1. Dynamics: Circular Motion
Physics 7C lecture 04
Dynamics: Circular Motion
Thursday October 10, 8:00 AM – 9:20 AM
Engineering Hall 1200

2. Dynamics of circular motion
If a particle is in uniform circular motion, both its acceleration and the net force on it are directed toward the center of the circle.
The net force on the particle is Fnet = mv2/R.

3. What if the string breaks?
If the string breaks, no net force acts on the ball, so it obeys Newton’s first law and moves in a straight line.

4. Avoid using “centrifugal force”
Figure (a) shows the correct free-body diagram for a body in uniform circular motion.
Figure (b) shows a common error.
In an inertial frame of reference, there is no such thing as “centrifugal force.”

5. Force in uniform circular motion
A sled on frictionless ice is kept in uniform circular motion by a rope.
Follow Example 5.19.

6. A. the sled will keep moving in a circle.
Q5.11
A sled moves on essentially frictionless ice. It is attached by a rope to a vertical post set in the ice. Once given a push, the sled moves around the post at constant speed in a circle of radius R.
If the rope breaks,
A. the sled will keep moving in a circle.
B. the sled will move on a curved path, but not a circle.
C. the sled will follow a curved path for a while, then move in a straight line.
D. the sled will move in a straight line.
Answer: D

7. A5.11
A sled moves on essentially frictionless ice. It is attached by a rope to a vertical post set in the ice. Once given a push, the sled moves around the post at constant speed in a circle of radius R.
If the rope breaks,
A. the sled will keep moving in a circle.
B. the sled will move on a curved path, but not a circle.
C. the sled will follow a curved path for a while, then move in a straight line.
D. the sled will move in a straight line.

8. A conical pendulum
A bob at the end of a wire moves in a horizontal circle with constant speed.
Follow Example 5.20.

9. D. is any of the above, depending on the bob’s speed v.
Q5.12
A pendulum bob of mass m is attached to the ceiling by a thin wire of length L. The bob moves at constant speed in a horizontal circle of radius R, with the wire making a constant angle  with the vertical. The tension in the wire
A. is greater than mg.
B. is equal to mg.
C. is less than mg.
D. is any of the above, depending on the bob’s speed v.
Answer: A

10. A5.12
A pendulum bob of mass m is attached to the ceiling by a thin wire of length L. The bob moves at constant speed in a horizontal circle of radius R, with the wire making a constant angle  with the vertical. The tension in the wire
A. is greater than mg.
B. is equal to mg.
C. is less than mg.
D. is any of the above, depending on the bob’s speed v.

11. A car rounds a flat curve
A car rounds a flat unbanked curve. What is its maximum speed?
Follow Example 5.21.

12. A car rounds a flat curve
If the car is moving at speed v, what is the coefficient of friction?

13. A car rounds a flat curve
f = m v2/R
f = μ n = μ mg
we have: μ = v2/ (R g)

14. A car rounds a banked curve
At what angle should a curve be banked so a car can make the turn even with no friction?
Follow Example 5.22.

15. A car rounds a banked curve
If the car is moving at speed v, what is the tilt of the track?

16. A car rounds a banked curve
n cos β = mg
n sin β = m v2/R
we have: tan β = v2/(R g)

17. Uniform motion in a vertical circle
A person on a Ferris wheel moves in a vertical circle.
Follow Example 5.23.

18. Q5.13
A pendulum of length L with a bob of mass m swings back and forth. At the low point of its motion (point Q ), the tension in the string is (3/2)mg. What is the speed of the bob at this point? P R
Answer: E Q

19. A5.13
A pendulum of length L with a bob of mass m swings back and forth. At the low point of its motion (point Q ), the tension in the string is (3/2)mg. What is the speed of the bob at this point? P R Q

20. The fundamental forces of nature
According to current understanding, all forces are expressions of four distinct fundamental forces:
gravitational interactions
electromagnetic interactions
the strong interaction
the weak interaction
Physicists have taken steps to unify all interactions into a theory of everything.

21. Dynamics of circular motion

22. Dynamics of circular motion
difference between time 0 and t

23. Dynamics of circular motion
difference between time 0 and t
when t is small,
Δv = v * ω t = v2 t / R
(since ω = v / R)
let t = dt :
dv = v2/R dt
a = dv / dt = v2/R Δv

24. Six artificial satellites
Six artificial satellites complete one circular orbit around a space station in the same amount of time. Each satellite has mass m and radius of orbit L. The satellites fire rockets that provide the force needed to maintain a circular orbit around the space station. The gravitational force is negligible.

25. Six artificial satellites
Six artificial satellites complete one circular orbit around a space station in the same amount of time. Each satellite has mass m and radius of orbit L. The satellites fire rockets that provide the force needed to maintain a circular orbit around the space station. The gravitational force is negligible.
ω is the same for all satellites.
T = ω2 m L, scaling as the product of m and L