The document discusses circular motion and centripetal acceleration. It explains that an object moving in a circle at constant velocity has acceleration because its velocity vector is constantly changing direction towards the center of the circle. It defines centripetal acceleration and describes how to calculate it using an object's velocity and radius of the circular path. Finally, it states that centripetal force is required to cause the centripetal acceleration and examples of this force include tension or friction.

1. Motion in Two Dimensions

2. Section 6.2
Circular Motion
 How do I explain why an object that is moving in a
circle at constant velocity has acceleration?
 How can I calculate centripetal acceleration?
 How can I identify the forces that cause centripetal
acceleration?

3. Uniform Circular Motion
 A mass being rotated at a constant velocity

4. Uniform Circular Motion
 Velocity is constant
 The velocity vector is
constantly changing
direction
 This change in direction
provides an acceleration
towards the center of the
circle

5. Centripetal Acceleration


6. Circular Velocity
 What if they do not give you velocity in m/s?
 How about 33 rpm?
 T = Period = the time needed to turn 1 complete
revolution
 What is the T for a 33 rpm record?

7. Centripetal Acceleration


8. Centripetal Force
 Since the acceleration points towards the center, the
centripetal force points towards the center
 F = ma, so Fc = ?
 Fc is the result of another force that turns the object in
a circle
 This force can be supplied by tension or friction

9. Centripetal Force
 Supplied by tension
 A 13 g stopper is attached to a 0.93 m string. The
stopper is swung in a horizontal circle, making one
revolution in 1.18 s. Find the tension force exerted by
the string on the stopper.

10. Centripetal Force


11. Homework
 Page 156
 # 12 - 15