Centripetal forces keep these children moving in a circular path.

Uniform circular motion is motion along a circular path in which there is no change in speed , only a change in direction .
F c v Constant velocity tangent to path.
Constant force toward center.
Question: Is there an outward force on the ball?

The question of an outward force can be resolved by asking what happens when the string breaks! v
Ball moves tangent to path, NOT outward as might be expected.
When central force is removed, ball continues in straight line.
Centripetal force is needed to change direction.

You are sitting on the seat next to the outside door. What is the direction of the resultant force on you as you turn? Is it away from center or toward center of the turn?
• Car going around a curve.
F c
Force ON you is toward the center.

F c
Reaction
The centripetal force is exerted
BY the door ON you. (Centrally)
There is an outward force, but it does not act
ON you. It is the reaction force exerted BY you
ON the door. It affects only the door.

Disappearing platform at fair.
R
F c
What exerts the centripetal force in this example and on what does it act?
The centripetal force is exerted BY the wall
ON the man. A reaction force is exerted by the man on the wall, but that does not determine the motion of the man.

How is the water removed from clothes during the spin cycle of a washer?
Think carefully before answering . . . Does the centripetal force throw water off the clothes?
NO.
Actually, it is the LACK of a force that allows the water to leave the clothes through holes in the circular wall of the rotating washer.

Example 1: A 3-kg rock swings in a circle of radius 5 m . If its constant speed is 8 m/s , what is the centripetal acceleration?
v
R m v
2 a c
 m = 3 kg
R
R = 5 m ; v = 8 m/s
F c
 ma c
 mv
2
R a
 c
(8 m/s)
2
5 m

F = (3 kg)(12.8 m/s 2 )
F c
= 38.4 N
2

Example 2: A skater moves with 15 m/s in a circle of radius 30 m . The ice exerts a central force of 450 N . What is the mass of the skater?
Draw and label sketch v = 15 m/s
F c
 mv
2
R
; m

F R c v
2
450 N
F c m=?
R
30 m m

(450 N)(30 m)
(15 m/s)
2 m = 60.0 kg
Speed skater

Example 3.
The wall exerts a 600 N force on an 80-kg person moving at 4 m/s on a circular platform. What is the radius of the circular path?
Draw and label sketch
F c m = 80 kg; v = 4 m/s 2
= 600 N r = ?
F
Newton’s 2nd law for circular motion:
 mv r
2
; r
 mv
F
2 r

(80 kg)(4 m/s)
2
600 N r = 2.13 m

v
F c
R
What is the direction of the force ON the car?
Ans.
Toward Center
This central force is exerted
BY the road ON the car.

v
F c
R
Is there also an outward force acting ON the car?
Ans.
No, but the car does exert a outward reaction force ON the road.

m
F c
R
The centripetal force F c that of static friction f s
: n f s
F c
= f s
R is v mg
The central force F
C and the friction force f are not two different forces that are equal. s
There is just one force on the car. The nature of this central force is static friction.

Finding the maximum speed for negotiating a turn without slipping.
n f s
F c
= f s m
F c
R
R v mg
The car is on the verge of slipping when F
C is equal to the maximum force of static friction f s
. mv 2
F c
= f s
F c
= f s
= m s mg
R

Maximum speed without slipping (Cont.) n f s mg
R
F c
= f s mv 2
R
= m s mg v = m s gR m v
F c
R
Velocity v is maximum speed for no slipping.

Example 4: A car negotiates a turn of radius 70 m when the coefficient of static friction is 0.7
. What is the maximum speed to avoid slipping?
v m
F c
R m s
= 0.7
F c
= mv 2 f s
= m s mg
R
From which: v = m s gR g = 9.8 m/s 2 ; R = 70 m v
 m s gR

(0.7)(9.8)(70 m) v = 21.9 m/s