1

Chapter (6)

Circular Motion

2

Circular MotionConsider an object moving at constant speed in a circle. The direction of motion is changing, so the velocity is changing (even though speed is constant).

Therefore, the object is accelerating.

The direction of the acceleration is toward the center of the circle and so we call it centripetal acceleration. The magnitude of the acceleration is

r

vac

2

3

Centripetal Acceleration

radiansin measured if

2 travelledDistance

ˆ)sin(ˆ)cos(

ˆ)sin(ˆ)cos(

2

1

r

yvxvv

yvxvv

• The best estimate of the acceleration at P is found by calculating the average acceleration for the symmetric interval 12.

4

r

va

r

v

θr/v

vva

θr/v

vva

θr/v

y

y

x

2

2 sin

2

sinsin

02

coscos

onAccelerati of Components

2 d/v t timeElapsed

0if

r

vac

2

Then,

5

Example: What is the centripetal acceleration of the Earth as it moves in its orbit around the Sun?

Solution:

rac

2

t

r 2But

smt

rac /1093.5

4 32

2

Then

yeart

mearthr

1

10496.1)( 11

6

Tangential acceleration

The tangential acceleration component causes the change in the speed of the particle. This component is parallel to the instantaneous velocity, and is given by

Radial acceleration

The radial acceleration component arises from the change in direction of the velocity vector and is given by

Tangential and Radial acceleration

dt

dat

raa cr

2

Note: If the speed is constant then the tangential acceleration is zero (uniform Circular Motion)

7

Total acceleration

The total acceleration vector a can be written as the vector sum of the component vectors:

rt

rt

aaa

aaa

22

Since the component perpendicular to other

8

Example: A car exhibits a constant acceleration of 0.300 m/s2 parallel to the roadway. The car passes over a rise in the roadway such that the top of the rise is shaped like a circle of radius 500 m. At the moment the car is at the top of the rise, its velocity vector is horizontal and has a magnitude of 6.00 m/s. What is the direction of the total acceleration vector for the car at this instant?

9

If the angle between

22

/072.0500

36sm

rar

222 /309.0 smaaa rt

5.13tan 1

t

r

a

a

10

Problem: A train slows down as it rounds a sharp horizontal turn, slowing from 90km/h to 50km/h in the 15s that it takes to

round the bend.  The radius of the curve is 150m.  Compute the acceleration at the train.

Problem: A particle moves in a circular path 0.4m in radius with constant speed.  If the particle makes five revolution in

each second of its motion, find (a) the speed of the particle and (b) its acceleration.

11

Centripetal Force

• A string cannot push sideways or lengthwise.

• A string in tension only pulls.

• The string pulls the ball inward toward the center of the circle

12

What if we cut the sting?The ball should move off with constant velocityThis means the ball will continue along the tangent to the circle.

13

Centripetal ForceIf there is a centripetal acceleration, then the net

force must also be a centripetal force:

r

vmmaF cc

2

14

• The Conical Pendulum• As the ball revolves

faster, the angle increases• What’s the speed for a

given angle?

Example:

15

tansin

)sin(

tan

tan

)2........(....................cos

)1.....(....................sin

2

2

Lg

Lrbut

rgv

rg

v

then

mgTr

mvT

16

Problem: I rotate a ball at an angle of 30o.  What is the centripetal acceleration?  If the string is 1 meter long, how fast is it rotating?

17

ProblemDriving in your car with a constant speed of 12 m/s, you encounter a bump in the road that has a circular cross section, as indicated in the Figure. If the radius of curvature of the bump is 35 m, find the apparent weight of a 67-kg person in your car as you pass over the top of the bump.

Nmg

a=v2/r