1. Circular Motion

2. What is circular motion?
Objects that move in a circle experience circular motion.
I know that’s tough.
Let’s take a moment and let it sink in…

3. Now that that is out of the way…
There are specific features of circular motion that make it different from linear or projectile motion

4. Constant speed
An object with a constant speed, which experiences no other forces, will travel in a straight line at that speed infinitely
Newton’s First Law

5. Constant speed

6. Constant speed
Objects can travel in a circle and maintain a constant speed

7. Is velocity constant? No Velocity is speed in a given direction
Those directions cannot be circular

8. What is acceleration? Acceleration is a change in velocity
We have so far defined acceleration as a change in speed but it can be a change in direction, also

9. Circular Motion
An object traveling in a circle travels at a constant speed but is accelerating

10. What causes acceleration?
All changes in velocity are caused by a force
F = ma
Newton’s Second Law

11. What is the force?
The force keeping the object in its circular path is called a centripetal force
Centripetal means “center seeking”

12. Centripetal force
It is a real force
It is a contact force

13. What direction does it point?
The centripetal force always points towards the center of the circle

14. What applies the force? It depends on the situation
In general, whatever keeps the item in it’s circular path applies the centripetal force

15. Example

16. Example

17. Example

18. Are there other forces?
When you make a turn in your car, what makes you pull to one side?
When you swing a bucket of water above your head, what keeps the water in the bucket?

19. What causes that?
In truth, it is a delicate interplay between the inertia of the item and the acceleration
It is another force

20. Centrifugal force
From the Latin, centrum, “center,” and fugere, “fleeing”
This is the force that pushes away from the center of the circle

21. Centrifugal Force
It is the reaction force that compliments the action of the centripetal force
Newton’s Third Law

22. Centrifugal The centrifugal force is a fictitious force
Is it also a contact force

23. Example
You have a bucket of water and you are swinging it around above you head. What forces are acting on it and what do they act on?

24. The two forces
Remember, we have two forces, the centripetal and the centrifugal
The centripetal acts on the bucket
The centrifugal acts on the water

25. The math You knew it was coming
Math is the language of physics and you need to learn to speak that language

26. Centripetal acceleration
There are two equations we can use depending on what we know

27. The first (and easiest)
v is the velocity of the object
r is the radius of the circle

28. The second T is the time it takes for one full revolution
r is the radius of the circle

29. Centrifugal acceleration
If the centrifugal force arises from Newton’s Third Law and is the equal but opposite reaction to the centripetal force, what is the equation going to be?

30. Centrifugal accelerationOr

31. Sample problem
A 1000 kg car enters an 80 meter radius curve at 20 m/s. What centripetal force must be supplied by friction so the car does not skid?

32. What do we know?
m = 1000 kg
r = 80 m
v = 20 m/s

33. Find the force F = ma = mv2/r F = 1000 × (202/80) F = 1000 × 400/80
F = 1000 × 5 = 5000 N

34. Sample problem
The centripetal force on a 0.82 kg object on the end of a 2.0 m massless string being swung in a horizontal circle is 4.0 N. What is the tangential velocity of the object?

35. What do we know?
m = 0.82 kg
r = 2.0 m
Fc = 4.0 N

36. Find the velocity F = ma = mv2/r 4.0 = 0.82 × v2/2.0 8.0 = 0.82v2
v = 3.12 m/s

37. Sample problem
A dragonfly is sitting on a merry-go-round 2.8 m from the center. If the centripetal acceleration of the dragonfly is 3.6 m/s2, what is the period of the merry-go-round?

38. What do we know?
r = 2.8 m
a = 3.6 m/s2

39. Find the period ac = (4π2r)/T2 3.6 = (4π2 × 2.8)/T2 3.6 = 110/T2
T = 5.5 s

40. Sample problem
A car moving at a 1.08 × 108 m/s (30 km/h) rounds a bend in the road with a radius of 21.2 m. What is the centripetal acceleration on the car and the centrifugal acceleration on the occupants?

41. What do we know?
v = 1.08 × 108 m/s
r = 21.2 m

42. Centripetal
a = v2/r
a = (1.08 × 108)2 / 21.2
a = 5.50 × 1014 m/s2

43. Centrifugal
a = -v2/r
a = -(1.08 × 108)2 / 21.2
a = × 1014 m/s2