Kinematics in Two Dimensions

PHY 130 FUNDAMENTAL PHYSICS 1

CHAPTER 3 K I N EM ATI C I N T WO D I M ENS IO N

LESSON OUTCOMES • M u l t i p l y o r d i v i d e a v e c t o r q u a n t i t y b y a s c a l a r q u a n t i t y.

• Use

the

direction

methods of

the

of

graphical

vector

resultant

analysis in

to

problems

determine involving

the

magnitude

vector

addition

and or

subtraction of two or more vector quantities.

• Use the trigonometric component method to resolve a vector components in the x and y direction and the vector resultant in problems involving vector addition or subt raction of t wo or more vect or quant ities.

• Use the kinematics equations along with the vector component method to solve problems involving two dimensional motion of projectiles.

VECTORS

Vectors • Physical quantities could be categorized to:

ď ą

Scalars – quantities which has only magnitude. (example: mass, volume, etc)

ď ą

Vectors – quantities which have both magnitude and direction. (example: velocity, acceleration, etc)

• Vector are usually written with their magnitude and directions

• Vectors are written in two ways: •

Method 1:

• Method 2:

đ?‘Ž (using an arrow above) đ?’‚ (using bold face print)

Example :

QUANTITIES

SYMBOLS

WRITTEN AS

Force

F

đ?‘­ or đ??š

Momentum

p

đ?’‘ or đ?‘?

Vector Representation • Graphically vector can be represented on paper with an arrowed straight line.

• Magnitude represented by the length of the line (scaled) and direction represented by the angle and arrow • Example: N

30

d = 100 m, East E

v = 20 ms-1, 30 north of east

Equality Of Vectors •

Equality of vectors – two vectors A and B are said to be equal if i. magnitude of A = magnitude of B ii. A and B are parallel or in the same direction.

A

A=B

B

A

B

AB Different directions

A

B

AB Different magnitudes

Multiplication Vector With Scalar •

Scalar multiplication – if v is a vector and k is a scalar, then when k is multiplied with v, a new vector, kv, is formed where this new vector has the same direction as v but a magnitude that is k times the initial value.

suppose k = 3 and v = 3 ms-1, North-east

•

Example:

•

Then the multiplication of k and v produces a new vector say w = 3v

•

Where w = 9 ms-1, North-east

w v

Vector Addition/ Subtraction In One Dimension

• For vectors in one dimension, simple addition and subtraction are all that is needed.

• If the motion is in two dimensions, the situation is somewhat more complicated.

• Methods of addition (considering only two vectors) 

geometric method



analytical method

Vector Addition/ Subtraction In Two Dimension •

Geometrical method: ď ą Triangle method ď ą Parallelogram method

•

Triangle method: i.

Sketch vector đ?‘Ž using appropriate scale.

ii.

Sketch vector đ?‘? using the same scale.

iii.

Placed the tail of đ?‘? at the tip of đ?‘Ž đ?‘? đ?‘? đ?‘Ž

đ?‘? =đ?‘Ž+đ?‘?

Vector Addition/ Subtraction In Two Dimension • Parallelogram method: the two vectors are joined at the tails, the resultant is the diagonal of a parallelogram formed with the two vectors.

Vectors Component • Component of vectors: Vectors can be described by its components • Consider horizontal (x) and vertical (y) components • A vector A can be written as A = Ax + Ay

A

Ay – vertical component or y-component

Ax- horizontal component or x - component

If the components are perpendicular, they can be found using trigonometric functions.

Vectors • The magnitude of the horizontal and vertical component can be obtained using Law of right-angled triangle, that is

F

Fy= Fsin θ θ Fx= Fcos θ

• Hence vector F can be written as

F  F cos ,   F sin , 

• Where the arrows indicate the direction of the component vector

Fig. 3.14, p.65

Unit Vectors • Unit vector of a vector: • Written as

vector

• Defined as

v̂ vector magnitude of vector v vˆ  v

Unit vector of a vector 

• Example: given a vector • Therefore, unit vector,

v = 10 ms-1, East

10 ms -1 , East vˆ  10  1 ms -1 , East

• A unit vector is a vector with magnitude equal to 1

Unit Vectors • Writing vectors in the i , j , k form • Let i be the unit vector in the x direction, j be the unit vector in the y direction, and k be the unit vector in the z direction. y

1 j k

1 z

i 1

x

Unit Vectors • A vector can be written in the unit vector notation, for example: • Suppose A is a velocity vector = 20 ms-1, in the x-direction

• Hence A can be written as A = 20 i ms-1 • Let B be force vector, B = 40 N, 30o from the horizontal

30o

Unit Vectors • The vector B = 40 N, 30o from the horizontal can be written in i and j form.

• Calculate the x-component and y-component of B • x-component = 40 cos 30 = 34.64 N

• y-component = 40 sin 30 = 20 N • Hence B = (34.64 i + 20 j ) N

B 20 j

34.64 i

Example 1 Component -x

Component -y

F1

6

0

F2

7 cos 120ď‚° = − 3.5

7 sin 120ď‚° = 6.062

F3

4 cos 220ď‚° = − 3.064

4 sin 220ď‚° = − 2.571

R

-0.564

3.491

The Figure shows three forces F1, F2 and F3 acting on a point O. Calculate the resultant force đ?‘… = −0.564đ?‘– + 3.491đ?‘— đ?‘ đ?‘…=

đ?‘…đ?‘Ľ2 + đ?‘…đ?‘Ś2 =

đ?œƒ = đ?‘Ąđ?‘Žđ?‘›âˆ’1

đ?‘…đ?‘Ś đ?‘…đ?‘Ľ

=−

(−0.564)2 +3.4912 = 3.536 đ?‘ 3.491 0.564

= 88.82° above negative-x axis

Example 2 An airplane trip involves three legs, with two stopovers. The first leg is due east for 620 km; the second leg is southeast (45°) for 440 km; and the third leg is at 53° south of west, for 550 km, as shown. What is the plane’s total displacement?

Component -x

Component -y

D1

620

0

D2

440 cos 45ď‚° = 311

− 440 sin 45ď‚° = − 311

D3

− 550 cos 53ď‚° = − 331

− 550 sin 53ď‚° = − 439

DR

600

− 750

đ??ˇđ?‘… = 600đ?‘– − 750đ?‘— đ?‘˜đ?‘š

đ??ˇđ?‘… =

đ??ˇđ?‘Ľ2 + đ??ˇđ?‘Ś2 =

đ?œƒ = đ?‘Ąđ?‘Žđ?‘›âˆ’1

đ??ˇđ?‘Ś đ??ˇđ?‘Ľ

=−

750 600

(600)2 +(−750)2 = 960 km

= 51° south of east

Projectile Motion • A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola.

• It can be understood by analyzing the horizontal and vertical motions separately

• Example: motion of ball kicked, golf ball hit by a club • Object given some initial velocity, then object moves under free fall or gravitational pull or free fall acceleration ( g = 9.8 ms-2) 21

Projectile Motion • In projectile motion the horizontal component and vertical component are independent of each other. y

v = vx ; vy = 0 vy

v0y

v

Vx

v0 θ v0x

x R = range (maximum horizontal distance)

22

Projectile Motion Generalizing the one-dimensional equations for constant acceleration:

Projectile Motion Projectile motion is motion with constant acceleration in two dimensions, where the acceleration is g and is down.

Projectile Motion 2

vo R sin 2 g

y

• Range equation:

θ x

R/2

R

Projectile Motion y

• Rmax when θ = 45 • H increases as θ increases θ = 45 0

H x Rmax

Projectile Motion 1.

Read the problem carefully, and choose the object(s) you are going to analyze.

2.

Draw a diagram.

3.

Choose an origin and a coordinate system.

4.

5.

Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration g.

Examine the x and y motions separately.

•

6. List known and unknown quantities. Remember that vx never changes, and that vy = 0 at the highest point.

•

7. Plan how you will proceed. Use the appropriate equations; you may have to combine some of them.

Projectile Motion – Example 1 y 20

ms-1

α vx

vy

40 R a) Find the velocity at t = 1 s: • vx = vocos θ = 20 cos 40 = 15.32 ms-1

• vy = vosin θ – gt = 20sin 40 – 9.8(1) = 3.06 ms-1 • v = vx i + vy j • v = √(15.32)2 + (3.06)2 = 15.62 ms-1 • α = tan-1 (3.06/15.32) = 11.30

o

x

Projectile Motion – Example 1 y 7.96 hmax

20 ms-1 40

15.32

R

x

b) Find the position of object at t = 1 s: x = (v0cos θ)t = (20 cos 40 )(1) = 15.32 m y = (v0sin θ)t – ½gt2 = (20sin 40)(1) - ½(9.8)12 = 7.96 m

c) Find the maximum height, hmax Max. height when vy = 0 , vy = vosin θ – gt = 20sin 40 – 9.8t = 0, t = 1.31 s

When t = 1.31 s , y = (v0sin θ)t – ½gt2 = (20sin 40)(1.31) - ½(9.8)(1.31)2 = 8.43 m

Projectile Motion – Example 2 A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance.

1 đ?‘Ś = đ?‘Łđ?‘Ś0 + đ?‘Žđ?‘Ś đ?‘Ą 2 2 1 −50 = 0 + − 9.8đ?‘Ą 2 2 đ?‘Ą = 3.19 đ?‘ đ?‘Ľ = đ?‘Łđ?‘Ľ0 đ?‘Ą

90= đ?‘Łđ?‘Ľ0(3.19) đ?‘Łđ?‘Ľ0 = 28.2 đ?‘š/đ?‘

known

unkown

x0 =y0 = 0

vx0

x= 90 m

t

y = - 50 m ax = 0 ay = - 9.8 m/s2

vy0 = 0