1. Uniform Circular Motion
Constant speed, but change direction  acceleration.
If acceleration net force.
If motion in circle at const speed, force towards the center.
Centripetal (center seeking) acceleration
ac = v2 / r

2. A puck is traveling at a constant speed around a circle on a table
A puck is traveling at a constant speed around a circle on a table. What would we have to know to find the tension in the string?

3. Clicker!
You are whirling a tennis ball on a string around in circles when the string suddenly snaps. What direction does the tennis ball fly? (the figure below is a top view).
The ball will fly along a tangential direction.

4. Clicker! v = 2pr/T = (2p * 1m )/1s = 2p m/s
You are whirling a 0.5 kg ball on a string around in circles. If the radius of the circle is 1m, and the ball takes 1s to complete a circular path, what is the speed of the ball?
1 m/s
2 m/s
4p m/s
2p m/s
4p2 m/s 1m
v = 2pr/T = (2p * 1m )/1s = 2p m/s

5. Clicker! a = v2/r = (2p2 m/s)/ 1m = 4p2 m/s^2
You are whirling a 0.5 kg ball on a string around in circles. If the radius of the circle is 1m, and the ball takes 1s to complete a circular path, what is the centripetal acceleration of the ball?
1 m/s^2
2 m/s^2
4p m/s^2
2p m/s^2
4p2 m/s^2 1m
a = v2/r = (2p2 m/s)/ 1m = 4p2 m/s^2

6. Clicker! Fc = m ac = 0.5 kg * 4p2 m/s^2 = 2p2 N
You are whirling a 0.5 kg ball on a string around in circles. If the radius of the circle is 1m, and the ball takes 1s to complete a circular path, what is the centripetal force experience by the ball?
1 N
2 N
2p2 N
2p N
4p2 N 1m
Fc = m ac = 0.5 kg * 4p2 m/s^2 = 2p2 N

7. Clicker! T = mg = 0.5 kg . 9.8 m/s^2 = 4.9 N
A puck of mass 0.25kg is tied to a string and allow to revolve in a circle of radius 1.0m on a frictionless tabletop. The other end of the string passes through a hole in the center of the table, and a mass of 0.5kg is tied to it. The suspended mass remains in equilibrium while the puck on the table revolves. What is the tension in the string?
0.25 g = 2.45 N (2) 0.75 g = 7.35 N
(3) g = 4.9 N (4) None of the above. T
T = mg = 0.5 kg m/s^2 = 4.9 N
Because the same rope is tied to the puck, the tension
is the same along the rope.

8. Clicker!
A puck of mass 0.25kg is tied to a string and allow to revolve in a circle of radius 1.0m on a frictionless tabletop. The other end of the string passes through a hole in the center of the table, and a mass of 0.5kg is tied to it. The suspended mass remains in equilibrium while the puck on the table revolves. What is the centripetal force experienced by the puck?
0.25 g = 2.45 N (2) 0.75 g = 7.35 N
(3) g = 4.9 N (4) None of the above. Fc T
Because the same rope is tied to the puck, the tension
is equal to the centripetal force of the puck.

9. Clicker! Fc = m ac . Therefore, ac = Fc/m = 4.9 N/ 0.25 kg
A puck of mass 0.25kg is tied to a string and allow to revolve in a circle of radius 1.0m on a frictionless tabletop. The other end of the string passes through a hole in the center of the table, and a mass of 0.5kg is tied to it. The suspended mass remains in equilibrium while the puck on the table revolves. What is the centripetal acceleration experienced by the puck? (Tension = 4.9 N)
19.6 m/s^ (2) 9.8 m/s^2
(3) 30 m/s^ (4) 2 m/s^2.
Fc = m ac . Therefore, ac = Fc/m
= 4.9 N/ 0.25 kg
= 19.6 m/s^2

10. Clicker! ac = v2/r v = a r = 19.6 m/s^2 * 1.0 m = 4.4 m/s
A puck of mass 0.25kg is tied to a string and allow to revolve in a circle of radius 1.0m on a frictionless tabletop. The other end of the string passes through a hole in the center of the table, and a mass of 0.5kg is tied to it. The suspended mass remains in equilibrium while the puck on the table revolves. What is the speed of the puck?
19.6 m/s (2) 4.4 m/s
(3) 30.0 m/s (4) 3.4 m/s.
ac = v2/r
v = a r = 19.6 m/s^2 * 1.0 m = 4.4 m/s

11. Clicker!
A motorcycle stunt driver performs a vertical loop the loop as shown. Which point will have the largest centripetal force ?
(5) The same centripetal force at all points. W FN FN W
At (3), Fc = FN +W
At (1), Fc = FN -W

12. Curved Paths
Curved paths can be considered as small parts of circles. ac

13. Satellites Fc = gravity is only force causing centripetal acceleration
r to center of circle
Independent of mass of satellite
Fc =

14. Example. What will be the speed of a satellite orbiting 100 km above the Earth surface?