Transcript Document

Uniform Circular Motion
Centripetal
forces keep
these children
moving in a
circular path.
Uniform Circular Motion
Uniform circular motion is motion along a
circular path in which there is no change in
speed, only a change in direction.
Fc
v
Constant velocity
tangent to path.
Constant force
toward center.
Question: Is there an outward force on the ball?
Uniform Circular Motion (Cont.)
The question of an outward force can be
resolved by asking what happens when the
string breaks!
Ball moves tangent to
v
path, NOT outward as
might be expected.
When central force is removed,
ball continues in straight line.
Centripetal force is needed to change direction.
Examples of Centripetal Force
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.
Fc
Force ON you is toward the center.
Car Example Continued
Reaction
Fc
F’
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.
Another Example
Disappearing
platform at fair.
R
Fc
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.
Spin Cycle on a Washer
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?
2
v
v
m
m = 3 kg
ac 
R
R
R = 5 m; v = 8 m/s
2
(8 m/s)
2
ac 
 12.8 m/s
5m
mv
Fc  mac 
R
2
F = (3 kg)(12.8 m/s2)
Fc = 38.4 N
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
Fc R
mv 2
Fc 
; m 2
v = 15 m/s
R
v
Fc R
450 N
30 m
m=?
Speed skater
(450 N)(30 m)
m
2
(15 m/s)
m = 60.0 kg
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
Newton’s 2nd law
for circular motion:
m = 80 kg;
v = 4 m/s2
Fc = 600 N
2
mv
mv
F
; r
r
F
r=?
(80 kg)(4 m/s)
r
600 N
2
r = 2.13 m
2
Car Negotiating a Flat Turn
v
Fc
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.
Car Negotiating a Flat Turn
v
Fc
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.
Car Negotiating a Flat Turn
The centripetal force Fc is
that of static friction fs:
m
Fc
R
n
fs
Fc = fs
R
v
mg
The central force FC and the friction force fs
are not two different forces that are equal.
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
fs
Fc = fs
m
v
R
Fc
R
mg
The car is on the verge of slipping when FC is
equal to the maximum force of static friction fs.
Fc = fs
Fc =
mv2
R
fs = msmg
Maximum speed without slipping (Cont.)
n
Fc = fs
fs
R
mv2
R
mg
v=
m
v
Fc
R
= msmg
msgR
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?
m
v
Fc
Fc =
R
ms = 0.7
mv2
R
fs = msmg
From which: v =
msgR
g = 9.8 m/s2; R = 70 m
v  ms gR  (0.7)(9.8)(70 m) v = 21.9 m/s