Quick Check 1. What is the centripetal acceleration of the Moon toward Earth? Given: R = 3.84 × 108 m and T = 2.36 × 106 s
2. A skater travels at 2.0 m/s in a circle of radius 4.0 m. What is her centripetal acceleration?
3. A 20.0 g rubber stopper is attached to a 0.855 m string. The stopper is spun in a horizontal circle making one revolution in 1.36 s. What is the acceleration of the stopper?
Centripetal Force
The net force that causes centripetal acceleration is called centripetal force. Acceleration and net force are related by Newton’s second law, which says that F = ma. So the magnitude of the centripetal force can be calculated by: centripetal force Therefore,
Fc = m
Fc = mac
v2 4π 2R or Fc = m 2 R T
Sample Problem 3.1.1 — Centripetal Force In a local playground a merry-go-round is turning at 4.50 m/s. If a 50.0 kg person is standing on the platforms edge, which is 5.80 m from the centre, what force of friction is necessary to keep her from falling off the platform?
What to Think About
How to Do It
1. This is a circular motion question and the force of friction between the person and the platform is the centripetal force.
ac =
2. Find the centripetal acceleration.
ac =
v2 R ( 4.50 m s)2
5.80 m ac = 3.49 m s 2
3. Find the centripetal force, which is the frictional force keeping her on the platform as it spins.
Ffr = Fc = mac Ffr = ( 50.0 kg)( 3.49 m s 2 ) Ffr = 175 N
158 Chapter 3 Circular Motion and Gravitation
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