Pseudo force

Pseudo Force

Dr Rajeev Tyagi

Activity 1: A simple pendulum moved with linear acceleration Activity 1b: A simple pendulum moved with two accelerations (ex. non-uniform circular motion) Activity 2: A duster on an accelerated inclined book Activity 3: A duster on an accelerated horizontal book

m

aT

To a person sitting in the train, it appears as if a force is applied to the bob in the backward direction (Note: there is no difference between truck at rest and the one moving with constant speed) a. This force is called Pseudo force (as there is actually no such force acting) b. It acts opposite to acceleration of system c. Its value for a mass M is (M × acceleration of system) d. Only the person in train (non-inertial frame) will observe it e. In FBD if you include pseudo force then acceleration obtained will be relative acceleration f. If you do not include pseudo force then acceleration obtained is actual. To obtain relative acceleration use aAB = aA – aB Q1. In the diagram given above, find a) angle that the string makes with vertical, b) tension in the string.

Q2. In a carriage, how do you differentiate whether the tilt is caused due to pseudo force or due to vehicle standing on incline? (Ans: The value of T will be different; T = Mg for tilt, T = M√(a2 + g2) for pseudo force.

Exercise related to Activity 2: Case (1) μ = 0 Q 1. Find value of aB if m is stationary aB = g tanθ Q 2. What is the value of normal reaction?

N = mg / cosθ

m M aB

Case (2) μ ≠ 0.

Q. Find the minimum and maximum value of aB for the block to remain at rest.

Note N ≠ m2g / cosθ, also N ≠ m2g cosθ, Ans: N = m2aB sinθ + m2g cosθ

Pseudo Force by Dr. Rajeev Tyagi. +91-9810244396, rajeev_tyagi@rediffmail.com

Exercise related to Activity 3: (Question on mass kept on accelerated truck, or force applied to lower block, with another kept on it) If the acceleration of truck (or of lower block) is given, then the situation is easier!

Q. In the case mentioned below, find a) time taken for the block to slip off the truck / lower block, b) the actual distance moved by block before it slips.

M1 1 M2

P

2 Case 1: 2 = 0, A force P is applied to M2 First assume as if the entire system moves as one piece [If later (after ‘check’) you find that it is not so, then re-do the question]. The acceleration of system will be – a2 = P / (M1 + M2) a) What is the condition for M1 to slide on M2? M1 a2 > 1S M1 g b) If it is NO, then M1 does NOT slide on M2 and a1 = a2 = P / (M1 + M2) If YES (sliding occurs) Then FBD for M1 is M1

1k(M1 g)

So, actual acceleration of M1 is, a1 = 1K g Acceleration of M1 relative to M2 is found by considering pseudo force too in its FBD M1 a2 M1 1k(M1 g) Else use aAB = aA – aB The FBD for M2 is

1k(M1 g) P

So a2 =

P  1K M 1 g M2

Case 2: 2 ≠ 0 Question 31, pg 99 HCV1 – Do it by both methods – considering pseudo force in the FBD of m2 and without it.

M1 has length L. Find time for smaller block to fall down from the bigger block.

m2 μ/2 μ

Questions from H C Verma 1 on Pseudo Force Ex. 4-pg 92; Ex. 9-pg 94; Q27, 28, 31 -pg 99

Pseudo Force by Dr. Rajeev Tyagi. +91-9810244396, rajeev_tyagi@rediffmail.com

m1

[IIT 2006, (+6,0)] A close circular disc with a (rectangular groove along its diameter is placed horizontally. A block of mass 1 kg is placed as shown. The co-efficient of friction between the block and all the surfaces of groove in contact is μ = 2/5. The disc has an acceleration of 25 m/s2. Find the acceleration of block with respect to disc. (The angle of groove with horizontal line is 37o).

θ=37o

[IIT 2000, 10M] In the figure masses M1, M2, and M are 20 kg, 5 kg and 50 kg respectively.

The coefficient of friction between M and ground is zero. The coefficient of friction between M1 and M and that between M2 and ground is 0.3. The pulleys and the strings are massless.

The string is perfectly horizontal between P1 and M1 and also between P2 and M2. The string is perfectly vertical P1 and P2. An external horizontal force F is applied to the mass M. Take g = 10. a) Draw free body diagram of mass M, clearly showing all the forces. b) Let the magnitude of the force of friction between M1 and M be f1 and that between M2 and ground be f2. For a particular force F it is found that f1 = 2 f2.

Find f1 and f2. Write equations of motion of all the masses. Find F, tension in the string and acceleration of the masses.

P1

M1

M

M2

F

P2

f1max= 60N, f2max= 15N, Case 1: Nothing moves - then f2 = 15, f1 = 15, so f1 = f2, not what is asked in question Case 2: M1 slides on M, M2 remains at rest - Not possible [If M1 slides f > 60, then M2 should also slide] Case 3: M1 does not slides on M, M2 slides – so f2 = 15, Case 4: Both M1, M2 slide – then f1 = 60, f2 = 15, so f1 = 4xf2, not what is asked in question Only case3 is possible i.e. f2 = 15, so f1 has to be 30 N.

Pseudo Force by Dr. Rajeev Tyagi. +91-9810244396, rajeev_tyagi@rediffmail.com

Additional Ques. 1: In the figure below, find the acceleration of block ‘m’.

m

R mA R

M mg

θ

(R+Mg)sinθ

If A is the acceleration of wedge M down the plane, then mA is the pseudo force on m ‫ ٭‬If we take Pseudo force, then force eqn will give ‘relative’ acceleration In vertical direction, relative acceleration of m wrt M is zero, so R + mA sinθ = mg  R = m(g – A sinθ) (1) In Horizontal direction acceleration of ‘m’ wrt ‘M’ is, a = mAcosθ / m = A cos θ For M, along the slope  (R + Mg) sinθ = MA ( M  m) g sin  Substituting R from (1) in (2) A  M  m sin 2 

(2)

θ

A cosθ

The block ‘m’ is having two accelerations A So net acceleration of ‘m’ is a

2 net

= A  A cos   2( A)( A cos  )cos(180   ) 2

2

2

What if both the surfaces have friction?

R + mA sinθ = mg  R = m(g – A sinθ) will remain same mAcosθ – μR = m x amMx (R + Mg) sinθ –μN1 = MA, where N1 = (R + Mg) cosθ Q2. The block of mass ‘m’ slides on wedge of mass ‘M’ which in turn slides backward on the horizontal surface. Find the acceleration of m w.r.t. M. What if all surfaces have friction?

N2 N1 mao

m

ao

N1

M

mg θ

Mg

For ‘m’, along the slope mao cosθ + mg sinθ = ma where ‘a’ is the acc of block w.r.t. wedge So, a = ao cosθ + g sinθ for ‘m’, perpendicular to slope,  N1 + mao sinθ = mg cosθ For wedge  N1 sinθ = Mao

Now there are 3 equations, and 3 variables N1, a0, a

If all surfaces have friction: mao cosθ + mg sinθ – μN1 = ma N1 sinθ – μN2 = Mao Pseudo Force by Dr. Rajeev Tyagi. +91-9810244396, rajeev_tyagi@rediffmail.com