cos

 2

1   cos 2 2

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cos 60  cos 60 

 2

  1200 3. Two forces of magnitude 2,

3 2 acting at a point have a resultant

10 . Find the angle between the

forces. If R be the resultant force

R 2  P2  Q2  2PQ.cos 

Then

Given

P  2, Q 

3 , R  10 2

 3   3  10  2     2.2.   .cos   2  2 2

 10  4  18  12 2.cos   12  12 2.cos  

1  cos  2

cos135  cos 

  135 4. The magnitude of the resultant of two forces P and Q acting at a point is Q. when Q is doubled, again the magnitude of the resultant is Q. Find P and the angle between P and Q.

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Let



be the angle between P and Q.

Let R be the resultant. Then,

R 2  P2  Q2  2PQ.cos 

 1

Given R = Q

 Q2  P2  Q2  2PQ.cos 

0  P2  2PQ.cos 

0  P  P  2Q.cos    P  2Q.cos   0  P  2Q.cos 

cos   

P 2Q

  2

1  Q2  P2   2Q   2.P.2Q.cos  2

 P  Q2  P 2  4Q2  4PQ.     2Q  Q2  P2  4Q2  2p2 P2  3Q2

P  3Q

  3

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 2  cos   

3Q 2Q

cos   

3 2

  4

  150

5. Two forces of magnitudes P and Q inclined at an angle  have a resultant  2n  1 P  Q . If the 2





angle between them in 90   , then their resultant is  2n  1 P  Q . Show that 2

tan  

n 1 . n 1

Let two forces P and Q act at a point o inclined at an angle  . Let R be their resultant. Then

R 2  P2  Q2  2PQcos  Given R   2n  1 P  Q 2

 2n 1

P

 Q2   P2  Q2  2PQcos 

P

2  Q2   2n  1  1  2PQcos   

P

 Q2  4n 2  4n  1  1  2PQcos 

P

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P

 Q2  2n  n  1  PQcos 

  2

Next, If the angle between P and Q is then R   2n  1 P  Q 2

1 

 2n 1

P

 Q2   P2  Q2  2PQcos  90   

P

2  Q2   2n  1  1  2PQsin   

P

 Q2  4n 2  4n  1  1  2PQsin 

P

 Q2  4n  n  1  2PQsin 

P

 Q2  2n  n  1  PQsin 

 2 ;  3

P P



n 1 n 1  n 1 n 1



tan  

2 2

 Q2  2n  n  1

 Q  2n  n  1 2



  3

PQsin  PQ cos 

n 1 n 1

6. The magnitude of the maximum and minimum resultant of two forces acting at a point are R and S respectively. If  be the angle between the forces, show that the magnitude of the resultant is

R 2 cos 2

 2   S sin 2 2

Let P and Q be the two forces acting at a pint o. Max resultant R = P + Q Min resultant S = P – Q

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P

R S 2

Q

R S 2

Let  be the angle between P and Q.

Let

R 1 be the new resultant.

R12  P2  Q2  2PQcos 

Then

 R  S   R  S   R  S  R  S  R12       2    2   2   2  2  2

R  2 1

R12 

R 2  S2  2RS  R 2  S2  2RS  2  R 2  S2  Cos  4 1  2R 2  2S2  2R 2 Cos  2S2 cos   4

1 R12  .2  R 2 1  cos    S2 1  cos    4

R12 

1 2   R .2cos 2  S2 .2sin 2   2 2 2

R12  R 2 cos 2

 2 2  S sin 2 2

 R1  R 2 cos 2

 2 2  S sin 2 2

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7. The resultant of two forces P and Q acting at a point o is R. when Q is doubled, R is doubled. When Q is reversed, again R is doubled. Prove that

P:Q:R  2 : 3 : 2 .

Let two forces P and Q act act a point O, at an angle  . Let R be the resultant. Then

 1

R 2  P2  Q2  2PQcos 

When Q is doubled, R is doubled

Q  Q ,

 2R 

R  2R

 P2   2Q   2.P.  2Q  .Cos  2

4R 2  P2  4Q2  4PQCos

 (2)

Again, when Q is reversed, R is doubled.

 2R 

 P2   Q   2.P.  Q  .Cos  2

4R 2  P2  Q2  2PQCos

 (3)

Again, when Q is reversed, R is doubled.

(1)  (3);

5R 2  2P2  2Q2 2P2  2Q2  5R 2  0

 (4)

(2)  2  (3)

12R 2  3P2  6Q2 P2  2Q2  4R 2  0 (4)  (5);

 (5)

P2  R 2  0

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PR

(5) 

 (6)

R 2  2Q2  4R 2  0 2Q2  3R 2

Q

(6) & (7) 

3 R 2

 (7)

P:Q:R  R :

 1:

3 R:R 2 3 :1 2

 2: 3: 2



P:Q:R  2 : 3 : 2

1.4 TRIANGLE LAW OF FORCES Statement If three forces acting at a point be represented in magnitude and direction by the three sides of a triangle taken in order , then they are in equilibrium. Proof:

Let three forces

P,Q, R act at a point O as shown in the figure. Let them be representation in magnitude

and direction by the three sides of ABC taken in order. ___

___

AB  P

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___

BC  Q

CA  R

Now, by the definition of vector addition, ___

___

AB BC  AC ___

AB BC   CA ___

AB BC CA  0 ___

___

P  Q R  0

The system is in equilibrium 1.4.1 Converse of Triangle law of forces If three forces acting at a point be in equilibrium then they can be represented in magnitude and direction by the three sides of a triangle taken in order. 1.4.2 Polygon law of forces. If n forces acting at a point be represented in magnitude and direction by the n sides of a polygon taken in order, then they are in equilibrium. LAMI’S Theorem if three forces acting at a point be in equilibrium, then the magnitude of each force baries as the suie of the angle beween the other two.

Let three forces

P,Q, R acting at a point O be in equilibrium.

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Then, by the converse of the triangle law of forces, they can be represented in magnitude and direction by the three sides of ABC taken in order. ___

___

AB  P , BC  Q, CA  R Now, in

ABC by sine formula

a b c   sin A sin B sin C

AB BC CA   sin      sin      sin      P Q R   . sin  sin  sin  The magnitude of each force is directly proportional to the sine of the angle of the angle between the other two forces. 1.4.3 PROBLEMS 1. Three forces acting at a point are in equilis. The angle between the first and second is angle between the second and third is

90 and the

120 . Find the ratio of the forces.

Let P, Q, R be the three forces acting at a point along OA, OB & OC. Given 

A O B  90 

BO C  120 

CO A  360   90  120   150 FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

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The forces are in equilibrium. By Lami’s theorem

P Q R   sin120 sin150 sin 90 P Q R   3 1 1 2 2 P Q R   3 1 2 2. I is the incentre of they are in the ratio

ABC . If three forces acting at I along IA, IB and IC are in equilibrium, show that

cos

A B C : cos : cos . 2 2 2

I is the incentre of ABC . Let three forces

P, Q, R acting at I along IA, IB and IC be in equilibrium.

By Lami’s theorem,

P 

sin B I C Now,







sin C I A

 1



sin A I B

   B C  sin B I C  sin 180       2 2  

B C  sin    2 2

 

A  B  C  180 ,

A B C     90  2 2 2 

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A   sin 90   2  

 sin B I C  cos

A 2



B 2



C 2

sin C I A  cos sin A I B  cos

 (1) 

cos

A 2



Q cos

B 2



cos

C 2

3. O is the orthocenter of ABC if three forces P, Q, R acting at O along OA, OB & OC are in equilibrium, prove that P: Q: R=a: b: c

O is the orthocenter of

ABC .

Three forces P, Q, R acting at o along OA, OB and OC are in equilibrium. By Lami’s theorem





sin B I C



sin C I A



R 

 (1)

sin A I B



Now, I BC  90  C 

I C B  90  B 

 B I C  180  90  C  90  B FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

ENGINEERING MECHANICS 

19 

BIC  B C

A  B  C  180 



B I C  180  A 

 sin B I C  sin 180  A  

sin B I C  sin A 

sin C I A  sin B 

sin A I B  sin C

 (1) 

P Q R   sin A sin B sin C

In ABC , by sine formula, we have

a b c   sin A sin B sin C

 (3) P Q R   a b c

(2) & (3) 

4. Three forces P, Q, R acting at the circumcentre S of ABC along SA, SB and SC are in equilibrium. Show that

i   ii 

P Q R   sin 2A sin 2B sin 2C P Q R  2 2  2 2 2 2 2 2 a  b  c  a  b  c  a  b  c  a  b2  c2  2

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S is the circumcentre of ABC . Three forces P, Q, R acting at S along SA, SB & SC are in equilibrium. By Lami’s theorem

P sin BSC



Q sin CSA



R sin ASB

P Q R   sin 2A sin 2B sin 2C

 1

The angle subtended at the centre of a circle in twice the angle subtended by a chord on the circumference. Next,

P Q R   2sin A.cos A 2sin B.cos B 2sin C. cos C But

cos A 

b2  c2  a 2 2bc

P.2bc Q.2ca R.2ab   2 2 2 2 2 2 2sin A.  b  c  a  2sin B.  c  a  b  2sin C. cos  a 2  b 2  c2  P.abc Q.abc R.abc   2 2 2 2 2 2 a sin A.  b  c  a  bsin B.  c  a  b  csin C. cos  a 2  b2  c2  P Q R   2 2 2 2 2 2 a sin A.  b  c  a  bsin B.  c  a  b  csin C. cos  a 2  b 2  c2 

  2

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By sine formula,

a b c   sin A sin B sin C

 2 ;  3

  3

P Q R  2 2  2 2 2 2 2 2 a  b  c  a  b  c  a  b  c  a  b2  c2  2

5. The two ends A and D of a string ABCD are fixed. A mass of 12 kgs is suspended from B and another mass of m kgs is suspended frm C. If AB is inclined at

60 , BC is horizontal and CD is inclined at 30 ,

prove that m = 4 kgs.

The forces are acting as shown in the figure By Lami’s theorem At B,

T1 T2 12g   sin120 sin 90 sin150 T T 12g  1 2  3 1 1     2 2  

 T1  T2 

24g 12g and T2  3 3

12g 3

 1

Considering the equilibrium at C,

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T3 T2 mg   sin150 sin120 sin 90 T2 mg  1  3     2  2 

 mg 

mg 

T2 3

12g 3. 3

 m  4 kgs 6. A, B are two fixed points in the same horizontal level. AB = c. Two strings AC and BC of lengths b, a are attached at A and B and a partied of mass m is suspended from c. show that the tensions in the



strings are in the ratio b c  a  b 2

 :a  b

 a2  a2 

A, B two fixed points AC, BC – two strings AB = a, AC = b, BC = c Let

T1 , T2 be the tensions in the strings.

A mass of m kgs is suspended from C. In the equilibrium position, the forces are acting as shown in the figure.

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By Lami’s theorem.

T3 T1 mg   sin c sin  90  B  sin  90  A 

T1 T  2 cos B cos A T1 T  2 22 2 2 2 c  a  b  b  c  a  2

2ca

2bc

T1.a T .b  2 22 2 2 2 c a b b c a 2



T1 T2  2 2 2 b  c  a  b  2  b  c2  a 2  2

1.4.4 Self assessment problems II 1. Three forces P, P, 2R acting at a point are in equilibrium . If the first two forces are perpendicular show that

P  2R .

1.4.5 To find the components of a force along two perpendicular directions.

Let ox and oy be two perpendicular lines. Let

i and j be unit vectors along ox and oy.

Let

F be a force acting along OA.

Let

XOA  

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F along ox  F.i

 F i cos 

 Fcos 

F along oy  F.j

/// the component of

 F j cos  90   

 Fcos  Note: For any force

F the component along direction perpendicular to F is zero.

1.6 To find the resultant of a system of coplanar forces acting at a point.

Let O be a given point Let ox and oy be two Let

 r lines through o.

i and j be two unit vectors along ox and oy. Let

F1 , F2 ,.....Fn be a system of n coplanar forces acting at o. Let them make angles

1 , 2 ,.........n with ox. Let R be the resultant force. Let it make an angle with ox. Now, Resultant force

R  F1  F2  .....  Fn

Taking dot product with

 1

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



R.i  F1  F2  .....  Fn .i

R.i  F1.i  F2 .i  .....  Fn .i R i cos   F1 i cos 1  F2 i cos 2  ....................  Fn i cos n

R cos   F1 cos 1  F2 cos 1  ............  Fn cos n

  2

R cos   X

X  F1 cos 1  F2 cos 1  ............  Fn cos n

Where

Similarly in (1), taking dot product with





R.j  F1  F2  .....  Fn .j

R.j  F1.j  F2 .j  .....  Fn .j R j cos  90    F1 j cos  90  1   F2 j cos  90  2   ....................  Fn j cos  90  n 

R sin   F1 sin 1  F2 sin 1  ............  Fn sin n

  3

R sin   Y Where

Y  F1 sin 1  F2 sin 1  ............  Fn sin n

 2    3 2

R 2  cos2   sin 2   X2  Y2

R 2  X2  Y2 R  X2  Y2

 3 ;  2

  4

R sin  Y  R cos  X

tan  

Y X

  5

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From (4) and (5), we get the magnitude and direction of the resultant force. 1.6.1 To find the necessary and sufficient conditions for the equilibrium of a system of coplanar forces acting at a point Statement The necessary and sufficient condition for the equilibrium of a system of co-planar forces acting at a point is that the sum of the components of the forces along any two

 r directions vanish separately.

Let O be a given point Let ox and oy be two Let

 r lines through o.

i and j be two unit vectors along ox and oy. Let

F1 , F2 ,.....Fn be a system of n coplanar forces acting at o. Let them make angles

1 , 2 ,.........n with ox. Let R be the resultant force. Let it make an angle with ox. Let X and Y be the sum of the components of the forces along ox and oy. Then,

and

R 2  X2  Y2

tan  

Y X

 1   2

necessary part if the system is in equilibrium, then,

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R2  0

X2  Y2  0 X = 0 and Y = 0 The sum of the components of the forces along two perpendicular directions vanish separately. Sufficient part Let X =0 and Y = 0

X2  0 and Y2  0  X2  Y2  0  0 R2  0 R 0 The system is in equilibrium Necessary and sufficient condition for the equilibrium of a system of coplanar forces acting at a point to be in equilibrium is that the sum of the components of the forces along any two perpendicular directions vanish separately. 1.6.2 Problems 1. ABCD is a square. Forces of magnitudes P, 2P, 3P, 4P and

2 2P act at A along AB, BC, CD, DA and

AC. Show that they are in equilibrium.

The forces are acting as shown in the figure. Let X and Y be the sum of the compon of the forces along AB and AD. Then,

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X  P.cos 0  2Pcos90  3Pcos120  4Pcos 270  2 2Pcos 45

 1  X  P.1  2P.0  3P  1  4P.0  2 2P.    2 X  3P  3P

1

X0 Next,

Y  P.sin 0  2Psin 90  3Psin120  4Psin 270  2 2Psin 45

 1  Y  P.0  2P.1  3P.0  4P.  1  2 2P.    2 Y  4P  4P

 2

Y0

If R be the resultant force, then

R 2  X2  Y2

R 2  0 2  02

R2  0 R 0 The system is in equilibrium 2. Forces of magnitudes

2, 3,5, 3, 2 and 2 act at are angular point of a regular hexagon towards other

vertices. Find the resultant.

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ABCDEF is a regular hexagon. The forces are acting at A, as shown in the figure. Take AB as x axis and AE as y axis. Let X and Y be the sum of the components If the forces along Ax and Ay. Then,

X  2cos 0  3 cos30  5cos 60  3 cos90  2cos120  3 1  1 X  2.1  3.    5.  3.0  2    2  2  2 

3 5 X  2    0 1 2 2

 1

X5 Next,

Y  2sin 0  3 sin 30  5sin 60  3 sin 90  2sin120  3 1 3 Y  2.0  3.  5.  3.1  2   2 2  2 

Y5 3

  2

Let R be the resultant

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Then,

R 2  X2  Y2



R 2  52  5 3



  3

R 2  10 Let R make an angle Then

tan  

 with AB.

Y X

5 3 5

tan   3   60 The resultant force is of 10 units along AD. 3. Three forces of magnitudes 3, 4, 5 units act at a point parallel to the three sides of an equilateral triangle. Show that the resultant is perpendicular to the second force and find its magnitude.

ABC is an equilateral triangle. Take AB as x axis and AY

 r to AB as y axis.

The forces are acting as shown in the figure. Let X and Y be the sum of the components of the forces along Ax and Ay.

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X  4cos 0  5cos120  3cos 240

 1  1 X  4.1  5.     3     2  2

 1

X0 Next,

Y  4sin 0  5sin120  3sin 240  3  3 Y  4.0  5.    3     2   2 

  2

Y 3

Let R be the resultant force. Then

R 2  X2  Y2

R2  0 

 3

  3

R 3

Let R make an angle

 with Ax

Then

tan  

Y X

tan  

3 0

tan  



 2

The resultant force is perpendicular to AB. Its magnitude is

3 units.

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1.7.1 To find the resultant of two like parallel forces

Let

F1 and F2 be two like parallel forces acting on a rigid body at A and B respectively. We will

find their resultant. Introduce two forces

P1 and P2 at A and B respectively. These two forces are equal in

magnitude and opposite in direction. So they will balance each other. Let

R 1 be the resultant of the forces F1 and P1 acting at A.

Let

R 2 be the resultant of the forces F2 and P2 acting at B.

Let

now, we can resolve Similarly

and

meet

R 1 at o into two forces P1 along OD and F1 along OC.

R 2 acting t o can be resolved into two forces P2 along OE and F2 along OC.

The forces

P1 along OD and P2 along OE are equal in magnitude and opposite in direction.

They balance each other. We are left with

F1 and F2 acting at o along OC.

Their resultant force is

R ď&#x20AC;˝ F1 ď&#x20AC;Ť F2 acting along OC.

Next,

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OAC and AGF are similar

AC OC  GF AF

AC OC  P1 F1 F1 P1

OC  AC. Similarly

 1

OCB and BKL are similar

CB OC  KL BK

CB OC  P2 F2 OC  CB.

F2 P2

  2

1   2  AC. But

F1 F  CB. 2 P1 P2

P1  P2

F1.AC  F2 .CB The resultant of two like parallel of

F1 at A and F2 at B is of magnitude F1  F2 . It acts at a point c

on AB such that

F1.AC  F2 .CB

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Note: Let

F1 and F2 be two unlike parallel forces.

The resultant force is of magnitude

F1  F2 .

It acts at a point C outside AB such that

F1.AC  F2 .CB

1.7.2 Problems 1. Two like parallel forces P, Q act at A and B. when the forces are interchanged, show that the point of action of the resultant is displaced through a distance

 PQ  P  Q  . AB  

Let P and Q be two like parallel forces acting at A and B respectively

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 P  Q  acts at a point C on AB such that

P.AC  Q.CB

AC CB AC  CB AB    Q P QP PQ

AC 

Q PQ

 1

When P and Q are interchanged, let the new resultant act at a point Then,

AC1 

P .AB PQ

C1 on AB.

  2

 2  1 AC1  AC  CC1 

P Q .AB  .AB PQ PQ

PQ .AB PQ

2. Three like parallel forces P, Q, R act at the vertices A, B, C of a triangle ABC. Show that their resultant will pass through the control of

ABC , if P = Q = R.

Let P, Q and R be three like parallel forces acting at A, B and C respectively. First, the resultant of two like parallel forces Q at B and R at C is another like parallel force

 Q  R  . It acts at a point D on BC such that FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

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 1

Q.BD = R.DC If Q = R Then BD = DC D is the mid point of BC Next, the resultant of P at A and

 Q  R  at D is another like parallel force  P  Q  R 

It acts at a point G on AD such that

P.AG   Q  R  .GD If P = Q = R

P.AG  2P.GD AG  2GD

AG 2  GD 1 G divides the median Ad in the ratio 2 : 1 G is the centroid of ABC . The resultant passes through the centroid. 1.7.3 Self assessment problems IV 1. Two like parallel forces 24 gmwt and 9 gmwt are acting at A and B where AB = 22 cms. If their resultant acts at C, show that AC = 6 cms. 2. Three like parallel forces P, Q, R act at the vertices of incentre, show that

ABC . If their resultant passes through the

P Q R   a b c

1.8 Moment of a Force Definition The moment of a Force

F acting at A about the point O is G  OA  F

Magnitude of the moment = OA  F

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 OA  F 

d 1 OA

 F d = Force × Perpendicular distance If the moment is in the anticlockwise direction it is taken as positive. If the moment is in the clockwise direction it is taken as negative. If the force

F passes through the point O, then the moment of F about O is zero.

1.9 Varignon’s Theorem on moments If two coplanar forces have a result the sum of the moments of the two forces bout any point in their plane is equal to the moment of their resultant about that point. Proof: Case 1:

Let

P and Q be two coplanar forces acting at a point A. Then their resultant R passes through

A. Also

 1

R  PQ

Let O be any point in their plane. Taking cross product with



OA  R  OA  P  Q

OA ,



OA  R  OA  P  OA  Q

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The moment of the resul tan t about O  Moment of P about O  Moment of Qabout O

Let

P and Q be two like parallel forces acting at A and B respectively.

Then their resultant is

  2

R  PQ

It acts at a point C on AB, such that

P.AC  Q.CB P.AC.sin   Q.CB.sin     

sin       sin 

P.AC.sin .n  Q.CB.sin      .n

P  AC  Q  CB

  3

Let O be any point in their plane. Sum of the moments of the forces about O

 OA  P   OB  Q









 OC  CA  P  OC  CB  Q

 OC  P  CA  P  OC  Q  CB  Q

 OC  P  OC  Q  P  CA  Q  CB





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 OC  R  0

39  

P  AC  Q  CB

= Moment of the resultant about O. Hence the proof Theorem: If the sum of the moments of system of coplanar acting on a rigid body about three non-collinear points vanish separately, then the system is in equilibrium. Proof:

Let

F1 , F2 ,.........Fn be a system of n coplanar forces acting at the points A1 , A2 ,..........., A n on

a rigid body. Let A, B, C be three non-collinear points in their plane. Let the sum of the moments of the forces vanish separately about A, B and C.

AA1  F1  AA2  F2  ..................  AA n  Fn  0

 1

BA1  F1  BA2  F2  ..................  BA n  Fn  0

 2

CA1  F1  CA2  F2  ..................  CA n  Fn  0

  3

1   2 

 AA1  BA1   F1   AA2  BA2   F2  ...........   AAn  BAn   Fn

0

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



AB  F1  F2  ..........  Fn  0

 n  AB    Fi   0  1   n  BC    Fi   0  1 

/// ly  2    3 

 4   AB is parallel to  Fi n

is parallel to

(or)

F  0 i

 Fi (or) 1

 5

 5  BC

  4

F  0 i

But A, B, C are three non-collinear points. n

F

cannot be parallel to both

AB and BC .

1 n

F  0 i

F1  F2  ..........  Fn  0 The system is in equilibrium 1.10 COUPLE Definition The two equal and unlike parallel forces not passing through the same point constitute a couple.

Moment of the couple =

AB  F

Theorem:

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A force acting at a point on a rigid body is equivalent to another equal and parallel force acting at some other point of the rigid body together with a couple. Proof:

Let a force

F act a t a point A along AC.

Let B be any other point on the rigid body. Let DBE be parallel to AC. At B introduce two forces

F , F along BD and BE. These two equal and

opposite forces balance each other. The old position remains unchanged. Now consider the force

F at A along AC and the force F at B along BD. These two equal and

unlike parallel forces not passing through the same point constitute a couple. Moment of the couple =

AB ď&#x201A;´ F

Then we are left with a force The force

F at B along BE.

F at A long AC is equivalent to an equal and parallel force acting at B along BE together with a

couple of moment

AB ď&#x201A;´ F

1.10.1 Theorem: A force and a couple acting in the same plane on a rigid body is equivalent to an equal and parallel force acting at some other point of the rigid body. Proof:

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Let a force

F act a point B along BE. Let G be the moment of the given couple. We will take an

equivalent couple with forces

F, F and arm AB .

G  AB  F

Then

Now, consider the two forces

F, F acting at B along BD and BE respectively. These two equal

and opposite forces acting at B balance each other. Then we are left with an equal and parallel force

F at

A along AC. The force

F acting at B along BE together with a couple of moment G  AB  F is equivalent to

an equal and parallel force

F acting at A along AC.

1.10.2 Theorem: If three forces acting on a rigid body are represented in magnitude and direction by the three sides of a triangle taken in order then they are equivalent to a couple whose moment is equal to twice the area of the triangle. Proof:

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Let three forces

P, Q and R act on a rigid body. Let them be represented in magnitude and direction by

the sides BC, CA and AB of

 ABC

Let DAE be parallel to BC. At A, introduces two forces

P along AE, Q along CA and R along AB. These three forces

acting at a point A are represented in magnitude and direction by the three sides of

 ABC taken in

order. By the triangle law of forces they are in equilibrium. Then we are left with a force

P at A along AD

and a force

P at B along BC. These two equal and unlike parallel forces not passing through the same

point form a couple. Magnitude of the couple  P  AL

 BC  h

1   2   BC  h  2   2  area of ABC 1.10.3 Problems 1. ABCD is a square of side 4 cms. Forces of magnitudes 2P, 3P, 4P, 6P act along AB, AC, AD and BC. Find the sum of moments of the forces about it s centre O.

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ABCD is a square of side 4 cms. Its centre is O. The forces are acting as shown in the figure. Taking moments about O, Sum of the moments of the forces =2.P.2 + 6.P.2 + 3.P.0 – 4.P.2 =8P 2. ABCDEF is a regular hexagon of side 2 cms. Forces of magnitudes 1, 2, 3, 4, 5 and 6 act along AB, BC, CD, DE, EF and FA. Find the sum of the moments of the forces about its centre O.

ABCD is a regular hexagon of side 2 cms. Its centre is O.

sin 60 

OL OA

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d  2.

3 2

d  3 cms The forces are acting as shown in the figure. Sum of the moments of the forces about O = 1.d + 2.d + 3.d + 4.d + 5.d + 6.d =21.d =21

1.10.4 Theorem If three coplanar forces acting on a rigid body keep it in equilibrium, then either all the three forces are either parallel or concurrent. Proof:

Let

P,Q, R be three coplanar forces act on a rigid body and keep it in equilibrium.

Then

P  Q  R  0  1 Since

P, Q and R are three coplanar forces, either all the three forces are parallel or they are

not parallel. If all of them are parallel, then the theorem is true. If all the three forces are not parallel, let two of them meet at A ie., Two forces

P and Q meet at

A. Then by the parallelogram law of forces, their resultant passes through the diagonal AC of the parallelogram ABCD.

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1 

P  Q  R

R passes through A

we get

R passes through A. All the three forces are concurrent at A. Hence the proof. Note 1: The weight of a rigid body always acts vertically downwards at the centre of gravity. Note 2:

Normal reactions

(a) A rod resting on a plane.

AB – rod R - normal reaction (a) Rod leaning against a wall

(b) (c) Rod resting on a peg

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(d) Rod restitng of a curved surface

Note 3:

(i)

Cot formula

 m  n  cot   mcot   n cot 

(ii)  m  n  cot   n cot B  mcot C 1.10.5 problems

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1. A rod completely within a smooth hemispherical lowl of radius r. its centre of gravity dicides it in the ratio a : b. If

 be the inclination of the rod with the horizontal, show that tan  

ba tan  where 2 ba

is the angle subtended by the rod at the centre of the lowl.

The three forces acting on the rod are

w  at the centre of gravity G.

(i)

It weight

(ii)

Normal reaction R at A

(iii)

Normal reaction S at B

Since these three forces acting on the rigid body keep it in equilibrium, all of them meet at O. In OAB by cot formula

 m  n  cot   n cot B  mcot C  a  b  cot 90

   b cot  90     a cot  90   

 a  b  tan    b  a  tan   tan  

ba tan  ba

2. A heavy uniform rod of length 2a rests partly within and partly without a smooth hemispherical lowl of radius r. If  be the inclination of the rod with the horizontal, show that 2r cos 2  a cos  .

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The three forces acting on the rod AB are

w  at G.

(i)

It weight

(ii)

Normal reaction R at A along AO.

(iii)

Normal reaction S at C

 r to AB

These three forces keep the body in equilibrium. They will concur at D.

AG  GB  a AO  OD  r

EAG   ACO  OAC   In ADE

cos EAD 

AE AD

cos 2 

AE 2r

 AE  2r cos 2 Next, in

 1

AGE

cos EAG 

AE AG

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cos  

AE a

  2

 AE  a cos 

1   2  a cos   2r cos 2 3. A uniform rod rests in equilibrium on two smooth inclined planes of inclinations  and equilibrium position, show that the rod is inclined at an angle

tan  

sin      2sin .sin 

 . In the

with the horizontal.

Let AB = 2a length of the rod. Let

 be the inclination of the rod with the horizontal

The three forces acting on the rod AB are

w  at G.

(i)

It weight

(ii)

Normal reaction R at A.

(iii)

Normal reaction S at B.

These three forces acting on the rod keep it in equilibrium. They will concur at a point O. In OAB , by cot formula,

 m  n  cot   mcot   n cot   a  a  cot 90

   a cot   a cot 

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 cos  cos   2a tan   a     sin  sin    sin .cos   sin .cos   2 tan     2sin .sin    tan  

sin      2sin .sin 

4. A uniform rod of length 2a is in equilibrium with one end against a smooth vertical wall and resting on a smooth peg at a distance b from the wall. Show that the inclination of the rod with the vertical is 1

 b 3 sin   . a 1

AB = 2a, length of the rod G – Centre of gracity AG = GB = a C – peg CD = b In the equilibrium position, let the rod be inclined at an angle In

 with the vertical wall.

ACD

sin  

b AC

 AC 

b sin 

1

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The three forces acting on the rod are

w  at G.

(i)

It weight

(ii)

Normal reaction R at C perpendicular to the rod

(iii)

Normal reaction S at A perpendicular to the wall,

These three forces concur at O. In OAG by cot formula,

 m  n  cot   n cot B  mcot C  AC  CG  cot 90

 CG cot 90    ACcot 

 0  AC  CG  tan   AC tan   ACcot  AC  tan   cot   AC tan  b  sin  cos   sin      a. sin   cos  sin   cos  b  sin 2   cos 2   sin     a. sin   cos .sin   cos 

1  a.sin  sin 2 

b   sin 3  a 1

 b 3  sin     a 1

 b 3   sin   a 1

5. A uniform rod of length a rests with one end against a smooth vertical wall and a string of length b tied at the other end of the rod, the string being attached at a point vertically above the rod. If the rod is inclined at an angle

 with the vertical, show that cos 2  

b2  a 2 3a 2

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AB = a, length of the rod. G – centre of gravity. AG=GB=

a 2

BC =l, length of the string Let the rod make an angle

 with the wall.

Let the string make an angle

 with the wall.

The three forces acting on the rod are

w  at G.

(i)

Its weight

(ii)

Normal reaction R at A.

(iii)

Tension T is the string.

These three forces meet at O. In OAB , by Cot formula,

 m  n  cot   mcot   n cot  a a a a    cot 180     cot 90  cot  2 2 2 2 a a cot   0  cot  2 1  cot   cot  2

 1

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Next, in OGB , by sine formula,

a b 2  2 sin  sin 180   

a sin   bsin 

 sin   Now

a sin  b

 2

1  cot   2cot 

b2  a 2 sin 2  cos   2. a sin  sin 

b2  a 2 sin 2   2a cos  

b2  a 2 sin 2   4a 2 cos2 

b2  a 2 1  cos2   4a 2 cos2 

b2  a 2  a 2 cos2   4a 2 cos2 



cos 2  

b2  a 2 3a 2

6. A solid hemisphere rests with its curved surface against a smooth vertical wall and a string tied to a point on the rim, the other end of the string being attached to a fixed point on the wall. In equilibrium, if the string and the plane surface of the hemisphere make angles

 and O with the vertical, prove that

3 tan    tan  8

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Let r be the radius of the solid hemisphere. In the position of equilibrium, the forces are acting as shown in the figure. In OFG ,

cos  

OF OG

cos  

OF 3 r 8

3 OF  r cos  8

 1

In OBF , by Sine formula,

OB OF  sin  90    sin     

3 r cos  r  8 cos  sin      3 sin       cos .cos  8 3 sin .cos   cos .sin   cos .cos  8

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 cos .cos 

sin .cos  cos .sin  3   cos .cos  cos .cos  cos .cos  8 tan   tan  

3 8

3 tan    tan  8 7. A solid cone of height h and semivertical angle  is placed with its base against a smooth vertical wall and is supported by a string attached to its vertex and to a point on the wal. Show that the greatest possible length of the string is h 1 

16 tan 2  . g

Let h be the height r be the radius and  be the semi-vertical angle of the cone. Let T be the tension in the string Let G be the centre of gravity of the cone. Then,

BG 

h 3h , AG  4 4

In the equilibrium position, the forces are acting as shown in the figure. In ABD

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AD2  AB2  BD2  AD2  h 2  BD2  1 Next, ABD and



AGO are similar

AB BD  AG OG

h BD  3 r h 4

r    tan   h ; r  h tan  

4 BD  r 3 4 BD  h tan  3 4 1  AD  h   h tan   3  2

 16  AD2  h 2 1  tan 2   9    AD  h 1 

16 2 tan  9

Greatest length of the string =

h 1

16 2 tan  9

1.11 COPLANAR FORCES ACTING ON A RIGID BODY BOOK WORK: A system of coplanar forces acting on a rigid body can be reduced to a single resultant force acting on the rigid body together with a couple. Proof:

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Let

F1 , F2 ,...........Fn be a system of n coplanar forces acting on a rigid body at A1 , A2 ,.........An .

let O be any arbitrary pt on the rigid body. Let BOC be parallel to

F1 acting at A1 .

At O introduce two equal and opposite forces F1 ,

F1 along OB and OC respectively. Since these

two direction, they cancel each other. The old position remains unchanged. Now, the force

F1 at O along OB and the force F1 acting at A1 from a couple of moment

OA1  F1 . We are best with a force F1 acting at O along OC. The force

F1 at A1 is equivalent to a force F1 at O together with a couple of moment OA1  F1 .

Similarly the forces

F2 , F3 ,.............Fn

F2 , F3 ,.............Fn acting at A2 , A3 ,.............A n are equivalent to forces

acting

together

with

couples

moments

OA2  F2 ,OA3  F3 ,....................,OAn  Fn . Now, the n forces

F1 , F2 , F3 ,.............Fn acting at the point O have a single resultant force.

R  F1  F2  F3  .............  Fn

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couples

are

59 equivalent

single

couple

moment

G  OA2  F2 ,OA3  F3 ,....................,OAn  Fn The system of n coplanar force acting on the rigid body is equivalent to a single resultant force n

i 1

R   Fi together with a single resultant couple of moment G   OAi  Fi Note: The equation of the line of action of the resultant is

xY  yX  G

1.11.1 Theorem: A system of coplanar forces acting on a rigid body is in equilibrium if and only if the sum of the components of the forces along any two perpendicular directions vanish separately and the sum of the moments of the forces about any point in their plane is zero. Proof:

Let

F1  F2  F3  .............  Fn be a system of n coplanar forces acting on a rigid body at

A2 , A3 ,.............A n . Let O be any arbitrary pt in their plane. n

Then, the given system of forces can be reduced to a single resultant force

R   Fi acting at i 1

the point O, together with a single resultant couple of moment

G   OAi  Fi . i 1

Let ox

and oy be any two perpendicular lines through O. let X and Y be the sum of the

components of the forces along ox and oy.

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Then,

 1

R 2  X2  Y2 n

G   OAi  Fi

  2

i 1

Necessary Part: Let the forces be in equilibrium. Then, the rigid body cannot move and cannot rotate. ~

R  0 and G  0

R 2  0 and

 OA  F  0 i 1

X2  Y2  0 n

X = 0 and Y = 0 and

 OA  F  0 i 1

Sufficient Part Let X = 0, Y = 0 and

G0

X2  Y2  0  0 and G  0

R 2  0 and G  0 R  0 and G  0 R  0  The rigid body cannot move.

G  0  The rigid body cannot rotate. The system is in equilibrium. 1.11.2 Problems 1. ABCD is a square of side a. Force of magnitudes P, 2P, 3P and 4P act along the sides AB, BC, CD and DA. Find the resultant and the equation of its like of action.

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ABCD is a square of side a cms. The forces are acting as shown in the figure. Take AB as x-axis and AD as y-axis Let X and Y be the sum of the components of the forces along x and y axis. Then,

X  P.cos 0  2P.cos90  3P.cos180  4P.cos 270

X  P.1  2P.0  3P.  1  4P.0  1

X  2P ly

///

Y  P.sin 0  2P.sin 90  3P.sin180  4P.sin 270

Y  P.0  2P.1  3P.0  4P.  1   2

Y  2P

Let R be the resultant force. Then

R 2  x 2  y2 R 2   2P    2P  2

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R  2 2P

  3

Let R make an angle

 with x axis.

Y X

Then,

tan  

2P 1 2P

  135

  4

The taking moments about A,

G  P.0  2P.a  3P.a  4P.a

G  5pa The equation of the line of action of the resultant in

xY  yX  G

x  2P   y  2P   5Pa 2x  2y  5a  0 2. ABCD is a square of side 4 cms. Forces of magnitudes 4, 3, 2, 5 and

2 2 act along BA, CD, DC, AD

and DB. Prove that they reduce to a couple and find its moment.

ABCD is a square of side 4 cms. The forces are acting as shown in the figure. Take AB as x axis and AD as y axis.

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Let X and Y be the sum of components of the forces along x and y axes. Then,

X  4.cos180  3.cos 270  2.cos 0  5.cos90  2 2.cos315

1 X  4  1  3.0  2.1  5.0  2 2   2 X  4  4

 1

X0

Y  4.sin180  3.sin 270  2.sin 0  5.sin 90  2 2.sin 315

 1  Y  4.0  3  1  2.0  5.1  2 2    2  Y  5  5 Y0 let R be the resultant force. Then,

R 2  X2  Y2

R2  0  0 R 0 Taking moments about A,

 4  G  4.0  3.4  2.4  5.0  2 2.    2 G  28 The given system reduces to a couple of moment -28 units. 3. If four forces acting along the four sides of a square are in equilibrium, show that their magnitudes are equal.

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Let ABCD be a square of side a. Let

F1 , F2 , F3 , F4 act along AB, BC, CD & DC.

Take AB as x-axis and AD as y-axis. Let X and Y be the sum of the Components of the forces along x and y axes. Since the force are in equilibrium, X= 0, Y = 0 and G = 0

X  F1.cos 0  F2 .cos90  F3.cos180  F4 .cos 270  0

 F1.1  F2 .0  F3 .  1  F4 .0  0 ~

 1

F1  F3

Y  F1.sin 0  F2 .sin 90  F3.sin180  F4 .sin 270  0

 F1.0  F2 .1  F3 .0  F4 .  1  0 ~

F2  F4

  2 G  F1.0  F2 .a  F3.a  F4 .0  0 ~

F2 .a  F3 .a

F2  F3

F2  F3

  3

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F1  F2  F3  F4 The magnitudes of the forces are equal. 4. ABCDEF is a regular hexagon. Forces of magnitudes 3, 4, 5, 6, 7 & 8 act along the sides of the hexagon in order. Find the resultant force and the equation of its line of action.

sin 60 

d

d a

a 3 2

5. ABCDEF be a regular hexagon of side a. forces of magnitudes 3, 4, 5, 6, 7 and 8 are acting along the sides AB, BC, CD, DE, EF & FA resp. Take the centre O as origin. Take OC as x axis and OY

 r to axis.

Let X and Y be the sum of the components of the forces along ox & oy.

X  3.cos 0  4.cos 60  5.cos120  6cos180  7.cos 240  8cos300

1  1 1  1 X  3.1  4.  5     6  1  7     8. 2  2 2  2 X  3

Y  3.sin 0  4.sin 60  5.sin120  6sin180  7.sin 240  8sin 300 Y  3.0  4.

Y  3

 3 3 3  3  5.  6.0  7     8    2 2  2   2 

3 2

  2

Let R be the resultant force. Then,

R 2  X2  Y2  3 3 R   3      2  2

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  3

R 6

Let R make an angle

 with ox.

Then,

tan  

Y X

 3 3   2    3

  240 Taking moments about O

G  3.d  4.d  5.d  6.d  7.d  8.d G  33d

G  33.

3 a 2

  5

The equation of the line of action of the resultant is

xY  yx  G  3 3 33 3 x   a   y  3  2 2  

2 3  y  11 3  0 5. A rod AB of length

a  b

and whose centre of gracity is at a distance a from A and weight W is

placed on two kinife edges C, d at a distance c apart. If equal lengths of the rod project outside C and D, prove that the pressures at C D are

W W  c  a  b  and  c  b  a  2c 2c

AB  a  b

AG  a GB  b

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ENGINEERING MECHANICS Let AC  DB  z Then CG  a  x and GD  b  x Let R and S be the reactions at C & D. Now

CD  C CG  GD  C a xbz  c

a bc  x

x

a bc 2x

 1

Taking Let W be the weight of the rod. Then, W  R  S Taking moments about G,

R.CG  S.GD  0

S.GD  R.CG

S R S R W W     CG GD CG  GD CD C S R W   ax bx c

 1

S W  ax c S

W a  x  C

S

w   a  b  c  a   c   2  

S

W  2a  a  b  c   c  2

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S

W c  a  b 2c

Next.

 2

1 

R w  bx c R

W b  x C

R

w   a  b  c  b  c   2  

R

W  2b  a  b  c   c  2

R

W c  b  a  2c

 3

The pressures on the reds are

W W  c  a  b  and  c  b  a  2c 2c

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UNIT – 2 2.0 Introduction 2.1 Objectives 2.2 Friction 2.3 Laws of friction 2.4 Coefficient of friction 2.5 Angle of friction 2.6 Common Catenary 2.7 Important Formulae 2.8 Suspension Bridge

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2.0 Introduction In this unit you will learn about a new concept known as friction. Friction and frictional force are defined and the laws of friction are given in detail. The coefficient of friction, angle of friction and cone of friction are also discussed. Different types of problems regarding the equilibrium of bodies under the action of frictional forces are given in detail. In the second half of the unit, the equilibrium of strings hanging under gravity are dealt with. Many standard formulae are derived. Many examples are given to make you understand about the common catenary. Lastly, the equation of the suspension bridege is derived. 2.1 Objectives After going through this unit, you would understood about 

Friction



Frictional force



Laws of friction



Coefficient of friction



Angle of friction



Cone of friction



Common catenary



Suspension fridege

Introduction In the first unit, we have discussed the equilibrium of smooth bodies. In this unit, we will discuss the action of forces on rough bodies. When we drag a heavy body along the ground by means of a horizontal force, a resistance to the motion of the body is experienced. The resistance is felt along the horizontal direction at the point of contact and prevents the body from sliding along the ground. This phenomenon is known as friction. In this unit we will see in detail about friction and its effect on the equilibrium of bodies with rough surface. 2.2.1 Definition:

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If two bodies are in contact with are another, the property of the two bodies because of which a force is exacted between them at their point of contact to prevent one body from sliding on the other is called friction. The force exerted is known as force of friction. Note:

Frictional force is a passive and self-adjusting force.

2.3 Laws of friction: Law I: When two bodies are in contact with one another, the friction on one of them acts at the point of contact in a direction opposite to that in which the point of contact would tend to move. Law II: When there is equilibrium, the magnitude of friction is just sufficient to prevent the motion of one body relative to the other. Law III: The magnitude of limiting friction always bears a constant ratio to the normal reaction between the two bodies. Law IV: The limiting friction is dependent only on the nature of materials of the bodies. It is independent of the shape or size of the bodies of contact. Law V: When one body slides on the other, then also the frictional force acts in a direction opposite to that of motion the magnitude of friction is slightly less then the limiting friction. It is independent of the velocity of the body in motion. Note: The last law is known as the law of dynamical friction. 2.4 Definition:

coefficient of friction

When two bodies are in contact with one another, in the position of limiting equilibrium, the ratio of the limiting friction to the normal reaction is called the coefficient of friction. It is denoted by Âľ.

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If R be the normal reaction and F be the limiting friction, then

F  R

F  R. Note: For all the bodies, µ varies from 0 to 1. For perfectly elastic bodies, µ = 1 For inelastic bodies µ = 0 The value of the frictional force varies from 0 to  R. 2.5 Definition

Angle of friction

When two rough bodies in contact are in limiting equilibrium, the frictional force is limiting. The angle which the resultant of the limiting friction and the normal reaction makes with the normal reaction is called the angle of friction. It is denoted by

.

Let R be the normal reaction and

R be the limiting friction. Let S be the resultant of R and Let

R .

 be the angle between R and S.

Then,

Scos   R and Ssin   R



Ssin  R  Scos  R

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tan    Note: The maximum value of

 is 45

The come with vertex at A, axis AD and semi-vertical angle

 is called the cone by friction.

2.5.1 Problems 1. A uniform ladder rests in equilibrium with one end on the rough horizontal ground and the other end against a smooth vertical wall. If

 be the inclination of the ladder with the vertical, show that tan   2 .

AB  2a, length of the ladder G - inclination of the rod with the wall R – normal reaction at A S – normal reaction at B

R - limiting friction at A W - weight of the ladder. In the postion of limiting equilibrium, the forces are acting as shown in the figure. In

OAB , by cot formula,

 a  a  cot 180

   a cot 90  a.cot 

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2a cot   a.0  a.

1 tan 

2a a  tan   2  tan   tan   2 2. A uniform ladder is ni equilibrium with one end resting on a rough horizontal ground and the other end against a rough vertical wall. If the ladder is about to ship, show that the inclination of the ladder with the horizontal is given by

tan  

1   where ,  are the coefficients of friction on the ground and the 2

wall.

AB  2a, length of the ladder G – Centre of gracity In limiting equilibrium, the forces are acting as shown in the figure. Let

 be the inclination of the rod with the horizontal ground.

OAB , Cot formula

 a  a  cot 90

   a.cot  90     a cot 

2a tan   a.tan  

2 tan    

a tan 

1 

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  1 

2 tan   tan  

1   2

3. A ladder rests in limiting equilibrium its lower end on a rough horizontal ground its upper end against a rough vertical wall. Its centre of gracits divides it in the ratio a : b. if the ladder is inclined at an angle with the wall, show that

tan  

a  b  a  b

where  and



 are the coefficients of friction at the lower and

upper ends.

AB = a + b, length of the ladder. AG = a, GB = b W – weight of the ladder. R, S – Normal reactions at A and B

R, 'S - Eimiting frictions at A and B. F1 - Resultant force of R and R

F2 - Resultant force of S and 'S  - Inclination of the ladder with the wall. In limiting equilibrium, the forces are acting as shown in the figure.

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In OAB by Cot formula

 a  b  cot 180

   b cot 90     a cot 

  a  b  cot   b tan     a  b  cot   b    a  b  cot  

a tan 

a 

b  a 

 cot  

 tan  

a  b ab

a  b  a  b

4. A uniform ladder of length l rests on a rough horizontal ground, with its upper end projecting slightly over a smooth horizontal red at a height h above the ground. If ladder is about to ship, prove that



h l2  h 2 l2  h 2

AB = l, length of the uniform ladder B – smooth horizontal red G – centre of gracity R – normal reaction at A

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F1 - resultant of R and R S – Normal reaction at B

 - angle of inclination of the red with the horizontal. In limiting equilibrium, the forces are acting as shown in the figure. In OAB , by Cot formula

 l 2  l 2  cot 90    2l cot   2l cot  l tan  

 tan  

l 1 l  2 tan  2 tan 

1 1  2 tan  2

2 tan 2   1 1  2 tan  2 

tan  1  2 tan 2 

h 

l  h2 h2 1 2 2 l  h2 2

h. l2  h 2  2 l  h 2  2h 2 

h l2  h 2 l2  h 2

5. A ladder of length 2l is in contact with a vertical wall and a horizontal floor, the angle of friction being



at each point of contact. If the weight of the ladder acts at a point distant k below the midpoint. Prove that

lcot   lcot 2  k cos ec2 where  is the inclination of the ladder with the well.

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AB  2l, length of the ladder G – Centre of graving C – Mid point CG – K; AG = l – k, BG = l + k R – Normal reaction at A

R - Limiting friction at A S – Normal reaction at B

S - Limiting friction at B

F1 - Resultant of R and R

F2 - Resultant of S and S  - Inclination of the ladder with the wall. In limiting equilibrium the forces are as shown in the figure. In

OAB , by Cot formula

 m  n  cot   mcot   n cot   l  k  l  k  cot 180

    l  k  cot  90      l  k  cot 

2lcot    l  k  tan    l  k  cot  2lcot   l  tan   cot    k  tan   cot  

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 sin  cos    sin  cos   2l cot   l   k    cos  sin    cos  sin    sin 2   cos 2    sin 2   cos 2   2l cot   l    k  cos  sin    cos  sin    

l cot   l cot  

l  cos 2   sin 2   2sin .cos 

 k.

1 2sin .cos 

l.cos 2 k  sin 2 sin 2

lcot   l.cot 2  k.cosec2. 6. a uniform red AB is am limiting equilibrium resting horizontally on two inclined planes at right angles one of which makes an angle  with the horizontal. If the coefficient of friction is the same for both ends, show that



1  tan  1  tan 

AB = 2a, length of the rod G – centre of gracity R – Normal reaction at A

R - limiting friction at A S – normal reaction at B

S - limiting friction at B

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F2 - resultant force of S & S w – weight of the red. In limiting equilibrium, the forces are acting as shown in the figure. Now,

CAG  90  ;CBG  

GAO    ; GBO  90    . In

OAG by Cot formula,

 m  n  cot   n cot B  mcot e  a  a  cot 90

 a.cot       a.cot 90      

0  a.

1  a.tan      tan     

tan 2       1 tan       1 tan   tan  1 1  tan .tan  tan     1  tan .    tan   1  tan 

 1  tan    1  tan   

1  tan  1  tan 

7. A uniform rod rests in limiting equilibrium within a rough hollow hemisphere. If the rod subtends an angle 2 at the centre and horizontal is

 be the angle of friction, show that the inclination of the red with the

sin 2   tan 1   cos 2  cos 2 

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AB = 2a, length of the rod G – centre of gracity R – Normal reaction at A

R - limiting friction at A S – normal reaction at B

S - limiting friction at B  - angle of friction  -inclination of the rod with the horizontal

2 - angle subtended by the red at the centre. In limiting equilibrium, the forces are acting as shown in the figure.

OAB

ˆ  OAB ˆ     90       90       OAB ˆ  OBA ˆ     90       90       OBA In OAB by Cot formula,

 m  n  cot   mcot B  mcot e.  a  a  cot 90

   a.cot 90        a.cot 90      

2a tan   a.tan       a.tan      2 tan  

sin     

cos     



sin     

cos     

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2 tan  

tan  

sin      .cos       sin      .cos      cos      .cos     

sin            

2cos      .cos     

sin 2 cos           cos           

sin 2     tan 1   cos 2  cos 2  2.5.2 TYPE: 2 1. A solid hemisphere rests on a rough horizontal floor and against a smooth vertical wall. Show that, is

 3

the hemisphere can rest in any position and if it be less, the least angle that the base of the

hemisphere can make with the vertical is

show that the angle is cos

1

 8  cos 1   . If the wall be rough with coefficient of friction   3 

 8 1        3 1    

First we will consider the case in which both the wall and the horizontal floor are rough. a – radius of the hemisphere G – Centre of gravity

3 OG  a. 8 R – Normal reaction at A

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R - limiting friction at A S – normal reaction at B

 'S - limiting friction at B  -inclination of the plane surface with the vertical. In limiting equilibrium, the forces are acting as shown in the figure. Resolving the forces horizontally and vertically,

 1

S  R

  2

W  R  S i.e., W  R  .R

W  R 1   R 

1 

W 1  

S

  3

W 1  

  4

Taking moments about O

R.a  W.Oe  S.a  0

3 W. a cos   a  R  S 8

3 W W W. cos   .  . 8 1   1  

 1    3  cos   8 1   8  1     cos   . 3 1   

 8  1       cos 1  .   3 1    

 5

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  0   5 

 8  1  0     cos 1  .   3 1  .0  

8    cos 1    3  If

3   , there exist real values for  . 8

Maximum value of

  6



8   is cos 1    . 3 

3 , there exists no real value for  . 8

The hemisphere can test in any position. 2. A solid hemisphere of weight W rests on a rough inclined plane with its curved surface in contact with the plane. Its plane base is kept horizontal by a weight P attached to a point on its rim. Show that



P W  W  2P 

r – radius of the hemisphere. G – centre of gravity. R – normal reaction at A

R - limiting friction at A W – weight of the hemisphere.

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In limiting equilibrium, the forces are acting as shown in the figure. Taking moments about A,

 1

W.OD  P.DB  0 now, in OAD

sin  

OD  OD  r sin  r

DB  OB  OD  r  r sin 

1  W.r sin   P  r  r sin    0 Wsin   P  Psin   0

sin   W  P   P  sin  

P WP

 tan  



 W  P

 P2

P W  2WP  P 2  P 2 2

P W   W  2P 

3. A circular disc of radius a is kept at on a rough inclined plane of inclination by a string attached to the rim of the disc parallel to the inclined plane. If W be the weight of the disc, find the tension in the string and the coefficient of friction.

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 - Angle of inclination of the inclined plane. T – tension in the string R – normal reaction at A

R - limiting friction at A W – weight of the disc. In limiting equilibrium, the forces are acting as shown in the figure. Resolving the forces along and perpendicular to the inclined plane.

 1

T  R  Wsin 

  2

R  W cos 

1  T  .Wco  Wsin  T  Wsin   Wcos 

T  W  sin    cos  

  3

Taking moments about A,

T.AB  WA.e  0 T.2a  W.a sin   0

2T  Wsin 

T 

W sin  2

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Tension in the string

 3 

W sin  2

T

  4

W sin   W  sin    cos   2 1 sin   sin    cos  2 1  cos   sin  2 

1 tan  2

4. Show that the maximum angle of inclination of an inclined plane so that a particle placed on it may not slide downwards is equal to the angle of friction. A particle is placed on a rough inclined plane and its angle is increased steadily. At the point of sliding, show that the angle of inclination is equal to the angle of friction.

Let a particle of it W be placed on the rough inclined plane. The inclination of the plane is increased steadily. When the particle is about to slide downwards let  be the angle of inclination of the inclined plane. Let R be the normal reaction The limiting friction The three forces

 r to the inclined plane.

R acts along the inclined plane, upwards as shown in the figure.

W, R, R acting on the particle at P, keep it in equilibrium.

By lami’s theorem

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W R R   sin 90 sin  90    sin 180    R R  sin  90    sin 180   

1   cos  sin  tan    tan   tan 

  Angle of inclination = Angle of friction 5. A cone of semi – vertical angle  rests on a rough inclined plane. The inclination of the plane is gradually increased. Show that the cone will slide before it topples are if

  4 tan  .

Let r be the radius, h be the height and

 be the semi vertical angle of the cone. G centre of gracity W weight of the cone R normal reaction

R limiting friction When the cone is about to slide downwards, the forces are acting as shown in figure (i).

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ENGINEERING MECHANICS Let

 be the angle of inclination of the inclined plane.

Resolving the forces along and perpendicular to the inclined plane, we get

R  W cos 

 1

R  Wsin 

  2

 2  ; R  W sin  1 R W cos    tan  tan   tan 

  3



when the cone is about to topple over, the forces are acting as shown in fig (ii). Let In

 be the angle of inclination of the inclined plane.

GBD ,

tan  

BD GD

tan  

r h   4

r tan   4.   h

tan   4.tan 

r   tan    h 

  4

if the cone slides before toppling,



tan   tan  tan   tan    4 tan 

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ENGINEERING MECHANICS 6. How high can a particle rest inside a follow sphere of radius a if   1

Let O be the centre and a be the radius of the hollow sphere. Let P be the position of the particle, in limiting equilibrium. Let R be the normal reaction and

R be the limiting friction In the position of limiting equilibrium the forces are acting as shown in the figure. By Lami’s theorem,

W R R   sin 90 sin  90    sin 180    R R  sin  90    sin 180   

1   cos  sin  sin   cos tan  

1 1 (Given   ) 2 2

Now, let h be the height of the particle P above the lowest position A. Let

PM  OA .

Then,

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 h  AM

h  OA  OM h  a  a cos 

2   h  a 1   5  The particle can rest upto a height

2   a 1   from the lowest position A. 5 

2.6. Equilibrium of strings and chains the common catenary A heavy, uniform, inextensible string hangs freely under gravity being suspended from two fixed points (not in the same vertical line). To find the equation of the curve in which it hangs.

Let A and B be the two fixed points from which a uniform, heavy, inextensible string is suspended. Let C be the lowest point of the curve and P be any point of the string. Let

CP  s . Let w be

the weight per unit length of the string. Let the tangent at P make an angle  with the horizontal. Take a point o at a vertical depth c below the lowest point C as origin and the horizontal and the upward vertical through o as x and y axes. The three forces acting on the portion of the string are (i)

The tension T at P along the tangent at P

(ii)

The tension

(iii)

The weight

T0 at C along the horizontal line ws  of the arc CP at its centre of gravity G.

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ENGINEERING MECHANICS The three forces concur at the point D. Resolving the forces horizontally and vertically, We get

T cos   T0  wc

 1

Tsin   ws

  2

 2  , T sin   ws 1 T cos  wc tan  

s c

  3

s  c tan 

This is the intrinsic equation of the catenary. Now, the slope of the tangent at P is

dy  tan  dx

 3  Put

p

s  c.

dy dx

s  c.p Differentiating w.r.t.x, 2  ds 2  dy    1        dx    dx 

ds dp  c. . dx dx

1  p 2  c. 

dp . dx

dx dp  c 1  p2

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Integrating



dp 1  p2



1 sinh 1 p  .x  k c

dx c

  4

where k is constant

x  0, p  0

at C,

sinh 1 0  0  k

 4  p  sinh

k  0

sinh 1 p 

x c

dy x  sinh dx c x dy  sinh .dx c Integrating,

y

cosh 1 c



x dy   sinh .dx c

x c k 1

Where At C,

  5

k1 is a constant

x  0, y  c

c  e.cosh 0  k1 c  c.1  k1  k1  0

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 5 

y  cosh

x c

 6

this is the Cartesian equation of the common catenary. 2.7 Some Important Formula 1.

x y  c cosh . c

dy x 1  c.sinh . dx c c 

dy x  sinh . dx c

s  c tan   c.

y2  c2  s 2

dy x  c.sinh dx c

Proof:

y  c.cosh

x c

y2  c2 cosh 2

x c

x  y2  c2 1  sinh 2  c  y2  c2  c2 sinh 2

x c

x  y2  c2   esinh  c 

y2  c2  s 2 5.

y  c.sec  Proof:

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ENGINEERING MECHANICS y2  c2   e tan  

y2  c2  c2 tan 2  y2  c2 1  tan 2  

y2  c2 sec2  y  csec .

x  c.log  tan   sec  

Proof:

y  csec  and s  c tan 

c cosh

ec  e 2

e

x x  csec  and csinh  c tan  c c



 cosh

x x  sec  and sinh  tan  c c

 cosh

x x  sinh  sec   tan  c c

x c

ec  e  2



x c

 sec   tan 

Taking log,

x  log  sec   tan   c

x  clog  sec   tan   7.

 ys  x  c log    c 

Proof: We have

x  elog  sec   tan  

 1

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but we know that

s  c tan  and y  csec 

 tan  

s y and sec   c c

y s x  c.log     c c

1 

 ys   x  c log    c  8.

T  wy

Proof: We have

T cos   wc and

Tsin   ws

 T cos     Tsin     wc    ws  2

T2  cos2   sin 2    w 2  e2  s 2 

T2 .1  w 2 .y2  T  wy. 2.7.1 Problems 1. Uniform chain of length 2l is to be suspended from two fixed points A and B in the same horizontal line so that either terminal tension is n times at the lowest

2l n 1 2



log n  n 2  1

point. Show that the span AB must be



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In a catenary, We have

y2  c2  s 2

 1

At A, s= l Given, Tension at

A  n  Tension at C

TA  n.T0 wyA  nwc yA  nc substituting in (1), we get

  nc   c2  l2 2

n 2 c2  c2  l2

n 2 c2  c2  l2 c2  n 2  1  l2

 c2 

l2 n2 1

c

l n2 1

we also have,

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 ys  x  c log    c  At A, s = l, y = nc

 nc  l   x A  c log    c 

l  x A  c log  n   c   l. n 2  1  log  n     l n2 1   l

xA 

xA  Span

n 1 2



log n  n 2  1



AB  2x A

AB  2

l n 1 2



log n  n 2  1



2. If a uniform chain of length 2l is suspended from two fixed points A and B or the same level and the depth of the middle point below is

l , Span AB is n

1 n 1  l  n   log n n 1 

Length of the chain = 2l Sag in the middle =

l h

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ENGINEERING MECHANICS Lowest point is C. Let span AB = 2a In a common catenary, We know that

 1

y2  c2  s 2 At A, s = l and

y  c

l n

l    a    c2  l2 n 

c2 

l2 l  2c  c2  l2 2 n n

l 2 l2 l  2 n n

 n  1 l 2c.  l2 n n2 2

l  n2 1  c    2 n  Next we have

 ys  x  c log    c  At A,

x  a,s  l, y  c 

l n

l   c n l  a  c log   c    

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100

  1   l  n  1   a  c log 1   c      l 1  n   a  c log 1   nc   2  l  n  1 2n  l  n  1  a log 1  2 2 n  n.l  n  1 

a

l 2   n 2  1 log 1   2n  n  1

l  n 2 1   n 1  2  a   log  2 n   n  1 

l 1   n 1  a   n   log   2 n   n 1  Span

AB  2a

l 1   n 1  AB  2.  n   log   2 n   n 1  1   n 1    AB   n   log   n   n 1   3. A uniform chain of length 2l is suspended from two points in the same horizontal level at a distance 2a

2a 2 la  log 2 2 apart. If a be the sag in the middle, show that l  a la

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Length of the chain = 2l Span AB = 2a Sag in the middle = a In a common catenary, We know that

 1

y2  c2  s 2 At A, s= l and

1 

y  ca

  c  a   c2  l2 2

c2  a 2  2ca  c2  l2 2ca  l2  a 2

c 

l2  a 2 2a

  2

Also we know that

 ys  x  c log    c  At A,

x  a; y  c  a;s  l

 la  a  c log 1   c  

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101

ENGINEERING MECHANICS

a

102

  l  a  .2a  l2  a 2 log 1  2 2  2a   l  a  

  l  a  2a  2a 2  log 1   2 2 l a   l  a  l  a  

2a 2 2a    log 1  2 2 l a  l  a  2a 2  l  a  2a   log  2 2 l a  l  a  

2a 2 l  a   log  2 2 l a  l  a 

4. A box kit is flying at a height h with a length l of the string paid out and with the vertex of the catenary on the ground. Show that at the like, the inclination of the string to the ground is

h 2 tan 1   and that its l

 l2  h 2   l2  h 2  tensions there and at the ground are w   and w    2h   2h 

A – kite C – Man on the ground L – length of the string h – Height of the kite above ground

 - Inclination of the string at A with the horizontal In a common catenary, We know that

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 1

y2  c2  s 2 At A,

103

s  l, y  c  h

1 

c  h 

 c2  l2

c2  h 2  2ch  c2  l2 2ch  l2  h 2

c 

l2  h 2 2h

Tension at C is

 1

T0  wc

 l2  h 2  T0  w    2h  Tension at A is

  2

TA  wyA

TA  w  c  h   l2  h 2  TA  w   h  2h   l2  h 2  2h 2  TA  w   2h    l2  h 2  TA  w    2h 

  3

Next, the tangent at A makes an angle  with x axis. In a catenary,

s  c tan  At A, s  l. and

c

l2  h 2 2h

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104

2hl  tan  l  h2 2

l2 ;

h  2  2 tan l  2 2 2  h 1    1  tan 2 l

h    tan l 2

h   tan 1    l 2 h   2 tan 1   l 5. A uniform heavy chain of length 2l hangs over two small smooth pegs in the same horizontal line at a distance 2a apart. If h be the say in the middle, show that the length of either part of the chain that hangs vertically is

h  l  2hl

Let of the chain = 2l A, B – pegs h – sag in the middle. Let

l1 be the length of the string that hangs vertically.

In a common catenary, We know that

y2  c2  s 2

 1

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105

y  l1;s  l  l1

1 

l12  c2   l  l1 

l12  c2  l2  l12  2ll1

O  c2  l2  2ll1

0  c2  l2  2l  c  h  0  c2  l2  2lc  2lh 2lh   c  l 

 2hl  c  l  c  l  2hl Length of the portion of the string hanging vertically = AD

 ch

 l  2hl  h  h  l  2hl 6. Show that the length of an endless chain which will hang over a circular pulley of radius a so as to be in

  3 4   contact with two thirds of the circumference of the pulley is a   log 2  3 3  





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106

a – radius of the circular pulley. The portion of the chain in contact with the circular pulley is Length of the portion

AEB =

AEB .

2  Circumference 3 2   2a 3 

4a 3

 1

The portion ACB hangs in the form of a catenary. At A, the tangent makes an angle

60 with x axis.

In a catenary, we know that

x  clog  tan   sec   At A,

x  AF  a sin 60  a.

  2 3 2

  60

 2 

3  c log  tan 60  sec 60 2



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3  c log 2

c 



32

2log 2  3

107

   3



Next, we know that

  4

s  c tan 

At A,

l1 

s  l1;   60 ; c 



2log 2  3



2 log 2  3

 l1 



2 log 2  3



.tan 60



. 3

2log 2  3



4a 2.3a  3 2 log 2  3





  4

Total length of the chain = Arc





AEB  2l



  4 3   a   3 log 2  3   





7. A heavy chain of length 2l has one tied at A and the other end is attached a small heavy ring which can slide on a rough horizontal rod which passes through A. if the weight of the ring is n times the weight of the chain, show that it greatest possible distance from A is



2l log   1   2 



where



1 .   2n  1

Here  is the coeff of friction.

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108

2l – length of the chain A – fixed point B – ring The forces acting at B are (i)

Tension T along the tangent

(ii)

Normal reaction R

(iii)

Limiting friction

(iv)

The weight of the ring 2nlw 

R.

Resolving the forces horizontally and vertically,

R  2nlw  Tsin 

 1

R  T cos 

  2

   2nlw  Tsin    T cos 

  3

But in a common catenary, We have and

T cos   T0  wc

Tsin   wl

 3 

  2nlw  wl   wc

C  l  2n  1 C

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109

Next, we have

x  clog  tan   sec  

  4

At A, s  c tan 

l

l tan  

  tan 

sec   1  tan 2   1   2 xA 

l log   1   2    

Span

AB  2.x A



l AB  2 log   1   2 



2.8 Suspension Bridge. In a suspension Bridge, two chains are hung up so as to be parallel, their ends being fixed to supports. From different points of these chains hang supporting rods which carry the roadway of the bridge. These supporting rods are usually at equal horizontal distances from one another. The chain is so loaded that the weight on each element of it is proportional to the horizontal projection of that element. A chain is suspended from two fixed points so that the weight of each element is proportional to its horizontal projection of that element. To show that it will havg in the form of a parabola. Proof: Let the chain the hung from two fixed points A and B. let C be the lowest point. Let W be the weight per unit length of the horizontal projection of the chain. Take the lowest pt C as origin and the horizontal and upward vertical through C as x and y axes. Let

P  x, y  be any point of the chain. Let the tangent at P make an angle  with x axis. The

three forces acting on the portion CP are

(i)

Tension T at P along the tangent at P

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Tension

(iii)

Weight

110

T0 at C wx 

These three forces concur at 0. Resolving the forces horizontally and vertically,

T cos   T0 wc

 1

Tsin   wx

  2

 2 ; 1

tan  

x c

dy x  dx c Integrating,

y

 dy   c dx

x2 k 2c

  3

Where k is constant At C,

x  0, y  0

 3  We get

0  0k y



k0

x2 2c

x 2  2cy This is the equation of the curve. This is a parabola.

UNIT – 3 3.0

Introduction

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ENGINEERING MECHANICS 3.1

Objectives

3.2

Displacement

3.3

Velocity

3.4

Parallelogram law of velocities

3.5

Motion in a straight line under constant acceleration

3.6

Simple harmonic motion

3.7

Impulsive force

3.8

Direct impact of two smooth spheres

3.9

Oblique impact of two smooth spheres

3.10

Impact of a sphere on a smooth fixed plane

3.0 Introduction

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111

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112

In this unit you will learn about the motion of a particle under the action of different kinds of forces. Some basic concepts like displacement, velocity and acceleration are introduced. The formulae regarding the motion of a particle moving with uniform acceleration are derived. Many examples are given to illustrate the above formulae. Next a special type of motion known as simple harmonic motion is discussed in detail. After that direct impact of two smooth spheres, oblique impact and the impact of a sphere on a fixed plane are discussed. Suitable examples are also given. 3.1 Objectives After studying this unit, you will be able to understand 

Displacement



Velocity



Acceleration



Motion of a particle under uniform acceleration



Simple harmonic motion



Direct impact



Oblique impact



Impact of a sphere on a fixed plane

DYNAMICS Dynamics is that part of mechanics, which deals with forces and their actions on bodies in motion. In this unit, we will discuss the motion of a particle along a straight line and along a plane curve. 3.2 Definition : Displacement A particle is said to be displaced when it changes its position. It is a vector quantity.

A When a particle moves from A to B, its displacement is

AB .

3.3 Definition : Velocity The velocity of a particle is its rate of displacement.

A ie, the displacement in one second is called velocity.

B P(r)

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113

r is the position vector of a particle at time t, then its velocity v  ddtr Its units are metre / sec or

Km/hr. 3.3.1 Definition: Resultant velocity If a particle has simultaneously many velocities

v1 , v2 ,............. vn then its resultant velocity is

V  v1  v2  ............  vn 3.4 Parallelogram law of velocities If a particle has two velocities

v1 and v2 which are represented in magnitude and direction by the

two adjacent sides of a parallelogram , then their resultant velocity is represented in magnitude and direction by the diagonal of the parallelogram through that point. To find the magnitude and direction of the resultant velocity

Let a particle at O have two velocities direction by Let Let

v1 and v2 . Let them be represented in magnitude and

OA and OB . ˆ  . AOB V be the resultant velocity.

By the definition of vector addition ,

OA + AC = OC

V1 + V2 = V The resultant velocity

V = V1 + v2 __(1)

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114

The diagonal OC represents the resultant velocity

V both in magnitude and direction.

Let the diagonal OC make an angle  with OA. (1) 

v  v1 + v2

v = v. v =

(v1 + v2 ). ( v1 + v2 )

v1 . v1 + v1 . v2 + v2 . v1 + v2 . v2 2

= v1 + v1 .

v2 + v1 . v2 + v2

v12  2v1.v2  v2 2

v12  v2 2  2v1v2 cos 

v12  v2 2  2v1v2 cos  ...........(1)

Let the diagonal oc make an angle



with OA.

Let CD  OA

ˆ  . CAD

In  ACD, AD=AC

cos   v2 cos 

CD= AC sin In

 = v2 sin 

OCD,

tan  

CD OD

tan  

CD OA  AD

tan  

V2 sin  ___(2) V1  V2 cos 

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115

V1 and V2 are perpendicular,

  90 Sin Cos (1)

 1

 0

v =

v =

(2) 

v12  v2 2

tan  

tan

3.4.1

v12  v2 2  2v1v2 .0



v2 .1 v1  v 2 .0

v2 v1

problems

1. A train moves due east with a velocity of 80 km/h and a passenger throws a ball with a velocity of 60

toards north. Find the resultant velocity.

The train is moving eastwards with a velocity of 80 km/h. The ball is thrown northwards with a velocity of 60 km/hr. The ball has simultaneously two velocities 80 km/hr along OA and 60 km/hr along OB. Let

V be the resultant velocity along OC.

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116

802  602

Then, V=

10000

V= 100 km/h. Let OC make an angle Then, tan



v2 V1

Tan



60 80



with OA.

3 4

  tan 1 ( ) . 2. A particle has two velocities of magnitudes v and 2v. If the resultant velocity is perpendicular to v, find the angle between the velocities. A particle has two velocities v and 2v. Let  be the angle between the velocities. If the resultant velocity makes an angle

tan   Given



with v, then

2v sin  v  2v cos 

  90

 tan 90  

2v sin  v  2v cos 

v  2v cos   0

v(1  2cos  )  0 1  2cos   0 cos   

1 2

  1200  The angle between the velocities is 120  FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

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3.4.2 Definition: Acceleration The acceleration of a moving particle is its rate of change of velocity. If a particle moves along a straight line AB, and if

dr dt

v

Velocity at P is

Acceleration at  is

a

r is the position vector of the particle at any time t, then

B P(r)

dv d 2 r  dt dt 2

3.5 B.W A particle moves along a straight line under uniform acceleration a. To derive the formulae i) 2 v=u+at ii) s  ut  1 at and iii)

Proof:

b2  u 2  2as.

B s

P(v)

Let a particle start from A with an initial velocity u

and move along the straight

line AB.

Let ‘a’ be its uniform acceleration. At any time t, let the particle be at P Where AP=S. Let v be the velocity of the particle at P. By the definition of acceleration,

dv a, dt

dv  a dt 

 dv   a dt v  at  k

Initially at A, 

 1 where k is a constant

t=0, v=u u = a.0 + k or k = u

1  v=at+u v=u+at

  2

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v

Also we know that,

118

ds dt

ds  u  at dt ds   u  at  dt



 ds 

  u  at  dt

at 2 s  ut   k1 2

  3 where k1 is a constant

Initially at A , t=0, s=0  0=0+0+ k1 

k1 =0

1 s  ut  at 2 2

 3 

  4

Eliminating t between (2) and (4), we will get the next formula. (2) 

v  u  at t =

vu a

Substituting in (4),

 vu  1  vu  s  u  a   a  2  a 

1  v  u   s  u   v  u   2  a    v  u  2u  v  u  s   2  a   s

 v  u  v  u 

2a 2as  v  u 2 2

 v 2  u 2  2as

 5

3.5.1 Problems

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1. A train moves in a straight line with a uniform acceleration and describes equal distances s,s in two Successive intervals of time

t 1 and t2 . Show that its acceleration is

2s(t1  t2 ) t1t2 (t1  t2 )

Let u be the initial velocity and a be the uniform acceleration of the train Given, the train describes equal distances s,s in two successive intervals of time

We know that

1 s  ut  at 2 2 When

 (1)

t  t1 , distance=s

1 s  ut1  at12 2 Next, When

t1 and t2 .

  2

t  t1  t 2 , distance = 2s

1 

2 s  u  t1  t 2  

1 2 a  t1  t 2  2

  3

 2    t1  t 2    3   t1 1 1 2 s  t1  t 2   2st1  at12  t1  t 2   a  t1  t 2  t1 2 2 1 s  t1  t 2  2t1   at1  t1  t 2   t1   t1  t 2   2 1 s  t1  t 2    at1t 2  t1  t 2  2 Acceleration

a

2s  t1  t 2  t1 t 2  t1  t 2 

2. A particle moves with a uniform acceleration along a straight line. If velocities in three Successive intervals of time

t1 , t2 and t3 prove that

v1 , v2 and V3 be its average

v 1 v2 t1  t2  v2  v3 t2  t3

Let u be the initial velocity and a be its uniform acceleration.

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120

v1 , v2 and v3 be the velocities at times t1 , t1  t2 and t1  t2  t3 respectively.

We know that v= u+ at

t  t1 , v  V1

 V1  u  at1

 2

t  t1  t 2 , v  V2

 V2  u  a  t1  t 2 

  3

t  t1  t 2  t 3 , v  V3  V3  u  a  t1  t 2  t 3  Given the average velocities are

v1 

 4

v1, v2 and v3

v  v3 u  v1 V  V2 ; v2  1 ; V3  2 2 2 2

 u  v1   V1  V2     v1  v 2 2   2    v 2  v3  V1  V2   v 2  v3      2   2 



u  v1  V1  V2 V1  V2  v 2  v3



u  v2 v1  v3



u   u  a  t1  t 2  

 u  at1    u  a  t1  t 2  t 3 

v1  v2 a  t1  t 2   v 2  v 3 a  t 2  t 3 



v1  v 2 t1  t 2  v 2  v 3 t1  t 2

3. The two ends of a train moving with a Constant acceleration pass a certain point with velocities u and v respectively.

Show that the Velocity with which the middle of the train passes the same point is

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A(u)

121

B(v)

C(v)

Let 2l be the length of the train, A be its Uniform acceleration, A, B be the two ends of the train and C be the middle point of the train. Let A and B pass a certain point with Velocities u & v. In moving a distance of 2l, the velocity of the train increases from u to v. We know that Here

v2  u 2  2as  1 v

s=2l

 u 2  2a.2l

4al  v 2  u 2 al 

v2  u 2 4

  2

Let V be the velocity of the train when the middle point c crosses the same fixed point. When s=l, v=V

V 2  u 2  2al  v2  u 2  V2  u 2  2    4  2u 2  v 2  u 2 V2  2 2 u  v2 2 V  2 V

u 2  v2 2

4. A train goes from one station A to another station B with an acceleration a during the first part of the journey and a retardation

a1 during the second part. The train starts from rest at A and comes to rest at

B. If T is the total time taken and S is the total distance described, Show that

T 2  2s ( 1  1 1 ) a a Let the train move from A to C with uniform acceleration a and from C to B with a uniform retardation From A to C

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a1 .

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122

Initial velocity = 0 Final velocity = V acceleration = a time =

distance=

v  u  at



1 s  ut  at 2  2 2 2 v  u  2as 

V  0  at1

 1

1 s1  0.t1  at12 2 2 v  0  2as1

  2

0  v  a 1t 2

  4

1 s 2  v.t 2  a1t 22 2 2 0  v  2a1s3

 5

  3

From C to B Initial velocity = v Final velocity = 0 acceleration = - a time =

distance=

v  u  at



1 s  ut  at 2  2 2 2 v  u  2as 

 6

1 &  4   t1 

v a



t1  t 2 

Total time

t2 

v a1

v v  a a1

T  t1  t 2

1 1  T  V    a a 

 7

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Next,

 3 &  6   v2 v2 & s2  2a 2a  2 v v2  s1  s 2   2a 2a  s1 

 Total distance s  s1  s 2

v2  1 1  s    2  a a  2

1 1  v    2 2 7   ; T   a a  8 s v2  1  1    2  a a  T2 1 1    2   s  a a  2

1 1   T 2  2s    .  a a  5. A lift ascends with a constant acceleration a, then with a constant velocity V and finally stops with a Constant retardation a. If the total distance traveled is S and the total time taken is T, show that the time for which the lift is moving with constant velocity is

T2 

4s . a

A to B , with constant acceleration a Initial velocity = o Final velocity = v time = t1 distance = s1 acceleration = a

v  u  at  v  0  at1

 1

v2  u 2  2as  v 2  0  2as1   2  FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

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124

B to C with constant velocity v Distance = velocity  time

s 2  v  t 2   3 C to D with constant retardation a Initial velocity = v Final velocity = o Time = distance =

acceleration = a

v  u  at  0  v  at 3

  4

v2  u 2  2as  0  v 2  2as3   5 

1 &  4   t1  Total time,

v  at1 & v  at 3

v v & t3  . a a

T  t1  t 2  t 3 v v  t2  a a 2v T  t2   6 a T

 2  &  5  v2 v2 s1  & s3  2a 2a Total distance

s  s1  s2  s3

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125

v2 v2 s  vt 2  2a 2a v   v s  v   t2   2a   2a v  s  v   t2  a   T  t2   T  t2  s  a  t2    2  2  a  T  t 2   T  t 2  2t 2  s   2 2   a s   T  t 2  T  t 2  4 4s  T2  t 22 a 4s t 22  T2  a t 2  T2 

4s a

Simple Harmonic Motion Definition: If a particle moves in a Straight line such that its acceleration is always directed towards a fixed point on the line and varies as its distance from the fixed point, then the motion of the particle is said to be a simple harmonic motion

3.6.1 Discuss the Simple harmonic motion of a particle: Let a particle move along the line BOA in a Simple harmonic motion. Let the acceleration of the particle be always directed towards the fixed point o on the line. The magnitude of the acceleration is directly proportional to its distance from o. At any time t, let the particle be at I where OP= x

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126

d2x  x. dt 2

Now, acceleration

d2 x  n 2 x dt 2

 1

where n is a Constant. The acceleration is negative because the direction of the acceleration is opposite to the direction of x increasing.

1 

d2 x  n2x  0 2 dt

D

 n2  x  0

  2  Ï§F D 

d . dt

This is a second order differential equation in x and t

 The auxiliary equation is

m2  n 2  0 m 2  n 2 m  in.  C.F  eot  A cos nt  Bsin nt  P.I = 0.  The solution of (2) is

  3

x  Acos nt  Bsin nt Differentiate,

dx  An sin nt  Bn cos nt dt At the extreme point A ,

t  0,

  4

dx 0 dt

 0  An sin 0  Bn cos 0 0  0  B.n.1 B  0

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 3  x  Acos nt At A,

127

 5

t  0; x  OA  a  a  A.cos 0 a  A.1 A  a

 5  x  a cos nt

  6

dx  an sin nt dt

Also,

v  an sin nt

v 2  a 2 n 2 sin 2 nt v 2  a 2 n 2 1  cos 2 nt  v 2  n 2  a 2  a 2 cos 2 nt  v2  n 2  a 2  x 2 

 7

 We get, i)

x  a cos nt

ii)

v2  n 2  a 2  x 2 

iii)

d2 x  n 2 x. dt 2

Amplitude 3.6.2 Definition: In a Simple harmonic motion, the Maximum displacement of the particle from the fixed point is called its amplitude. To find the amplitude, Put velocity = 0

v0

v2  0

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n2 a2  x2   0

a2  x2  0 x2  a2 xa  amplitude  a Oscillation 3.6.3 Definition In a Simple harmonic motion, an Oscillation is a complete to and fro motion ie, The motion of the particle from one end A to the other end B and than from B to A is called one Oscillation.

Period of Oscillation 3.6.4 Definition In a Simple harmonic motion, the time taken for one complete oscillation is called the period of oscillation.

x  a cos nt

Displacement

x  a cos  nt  2   2  x  a cos n  t   n   i.e, At time t, the particle is at P. Again at time (t  2 ) , the particle is at P, moving along the same direction.

Interval between these two instants= 2

 Period of Oscillation T 

2 sec. n

Note: In a Simple harmonic motion, the maximum acceleration will be at A. Magnitude of the max 2

acceleration= n a At the point o, the velocity will be maximum.

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Max Velocity=na.

3.6.5

Problems

1. A particle moves in a straight line such that, its velocity v at a distance x is given by where



and



v2     x2

are constants. Show that the motion is simple harmonic and find the amplitude and

period of oscillation. Given, a particle moves along a straight line such that the velocity v at a distance x is given by

v 2     x2 ___ (1) 2v.

differentiating w.r.t. to

dv dx  0  .2x. dt dt

dv  .2x.v dt dv    x dt d2x  2   x dt 2v.

This equation is of the form

d2 x  n 2 x dt 2

 The motion of the particle is a simple harmonic motion.

T

Period of oscillation

2 n

n2  

n  . T

2 

To find the amplitude , Put velocity = 0

v0

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v2  0   x 2  0 x 2  

 amplitude 

x2 

 

x

 

 . 

2. The displacement of a particle in t seconds is given by x=a constant + b sin nt. Show that the motion is simple harmonic and find its period and amplitude.

 1

Displacement

x  a cos nt  bsin nt

differentiating,

dx  an sin nt  bn cos nt   2  dt

Again differentiating,

d2x  an 2 cos nt  bn 2 sin nt 2 dt d2x  n 2  a cos nt  b sin nt  2 dt d2x  n 2 x   3 2 dt  The motion of the particle is simple harmonic. Period of oscillation

T

2 seconds n

To find the amplitude, Put velocity = 0

v0 dx 0 dt

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an sin nt  bn cos nt  0 an sin nt  bn cos nt tan nt 

Put

tan nt 

Amplitude

b a

b in 1 a

 a.



a a b 2

 b.

b a  b2 2

a 2  b2 a 2  b2

 Amplitude  a 2  b2 3. A particle moving in a simple harmonic motion has velocities u and v at P and Q.

v2  u 2 If  and  be the accelerations at P and Q, show that the distance PQ is   In a simple harmonic motion,

d2 x  n 2 x. 2 dt

We know that

At P , let

x  x1;

d2 x  dt 2

  n 2 x1

 1

d2 x  At Q, let x  x 2 ; dt 2

  n 2 x 2

  2

Next, the velocity is given by,

v2  n 2  a 2  x 2  At P,

  3

x  x1; v  u

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ENGINEERING MECHANICS  u 2  n 2  a 2  x12 

132   4

x  x2; v  u

At Q,

 v2  n 2  a 2  x 2 2 

  5

Now , 2 2 2 2 2 2 v 2  u 2 n  a  x 2   n  a  x1     n 2 x1  n 2 x 2

 

  x 2 2  x12    x1  x 2 

  x 2  x1  x 2  x1    x 2  x1 

 x 2  x1 =

PQ 



PQ . v2  u 2   th

4. Show that a particle moving in a simple harmonic motion requires

1 of its period to move from one 6

extreme position to one in which the displacement is half its amplitude. In a Simple harmonic motion, Displacement x= a cos nt ____(1) Here, amplitude  a Period



2 . n

At one end A, x  a

1 

a  a cos nt

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1  cos nt cos 0  cos nt 0  nt t  0 Let P be the mid point of OA. At P,

a x  ; let t  t1 . 2 a  a cos nt1 2

1 

1  cos nt1 2  cos  cos nt1 3   nt1 3  t1  3n 1 2 t1   6 n 1 t1   T 6

1 t1   period. 6 5. A particle is executing a simple harmonic motion with centre o and period T. It passes through the

point P with a Velocity V in the direction OP. Show that it will return to I after a time

VT T [ ] tan 1 2 .OP 

seconds. In a simple harmonic motion Displacement

x  a cos nt

 1

At one end A, x  a

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a  a cos nt 1  cos nt cos 0  cos nt 0  nt t  0. At the point P, x  x; t

t

P  š, x  x; t  t  x  a cos nt

  2

i.e.,OP  a cos nt differentiating w.r.t.t, ,

dx  an sin nt dt

Magnitude of the velocity

  3

V  an sin nt

 3 ;  2 

Now, period

T

V an sin nt  OP a cos nt

V  tan nt n.OP

2 n

  4

  5

 4    5 ;

VT 2  .tan nt n.OP n

T V .  tan nt 2 OP  VT   nt  tan 1   2.OP  

t 

1  VT  tan 1  n  2.OP 

 Time taken to move from A to P is

t

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 Time taken to move from P to A and then to move from A to P is



2T  VT  tan 1  2  2.OP 



T  VT  tan 1    2.OP 

6. A particle moves along a circle with uniform Velocity.

Show that its projection an any diameter

executes a simple harmonic motion.

Let a particle move along a circle with centre o and radius a. Let V be the uniform velocity. At any time t, let the particle be at P. Let M be the projection of P on the diameter AB.

ˆ  Let MOP Let OM  x Since the particle is moving along the circle with a uniform velocity v, its tangential acceleration

dv  0. dt

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Its normal acceleration



136

v2 . along PO a

 The component of acceleration along OA  

 The acceleration of M is

v2 cos  a

d2 x v2 x   . dt 2 a a

d2x v2   2 x. dt 2 a

d2x v2   2 x. dt 2 a This equation is of the form

d2 x  n 2 x . 2 dt  The motion of M is a simple harmonic motion. Its period of oscillation is

T

2 n

2 v a 2 T w T

where

 

v w  , angular velocity of P. a

. If the distance x of the point moving in a straight line and its velocity V are connected by the relation

4V 2  25  x2 ,show that the motion is simple harmonic. Find its amplitude and period. Let V be the velocity of the particle at a displacement x. Given,

4v2  25  x 2

 1

Differentiating w.r.t.t,

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4.2.v.

137

dv dx  0  2x. dt dt

dx  2x.v dt d2x 1   x. 2 dt 4

8v.

This equation is of the form

d2 x  n 2 x 2 dt

Therefore the motion of the particle is simple harmonic. Period of oscillation

T T

2 n

n2 

2 1 2

n

 

1 4

1 2

T  4 Seconds To find the amplitude, put velocity = 0

v0 v2  0 1 25  x 2   0  4 25  x 2  0 x 2  25 x 5 Therefore Amplitude = 5 Units

Impulsive Force 3.7 Definition A very large force acting for a very small internal of time is called an impulsive force. Impulse of the impact= Change in momentum.

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Note: Momentum = mass ď&#x201A;´ Velocity. 3.7.1 Principle of Conservation of momentum When the bodies collide, the total momentum after impact is equal to the total momentum before impact. 3.7.2 Newtonâ&#x20AC;&#x2122;s experimental law on impacts. 1. When two Smooth Spheres impinge, along the Common normal, the relative velocity of one w.r.t. the other after impact bears a constant ratio to their relative velocity before impact and is of the opposite sign. 2. Along the common tangent, the velocity of each sphere remains unchanged. 3. Total momentum after impact is equal to the total momentum before impact.

Coefficient of restitution 3.7.3 Definition When two Smooth Spheres impinge, along the common normal, the relative velocity of one w.r.t. the other after impact bears a constant ratio to their relative velocity before Impact and is of the opposite sign. This constant ratio is called the coefficient of restitution. It is denoted by e. Note: If a body is perfectly elastic, e =1. otherwise e <1.

3.8 Direct Impact of two Smooth Spheres Two Smooth spheres

m1 , m2 impinge directly. Discuss the motion after impact.

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Let the smooth spheres direction with velocities If

139

A & B of masses

m1 & m2 move along the same line in the same

u1 & u2 respectively.

u1  u2 , A and B will impinge directly.

After impact, A and B will move along the same line. Let their velocities be

Let e be the Coefficient of restitution. By Newton’s law on impacts, Relative velocity after impact}=-e  Relative velocity before impact

v1  v2  e  u1  u 2 

 1

By the principle of conservation momentum

m1v1  m2v2  m1u1  m2u2 ____ (2)

1  m2   2 ; m 2 v1  m1v1  em 2  u1  u 2   m1u1  m 2 u 2 v1  m 2  m1    m1  em 2  u1  m 2 u 2 1  e   v1 

 m1  em2   m2 u 2 1  e  m1  m 2

  3

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 2  1  m1 m 2 v 2  m1v 2  m1u1  m 2 u 2  m1e  u1  u 2  v 2  m 2  m1   m1u1 1  e    m 2  em1  u 2  v2 

 m2  em1  u 2  m1u1 1  e  m1  m 2

  4

 3 ,  4  give the velocities of A and B after impact. 3.8.1 To find the impulse and loss in kinetic energy in a direct impact

Before impact Mass of A= m1 Mass of B=

Velocity of A=

Velocity of B=

u1  u2 , A and B will impinge directly.

After impact

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Let velocity of A  v1 & Velocity of B  v 2 Let e be the Coefficient of restitution. By Newton’s law on impacts,

v1  v2  e  u1  u 2 

 1

By the principle of Conservation of momentum,

m1v1  m2 v2  m1u1  m2u 2 (i)

  2

To find the impulse Let I be the impulse of the impulsive force exerted by A on B. Then I = Change in the momentum of B.

I  m2 v2  m2 u 2 I  m2  v2  u 2 

  3

By Newton’s third law, B exerts an equal and opposite force on A. -I = Change in the momentum of A

I  m1v1  m1u1 I  m1  v1  u1 

  4

 3  m1   4  m2  m1  m2  I  m1m2  v2  u 2   m1m2  v1  u1   m1m2  v2  u 2  v1  u1 

 m1  m2  I  m1m2  v2  v1    u1  u 2   m1m2  e  u1  u 2    u1  u 2    m1m2  u1  u 2  e  1

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I 

142

m1m2  u  u 1  e   m1  m2  1 2

  5

ii) To find the less in Kinetic energy Loss in K.E = initial K.E – Final K.E

1 1 1  1    m1u12  m 2 u 2 2    m1v12  m 2 v 2 2  2 2 2  2  1 1  m1  u12  v12   m 2  u 2 2  v 2 2  2 2 1 1  m1  u1  v1  u1  v1   m1  u 2  v 2  u 2  v 2  2 2 1 1  I  u1  v1   I  u 2  v 2  2 2 1  I  u1  v1    u 2  v 2   2 1  I  u1  u 2    v1  v 2   2 1  I  u1  u 2   e  u1  u 2   2 1  I  u1  u 2 1  e  2 1 m1m 2   u1  u 2 1  e  u1  u 2 1  e  2  m1  m 2  Therefore loss in K.E



1 m1m2 2 2  u1  u 2  1  e  2  m1  m2 

3.8.2 Problems 1. If two equal and perfectly elastic spheres impinge directly, show that after impact they interchange their velocities.

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Before impact Let mass of A=m Mass of B= m Velocity of A=

Velocity of B=

After impact Let Velocity of A= & Velocity of B=

Since the falls are perfectly elastic, e=1. By Newton’s law on impacts,

v1  v 2  e  u1  u 2  v1  v 2  1 u1  u 2  v1  v 2  u1  u 2

 1

By the principle of conservation of momentum

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mv1  mv 2  mu1  mu 2 v1  v2  u1  u 2

  2

1   2 ; 2v1  2u 2  v1  2u 2

  3

1   2 ; 2v 2  2u1  v 2  u1

  4

 3 &  4  v1  u 2 ; v2  u1 Therefore after impact their velocities are interchanged. 2. A ball of mass 2m impinges directly on another ball of mass m at rest. If the velocity of the former before impact is equal to that of the former before impact is equal to that of the letter after impact, prove that

e

1 . 2

Before impact Let mass of A = 2 m

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Mass of B= m Velocity of A= u Velocity of B= 0 After impact Let Velocity of A  v1 & Velocity of B  v 2 By Newton’s law on impacts,

v1  v 2  e  u  0  v1  v 2  eu

 1

By the principle of conservation of momentum

2mv1  mv2  2m.u  m.0 2v1  v 2  2u

  2

1  2   2 2v2  v 2  2eu  2u

3v 2  2u  e  1 Given

v2  u  3u  2u 1  e  3  1 e 2 1 e  2

3. A ball of mass m impinges directly on another ball of mass 2m which is moving in the same direction with

1 3 of its velocity. If e  show that after impact, the first ball reduces to rest. 7 4

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Before impact Let mass of A = m Mass of B = 2m Velocity of A = u Velocity of B



u 7

After impact Let Velocity of A  v1 & Velocity of B  v 2

Given

3 e . 4

By Newton’s law on impacts,

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v1  v 2  e  u1  u 2  3 u v1  v 2    u   4 7 3 6u v1  v 2   . 4 7 9u v1  v 2    1 14 By the principle of conservation of momentum

mv1  2mv 2  mu  2m 9 v1  2v 2  u 7

u 7

  2

1  2   2 2v1  v1  2 

9u 9u  14 7

9 9 3v1   u  u 7 7 3v1  0  v1  0 Therefore after impact, the first ball comes to rest. 4. A ball impinges directly on another equal ball at rest. Show that after impact, their velocities are in the ration

1 e . 1 e

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148

Before impact Let mass of A = m Mass of B = m Velocity of A = u Velocity of B = 0 After impact Let Velocity of A  v1 & Velocity of B  v 2 By Newton’s law on impacts,

v1  v 2  e  u1  u 2  v1  v 2  e  u  0  v1  v 2  eu

 1

By the principle of conservation of momentum

mv1  mv2  mu  m.0  v1  v2  u

  2

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149

2v1  eu  u v1 

u 1  e  2

  3

1   2 ; 2v 2  eu  u v2 

 4 ;  3 

u 1  e  2

  4

v2 u 2   1  e   v1 2 u 1  e 

v2 1  e  v1 1  e

5. A ball impinges directly on another ball on times its mass which is moving with

same direction. If the impact reduces the first ball to rest, prove that

e

1 of its velocity in the n

mn n and m  . m  n  1 n2

Before impact Let mass of A = m Mass of B = m Velocity of A = u Velocity of B = 0 After impact Let Velocity of A  v1 & Velocity of B  v 2 By Newton’s law on impacts,

v1  v 2  e  u1  u 2  v1  v 2  e  u  0  v1  v 2  eu

 1

By the principle of conservation of momentum

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150

Before impact Let mass of A = M Mass of B = mM Velocity of A = u Velocity of B



u n

After impact Let Velocity of A  v1 & Velocity of B  v 2 By Newton’s law on impacts,

u  v1  v 2  e  u   n   1 v1  v 2  eu 1    n

 1

By the principle of conservation of momentum

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151 Mv1  mMv 2  M.u  M.m. mu n  m v1  mv 2  u 1   n 

u n

i.e., v1  mv 2  u 

  2

1  m   2  1  m mv1  v1  emu 1    u 1   n  n  Given

v1  0  1  m  0  emu 1    u 1   n  n   n 1   n  m  em     n   n  mn e    3 m  n  1

Also, we know that



mn 1 m  n  1

m  n  mn  m 2m  mn  n n  mn  2m n  m  n  2 m 

n n2

6. Three spheres A,B,C of masses m,2m,m lie in a straight line on a smooth table. A is projected towards B. If e = 1 , show that after two impacts, their velocities are in the ratio 0:1:2 and there will be no

further impacts.

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Before impact Let mass of A = m Mass of B = 2M Velocity of A = u Velocity of B = 0 After impact Let Velocity of A  v1 & Velocity of B  v 2 By Newton’s law on impacts,

v1  v 2  e  u1  u 2  1  u  0 2 u v1  v 2    1 2 v1  v 2  

By the principle of conservation of momentum

m.v1  2m.v2  m.u  2m.0 v1  2v 2  u

  2

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153

u 3v1  2.  u 2 3v1  0 v1  0

  3

1   2 ; 3v 2   u 2 u  v2  2

u u 2

3v 2  3

  4

After the first impact A comes to rest and B moves with a velocity

u . 2

Second Impact B and C

Before impact Let mass of B = 2m Mass of C = m

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Velocity of B



154

u 2

Velocity of C = 0 After impact Let Velocity of B  v3 & Velocity of C  v 4 By Newton’s law on impacts,

1 u v3  v 4   . 2 2 u  v3  v 4   4

  5

By the principle of conservation of momentum

u 2m.v3  m.v 4  2m.  m.0 2 2v3  v 4  u  6

 5   6  ;

3v3  v3 

 5  2   6  ;

3u 4

u 4

 7

u 2v4  v4  2.  u 4

3v 4    v4 

u 2

3u 2  8

After the second tmpact Velocity of A : Velocity of B : Velocity of C

u u  0: : 4 2

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1 1  0: : 4 2  0 :1: 2 Since, Velocity of A < Velocity of B < Velocity of C, there will be no further impacts.

3.9 OBLIQUE IMPACT Two smooth spheres of masses

m1 and m2 moving with velocities u1 and u2 impinge obliquely.

Find their velocities after impact. Before impact Let mass of A = m 1 Mass of B = m2 Velocity of A = u1 Velocity of B = u2 Let

u1 , u 2 make angles  and  with the common normal AB.

After impact Let Velocity of A  v1 & Velocity of B  v 2 Let

v1 , v2 make angles  and  with the common normal AB.

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By Newton’s law on impact Along the common tangent,

 v1 sin   u1 sin 

 1

v2 sin   u 2 sin 

  2

Along the common normal

v1 cos   v2 cos   e  u1 cos   u 2 cos 

  3

By the principle of conservation of momentum

m1v1 cos   m2 v2 cos   m1u1cps  m2u 2 cos 

  4

 3  m 2   4  m2 u1 cos   m1u1 cos   em2  u1 cos   u 2 cos   m1u1 cos   m2 u 2 cos 

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v1 cos   m2  m1   u1 cos   m1  em2   m2u 2 cos  1  e   v1 cos  

u1 cos   m1  em2   m2 u 2cos 1  e  m1  m2

  5

 4   3  m1 m2 v2 cos   m1v2 cos   m1u1 cos   m2u 2 cos  em1  u1 cos   u 2 cos  u 2 cos   m1  m2   m1u1 cos  1  e   u 2 cos   m 2  em1   v 2 cos   From

m1u1 cos  1  e   u 2 cos   m 2  em1  m1  m 2

1 ,  2 , 5 ,  6 , we get the values of

v1 , v2 , ,  .

3.9.1 To find the impulse and loss in kinetic energy due to the oblique impact To find the impulse Let I be the impulse of the impulsive force exerted by A on B. Then I= Change in the momentum of B (along the common normal)

I  m2 v2 cos   m2 u 2 cos  I  m2  v2 cos   u 2 cos  

 7

By Newton’s third law, B will exert an equal and opposite force.

I  chavgein the momentumof A I  m1v1 cos   m1u1 cos  I  m1  v1 cos   u1 cos  

 8 

 7   m2  8  m1  m1  m2  I  m1m2  v2 cos   u 2 cos   v1 cos   u1 cos    m1m2  v 2 cos   v1 cos     u1 cos   u 2 cos     m1m2 e  u1 cos   u 2 cos     u1 cos   u 2 cos   

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I 

m1m2  u1 cos   u 2 cos  1  e  m1  m2

158

 9

To find the loss in K.E Loss in K.E = Initial K.E – Final K.E

1 1 2 2 2 2 1 1   m1  u1 cos    m 2  u 2 cos      m1  v1 cos    m 2  v 2 cos    2 2 2  2  1 1 2 2 2 2  m1  u1 cos     v1 cos     m 2  u 2 cos     v 2 cos     2   2  1  m1  u1 cos   v1 cos   u1 cos   v1 cos   2 1  m 2  u 2 cos   v 2 cos   u 2 cos   v 2 cos   2 1 1 I  u1 cos   v1 cos    I  u 2 cos   v 2 cos   2 2 1  I  u1 cos   v1 cos    u 2 cos   v 2 cos    2 1  I  u1 cos   u 2 cos     v1 cos   v 2 cos    2 1  I  u1 cos   u 2 cos    e  u1 cos   u 2 cos    2 1  I  u1 cos   u 2 cos  1  e  2 1 m1m 2   u1 cos   u 2 cos  1  e   u1 cos   u 2 cos  1  e  2 m1  m 2 

 Loss in K.E  3.9.2

1 m1m2 2  u1 cos   u 2 cos  1  e2  2 m1  m2

Problems

1. A Smooth sphere impinges obliquely on another sphere at rest. Show that, after impact the lather moves along the common normal.

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Before impact Let mass of A = m 1 Mass of B = m2 Velocity of A = u1 Velocity of B = 0 Let

u1 make angles  with the common normal AB.

After impact Let Velocity of A  v1 & Velocity of B  v 2 Let

v1 , v2 make angles  and  with the common normal AB.

By Newton’s law on impact Along the common tangent,

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160

 v1 sin   u1 sin  v 2 sin   0 sin   0   0 Therefore after impact, B moves along the common normal. 2. A perfectly elastic sphere impinges obliquely on another equal sphere at rest. Show that, after impact they will move along perpendicular directions.

Before impact Let mass of A = m Mass of B = m Velocity of A = u1 Velocity of B = 0 Let

u1 make angles  with the common normal AB.

After impact

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Let Velocity of A  v1 & Velocity of B  v 2 Let

v1 , v2 make angles  and  with the common normal AB. Given e = 1.

By Newton’s law on impact Along the common tangent,

v1 sin   u1 cos 

 1

v 2 sin   0 sin   0 0 Therefore after impact, B moves along AB. Along the common normal,

v1 cos   v 2 cos   e  u1 cos   0  v1 cos   v 2 cos   u1 cos 

 e  1   2

By the principle of conservation of momentum

mv1 cos   mv 2 cos   mu1 cos   m.0

  3

v1 cos   v 2 cos   u1 cos 

 2    3 ;

2v1 cos   0 cos   0 

 2

Therefore after impact, B moves in a direction perpendicular to AB. Therefore after impact, A and B moves along perpendicular directions. 3. A smooth sphere impinges on another equal sphere moving with the same speed in a perpendicular direction. The line joining the centers is

r

to the direction of motion of the second

ball. Show that due to the impact, the second ball turns through an angle

 1 e  tan 1  .  2 

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Before impact Let mass of A = m Mass of B = m Velocity of A = u Velocity of B = u Let

u1 make angles  with the common normal AB.

After impact Let Velocity of A  v1 & Velocity of B  v 2 Let

v1 , v2 make angles  and  with the common normal AB.

By Newton’s law on impacts, Along the common tangent,

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 v1 sin   0 sin   0  1

0 Also v2 sin   u

  2

Along the common Normal,

v1 cos   v2 cos   e  u  0 

  3

By the principle of conservation of momentum

mv1 cos   mv2 cos   m.u  m.0 v1 cos   v2 cos   u

 3   4  ;

2v 2 cos   eu  u  v 2 cos  

 2 ;  5

  4

 e  1 u

2 v 2 sin  u.2  v 2 cos   e  1 u

 tan  

2 1 e

  5

  6

Let  be the angle of deflection in the direction of motion due to the impact. Then,

  90   tan   cot  1 e 2  1 e    tan 1    2  tan  

4. If two equal and perfectly elastic spheres impinge obliquely along two perpendicular Directions show that after impact, they will move along perpendicular directions.

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Before impact Let mass of A = m Mass of B = m Velocity of A = u1 Velocity of B = u2 Let

u1 and u2 make angles  and  with the common normal AB.

After impact Let Velocity of A  v1 & Velocity of B  v 2 Let

v1 , v2 make angles  and  with the common normal AB.

Given e = 1. By Newton’s law on impact

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Along the common tangent,

v1 sin   u1 sin 

 1

v2 sin   u 2 sin 

  2

Along the common Normal,

v1 cos   v 2 cos   e  u1 cos   u 2 cos   v1 cos   v 2 cos   1 u1 cos   u 2 cos   v1 cos   v 2 cos   u1 cos   u 2 cos 

  3

By the principle of conservation of momentum

m.v1 cos   m.v 2 cos   m.u1 cos   m.u 2 cos 

 4

v1 cos   v 2 cos   u1 cos   u 2 cos 

 3   4  ;

2v cos   2u 2 cos   5

v1 cos   u 2 cos 

1 ;  5

v1 sin  u1 sin   v1 cos  u 2 cos  tan  

u1 sin  u 2 cos 

 7

Given A and B are moving along perpendicular directions before impact.

 

  2

7 

u1 sin    u 2 cos     2  u1 sin  tan   u 2   sin  

tan  

tan   

u1 u2

 8

Next,

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 2 ; 6

166

v 2 sin  u 2 sin   v 2 cos  u1 cos    u 2 sin     2  tan   u1 cos  tan  

u 2 cos  u1 cos 

tan  

u2 u1

1 tan  tan    cot  tan   

  tan   tan     2     2    2 Therefore after impact, A and B move along perpendicular directions. Impact of a sphere on a fixed plane A smooth sphere of mass m impinges obliquely on a fixed plane. Find the motion of the sphere after impact.

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Let a Smooth sphere of mass m impinge obliquely on a smooth fixed plane. Let u be the velocity before impact. Let u be inclined at an angle



with the common normal OA.

After impact, let v be the velocity, along a direction inclined at an angle



with OA.

Along the common tangent,

 vsin   u sin 

 1

Along the common normal,

v cos   0  e  u cos   0  v cos   eu cos 

  2

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1   2  2

;

168

v 2  sin 2   cos 2    u 2  sin 2   e 2 cos 2    v 2  u 2  sin 2   e2 cos 2    v  u sin 2   e 2 cos 2 

1 ;  2

v sin  u sin   v cos  eu cos  1 tan   tan  e

From

  3

  4

 3 ,  4  , we get the magnitude and direction of the velocity after impact.

To find the impulse Impulse I = Change in the momentum (along the common Normal)

I  mv cos   m  u cos   I  m.e.u cos   mu cos  I  mu cos  1  e  To find the loss in K.E Loss in K.E = Initial K.E – Final K.E

1 1 2 2  m  u cos    m  v cos   2 2

1 1 2  mu 2 cos2   m  eu cos   2 2 1  mu 2 cos 2  1  e2  2 Note: If the sphere implugs the plane at right angles, then

  0.

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 4 

169

1 tan   .tan 0 e 1 tan   .0 e tan   0 0

 3 

v  u sin 2 0  e 2 cos 2 0 v  u 0.  e 2 .1 v  eu

Therefore if a sphere at right angles a fixed plane with a velocity u, it will refound at right angles with a velocity eu. 3.10.1 Problems 1. A Smooth ball falls from a height h and hits the ground after a time t seconds. Show that it rebounds and reaches a maximum height

e2 h in et seconds.

Let a ball from rest at A. Let h be the height of A above the ground. Let it hit the ground at B with a Velocity u. Now,

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v2  u 2  2as  

u 2  0  2gh u 2  2gh

 1

The time taken from A to B is t secs.

v  u  at



u  0  gt



u  gt

  2

Before impact, the velocity of the ball is u. It will rebound vertically after impact with a velocity eu. After impact, Let h1 be the height reached and let t1 be the time taken for it.

0   eu   2gh1

v2  u 2  2as 

Next,

2gh1  e 2 u 2 2gh1  e 2 .2gh  h1  e 2 h

  3

Next,

v  u  at



0  eu  gt1 gt1  e.gt

  4

 t1  at Therefore after impact, the max height reached

 e2 h.

Time taken = et.

2. A ball falls from a height h on a horizontal plane. Show that the total distance traveled before the ball

 1  e2  comes to rest is h  and the time taken is 2   1 e 

2h  1  e2   . g  1  e2 

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Let a ball fall from rest at A. Let h be the height of A above the horizontal plane. Let the ball reach the horizontal plane with a velocity u. Now,

v 2  u 2  2as 

u 2  0  2gh u 2  2gh h

u2 2g

 1

After the first, second, third…… …… impacts, the ball will move upwards

with velocities

eu,e2 u,e3u.... Let

h1 , h 2 , h 3 .... be the maximum heights reached.

Then,  h1

 eu   2g

e u   2

, h2

Therefore total distance traveled

e u   3

, h3

,....

 h  2h1  2h 2  2h 3  .....

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u2 e2 u 2 e4 u 2 e6 u 2   2.  2.  2.  .... 2g g 2g 2g e2 u 2 1  e 2  e 4  .... h g  2e 2   h 1  2  1 e  1  e 2  2e 2  h  2  1 e   1  e2   h 2   1 e  Next, Let t be the time taken from A to B.

v  u  at



u  0  gt u  gt t 

u g

  2

After the first, second, third…. ……….. impacts let

t1 , t 2 , t 3 ....... be the times taken to reach the

maximum heights. Now,  t1



eu e2 u e3 u , t2  , t3  ...... g g g

Therefore total time taken T

 t  2t1  2t 2  ....

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173 u eu e2u 2 2  .... g g g eu  t  2 1  e  e 2  .... g 

 1   t  2et    1 e  2e    t 1   1  e  1  e  2e  t  1  e   1 e   t   1 e  3. A smooth circular table is surmounted by a vertical rim. Show that a ball projected along the table from a point A on the rim, in a direction making an angle



with the radius through A will return to A after two

e3 impacts, if tan   . 1  e  e2 2

Let o be the center of the circle. A ball is project from A along AB.

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After two impacts at B an c, it returns to A. Let

ˆ  OBA ˆ  OAB ˆ  ˆ  OCB OBC

ˆ  OAC ˆ  OCA Considering the impact at B,

1 tan   tan  e

 1

Similarly considering the imact at c,

1 tan   tan  e 1 tan   2 tan  e Next, in

  2

ABC A  B  C  1800 2  2  2   180       90     90   tan       tan  90    tan   tan   cot  1  tan .tan  1 tan   tan  e2 e  1 1  tan . tan  tan  e 1 tan 2   tan 2   e 2  e tan 2  e e e tan 2   tan 2   e3  e 2 tan 2  e tan 2   tan 2   e 2 tan 2   e3 tan 2   e  1  e 2   e3 e3  tan   1  e  e2 2

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175

UNIT-4 4.0

Introduction

4.1

Objectives

4.2

Projectile

4.3

Equation of the trajectory

4.4

Time of flight

4.5

Range on an inclined plane

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4.0 Introduction In this unit, we will discuss about the motion of

a projectile. You will learn that all the particles

projected with any velocity along any direction trace a parabola.

The horizontal range, maximum

horizontal range, time of flight and the greatest height attained by a projectile are calculated. Numerous examples are given to learn in detail all about projectiles. The range on an inclined plane is also derived.

4.1

OBJECTIVES After going through this unit, you would have learnt about 

The path of a projectile.



The time of flight



The horizontal range



The maximum horizontal range



The greatest height



The velocity at any instant.



The range on an inclined plane.

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PROJECTILES In this unit we will discuss the motion of a particle projected from a point on the horizontal ground with any velocity along any direction.

A particle is projected from a point o on the horizontal ground. It is projected with a velocity u along a direction inclined at an angle

ď Ą with the horizontal.

It falls on the

ground at A. O is called the point of projection, U is called the initial velocity,

ď Ą

is called the angle of projection,

OA is called the horizontal range, The time taken to move from O to A, is called the time of flight. The path of the particle is called a trajectory. 4.3 B.W To prove the path of a projectile is a parabola

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Let o be the point of projection, u be the initial velocity &



be the angle of projection.

Take O as origin and the horizontal and the vertical as x and y axes. Let

i and j be unit vectors along ox and oy.

At any time t, let the particle be at P(x,y). Let

v be the velocity at P.

The only external force acting on the particle is its weight mg

.

The equation of motion of P is

F  ma d2 r  mg   j  dt 2 d2 r  gj  1 dt 2

Integrating,



OP  r   

d2 r .dt   g jdr dt 2

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dr   gt j  c1 dt

 (2)

Initially

at o,

t  0,

dr  u  u cos  i  u sin  j dt

 2  u cos  i  u sin  j  O  e1

 e1  u cos  i  u sin  j  2 

dr  gt j  u cos  i  u sin  j dt

dr  u cos  i   u sin   gt  j dt Integrating, r 



  3

u cos  i   u sin   gt  j dt

 gt 2  r  u cos t i   u sin t   j  e2 2   Initially at 0,

  4

e 2 - kh¿è

t  0; r  0 0  0 i  0  0 j   e2  e2  0

 4   r  u cos t i   u sin t  

1 2 gt  r 2 

1   x i  yj  u cos t i   u sin t  gt 2  r 2   x  u cos t  6 1 y  u sin t  gt 2 2 Eliminating t between

 6 

t

 5 

 7

 6 ,  7   , we get the equation of the path x u cos 

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Substituting in

7

y  u sin .

180 x 1  x   g.   u cos  2  u cos  

y  x tan  

gx 2 2u 2 cos 2 

 8

This is the equation of the path

8  g .x 2  x tan   y  0 2 2u cos  2

Here

"a " 

g , "b"  0; "h"  0 2u cos 2  2

"h 2  ab"  0 

g .0 2u cos 2  2

"h 2  ab"  0  The equation (8) represents a parahola.  The path of a projectile is a parahola. Note: The only force acting on the particle at I is Mg. Since it is acting in the vertical direction, there is no force in the horizontal direction  In the horizontal direction, the component of velocity u cos



remains

constant through out the motion. 4.4.1 To find the time of flight Let the particle projected from a point O reach the horizontal ground again at The point A. The time taken to travel from 0 to A is the time of flight.

1 y  u sin t  gt 2 . 2

Now,

At A ,

y  0; t  T 1  0  u sin .T  gT 2 2 1   0  T  u sin   gT  2  

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1  T  0 (or) u sin   gT  0 2 T  0 (or)

1 gT  u sin  2

T  0 (or) T   Time of flight T 

2u sin  g

4.4.2 To find the horizontal range The particle projected from o again reaches the ground at A. Then OA is called the horizontal range. We have,

x  u cos  .t

t

At A,

2u sin  g

2u sin  g 2sin  cos  x  u2 g

 x  u cos .

x 

u 2 sin 2 g

 Horizontal Range R 

u 2 sin 2  g

4.4.3 To find the maximum horizontal range u is the initial velocity and

 is the angle of projection. Horizontal range R=

sin 2 ___(1) g

If a particle is projected from O with the same initial velocity



, along different

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Directions We will find, when we will get the maximum horizontal range. Therefore u is constant. Also g is a constant. Therefore R will be maximum, when But maximum value of

sin 2 is maximum.

sin 2 =1 sin 2  sin

 2

 2   4 2 

Put



 in (1) 4

u2 .1 Maximum horizontal range R1  g  R1 

u2  and   g 4

4.4.4 To find the greatest height Let B be the highest point in the path of the projectile.

We have

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1 y  u sin t  gt 2 2 dy  u sin   gt dt

 1 .   2

At B, the vertical component of velocity is zero

i.e.,

dy 0 dt

u sin   gt  0 t

u sin  g

substituting in (1),

u sin  1  u sin   y  u sin .  g  g 2  g  y

u 2 sin 2  u 2 sin 2   g 2g

y

u 2 sin 2  2g

u 2 sin 2   Greatest height h  2g 4.4.5 To find the velocity at any instant We have,

dr  u cos  i   u sin   gt  j ......(1) dt dr   (u cos  ) 2  (u sin   gt ) 2 dt v  u 2 cos 2   u 2 sin 2   2u sin  .gt  g 2t 2 v  u 2 (cos 2   sin 2  )  2ugt sin   g 2t 2 v  u 2  g 2t 2  2ugt sin  .........(2) If v is inclined at an angle



with the horizontal, then

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tan  

184

u sin   gt u cos 

4.4.6 To prove that there are two directions of projection with a given velocity for a given horizontal range.

Let u be the initial velocity and



be the angle of projection.

Then, horizontal range

u 2 sin 2 R g gR  sin 2  2 u If

 1

gR  k  1 , there exist real values for  u2

Let

sin   k

1  sin 2  sin  Also

sin 2  sin   sin    

 2   (or)   

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 

185

   (m)  2 2 2

i.e, For a given velocity and for a given horizontal range their exist two angles of projection



and

  . 2 2 i.e. The angles are Supplementary. Some important formulae. 1. The equation of the path is

y= x tan

-

gx 2 . 2u 2 cos 2 

2. Time of flight T=

2u sin  g

u 2 sin 2 Horizontal range R= g

4. Greatest height H=

u 2 sin 2  2g

5. Maximum horizontal range

6. x= ucos

 .t

7. y= usin



8. 4.4.7

R1 

u2 ;   4 g

1 2 gt 2

v  u cos i  (u sin   gt ) j Problems

1. In a projectile motion, prove that horizontal range and

gT 2  2R tan  Where T is the time of flight, R is the

 is the angle of projection.

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Let u be the initial velocity and Time of flight

T

 be the angle of projection.

2u sin  g

R

Horizontal range

186

u 2 sin 2  g

Now, L.H.S

 2u sin   gT  g   g  

4u 2 sin 2   g. g gT 2 

4u 2 sin 2  g

 1

R.H.S

2R tan   2.

4u 2 sin 2  .tan  g

 2.u 2 .

2R tan  

2sin .cos  sin  . g cos 

4u 2 sin 2  g

  2

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gT2  2R tan . 2. If the greatest height attained is one-fourth its horizontal range, find its angle of projection.

Let u be the initial velocity and

Horizontal range

R

 be the angle of projection.

u 2 sin 2 g

u 2 sin 2  Greatest height h  2g Given

1 h R 4

u 2 sin 2 1 u 2 sin 2  . 2g 4 g 1 sin 2   .2sin .cos  2 sin   cos  tan   1 

 4

3. A particle is projected from a point O with an initial velocity u. If

h 1 and h2 are the heights

attained in two different paths for the same horizontal range R, show that

R  4 h1h2

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Let u be the initial velocity and

Horizontal range

R

188



u 2 sin 2 g

 1

gR u2

  2

 sin 2 

be the angle of projectile.

gR  1 there exist real values for  u2 Let

gR  sin  u2

 2  sin 2  sin  sin 2  sin   sin     2   (or)     

    (or) 2 2 2

  1 (or)

   1. where  1 2 2

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 For a given horizontal range R, there are two angles of projection 1 and   1 . 2 The greatest height attained by the particle is

h

u 2 sin 2  2g

when

  1 ; h  h1

u 2 sin 2   h1  2g when



  3

  1 ; h  h 2 2

  u 2 sin 2   1  2  h2  2g

u 2 cos 2 1 h2  2g

  4

Now,

u 2 sin 2 1 u 2 cos 2 1 4 h1h 2  4 . 2g 2g



4 2 .u sin 1.cos 1 2g



u 2 .2sin 1 , cos 1 g



u 2 sin 2 1 R g

 R  4 h1h 2

4. Two particles are projected from a given point on the horizontal ground with the same velocity u at angles

 and  , aimed at a target on the horizontal plane.

Once falls a metres too short and

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190

the other b metres too long. If



be the correct angle of projection so as to hit the target, prove

that

 a  b  sin 2  a sin 2  bsin 2

Let u be the initial velocity &

 be the angle of projection.

u 2 sin 2  Horizontal range R  g First,

angle of projection   Horizontal range

R  a  Next,

u 2 sin 2  g

angle of projection Horizontal range

R  b  Thirdly,

 R a

 1



Rb

u 2 sin 2 g

angle of projection

  2 

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ENGINEERING MECHANICS Horizontal range

R 

191 R

u 2 sin 2 g

  3

1  b   2  a bu 2 sin 2  au 2 sin 2  bR  aR   g g

a  b R  a  b

u2  b sin 2  a sin 2  g

u 2 sin 2 u 2   b sin 2  a sin 2  g g

  a  b  sin 2  b sin 2  a sin 2. 5. A particle projected from a point O just clears a wall at a distance a and height h. If angle of projection and R be the horizontal range, prove that tan

Let O be the point of projection u be the initial velocity of

Horizontal range

R

u 2 sin 2 g



=

 be the

Rh a( R  a)

be the angle of projection.

 1

The equation of the path is,

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192

y  x tan  

gx 2 2u 2 cos 2 

  2

The projectile clears a wall of a height ‘h’ and at a distance ‘a’ from 0

 P  a, h  lies on (2)  h  a tan  

h  a tan  

ga 2 2u 2 cos 2 

  3

u 2 sin 2 a2 . 2 R 2u cos 2  h  a tan  

a 2 .2sin .cos  R.2 cos 2 

h  a tan  

a2 .tan  R

 u 2 sin 2  1  g     R  

a   h  a tan  1    R  R a  h  a tan     R  

Rh  tan  a R  a 

 tan  

Rh a R  a 

6. A particle projected with a velocity u at an angle

45 falls on the ground at A. Show that in order

to hit a target at a highest h above A at the same angle of projection, the initial velocity must be increased to

u2 u 2  gh

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193

Let O be the point of projection, u be the initial velocity, &



be the angle of projection.

Horizontal range

Given

R

u 2 sin 2 g

  45

R 

u 2 sin 90 g

R 

u2 g

 1

Let B be a point at a height h above A. Let a second particle be projected from O with a velocity v. The equation of the path is

y  x tan   Here ,

gx 2 2u 2 cos 2 

  45 , u  v

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 y  x tan 45 

194

gx 2 2v2 .cos 2 45

B  R, h  lies on it h  R.1 

g.R 2 1 2.v 2 . 2

gR 2 hR 2 v h

u 2 g.u 4  g v 2 .g 2

u4 u2  h v 2 .g g u4 u 2  gh  v 2 .g g 

v2 1   gh u4 u2

v2 

u4 u 2  gh

v 

u2 u 2  gh

7. Show that the greatest height attained by a particle on a wall at a distance a from the point of

u 2 ga 2  projection is where u is the initial velocity. 2 g 2u 2

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195

Let O be the point of projection, u be the initial velocity, &



be the angle of projection.

The equation of the path is

y  x tan  

gx 2 2u 2 cos 2 

 1

Let the particle reach a height h on the wall at a distance a from the wall. Then, B(a,h) lies on (1)

ga 2 2u 2 cos 2  ga 2 sec 2  h  a tan   2u 2 ga 2 h  at  2 1  t 2    2 2u

 h  a tan  

 t  tan  

For the given velocity u, we will find the maximum height on the wall. Differentiality w.r.to t,

dh ga 2  a.1  2 .2t dt 2u

  3

Again differentiating w.r.to t,

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196

d 2h ga 2   .1 dt 2 u2

  4

For maximum or minimum value of ‘h’,

dh  0. dt a

ga 2 .t  0 u2

ga 2 .t  a u2 u2 t  ag When

t

u2 d2h ga 2 ,  4  2   2  0 dt u ag

 h is maximum when t 

u2 ag

u2 Put t  in (2) ag u 2 ga 2  u4  h1  a.  2 1  2 2  ag 2u  a g 

u 2 ga 2 ga 2 u 4 h1    . g 2u 2 2u 2 a 2g 2 h1 

u 2 ga 2 u 2   g 2u 2 2g

h1 

u 2 ga 2  2g 2u 2

8. A particle is projected over a triangle from one end B of its horizontal base, to graze the vertex and then falls at the other and C of the base. Prove that tan

 =tan

B+tan C where

 is

angle of projection.

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the

ENGINEERING MECHANICS

197

Let B be the point of projection, u be the initial velocity, &

 be the angle of projection.

The equation of the path is,

y  x tan  

gx 2 2u 2 cos 2 

 1

A  h cot B, h  lies on it h  h cot B.tan  

g.h 2 cot 2 B 2u 2 cos 2 

  2

C  h cot B  h cot C,0  1  lies on it  0   h cot B  h cot C  .tan   h  cot B  cot C  tan    2u 2 cos 2  

g  h cot B  h cot C 

2u 2 cos 2 

gh 2  cot B  cot C 

2u 2 cos 2  gh  cot B  cot C  tan 

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198

substituting in (2), we get

gh 2 cot 2 B.tan  h  h cot B.tan   gh  cot B  cot C 

cot 2 B.tan  1  cot B.tan   cot B  cot C cot B tan   cot B  cot e   cot 2 B.tan  1 cot B  cot C cot B  cot C  cot 2 B.tan   tan .cot B, cot C  cot 2 B.tan  cot B  cot C  tan .cot B.cot C. 1 1 tan    tan B tan C tan B.tan C tan e  tan B tan   tan B.tan C tan B.tan C  tan   tan B  tan C. 9. A particle reaches a point P of its path in t seconds and

t  is the time taken to move from P to the

horizontal plane. Show that the height of P above the horizontal plane is

1 gtt  . 2

Let O be the point of projection

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199

u be the initial velocity &  be the angle of projection At any time t, let the particle be at P.

1  y  u sin t  gt 2 2

 1

The time taken to move from P to the horizontal plane is

t .

 Total time taken to move from O to A is T  t  t

2u sin   t  t g 2u sin   t  t g

  2

1 1  2u sin   gtt  g  tt 2 2  g  1 2u sin  1 2  gt.  gt 2 g 2 1  u sin t  gt 2  y 2

1  y  gtt . 2  Height of P above the horizontal plane =

1 gtt  . 2

10. A particle is projected with a velocity u at an angle inclined at angle

.

A time t seconds, it moves in a direction

 to the horizontal. Prove that gtcos  = u sin (  -  )

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200

Let O be the point of projection, u be the initial velocity, &  be the angle of projection. At any time t, let the particle be at P. Let

be the velocity at P.

v  u cos  i   u sin   gt  j Let

make an angle



 1

with x-axis.

v  vcos  i  vsin  j

 2

1 &  2 

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201

v cos  i  v sin  j  u cos  i   u sin   gt  j  v cos   u cos 

  3

v sin   u sin   gt

 4

 3 ;  4

v cos  u cos   v sin  u sin   gt

 u sin   gt  cos   u cos .sin . u sin  cos   gt cos   u cos .sin  u sin .cos   u cos  sin   gt cos  u sin .cos   cos .sin   gt cos  u sin       gt cos . 11.

A particle is projected with a velocity u at an angle the initial direction after a time

.

Show that it will move at right angles to

u cos sec  . g

Let O be the point of projection, u be the initial velocity &



be the angle of projection.

At any time t, let the particle be at P.

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Let

Then, Let

be the velocity at P.

Then

202

= u cos

 i

+ (u sin

be inclined at an angle

 -gt) j



___ (1)

with the horizontal.

v = u cos  i + u sin  j ___ (2)

(1) & (2)  v cos  i  v sin  j  u cos  i  (u sin   gt) j  v cos   u cos  v sin   u sin   gt

 (4)

(3) v cos  u cos  ;  (4) v sin  u sin   gt When

the

particle

moves

right

angles

the

initial

direction,



  cos     2   u cos     u sin   gt sin      2   sin  u cos   cos  u sin   gt u sin 2   gt sin   u cos 2  gt sin   u cos 2   u sin 2  gt sin   u  cos 2   sin 2   gt sin   u

12.

t

u g sin 

t

u cos ec g

Show that the height of the directrix of the paracole above x-axis is

u2 2g

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  2

ENGINEERING MECHANICS

203

Let O be the point of projection, u be the initial velocity, &

ď Ą

be the angle of projection.

Taking O as origin, the horizontal and the upward vertical through O as x and y- axis, the equation of the path is

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204

gx 2 y  x tan   2 2u cos 2  g.x 2  x tan    y 2u 2 cos 2   2 2u 2 cos 2   g x  .tan .x    y  2 2 2u cos   g  2u 2 cos 2  sin  2u 2 cos 2  x  . .x   .y g cos  g 2

x2 

2u 2 sin .cos  2u 2 cos 2  .x   y g g 2

   u 2 sin .cos   u2 2u 2 cos 2  y    x  g sin .cos     g g     2

 u 2 sin .cos   2u 2 cos 2  u4 x    y  sin 2 .cos 2    2 g g g   2

 2u 2 sin .cos   2u 2 cos 2   u 2 sin 2   x    y     2g g 2g     2

 u 2 sin 2  2u 2 cos 2   u 2 sin 2    x   y  2g  2g  g   

 (1)

This equation is of the from  x  h   4a(y  k) 2

Where

h

u 2 sin 2 u 2 sin 2  ; k 2g 2g

4a  a

2u 2 cos 2  g

u 2 cos 2  2g

 The equation (1) represents a parabola with vertex at B (h,k)

 u 2 sin 2 u 2 sin 2   i.e., B  , 2g   2g

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205

 Height of the direction above x- axis = DB  BC = y co-ordinate of B + a

u 2 cos 2  u 2 sin 2    2g 2g u 2 cos 2   u 2 sin 2   2g  13. If

u2 2g

v1 and v2 are the velocities of a projectile at the ends of a focal chord and if v0 be the velocity at

the vertex of the parabola, show that

1 1 1  2 2. 2 v0 v1 v 2

Let O be the point of projection. u be the initial velocity. &



be the angle of projection.

Let PQ be the focal chord. Let Let

v1 & v2 be the velocities at P & Q.

v0 be the velocity at the vertex B.

We know that, the velocity at any point on the path of a projectile is equal to the velocity acquired by it in falling freely from the directrix to that point.

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206

"v2  u 2  2as" At P,

v12  0  2g.PL  2g.PL

At Q,

v22  0  2g.QM  2g.QM

At B,

v02  0  2g.BC  2g.BC

 (1)  (2)  (3)

In a parabola, we have

1 1 2   SP SQ l

(Latus tectum 2l=4a , l=2a)

1 1 1   SP SQ BC 1 1 1   SP  PL ; SQ  QM  PL QM BC 1 1 1   2g.PL 2g.QM 2gBC 1 1 1  2 2 2 v1 v 2 v 0 14. Show that the velocity at any point on the path of a projectile is equal to the velocity attained by it in falling freely from the directrix to the point.

Let O be the point of projection,

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207

u be the initial velocity and



be the angle of projection.

At any time t, let the particle be at P. Let

v be the velocity at P.

 v 2  u cos     u sin   gt  2

v 2  u 2 cos 2   u 2 sin 2   g 2 t 2  2ugt sin  v 2  u 2  cos 2   sin 2    g 2 t 2  2ugt sin  v 2  u 2  g 2 t 2  2ugt sin 

 (1)

Next, let Q be the point on the directrix vertically above P. If a particle falls freely from Q under gravity, let

v1 be its velocity on reaching the point P.

"v2  u 2  2as" At P,

v12  0  2g.PQ v12  2g  QM  PM   u2  v  2g   y P   2g  2 1

 u2  1  v12  2g    u sin t  gt 2   2   2g  2 u 1 v12  2g.  2gtu sin   2g. gt 2 2g 2 v12  u 2  g 2 t 2  2ugt sin 

 (2)

(1) & (2)  v2  v12 v  v1 .  The velocity at any point on the path of a projectile is equal to the velocity acquired by it in falling freely from the directrix to that point. B.W. 4.5 To find the range on an inclined plane. A particle is projected with a velocity u at an angle an inclined plane of inclination



with the horizontal. To find the range on

.

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Let o be the point of projection, u be the initial velocity &



Let



be the angle of projection. be the angle of inclination of the inclined plane.

Let the particle reach the inclined plane at the point A. Let the range on the inclined plane be OA=R. Take o as origin and the horizontal and the upward vertical through o as x and y axes. Let AB = OX.

OB  OB  R cos  R AB sin    AB  R sin . R cos  

 OAB,

 A  R cos , R sin  The equation of the path is

y  x tan  

gx 2 2u 2 cos 2 

A  R cos , R sin  lies on it.

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208

ENGINEERING MECHANICS

g.R 2 cos 2   R sin   R cos .tan   2 2u cos 2  sin  g.R cos 2  sin   cos .  cos  2u 2 cos 2  g cos 2  sin .cos  R. 2   sin . 2 2u cos  cos  g cos 2  sin .cos   cos .sin  R. 2  2 2u cos  cos  sin      g cos 2  R. 2  2 2u cos  cos  2u 2 cos .sin      R g cos 2  2u 2 sin      .cos  R g cos 2 

2u 2 sin(   ).cos   Range on the inclined plane R= g cos 2  4.5.1 Note.1. To find the time of flight Let T be the time taken to move from O to A. In a projectile motion, We have x= R cos

 ; t=T

2u 2 sin(   ).cos  .cos   u cos  .T g cos 2   Time of flight T=

2u sin(   ) g cos 

4.5.2 Note.2. Maximum range on the inclined plane

2u 2 sin(   ).cos  Range on the inclined plane R= g cos 2  u2 =  2sin(   .cos  ) g cos 2  FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

209

ENGINEERING MECHANICS

210

u2 [2sin(   )  sin(     )] g cos 2 

u2 [sin(2   )  sin  ]  (1) = g cos2 

R will be maximum when sin (2 

  ) is maximum.

But maximum value of sin ( 2   )  1 sin ( 2

  ) = sin 

2     2  



     4 2 (1)  Max. range on the inclined plane R1 

4.5.3

u2 1  sin  g cos 2 

R1 

u2 1  sin   g 1  sin 2  

R1 

u 2 1  sin   g 1  sin  1  sin  

R1 

u2 g 1  sin  

Problems

1. A gun is situated on an inclined plane and the maximum ranges up and down the inclined plane are

R1 and

l2 . If l be the maximum range on the horizontal plane with the same velocity, prove that

1 1 2   . l1 l2 l

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211

Let u be the initial velocity. Let



the inclination of the inclined plane.

The Maximum range up the inclined plane is

l1 

u2 g 1  sin  

III ly The maximum range down the inclined plane is l2 

Max range on the horizontal plane is

l

u2 g

u2 g 1  sin  

 1

  2

  3

Next,

1 1 g 1  sin   g 1  sin      l1 l2 u2 u2 1 1 g   1  sin   1  sin   l1 l2 u 2 1 1 1   .2 l1 l2 l 1 1 2    l1 l2 l 2. A gun is situated on an inclined plane of inclination the inclined plane are in the ratio

 . Prove that the maximum ranges up and down

1  sin  :1  sin  .

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Let u be the initial velocity. Let



be the irelination of the irclined plane.

Range on the inclined plane is

R

2u 2 sin      .cos  g cos 2 

u2 R  2sin      .cos   g cos 2   u2 R sin  2     sin   g cos 2   R will be maximum when

 (1)

sin  2    is maximum.

But maximum value of

sin  2    1

1  Maximum range on the irelined plane} l1 

u2 1  sin   g 1  cos 2  

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l1 

u2 1  sin   g 1  sin 2  

l1 

u 2 1  sin   g 1  sin  1  sin  

l1 

u2 g 1  sin  

 (1)

u2 III Maximum range down the inclined plane l2  g 1  sin   by

 (2)

g 1  sin   (1) l1 u2 ;   (2) l2 g 1  sin   u2 

l1 1  sin   l2 1  sin 

3. Show that for a projectile to reach a point P at a distance a and height b, the velocity of projection must not be less than c



g b  a 2  b2



Let o be the point of projection & u be the initial velocity. Let P be a point at a distance a from o and at a height b above the horizontal.

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ˆ  xoP The maximum range on the inclined plane is R1 

u2 g 1  sin  

R1.g 1  sin    u 2  b  u 2  R1.g 1    R1  u 2  g  R1  b  u2  g



a 2  b2  b





u  g b  a 2  b 2



This is the minimum value of u.

 The particle cannot reach P if the initial velocity of projection is less than 4. A particle is projected with a velocity u at an angle inclination

.





g b  a 2  b2



and it strikes at right angles an inclined plane of

Prove that

(i) cot   2 tan

  

(ii) time of flight is

2u g 1  3sin 2 

(iii) range on the inclined plane is

2u 2 sin  g (1  3sin 2  )

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Let o be the point of projection, u be the initial velocity &



be the angle of projection.

The component of u Along the inclined plane is Along a direction

u cos     

 r to the inclined plane is g cos  .

Given, the particle strikes the inclined plane at right angles to the inclined plane. At A, the component of velocity along the inclined plane is zero. Along the inclined plane,

"v  u  at"  0  u cos       g sin .T g sin .T  u cos      T 

u cos      g sin 

 (1)

But we know that

Time of flight

T

2u sin     

 (2)

g cos 

(1) & (2) 



u cos      g sin 



2u sin      g cos 

cos  2sin       sin  cos     

 cot   2 tan(  )  (3) cot  2 cos  tan       2sin  tan      

4sin 2   cos2 

cos

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1  Time of flight T T T

T

u cos      g sin 

216

u.2sin  4sin   cos 2 .g sin  2

2u 3sin 2   sin 2   cos 2  2u

  4

3sin 2   1

Next , Along the inclined plane

"V 2  u 2  2as" 0   u cos        2g sin .R 2

 R.2g sin   u 2 cos 2      u 2 .(2sin ) 2 1 R . 2 2 (cos   4sin ) 2g sin  R

u 2. 4sin 2  . [(cos 2   sin 2 )  3sin 2 ].2g sin 

 R

2u 2 sin  g 1  3sin 2  

5. From a point on a horizontal plane, a ball is projected with a velocity u at an angle Show that it will keep rebounding for a time



to the horizontal.

2u sin  u 2 sin 2 and will have a range g 1  e  g (1  e)

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Let o be the point of projection, u be the initial velocity, &



be the angle of projection

In a projectile motion, the horizontal component of velocity u cos



remains constant throughout

the motion. Considering the impact on A, The component of velocity along the common tangent is ucos and so it is not affected by the impact. Before impact at A, the vertical component of velocity is usin and so after impact the vertical component is eusin. Similarly, after the second , third , ……….impacts, the vertical components of velocity are

e2 u sin ,e3u sin .... Time taken from O to A is

t1 

2u sin  g

Similarly Time taken from A to B is

Time taken from B to C is

t2 

2eu sin  g

2e2 u sin  g

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…. ….

 Total time taken

T  t1  t 2  t 3 ....... 2u sin  2eu sin  2e 2 u sin     ..... g g g 2u sin  1  e  e 2  .....  T g T

2u sin   1    g  1 e  2u sin  T g 1  e  T

In the horizontal direction, the component of velocity ucos remains constant. Therefor

Total horizontal range = velocity x time

R  u cos   R

2u sin  g 1  e 

u 2 sin 2 g 1  e 

6. A particle is projected from O with a velocity u at an angle  and it impinges on a smooth vertical wall

 u 2 sin 2    d  from the wall. If it returns to g  

at a distance d. Show that it returns to a point at a distance e  the point of projection after the impact, show that

eu 2 sin 2   gd 1  e  .

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Let O be the point of projection, u be the initial velocity, and  be the angle of projection. A ball projected from O, hits the wall at A and falls on the ground at B. Let OC = d and CB= d1 From O to A, the horizontal component of velocity ucos remains constant. Let

t1 be the time taken from O to A. d  u cos .t 1  t1 

d .......(1) u cos 

Before impact at A, the balls hits the wall at right angles with a velocity ucos.  After impact, it will rebound at right angles to the wall with a velocity eucos. From A to B, the horizontal component of velocity eucos remains constant If

t2 be the time taken from A to B, then d  u cos .t 2  t2 

d1 .......(2) eu cos 

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219

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220

Total time of flight

T  t1  t 2 d1 2u sin  d   g u cos  eu cos  2u sin  de  d1  g eu cos  2u sin .eu.cos   de  d1 g d1 

eu 2 sin 2  de g

 u 2 sin 2  d1  e   d g  

 (1)

After impact if the ball returns to the point of projection O, then

 u 2 sin 2  e  d  d g    u 2 sin 2  gd  e d g   eu 2 sin 2  egd  gd eu 2 sin 2  gd  egd  eu 2 sin 2  gd 1  e 

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221

UNIT â&#x20AC;&#x201C; 5

5.0

Introduction

5.1

Objectives

5.2

Moment of Inertia

5.3

Tjeore, of parallel axes

5.4

Theorem of perpendicular axes

5.5

Moment of Inertia of Standard bodies.

5.6

Motion of a rigid body

5.7

Conservation of angular momentum

5.8

Compound pendulum

5.9

Centre of Suspension and centre of oscillation

5.10

Motion of a circular disc down an inclined plane

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222

5.0 Introduction In this unit, a new concept known as Moment of inertia is introduced. Two important theorems on moment of inertia are explained. The moments of inertia of many standard bodies are derived. Some problems are also worked out. Later, the motion of a rigid body rotating about a fixed horizontal axis is discussed in detail. The kinetic energy, angular momentum and the moment of the effective forces about the axis of rotation are derived. Many examples are given. Compound pendulum and its period of oscillation are explained. The lengths of the simple equivalent pendulum are calculated for many rigid bodies. The motion of a circular disc rolling down an inclined plane is also discussed. 5.1 OBJECTIVES After studying this unit, you will know about 

The moment of inertia



The parallel axes theorem.



The perpendicular axes theorem.



The moments of inertia of many standard bodies.



The motion of a rigid body rotating about a fixed horizontal axes.



The compound pendulum.



The period of oscillation of a compound pendulum.



The length of a simple equivalent pendulum.



The motion of a circular disc rolling down a rough inclined plane.

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223 Moment of Inertia

5.2 Definition The moment, of inertia of a particle of mass m, about a line l is

mr 2 where r is the perpendicular

distance of the particle from the line.

. The moment of inertia of a rigid body about a line l is

 mr



r 2dm Where dm is the

elementary mass of the rigid body. 5.3 Theorem of Parallel axes: If I be the moment of inertia of a rigid body of mass M about a line l and if inertia about a parallel line through its centre of gravity G, then

I G be the moment of

I  IG  Md 2 where d is the distance

between the parallel axes.

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5.4 Theorem of Perpendicular axes

I x and I y be the moments of inertia of a rigid body about two perpendicular lines ox and oy in

the plane of a lamina and if

I z be the moment of inertia about a line which is perpendicular to both ox and

oy, then

Iz  Ix  I y Note: The parallel axes theorem is applicable to both two dimensional and three dimensional bodies. The perpendicular axes theorem is applicable to only two dimensional bodies. Moment of Inertia of some standard bodies. 1. To find the moment of inertia of a rod.

Let BOA be a uniform rod of length 2  . Let o be the midpoint of the rod. Let oy

 r BOA.

We will find the M.I about OY We will take o as origin and OA as x axis. Let PQ be an elementary mass of the rod. It will be a particle of thickness 8x at a distance x from o. Elementary mass of PQ is

 m   x.

Where  is the density of the rod.

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225

oy  ( x ).x 2 a

 x3  Iy      3  a  a 3  a 3  Iy         3  3  2a 3 I y  .  1 3 Let M be the mass of the rod

 M  2a.  

M 2a

M 2a 3 . 2a 3 1  2 a I y  M. 3 Iy 

Note: If AY is a line 11

to oy, then by the parallel axes theorem,

M.I about AY is I=

IG  Md 2

I= M .

a2  M .a 2 3

I y  M.

4a 2 3

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2. To find the M.I of a rectangular lamina

Let ABCD be a rectangular lamina of sides 2a and 2b. Let o be the centre of the lamina. Let ox and oy be two

 r lines parallel to the sides as shown in the figure.

We will find the M.I about ox Let PQ be an elementary mass of the lamina. It will be a thin rod of length wb and thickness at a distance x from o. Elementary mass of PQ is

 m  2b. x.

Where  is the density of the lamina.

M.I of the elementary mass about ox =

(2b x ).

b2 3

b3  M.I of the lamina about ox = I x   2 dx x  a 3 a

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x

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227

b3 I x  2 2 dx x 0 3 3 a b I x  4  dx 3 x 0 b3 I x  .4 .a. 3 a

 1

Let M be the mass of the lamina.

 M  2a.2b. Then,

 

M 4ab

M 4b3 Ix  . .a 4ab 3

1 

I x  M.  M.I about ox is I x  M.

111 M.I about oy is

If oz is an axis

b3 3

I y  M.

a2 3

 r to both ox and oy,

Iz  Ix  I y b2 a2  M. 3 3 2 2 a  b 

I z  M. Iz  M

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5.5.3 To find the M.I of a triangular lamina.

Let ABC be a triangular lamina of base a and height h. We will the M.I about the base BC Let PQ be an elementary mass of the lamina.It will be a thin rod of length PQ and thickness



distance x from BC. Elementary mass of PQ is Where





m =PQ.  x. 

is the density.

M.I of the elementary mass about BC = (PQ.  x.  ).x

Therefore, M.I of the triangular lamina about BC =

 PQ..x dx 2

x 0

ABC III r APQ. 

BC h  PQ h  x

BC  h  x  h a h  x PQ  . h PQ 

 1 

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x at a

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x 0

I

229

a h  x h

.x 2 .dx.

a h  x3 x 4  h.   dx h x 0  3 4 h

a  x3 x 4  I    h.   h 3 4 0

a  h3 h 4  I    h.   h 3 4  a h4 I  . . h 12 a I  . h 3   2 12 Let M be the mass of the triangular lamina.

1  M  a.h. 2 2M   ah

 2  2M a 3 . .h ah 12 h2 I  M. 6 I

4. To find the M.I of a circular ring

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Let O be the centre and a be the radius of a circular ring. Let OX and OY be two perpendicular lines in the plane of the ring. Let OZ be perpendicular to both ox and oy. We will find the M.I about oz Let PQ be an elementary mass of the ring. Elementary mass of PQ=  m. M.I of the circular ring about OZ 

 m.a

I z  a 2   m  I z  a 2 .M Where M is the mass of the ring.

 Iz  M.a 2

 1

I x , I y be the M.I of the circular ring about ox and oy. Since the ring is symmetrical about ox and oy, we get

Ix  Iy . By the perpendicular axes theorem,

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Iz  Ix  I y M.a 2  I x  I x  2I x  M.a 2 a2 III I y  M. . 2 by

 M.I of the circular ring about a radius  M.

a2 . 2

M.I of the circular ring about a line through the centre perpendicular to the plane of the ring

 M.a 2

5.5.5 To find the M.I of a circular lamina Let O be the centre and ‘a’ be the radius of the circular lamina. Let ox and oy be two perpendicular lines in the plane of the lamina.

We will find the M.I about ox Let PQ be an elementary mass of the circular lamina. It will be a thin rod of length 2y and thickness Elementary mass of PQ is

x

at a distance x from O.

m  2y.x.

Where  is the density

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M.I of the elementary mass about ox

M.I of the circular lamina about ox

232

  2y.x.  .



x  a

y2 3

2 3 .y .dx. 3

2 a I x   y2dx 3 x  a

 1

On a circle,

x 2  y2  a 2 x  a cos  y  a sin  dx  a sin d x  a, a  a cos 

1  cos    x = a;

a= a cos

  0



2 0 3  1  I x    a sin    a sin   d 3 

   2 I x  a 4  sin 4 d  0 3  2 I x  a 4 .2  2 sin 4 .d  0 3 4 3 1  I x  . .a 4 . . . 3 4 2 2 4 a I x  .   2 4

Let M be the mass of the lamina. Then,

M  a 2 .  

M a 2

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ENGINEERING MECHANICS   2  Ix 

M a 4 . a 2 4

I x  M.

a2 4

 M.I of a circular lamina about a radius  M. Let oz be a line

233

a2 . 4

 r to both ox and oy

Then M.I about oz is

Iz  Ix  I y a2 a2  M. 4 4 2 a I z  M. 2 I z  M.

6. To find the M.I of an elliptical lamina.

Let o be the centre of the elliptical lamina. Let 2a, 2b be the lengths of the major and minor axes. Let ox and oy be two perpendicular lines in the plane of the lamina as shown in the figure. We will find the M.I about ox.

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Let PQ be an elementary mass of the lamina. It will be a thin rod of length 2y and thickness

 x , at a

distance x from O. Elementary mass of PQ is

m  2yx..

Where  is the density.

y2 M.I of the elementary mass about ox   2yx  . 3 M.I of the elliptical lamina about ox 

2y3 x a 3 dx a

2 a I x   y2dx 3 x  a

 1

On the ellipse,

x 2 y2   1. a 2 b2 x  a cos  y  b sin  dx  a sin d x  a gives

a  a cos 

1  cos   x  a gives

a  a cos 1  cos  0

1  I x  .

2 0 3  bsin   a sin  d   3

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ENGINEERING MECHANICS  2 I x  . ab3  sin 4 d 0 3  2 I x  . ab3 .2 2 sin 4 .d 0 3 4 3 1  I x  . ab3 . . . 3 4 2 2 a I 2  .. b3   2 4

Let M be the mass of the elliptical lamina,

M  ab.  

M ab

 2  M a .. .b3 ab 4 b2  I x  M. . 4 Ix 

IIIby , M.I about oy is I y  M.

a2 . 4

Let oz be a line perpendicular to both ox and oy. Then, M.I about oz is

Iz  Ix  I y b2 a2  M. 4 4  a 2  b2  Iz     4  I z  M.

5.5.7 To find the M.I of a solid sphere Let O be the centre and ‘a’ be the radius of a solid sphere. Let ox and oy be two perpendicular lines through o. We will find the M.I about ox.

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235

ENGINEERING MECHANICS

236

Let PQ be an elementary mass of the solid sphere.

x

It will be a circular lamina of radius y and thickness Elementary mass of PQ is

at a distance x from o.

m  y2 .x. .

Where  is the density

M.I of the elementary mass about ox

M.I of the solid sphere about ox

Ix 



  y2x  .

x  a



 a  y4dx 2 x  a

y2 2

y4 dx 2  1

x  a cos  y  a sin  dx  a sin d x  a gives

a  a cos 

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1  cos   a  a cos 

x  a gives

1  cos  0

 1   0 4  a sin   .  a sin d    2  5   .a  sin 5 d  0 2  5  2 5  .a .2 sin d 0 2 4 2  a 5 . . .1 5 3 8  .a 5   2 15

Ix  Ix Ix Ix Ix

Let M be the mass of the solid sphere . Then,

4 3 a  3 3M   4a 3 M

 2  Ix 

3M 8 . .a 5 3 4a 15

2 I x  M. .a 2 5

2  M.I of a solid sphere about a radius  M. .a 2 5 To find the M.I of a hollow sphere Let o be the centre and ‘a’ be the radius of a hollow sphere.

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238

s Let ox and oy be two perpendicular lines through o. We will find the M.I about ox Let PQ be an elementary mass of the hollow sphere. It will be a circular ring of radius y and thickness Elementary mass of PQ is

s

at a distance x from o.

 2ys.

Where  is the density.

  2ys  .y2

M.I of the elementary mass about ox M.I of the hollow sphere about ox 

I x  2



x  a

2y3ds

y3ds

 1

x  a cos  y  a sin  s  a ds  ad x  a gives

a  a cos 

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239

1  cos   a  a cos 

x  a gives

1  cos  0  1  I x  2

 a sin   0

.ad

I x  2.a 4 



 0

sin 3 d



I x  2.a 4 .2  2 sin 2 d 0

2 I x  2.a 4 .2. .1 3 8 4 I x  a    2 3 Let M be the mass of the hollow sphere Then, .

M  4a 2 .  

M 4a 2

 2  Ix 

8 4 M a . 3 4a 2

2 I x  M. a 2 3 2  M.I of the hollow sphere about a radius  M. a 2 3 To find the M.I of a solid cone

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240

Let O be the vertex Ox be the axis  be the semi-vertical angle r be the radius and

h be the height of the solid cone

We will find the M.I about ox Let PQ be an elementary mass of the cone. It will be circular disc of radius y and thickness a distance x from o. Elementary mass of PQ is

m  y2 .x.

Where  is the density

M.I of the elementary mass about ox

M.I of the solid cone about ox

Ix 



x 0

  y2x  . 

y2 2

y4 dx 2

 h 4 y dx 2 x 0

y    tan   x y  x tan  

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x

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241

 h 4 x tan   dx   2 x 0 h  Ix  tan 4   x 4dx x 0 2 Ix 

 x5   Ix  tan 4    2  5 0  h5 tan 4 . 2 5  I x  . .h 5 tan 4  10 Ix 

 1

Let M be the mass of the cone

1 M  r 2 h. 3 3M  2 r h

1  3M  5 . h tan 4  2 r h 10 3 4 I x  M. h tan 4  10.r 2 3 4 I x  M. h tan   2  10.r 3 I x  M. 2 r 4 10r Ix 

r    tan   h h tan   r 

Next, we will find the M.I about oy Elementary mass of PQ is

m  y2x.

M.I of the elementary mass about oy

 y2x..

y2  y2x.x 2 4

 y4     x 2 y2 dx x 0   4 

 M.I of the cone about oy I y  

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Iy  

  tan

242 4

  4 tan 2  

x 0

x 4 dx h

 x5  1 I y  . tan 2   tan 2   4  .   4  5 0

 3M h5 I y  .tan 2 . 2  4  tan 2   . 4 r h 5 3 Iy  M h 2 tan 2   .  4  tan 2   .h 2 2  20.r 3 2 2 I y  M. r .h  4  tan 2   2 20.r 3 I y  M. h 2  4  tan 2   20 5.6 Motion of a rigid body A rigid body rotates a fixed horizontal axis passing through the body. To find i) the kinetic energy of the rigid body ii) the angular momentum and iii) the moment of the effective forces about the axis of rotation Let a rigid body of mass M rotate about a fixed horizontal axis oz. Let OA be the fixed vertical line through O Let G be the centre of gravity of the rigid body.

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Let m be an elementary mass of the rigid body at P where OP = r. In a rigid body, the distance between any two fixed points remains constant throughout the motion. Therefore for all the positions of the rigid body,

ˆ   is constant. GOP

When the rigid body rotates about the axis oz at any time t, let

ˆ  AOG

OZA is a fixed plane in space and OZG is a fixed plane in the rigid body. 

ˆ  is the angle between these two fixed planes.. AOG As the rigid body rotates about oz, at any time t, the angular velocity of P 

  The angular velocity of P 

d    dt

d d  dt dt

d d 0 dt dt

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ENGINEERING MECHANICS [Since  is constant,

244

d  0] dt

d2 The angular acceleration of P  2      dt d 2 d 2  2  2 dt dt 2 d  2 0 dt d 2  2 dt  All the particles of the rigid body rotate with the same angular velocity

acceleration

d and same angular dt

d 2 dt 2

i) To find the Kinetic energy

When the rigid body rotates about the axis oz,P will move along a circle with centre o and radius r.

 It will have a velocity r 

d , along the tangent at . dt

 Kinetic energy of the elementary mass at }=

1 d 2 m( r ) 2 dt

[If a particle of mass m moves with a velocity v, then its K.E=

1 2 mv 2

1  d   Kinetic energy of the rigid body}   mr 2   2  dt 

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245

1  2   mr 2  2 1  2 MK 2 2  K.E of the rigid body 

1 MK 2 2 2

Where MK is the moment of inertia of the rigid body about the axis of rotation oz. Here K is called the radius of gyration. ii) To find the angular momentum. When the rigid body rotates about the axis oz, the elementary mass at  moves along a circle with centre o and radius r.

 It has a velocity r

d along the tangent at P. dt

 Momentum of the elementary mass at P = mr

d dt

[If a particle of mass m moves with a velocity v, then its momentum is mv]. Taking moments about o, the moment of momentum of the elementary mass about oz} =”Force  distance” =r.mr

 Moment of momentum of the rigid body about oz=  mr 2 =

d dt

 mr 2 2 =  ( mr

)

=  MK

 Angular momentum of the rigid body = MK 2 iii) To find the moment of the effective forces about the axis of rotation. When the rigid body rotates about the fixed horizontal axis oz,  moves along a circle with centre o and radius r.

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d ) 2 d dt  It has a tangential acceleration r 2 and normal acceleration r ( dt (

We will find the moment about oz. Since the normal acceleration

r .2 passes through o, its moment about oz=0.

 d 2   Moment of the effectice forces acting on the elementary mass about oz  r.  mr 2  dt  

  mr 2

 Moment of the effective forces acting on the rigid body about oz

d 2 dt 2

d 2   mr 2  dt 2  Mk 2 

 Mk 2 .

 Mk 2.

 Moment of the effective forces about oz i) Kinetic Energy 



1 Mk 2 2

ii) Angular momentum  Mk



iii) Moment of the effective forces about oz = Mk 5.7



Conservation of Angular Momentum. If a rigid body rotates about a fixed horizontal axis, and is the sum of the moments of the

external forces about its axis of rotation is zero, then its angular momentum about its axis of rotation is constant. Proof:

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Let a rigid body of mass M rotate about a fixed horizontal axis oz passing through the body. Then, the moment of the effective forces acting on the rigid body about its axis of rotation is

MK 2 . Here

MK 2 is the moment of inertia of the rigid body about oz.

 The equation of motion of the rigid body is

MK 2  L...........(1) Where L is the sum of the moments of the external forces about oz.

d ( MK 2 )  L.......(2) dt

(1) 

Here

MK 2 is the angular momentum of the rigid body about oz.

i.e 1 When a rigid body rotates about a fixed horizontal axis, its rate of change of angular moments of the external forces about the axis of rotation oz. If L=o,then

(2) 

d ( MK 2 )  o dt

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 MK 2  Constant.  When a rigid body rotates about a fixed horizontal axis oz, if the sum of the moments of the external forces acting on it about its axis of rotation is zero, then its angular momentum about its axis of rotation remains constant. This is known as the principle of Conservation of angular momentum. 5.7.1

Problems

1. A uniform rod of length 2a is free to turn about its one end, which is fixed. When it is hanging vertically, it is projected with an angular velocity w. In the horizontal position, if it comes to rest, prove that

w

3g 2a

Let OA= 2a, the length of the rod G- centre of gravity. M- mass of the rod. Let the rod rotate about a fixed horizontal axis oz. When the rod is hanging vertically, in the position OA, it is projected with an angular velocity w. At any time t, let

OA be the position of the rod. Let G be the position of the centre of gravity at that

instant. Let

GM r OA .

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GM  h and A^OG   .

Let Now,

Loss in kinetic energy= Work done. Initial K.E-Final K.E=Force  distance moved.

1 1 MK 2 2  MK 2 2  Mg  h.....(1) 2 2 Where

MK 2 is the moment of inertia of the rod about oz. 4a 2  Mk  M . 3 2

 (1)  1 MK 2 ( 2   2 )  Mgh 2

[In

OMG,cos  

OM a

OM  a cos  ]

1 4a 2 2 .M . (   2 )  Mgh 2 3 2a 2 2 (   2 )  g (OG  OM ) 3

2a 2 2 (   2 )  g (a  a cos ) ___(1) 3 When the rod reaches the horizontal position, its angular velocity becomes zero. i.e, then

  90 ,.   0

 (1)  2a 2 2 (  o)  g (a  a cos90 ) 3 2a 2 2   g.a 3. 3g  2  2a

(cos90  0)

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 

250

3g 2a

2. A uniform rod of length 2a is at rest hanging from one end. An angular velocity w about a horizontal axis through the fixed and is communicated to it.



If it just makes complete revolutions, prove that

3g a

Let OA= 2a, length of the rod. G- centre of gravity. M- mass of the rod. Let the rod rotate about a fixed horizontal axis oz. When the rod hangs vertically in the position OA, it is projected with an angular velocity .

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251

OA be the position of the rod. Let G be the position of the centre of gravity at that

instant. Let

GM r OA.

Let GM= h and

A^OG   .

Now, loss in K.E= work done. Initial K.E-Final K.E=Force  distance moved

1 MK 2 2  1 MK 2 2  Mg  h......(1) 2 2 Where

MK 2 is the moment of inertia of the rod about oz.

4a 2  MK  M . 3 2

(1) 

1 4a 2 2 1 4a 2 2 M. w  M.   Mg.h 2 2 3 3 2a 2 2 ( w   2 )  g.GM 3

2a 2 2 ( w   2 )  g (OG  OM ) 3 2a 2 2 ( w   2 )  g (a  a cos  ) 3 2a 2 2 ( w   2 )  ga(1  cos  ) 3 3g w2   2  (1  cos ) ___(2) 2a

If the rod has to make a complete revolution, when the rod reaches the upward vertical position , ie in the position OB,

 0

   ,  0.

 (2) 

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3g 1  cos   2a 3g  2  1  (1)  2a 3g  2  .2 2a 3g 2  a

2  0 

 

3g a

3. A uniform rod of length 2a can turn freely about one end. If it is released from the horizontal position, prove that its angular velocity when it is first vertical is

Let

3g 2a

OA  2a, be the length of the rod

G  centre of gravity M  Mass of the rod Let the rod rotate about a fixed horizontal axis oz. When the rod is in the horizontal position OA, it is held at rest and then released. At any time t, let Let

OA be the position of the rod.

G be the position of the centre of gravity at that instant.

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Let OB be the downward vertical position of the rod. Let

GM  OB .

Let

ˆ  . BOA

Let

OM  h .

Now, Loss in K.E = Work done Initial K.E – Final K.E = Force  distance moved

1 0  MK 22  Mg  h  2 1 MK 2 2  Mgh  1 2 where

MK 2 is the moment of inertia of the rod about oz.  MK 2  M.

 1 

4a 2 3

1 4a 2 2 .M.   Mg.a cos  2 3 2 

3g cos  2a

When the rod becomes vertical,

  0,let    3g .cos 0 2a 3g 2  .1 2a

2 

 

3g 2a

4. A uniform circular disc of radius ‘a’ and mass m can revolve about a fixed horizontal axis perpendicular to its plane at a distance ‘b’ from the centre. Find its Kinetic energy.

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Let C be the centre and a be the radius of the circular disc. Let the disc rotate about a fixed horizontal axis oz. Let oc = b. When the rod rotates about the fixed horizontal axis oz, Kinetic energy of the disc

Where

1  MK 22 2

 1

MK 2 is the moment of inertia of the disc about oz. Let

Cz be an axis parallel to oz.

We know that M.I of the disc about

Cz  M.

a2 . 2

By parallel axis theorem, M.I of the disc about OZ= M.I about

Cz + M.b2 a2  M.b 2 2  a 2  2b2 

MK 2  M. MK 2  M

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a Kinectic energy of the disc  M

5.8

255 2

 2b 2  2

2

COMPOUND PENDULUM A rigid body oscillating freely under gravity about a fixed horizontal axis is called a compound

pendulum. B.W To find the period of oscillation of a compound pendulum

Let a rigid body of mass M oscillate about a fixed horizontal axis oz. Let G be the centre of gravity. Let OA be the fixed vertical line through o. Let OQ=h and

GM  OA.

When the rigid body oscillates about oz, at any time t, let

ˆ  . AOG

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The only external force acting on the rigid body is its weight

Mg  atG.

Taking moments about oz, Where

MK 2 is the moment of inerti of the rigid body about oz.

MK 2  Mg.GM Where In

 1

MK 2 is the moment of inertia of the rigid body about oz.

OGM, GM OG GM sin   h  GM  h sin  sin  

1 

MK 2  Mgh sin 

For small values of

  2

 in radiands,

sin    nearly

 2  

MK 2  Mgh gh  K2 d 2 gh  2  2 dt K



  3

This equation is of the form

d2 x  n 2 x 2 dt

 This is a simple harmonic motion.

 Its period T 

2 n

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T

257

2 gh K2

K2 period T  2 gh 5.8.2

Simple equivalent Pendulum

A simple pendulum whose period of oscillation is equal to that of a compound pendulum is called a simple equivalent pendulum. For a simple pendulum of length l, Period T  2

l g

 1

K2 For a compound pendulum T  2 gh comparing

  2

1 ,  2 

we get

K2 l h

 Length of the simple equivalent pendulum

l

K2 . h

5.9 Centre of suspension and centre of oscillation Let a rigid body oscillate about a fixed horizontal axis oz.

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Let G be the centre of gravity. Let OG=h Let Let

MK 2 be the moment of inertia of the rigid body about oz. O be a point on OG, such that

i.e., OO 

K2 . h

Then, o is the centre of suspension and

O is the centre of oscillation. 5.9.1 To show that the centre of suspension and centre of oscillation are interchangeable. Let a rigid body oscillate about a fixed horizontal axis oz..

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ENGINEERING MECHANICS

259

Let o be the centre of Suspension, G be the centre of gravity And

O be the centre of oscillation.

Let OG=h

OO 

K2 h

 1

Mk 2 is the M.I of the rigid body about oz.

Where Let

MK 2 br yhr M.I of the rigid body about a parallel axis GZ 

Then, by the parallel axes theorem,

"I  IG  Md 2 " MK 2  MK 2  Mh 2  K2  K2  h2

 1 

OO 

  2

K2  h2 h

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ENGINEERING MECHANICS K2 OO  h h K2 OO   OG h K2 OO  OG  h 2 K OG  OG  OG.OG  K 2 Next, when

260

  3

O becomes the centre of suspension, let O be the centre of oscillation.

Then from (2), we get

OG.OG  K 2

  2

 3 &  4   OG.OG K 2  OG.OG K 2 OG 1 OG  OG  OG

 O and O coincide. i.e, When O becomes the centre of suspension, O becomes the centre of oscillation.

 The centre of oscillation and the centre of suspension are interchangeable. 5.9.2

To find the minimum period of oscillation of a compound pendulum.

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Let a rigid body oscillate about a fixed horizontal axis oz.

Period of oscillation T  2

T  2

Where

K2 gh

l g

 1

MK 2 is the M.I of the rigid body about oz.

Let OG=h. Let

O be the centre of oscillation.

Then  OO 

K2 h

K2 __(2) Let l  h Let

MK 2 be the M.I of the rigid body about a paralled axis GZ  .

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262

Then, by paralled axes theorem,

I  IG  Md 2 MK 2  MK 2  Mh 2 K2  K2  h2

l

 (2) 

  2

K2  h2 h

K2 l h h Differentiating w.r.t.h

dl K2   2 1 dh h Again differentiating w.r.t. h

d 2l 2K 2  3 dh 2 h For maximum or minimum value of l

dl 0 dh K2  2 1  0 h 2 K 0 h2 K2  h2 Kh h  K

d 2l 2K 2  3 when h  K , dh 2 K 

2 0 K

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Minimum value of l 

263

K2  h2 h K2  K2 K  2K. 

 Minimum period of oscillation

5.9.3

T1  2

K2 gh

T1  2

l g

T1  2

2K g

PROBLEMS

1. A weightless red ABC of length 2a is capable to rotate about a horizontal axis through A. A particle of mass m is attached at B and another of mass m is attached at C. Find the length of the simple equivalent pendulum. ABC=2a, length of the rod. Let the rod oscillate about the fixed horizontal axis AZ.

A particle of mass m is attached at B and another particle of mass m is attached at C.

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Let G be the common centre of gravity. Let AB=h. Taking moments about A,

2mg.AG  mg.AB  mg.AC 2mg.h  mg.a  mg.2a 2h  3a h

3a 2

 1

Considering the moment of inertia about AZ,

2m.K 2  m.a 2  m.  2a 

2K 2  a 2  4a 2 K2 

5a 2 2

  2

 The length of the simple equivalent pendulum is

K2 h  5a 2    2  l  3a     2 5a l 3 l

2. A heavy uniform rod of length 2l and mass M has a mass m attached to its one end. The whole system oscillates about a fixed horizontal axis through the other end. Show that the period of oscillation is 4

 M  3m  l vd ãWÎf. 3  M  2m  g

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Let

AC  2l, be the length of the rod.

The weight of the rod Also a weight

Mg  acts at the mid point B.

mg  acts at C.

Let AG=h. Taking moments about A,

 M  m  g.OZ  Mg.l  mg.2l  M  m  .g.h   M  2m  gl  M  2m  l h   1 M  m g Let the rod oscillate about the fixed horizontal axis AZ. Considering the moments about AZ,

4l2 2  M  m  .K  M.  m.  2l  3 2 4l  M  m  K 2   M  3m  3 2 4l  M  3m   K2    2 3 M  m 2

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ENGINEERING MECHANICS

Period of oscillation T  2

266

K2 gh

4l2  M  3m  M  m  T  2 3  M  m  .g  M  2m  .l T  4

 M  3m  l 3  M  2m  g

3. A square lamine of side 2a oscillates about a horizontal axis through one of its corners which is perpendicular to the plane of the lamina. Find the length of the simple equivalent pendulum.

Let

ABCD be a square lamina of side 2a.

Let it oscillate about fixed horizontal axis AZ. Let G be the centre of gravity. Let AG=h. Then,

AG  cos 45 AD h 1  2a 2 2a h 2 h  2a

 1

Let the Moment of inertia of the lamina about AZ. be

MK 2 .

By the theorem of parallel axes,

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267

I  IG  Md 2



2a 2 MK  M.  M. 2a 3 2a 2  6a 2 K2  3 2 8a K2    2 3 2



The length of the simple equivalent pendulum is

K2 h 8a 2 1 l  3 2a l

l

4 2a 3

4. A solid homogeneous right circular cone of height h and semivertical angle horizontal axis through its vertex.

 oscillates about a

Show that the length of the simple equivalent pendulum is

h 4  tan 2   .  5

Let o be the vertex H be the height R be the radius And



be the semivertical angle of the cone.

Let G be the centre of gravity of the cone.

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3 OG  h 4

268

 1

The cone oscillates about the fixed horizontal axis oy. We will find the M.I of the cone about oy. Let PQ be an elementary mass of the cone. It will be a circular disc of radius y and thickness Elementary mass of PQ is

x

at a distance x from o.

 m   y 2 x. p

Where  is the density. By the theorem of parallel axes,

I  IG  Md 2  y2  M.I of the elementary mass about oy}= I y     x 2 dm x 0  4  h

 y2 2 2   x  y dx x 0  4 

Iy  

 x 2 tan 2   2  x 2  ..  x tan   ..dx.  x 0 4  

Iy  

y   tan   y  x tan    x 

 tan 2   4  2 2 x2   ..x tan ..dx. x 0 4  

Iy  

 4  tan   .tan  2

.. 

x 0

x 4 dx h

 x5   I y  .  4  tan 2 x  .tan 2 .   4  5 0 I y  .

 h5 4  tan 2 x  .tan 2 .  4 5

  2

Next, let M be the mass of the cone.

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ENGINEERING MECHANICS 1  M  r 2 h. 3 1 M  .h 2 tan 2 .h. 3 3M   3 h tan 2 

269

h    tan   r h  r tan     3

 2  3M  . h 5 tan 2   4  tan 2   2 h tan  20 3 i.e., MK 2  M. h 2  4  tan 2   20 3 2  K2  h  4  tan 2     4 20 Iy 

 Length of the S.E.P. =

l l l

5.10

K2 h

K2 h 3h 2  4  tan 2   20



4 3h

h 4  tan 2   .  5

Motion of a uniform circular disc rolling down an inclined plane. A uniform circular disc of radius a rolls down a rough inclined plane. If it starts from rest, find its acceleration and the distance traveled in t seconds.

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270

a  radius o  centre

M  mass of the circular disc   angle of inclination of the inclined plane

A  Initial point of contact of the disc with the inclined plane x- distance moved in time t seconds

 - angle of rotation R  normal reaction F  frictional force B-point of contact after t seconds

A  position of the point A after t seconds

x  AB  a. The forces are acting as shown in the figure. Resolving the forces along and perpendicular to the inclined plane, we ge t

"F  ma " Mx  Mg sin   F

 1

O  R  Mg cos 

  2

The moment of the effective forces about the axis of rotation is

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  3

MK 2  F.a

MK 2 is the M.I of the circular disc about the axis of rotation.

Where

a2  MK  M. 2 2

 3  a2 M   Fa 2 a M   F 2

  4

Next,

x  a x  a x  a  

x a

 4  ax F 2a x F  M 2 M

  6

1  Mx  Mg sin   M x 

x 2

x  g sin  2

3 x  g sin  2 2g x  sin  3

 7

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271

ENGINEERING MECHANICS 1 "s  ut  at 2 " 2 1 2g x  0.t  . sin .t 2 2 3 g 2 x  t sin . 3  Acceleration x 

2g sin . 3

Distance moved in t seconds

x

g 2 t sin . 3

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273

UNIT-1 QUESTIONS Self assessment problems I 1. The magnitude of the resultant of two forces P of Q acting at a point is equal to P in magnitude. When P is doubled, prove that the new resultant is perpendicular to Q. 2. Two forces act at a point such that the vector sum and difference are perpendicular to each other. Show that the forces have equal magnitude. 3. Two forces P and Q act at a point inclined at an angle

 . When the forces are interchanged, show that

the new resultant turns through an angle  given by

 PQ   .tan . 2 PQ 2

tan

Self assessment problems II 1. Three forces P, P, 2R acting at a point are in equilibrium . If the first two forces are perpendicular show that P  2R .

Self assessment problems III 1. ABCD is a square. Forces of magnitudes 4, 3 2, 5 and 2 2 act at a point o along BA, CD, DC, AD and DB. Show that they are in equilibrium. 2. ABCDEF is a regular hexagon. Forces of magnitudes 2, a,5, b and 3 act at A along AB, CA, AD, AE and FA. If we forces are in equilibrium find a and b.

Self assessment problems IV 1. Two like parallel forces 24 gmwt and 9 gmwt are acting at A and B where AB = 22 cms. If their resultant acts at C, show that AC = 6 cms. 2. Three like parallel forces P, Q, R act at the vertices of ABC . If their resultant passes through the incentre, show that

P Q R   a b c

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274

1. A uniform bar of weight W and length 2a can turn freely about a smooth hinge at its upper end A and the lower end B is kept at a distance 2b from the vertical by a horizontal force applied at

W 4a 2  3b 2 B. show that the reaction at the hinge is 2 a 2  b2 2. A solid sphere of weight W and radius a placed on a smooth inclined plane of inclination or and a string of length l is attached to a point on the surface of the sphere, the other end of the string being fixed to a point on the inclined plane. Show that the tension in the string is

W  a  l  sin  l2  2al

Self assessment problems VI 1. A rod ABCD is supported at the points B and C such that AB = BC = CD. It tills when a weight of P kgs is suspended from A or a weight of q kgs is suspended from D. Find the weight of the rod and show that the centre of gracity divides the rod in the ratio 2p+q : p+2q. 2. ABCD is a square of side a force of magnitudes 3P, 2P, P, 2P act along AB, CB, DC and AD respectively. Find the resultant and its line of action. Answers Self assessment problems I 1. Let two forces P & Q act at a point O. then their resultant

R  PQ

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ENGINEERING MECHANICS  1

 R  PQ

Given R  P

 1 

P  PQ 2

P  PQ





P.P  P  Q . P  Q



P.P  P.P  P.Q  Q.P  Q.Q

0  P.Q  P.Q  Q.Q

0  2P.Q  Q.Q

 2P  Q.Q  0 2P  Q is perpendicular to Q When P is doubled, the new resultant 2P  Q is perpendicular to Q . 2. let two forces P & Q act at a point O.

Then Their vector sum is P  Q

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275

ENGINEERING MECHANICS Their vector difference is P  Q Given P  Q and P  Q are perpendicular







 PQ . PQ  0

P.P  P.Q  Q.P  Q.Q  0 2

P  P.Q  Q.P  Q  0 2



P Q

PQ

The magnitudes of the forces are equal 3.

Let two forces P and Q act at a point O. Let Q be the angle between them. Let OD be the sector of AOC . When the forces are interchanged, let the new resultant turn through an angle  . In AOC

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ENGINEERING MECHANICS

AOC 

     and ACO   2 2 2 2

OA AC      sin    sin    2 2 2 2

P Q      sin    sin    2 2 2 2   sin    P 2 2  Q   sin    2 2     sin     sin    PQ 2 2 2 2    PQ     sin     sin    2 2 2 2

        2 2 2 2  2 2 2 2  2sin  .cos    2 2     PQ      PQ         2 2 2 2  2 2 2 2  2.cos  .sin    2 2        

  sin .cos PQ 2 2    P  Q cos .sin 2 2  PQ 2  P  Q tan  2 tan

 tan

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278

Self assessment problems II 1. Let three forces P, P, 2R acting at a point O be in equilibrium. Given, the first two forces one equal in magnitude and  r to each other.

ABCD is a square. Take AB as x axis and AD as y axis. The forces are acting as shown in the figure. Let X and Y be the sum of the components of the forces along AB and AD. Then,

X  4cos180  3cos 270  2cos 0  5cos30  2 2 cos315

 1  X  4  1  3.0  2.1  5.0  2 2    2 X  4  4

X0

 1

Next,

Y  4sin180  3sin 270  2sin 0  5sin 30  2 2 sin 315

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 1  Y  4.0  3.  1  2.0  5.1  2 2    2  Y  0 3 0  5 2

  2

Y0

Let R be we resultant Then,

R 2  X2  Y2

R2  0  0 R 0

the system is in equilibrium. 2.

ABCDEF is a regular hexagon. The forces are acting at A as shown in the figure. Take AB as x axis and AE as y axis. Let X and Y be the sum of the components of the forces along x and y axis. Then,

X  2cos 0  a.cos 210  5cos 60  bcos90  3cos300 FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

279

ENGINEERING MECHANICS  3 1 1 X  2.1  a.     5.  b.0  3. 2 2  2 

X

12  3a 2

 1

Next,

Y  2sin 0  a sin 210  5sin 60  bsin 90  3sin 300  3 3  1 Y  2.0  a     5.  b.1  3    2  2  2 

Y

a  2b  2 3 2

  2

Since the system is in equilibrium, X =0 and y = 0

12  3a 0 2

12  3a  0

3a  12 a

12 3

a4 3 Next, Y = 0

a  2b  2 3 0 2

4 3  2b  2 3  0

2b  2 3

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ENGINEERING MECHANICS

281

b 3 ~ a4 3 & b 3 Self assessment problems IV 1. Let two like parallel forces 24 gmwt and 9 gmwt act at A and B respectively.

Let three like parallel forces P, Q, R act at A, B, C of ABC . First, the resultant of the two like parallel forces Q at B and R at C is another like parallel force Q + R. It acts at a point D on BC such that

Q.BD  R.DC

BD R  DC Q

 1

Similarly, the resultant of the two like parallel force R at C and P at A is R & P. It acts at a point E on CA such that R.CE  P.AE

CE P  AE R

  2

AF Q  FB P

  3

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282

If the resultant force passes through the incentre I of ABC , then AD, BE and CF will be the internal bisectors of the angles.

BD BA c   DC AC b BD c  DC b CE a  EA c AF b  FB a

1 ,  2 , 3  c R  ; b Q

a P  ; c R

b Q  a P

Q R R P  ;  ; b c c a

P Q  a b

P Q R   a b c Self assessment problems V

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ENGINEERING MECHANICS CB = 2b G – centre of gracity The three forces acting on the rod are (I)

Its Weight W  at G

(II)

Horizontal force P  at B

(III)

Reaction R at the hinge at A.

These three forces meet at O. By Lami’s theorem R W P   sin 90 sin  90    sin 180   

R W  1 cos 

 1

In ABC ,

AC2  AB2  BC2 AC2   2a    2b  2

AC  2 a 2  b2

  2

In OAC

CA2  OC2  AC2 CA2  b2  4  a 2  b2 

CA2  4a 2  3b2 CA  4a 2  3b2

  3

In OAC

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283

ENGINEERING MECHANICS

cos  

1 

AC OA 2 a 2  b2 4a 2  3b 2 R

W. 4a 2  3b 2 2 a 2  b2

W. 4a 2  3b 2 R 2 a 2  b2 2.

Let a be the radius and w be the weight of the sphere. Let  be the inclination of the inclined plane. Let l be the length and T be the tension in the string. In the equilibrium position, the forces are acting as shown in the figure. By Lami’s theorem T W R   sin 180    sin 180    sin     

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284

ENGINEERING MECHANICS

T W  sin  sin 

1

In AGC ,

sin  

AC GC

sin  

GC2  AG 2 al

a  l

 a2

al

l2  2al al

 1 

T

W sin   a  l  l2  2al

1. Self assessment problems VI

ABCD rod Let AB = BC = CD = x

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285

ENGINEERING MECHANICS

286

Let AB = BC = CD = x Let W be the wt of the rod Let G be the centre of gracity. When a weight of p kgs is suspended from A, the forces are acting as shown in the fig (i) Taking moments about B, P.AB  W.BG  0

~ P.x  W.BG

 1

Next when a weight of q kgs is suspended from B, the forces are acting as shown in fig (ii) Taking moments about C,

W.GC  q.CD  0

  2

qx  W.GC

1   2  P.x  q.x  W.BG  W.GC

 p  q  x  W.  BG  GC  p  q  x  W.x p  q  W 1 

  3

p.x   p  q  .BG BG 

px pq

  4

Next, AG  AB  BG

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ENGINEERING MECHANICS

AG  x 

AG 

px pq

 p  q  x  px pq

 2p  q  .x p  q

  5

Also GD  AD  AG GD  3x 

GD 

 5 ;  6

 2p  q  .x p  q

3  p  q  x   2p  q  .x

p  q

p  2q .x pq

  6

AG  2p  q  p  q  .x .x GD  p  q  p  2q

AG 2p  q  GD p  2q 2.

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287

ENGINEERING MECHANICS ABCD is a square of side a. The forces are acting as shown in the figure. Take AB as x axis and AD as y axis Let X and Y be the sum of the components of the forces along x and y axis.

X  3P.cos 0  2P.cos 270  P.cos180  2P.cos90

X  3P.1  2P.0  P  1  2P.0  1

X  2P

Y  3P.sin 0  2P.sin 270  P.sin180  2P.sin 90

Y  3P.0  2P.  1  P.0  2P.1   2

Y0

Let R be the resultant force Then

R 2  X2  Y2 R 2   2P   02 2

  3

R  2P

Let R make an angle  with x axis. Then

tan  

Y X

tan  

2P 0

tan   

0

 2

 4

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289

Taking moments about A, G=3.P.0-2P.a + P.a + 2P.0 G=-P.a The equation of the line of action of the resultant force is xy – yx = G x.0 – y.2P = -P.a 2y – a = 0 UNIT-2 QUESTIONS

Self assessment problems I 1. A uniform ladder rests in limiting equilibrium with its lower end on a rough horizontal floor and its upper end against an equally rough vertical wall. If  be the inclination of the ladder with the vertical, show that tan  

2 . 1  2

2. A ladder AB rests with A resting on the grand and B against a vertical wall, the coefficients of friction on the ground and the wall being  and  respectively. The centre of gravity of the ladder divides AB in the ratio 1 : n. If the ladder is on the point of shipping at both ends, show that the inclination to the ground is given by tan  

1  n .  n  1 

3. A uniform red AB rests within a fixed hemispherical lowl whose radius is equal to the length of the rod. If  be the coefficient of friction between the rod and the lowl, show that in limiting equilibrium, the inclination  of the rod to the horizontal is given by tan  

1  n  n  1 

4. A solid cylinder of radius r and height h rests on a rough inclined plane and the inclination of the plane is gradually increased. Show that the cylinder will ripple over before sliding, if the ratio of the diameter of the base of the cylinder to its height is less than the coefficient of friction.

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290

Self assessment problems II 1. A uniform chain of length l is suspended from two fixed points A and B in the same horizontal line so that either terminal tension is 5 times that at the lowest point. Show that the span AB must be

l 2 6



log 5  2 6



2. If T be the tension at any point P of a catenary and T0 at the lowest point C, prove that

T2  T02  W2 , where W is the weight of the arc CP of the string. 3. A telegraph wire stretched between two fixed points, distant a apart sags n in the middle.  a2 7  Prove that the tension at the ends is approximately w  n  n  where W is the weight per unit 8 6 

length of the string.

Answers

AB = 2a, length of the uniform ladder. G – Centre of gravity. R – Normal reaction at A

R - limiting friction at A

F1 - resultant of R and R

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ENGINEERING MECHANICS S – Normal reaction at B

S - limiting friction at B

F2 - resultant of S and S  - angle of friction

W – weight of the ladder   inclination of the ladder with the wall.

In limiting equilibrium, the forces are acting as shown in the figure. In OAB , by cot formula,

 m  n  cot   mcot   n cot .  a  a  cot 180

   a cot 90     a cot 

2a cot   a tan  

2cot    

a tan 

1 

2  1 2cot     cot  

1  2 2

 tan  

2 1  2

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ENGINEERING MECHANICS

AB = ladder G – Centre of gravity. R – Normal reaction at A

R - limiting friction at A

F1 - resultant of R and R S – Normal reaction at B

 'S - limiting friction at B

F2 - resultant of S and S W – weight of the ladder   inclination of the ladder with the wall.

In limiting equilibrium, the forces are acting as shown in the figure. In OAB , by cot formula

 m  n  cot   mcot   n cot  1  n  cot 90

   n cot  90     1.cot 

  n  1 tan   n tan  

  n  1 tan   n. 

1 tan 

1 

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ENGINEERING MECHANICS

  n  1 tan   n  1  tan  

n  1   n  1

 tan  

1  n  n  1

AB =a, length of the rod. G – Centre of gravity. W – weight of the rod. R – Normal reaction at A

R - limiting friction at A

F1 - resultant of R and R S – Normal reaction at B

 'S - limiting friction at B

F2 - resultant of S and S   inclination of the ladder with the wall.

Given, OA  AB  OB OAB is equilateral. FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

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ENGINEERING MECHANICS

294

ˆ  OBA ˆ  60 ˆ  AOB OAB

ˆ  60    OAB ˆ  60    OBA In OAB , by cot formula,

 m  n  cot   n cot B  mcot e a a a a    cot  90     cot  60     cot  60    2 2 2 2 a a tan    cot  60     cot  60    2  tan  

 1 1 1    2  tan  60    tan  60      

tan  

1 1  tan 60 tan  1  tan 60 tan    2  tan 60  tan  tan 60  tan  

tan  

1 1  3. 1  3     2  3  3  



 1  1  3 tan   2 

 3      3   1   3    3   



3    

tan  

1  3    3  3 2  3  3    3 2    2 3  2 

tan  

1  8  2  3   2 

 tan  

4 3  2

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ENGINEERING MECHANICS

Let r be radius, h be the height, w be the weight and G be the C, G of the cylinder. R be normal reaction and R be the limiting friction. When the cylinder is about to slide, the forces are acting as shown in the figure (ii) Let  be the angle of inclination of the inclined plane. Resolving the forces along and perpendicular to the clined plane, we get

R  W cos 

 1

R  Wsin 

  2

 2 ; 1

R W sin   R W cos 

  tan  tan   tan  



  3

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295

ENGINEERING MECHANICS when the cylinder is about to topple over, the forces are acting as shown in fig (i). let  be the angle of inclination of the inclined plane. In G AB , tan  

AB r 2r   h GA h 2

tan  

2r h

 

  4

In the cylinder slides before toppling over,

 tan   tan 

tan   tan 



2r h

In the cylinder tipples before sliding,

 tan   tan 

tan   tan 

2r  h Self assessment problems II 1.

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297

In a common catenary, Tension T  wy Given, Tension at A = 5 × Tension at C

TA  5  T0 wyA  5wc

yA  5c

 1

We know that

  2

y2  c2  s 2 At A, y  5c, s  l

 2 

24c2  c

 5c 

l  c   2

l2 4

l 4 6

 ys  Next, x  c log    c 

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ENGINEERING MECHANICS At A, y  5c, s 

298

l 2

 5c  l  2  x A  c log   c   

l   x A  c log  5   2c   xA 

 l.4 6  log  5   2.l  4 6  l

Span AB  2x A

AB  2.

AB 

l 4 6



log 5  2 6



l 52 6 2 6 log



Let A and B be the two fixed points from which a heavy, uniform chain is suspended. Let C be the lowest point of the curve and P be any point of the string. Let cp  s . Let W ne the weight per unit length of the string. Let the tangent at P make an angle  with the horizontal. The three forces acting on the portion CP of the string are (i)

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299

(ii)

Tension T0 at C along the horizontal like

(iii)

The weight ws  of the arc CP.

These three forces conur at D. Resolving the forces horizontally and vertically

T cos   T0

 1

Tsin   ws

  2

1   2 2

 T 2  cos2   sin 2    T0 2   ws 

T 2  T0 2   ws 

T2  T02  W2 Where W  ws , the weight of the portion cp 3. Span AB = a Sag = n The equation of the common catenary is

y  c cosh

x c

At A, x 

a 2

yA  c.cosh

a 2c

n  yA  c n  c cosh

a c 2c

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ENGINEERING MECHANICS

a   n  c cosh  1 2c     a 2  a 4        2c 2c n  c a        ........  1   2 4      a2  a4 n  c 2   ......... 4  8c 16  24c 

n

a2 a4  8c 384c3

 1

Since the wire is stretched, c is large and n is small. To the first approximation,

n

a2 8c

c

a2 8n

Put c 

n

a2 in the second term of (1) 8n

a 2 a 4 83 h 3   8c 384 a 6

a2 4n 3 n 2 8c 3a  4n 2  a2  n 1  2  8c  3a 



a2  4n 2  8n 1  2   3a 

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ENGINEERING MECHANICS

c

301

a 2  4n 2  1 1 8n  3a 2 

a 2  4n 2  c 1 8n  3a 2 

c

a 2 a 2 4n 2   8n 8n 3a 2

a2 n c  8n 6 Tension at A  w.yA

TA  w  c  n   a2 n  TA  w    n   8n 6   7n a 2  TA  w    approximately  6 8n 

UNIT-3 QUESTIONS

Self assessment Problems I 1. A particle moving in a Straight line with uniform acceleration describes distances a and b in successive intervals of time t1 and t 2 . Show that its acceleration is

2. For 1

2(bt1  at2 ) t1t2 (t1  t2 )

of the distance between two stations a train is uniformly accelerated and for 1

Of the distance it is uniformly retarded. It starts from rest at one station and comes to rest at another. Show that its greatest velocity is (1  1

 1 ) times its average velocity. n

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Self assessment problems II 1. A particle moves in a simple harmonic motion. Prove that f 2T2  42 v2 is constant where f is the acceleration, v is the velocity and T is the period of oscillation. 2. A particle moving in a straight line in Simple harmonic motion has displacements x1 , x2 and x3 in three Consecutive Seconds. Show that its period of oscillation is

2 seconds. 1  x1  x 3  cos    2x 2  2. A particle moves in a straight line in a simple harmonic motion with period

When it is at a distance

2 and amplitude a. n

a 3 from the mean position its velocity is increased by na. Show that 2

its new amplitude is a 3 .

Self assessment problems III 1. Two balls impinge directly and they interchange their velocities after impact. If they are perfectly elastic, show that their masses are equal. 2. Two Spheres A and B of same size lie on a Smooth horizontal circular groove at the opposite ends of a diameter. A is projected along the groove and after a time t linpuiges on B. Show that a Seconds impact will occur after a time

2t seconds. e

Self assessment problems IV 1. Two equal spheres moving with equal velocities impinge obliquely, their directions of motion making angles 30 and 90 with the line of centers at the instant of impact. If e  1 , show that after impact,

the spheres move in parallel directions. 2. A ball is dropped from the ceiling of a room. After rebounding twice from the floor, it reaches a height equal to one half that of the room. Show that the coefficient of restitution is ( 1 )

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3.11 Answers Self assessment problems I Let u be the initial velocity and f be the uniform acceleration. Given, the train moves distances a and b in two successive intervals of time t1 & t2 respect We know that

1 s  ut  ft 2 2

 1

When t  t1 , s  a

1 a  ut1  ft12 2

  2

When t  t1  t 2 , s  a  b

1 2  a  b  u  t1  t 2   f  t 1  t 2  2

  3

 2    t1  t 2    3   t 1 1 1 2 a  t1  t 2    a  b  t1  ft12  t1  t 2   f  t1  t 2  t1 2 2 1 at 2  bt1  ft1  t1  t 2   t1   t1  t 2   2  2  bt1  at 2   ft1t 2  t1  t 2   f

2  bt1  at 2  t1 t 2  t1  t 2 

2. A to B moving under constant acceleration a. Initial velocity = 0 Final velocity = v Time = t1 Distance = s1

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v  u  at

 v  0  at1

v  u  2as  v  0  2as1 2

 1  2

B to C moving with uniform velocity V. Distance = velocity X time

  3

s2  vt 2

C to D moving under constant reterdation a. Initial velocity = 0 Final velocity = v Acceleration = -a1 Time = t3 Distance = s3

v  u  at

 v  v  a 1t 3

v2  u 2  2as  0  v 2  2a1s3

  4  5

Total distance s  s1  s2  s3 Given s1 

s , m

s3 

s s  s2  m n s s s   vt 2  m n s s vt 2  s   m n

s h

 s

 6

Next, Total time T  t1  t 2  t 3

v v  t2  1 a a v v t2  T   1 a a

T

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Substituting in (6)

s s  v v v t   1   s   m n  a a  v2 v2 s s VT   1  s   a a m n s s VT  2s1  2s3  s   m n s s s s VT  2  2  s   m n m n s s VT  s   m n 1 1  VT  s 1     m n V

s 1 1 1    T m n

1 1  V  v 1     m n Self assessment problems II 1. In a simple harmonic motion, Displacement x = a Constant. If v be the velocity at a distance x, then

v2  n 2  a 2  x 2 

d2 x Acceleration 2  n 2 x dt Period T 

2 n

2  2   f 2 T 2  4  2 v 2    n 2 x     4 2 n 2  a 2  x 2   n  2 4  n 4 x 2 . 2  4 2 n 2  a 2  x 2  n 2 2 2  4 n x  4 2 n 2 a 2  4 2 n 2 x 2

 4 2 n 2 a 2 = constant. FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

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2. In a simple harmonic motion, Displacement x = a const

---(1)

At P, t  t1 , x  x1



x1  a cos nt

At Q, t  t  1,

  2 x  x2

x 2  a cos  t  1



At R, t  t  2,

x  x3

x3  a cos  t  2 



  3

  4

Next,

x1  x 3 a cos nt  a cos n  t  2   2x 2 2a cos n  t  1 a cos nt  cos  nt  2n     2a cos  nt  n   nt   nt  2n    nt  nt  2n  2 cos  .cos    2 2      2.cos  nt  n    

Period T 

cos  nt  n  .cos  n  cos  nt  n 

x1  x 3  cos n 2x 2  x  x3  n  cos 1  1   2x 2 

  5

2 n T

2  x  x3  cos 1  1   2x 2 

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3. In a simple harmonic motion, if v be the velocity at a distance x, then,

v2  n 2  a 2  x 2  When x  a

 1

3 , let vel = v. 2 2  a 3  2 1  v  n a     2      3   v 2  n 2 a 2  a 2  4   2

a2 v n . 4 2

Given, when it is at a distance a

  2

3 , its velocity is increased by na. Let a1 be the new amplitude. 2

1   v  na 

2  a 3  2  n a1     2      2

 3a 2  v 2  n 2 a 2  2vna  n 2  a12   4   n 2a 2 3a 2  2 2 2 2  n a  2vna  n  a1   4 4   a 2 2 3n 2 a 2 na   n 2 a12 4 4 3n 2 a 2  n 2 a12 a12  3a 2 a1  3a Therefore new amplitude a1  3a Self assessment problems III

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Before impact Let mass of A = m1 Mass of B = m2 Velocity of A = u1 Velocity of B = u2 After impact Let Velocity of A  v1 & Velocity of B  v 2 Given e = 1 By Newton’s law on impact

v1  v 2  e  u1  u 2  v1  v 2  1 u1  u 2  v1  v 2  u1  u 2

 1

By the principle of conservation of momentum

m1.v1  m2 .v2  m1.u 1  m2 .u 2 FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

ENGINEERING MECHANICS Given, v1  u 2 ;

v2  u 2 m1u 2  m 2 u1  m1u1  m 2 u 2 m1u 2  m 2 u1  m1u1  m 2 u1 m1  u 2  u1   m 2  u 2  u1 



m1  m 2

Therefore the masses are equal. 2.

Let O be the centre and r be the radius of the circular groove. Two balls A and B are at rest at the opposite ends of a diameter. A is projected along the groove with a velocity u. After a time t, it impinges on B. The distance traveled is r Distance = vel  time

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r  u  t u 

r t

 1

After impact, let A and B move with velocities v1 and v 2 By Newton’s laws on impacts,

v1  v 2  e  u1  u 2  v1  v 2  e  u  0  v 2  v1  eu  Relative velocity of B w.r.to A = eu. Let t1 be the time taken before the second impact. B has to make one complete revolution with this excess velocity before the second impact. Distance = vel  time

2r   v 2  v1   t1 2r  eu  t1 2r  e.

r  t1 t

2t  t1 e

t1 

t secs. e

Self assessment problems IV 1.

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Before impact Let mass of A = m Mass of B = m Velocity of A = u Velocity of B = u The directions of motion of A and B are inclined at 30 &90 with AB. After impact Let Velocity of A  v1 & Velocity of B  v 2 Let v1 , v 2 make angles  and  with AB. By Newton’s law on impact Along the common tangent,

v1 sin   u sin 30 v1 sin   u

 1

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Along the common Normal,

v1 cos   v 2 cos   e  u cos 30  0  1 3 v1 cos   v 2 cos    .u. 3 2

  3

By the principle of conservation of momentum

mv1 cos   mv 2 cos   m.u cos 30  m.0 v1 cos   v 2 cos   u.

 3 &  4  ;

 4    3 ;

1 ;  5

3 2

  4

2v1 cos   

3 3 u u t 2

2v1 cos  

3 u 3

2v 2 cos  

3 3 u u 2 6

2v 2 cos  

2 3 u 3

v 2 cos  

3 u 3

 6

v1 sin  u 6   v1 cos  2 3u tan   3   60

 8

    60 Therefore after impact A and B will move along parallel directions.

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Let h be the height of the ceiling. A ball is dropped from A. Let it hit the horizontal floor with a velocity u.

v 2  u 2  2as 

u 2  0  2gh u2 h  2g

 1

Since the ball impinges normally with a velocity u, it will rebound vertically with a velocity eu. After the second impact it will rebound vertically with a velocity e2u. Let h2 be the height reached after the second impact.

v 2  u 2  2as 

0   e 2 u   2g.h 2 2

2gh 2  e 4 u 2 2gh 2  e 4 .2gh 2h 2  e 4 h Given h 2 

h 2

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314

h  e 4 .h 2 1  e4  2 

 1 4 e  2

UNIT-4 QUESTIONS

Self assessment problems I 1. A particle is projected so as to have a horizontal range R and greatest height h. Prove that the maximum horizontal range with this velocity of projection is 2h +

R2 . 8h

2. A particle is projected from a point o and reaches a point P in t seconds. If t  be the time taken to move from P to the horizontal ground, show that the greatest height of the projectile is

1 g (t  t )2 . 8

3. A particle is thrown over a triangle from one end of a horizontal base and grazing the vertex falls on the other end of the base. If  and  be the base angles and the angle of projection is 45 prove that

tan   tan   1. Next we will see the motion of a particle projected on an inclined plane.

4.5.4

Self assessment problems II

1. The angle of elevation of a point P at a height h is . Show that in order to reach P, the initial velocity of projection must not be less than

gh 1  cos   .

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Let O be the point of projection, u be the initial velocity and  be the angle of projection. Let P be a point at a distance a and a height h from O.

ˆ  xOP The maximum range on the inclined plane is

R1 

u2 g 1  sin  

OP 

u2 g 1  sin  

h cos ec  

u2 g 1  sin  

h cos ec.g 1  sin    u 2 gh  cos ec  cos ec.sin    u 2 u 2  gh  cos ec  1  u  gh 1  cos ec  . 4.6

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Self assessment problems I 1.

Let O be the point of projection, u be the initial velocity and  be the angle of projection Horizontal range R 

Greatest height h 

u 2 sin 2 g

u 2 sin 2  2g

Maximum horizontal range R1 

u2 g

Now, 2

R2 u 2 sin 2   u 2 sin 2  1 2g 2h   2.   . . 2 2 8h 2g g   8 u sin 

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2 u 2 sin 2   u .2sin .cos   g   . 2 2 2 g g 4u sin  2



u 2 sin 2  u 4 .4sin 2 .cos 2   g g.4sin 2 



u 2 sin 2  u 2 cos 2   g g

2h 

R2 u2   sin 2   cos 2   8h g

2h 

R2 u2  .1 8h g

R2 2h   R1 8h

Let O be the point of projection, u be the initial velocity and  be the angle of projection. At any time t, let the particle be at P(x,y) Then,

1 y  u sin  t  gt 2 .........(1) 2

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T  t  t'

 Time of flight



Greatest height 

318

2u sin   t  t' g

u 2 sin 2  2g

 (1)

 (2)

(1) + (2)

1 1  2u sin   g  t  t '   .g.   8 8  g 

1 4u 2 sin 2   .g. 8 2g u 2 sin 2   2g 1 8



Greatest height  g t  t '



Let O be the point of projection and u be the initial velocity. The equation of the path is FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

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y  x tan  

319

gx 2 2u 2 cos 2 

  45 g.x 2 2u 2 cos 2 45 gx 2

y  x.tan 45  y  x.1 



2u 2 1/ 2 yx

gx 2 2u 2 .

yx



1 2

gx 2 u2

 (1)

B  h cot , h  lies on it. g  h cot   h  h cot   u2 gh cot 2  1  cot   u2

 (2)

A  h cot   h cot ,0   0   h cot   h cot   

g.  h cot   h cot  

gh 2  cot   cot    h  cot   h cot   u2 gh  cot   cot    u 2 2

 u 2  gh  cot   cot  

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gh cot 2  1  cot   gh  cot   cot   cot 2  1  cot   cot   cot  1

cot   cot   cot    cot 2  cot   cot 

cot   cot   cot 2   cot .cot   cot 2  1 1 1   tan  tan  tan .tan  tan   tan   1  tan   tan   1 UNIT- 5 QUESTIONS

Self assessment problems I 1. If a and b be external and internal radii of a hollow sphere, prove that its moment of inertia about a diameter is M.

5 5 2 a  b  . 5  a 3  b3 

2. Find the moment of inertia of a solid cube about an edge Until now , we have seen the motion of a particle, under the action of different kinds of forces. Next we will discuss the motion of a rigid body

Self assessment problems II 1. A uniform rod of length 2a can turn freely about one end. When it is in the upward vertical position, it is released. When it comes to the horizontal position, find its angular velocity. 2. A square lamina of side 2a can rotate about a fixed horizontal axis passing through one vertex perpendicular to the plane of the lamina . Find the kinectic energy of the lamina

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Self Assessment Problems III 1. An elliptic lamina is such that when it swings about one latus rectum as a horizontal axis, the other latus rectum passes through the centre of oscillation. Show that e 

1 . 2

2. A rod of mass m has a mass m attached to its one end. If the rod can oscillate about an axis passing through the other end, find the period of oscillation

Answers Self – assessment problems I 1.

Let o be the centre and a,b be the outer and inner radii of the hollow sphere. We will find the M.I about a radius ox Let us consider an elementary mass of the hollow sphere. It will be a hollow sphere of centre o, radius r and thickness  r . As r varies from b to a we will get the hollow sphere FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +91-9999554621

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Elementary mass m  4r 2r..



 23

M.I of the elementary mass about ox = 4r 2r. . r 2

M.I of the whole body about ox=

8 4 r dr r b 3



a 8 I x   r 4 dr r b 3 a

8  r5  I x     3  5 b

5 5 8 a  b  I x   3 5

 1

Let M be the mass of the hollow sphere . Then

4 4   M   a 3  b3  . 3 3  4 3 3 M a  b   3 3M     2 4  a 3  b 3 

1  5 5 8 a  b  3M Ix    3 5 4  a 3  b 3  5 5 2 a  b  I x  M. 5  a 3  b3 

2. Let 2a be the side of the cube. Let ox and oy be two perpendicular axes through the centre o.

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First we will find the M.I about ox. Let PQ be an elementary mass of the cube. It will be a square lamina of side 2a. Elementary mass of PQ is m  2a.2a.x. . Where  is the density

2a 2 M.I of the elementary mass about ox   4a x  . 3 2

M.I of the cube about ox. I x 



8 a I x  a 4  x   a 3 8 I x  . a 4 .2a 3

x  a

8 4 a dx 3

 1

Let M be the mass of the cube. Then, M  2a.2a.2a.

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 

324

M 8a 3

1  M 16 5 . a 8a 3 3 2a 2  I x  M. 3 Ix 

2  2a   M.I of the cube about the edge AB  M. a 2  M.   3  2

 M.

8a 2 3

Self assessment problems II 1.

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OA  2a, length of the rod G  Centre of gravity M  mass of the rod Let the rod rotate about a fixed horizontal axis oz. Initiallly the rod is held at rest in the upward vertical position and then released. At any time t, let OA be the position of the rod. Let the rod rotate an angle  in t secs. Let G be the position of the C.G at time t secs. Let GM  OA Let GM  h

ˆ  Let AOA Now , Loss in kinetic energy = Work done Initial kinetic energy – Final kinetic energy = Work done

1 0  MK 22  Mg.h 2

 1

Where Mk 2 is the M.I of the rod about oz

4a 2  MK  M. 3 2

 1  1 4a 2 2 .M.   Mg.  a  a cos   2 3 3g 2  1  cos   2a When the rod reaches the downward vertical position,

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  ,   w 3g 1  cos  2a 3g w2  1   1  2a  3g w 2  .2 2a

w2 

w 

3g a

Let ABCD be a square lamina of side 2a. Let the lamina rotate about a fixed horizontal axis AZ. When the lamina rotates about the axis AZ, at any time t, let  be the angle of rotation. Kinetic energy of the lamina 

1 MK 22 2

 1

Where MK 2 is the M.I of the lamina about AZ. By the parallel axes theorem,

I  IG  Md 2 2a 2  3 8a 2 2 MK  M. 3 MK 2  M.





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K.E of the lamina 

327

1 8a 2 2 M.  2 3  M.

4a 2 2  3

Self assessment problems III 1.

Let the equation of the ellipse be

x 2 y2   1. a 2 b2 Let the lamina oscillate about the latus rectum sz as a fixed horizontal axis. Let G be the ceentre of gravity. Then SG=h=ae

 1

Let Mk 2 be the M.I of the lamina about sz.

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ENGINEERING MECHANICS By parallel axis theorem,

I  IG  Md 2 a2 2  M  ae  4 1  k 2  a 2   e2    2 4  Mk 2  M.

Given that the other focus

ss1 

k2 h

1  a 2   e2  4  2ae   ae 1  2a 2 e 2  a 2   e 2  4  1 2 e 4

2e 2  1 4 1 e 2 e2 

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329

B be the midpoint of the rod. The weight of the rod mg  acts at B. Another weight 2mg  is attached to C Let Gbe the common centre of gravity. AG=h Taking moments about A, 3mg.h=mg.a+2mg.2a 3h=a+4a

h

5a 3

 1

Let the rod oscillate about the fixed horizontal axis AZ.

Considering M.I about AZ,

4a 2 2  2m  2a  3 28a 2 3mk 2  m. g 3m.k 2  m.

k2 

28a 2 g

  2

Period of oscillation T  2

T  2

28a 2 3  gg 5a

T  4

7a 15g

k2 gh

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