1. Example 1: A 3-kg rock swings in a circle of radius 5 m
Example 1: A 3-kg rock swings in a circle of radius 5 m. If its constant speed is 8 m/s, what is the centripetal acceleration? R v m
m = 3 kg
R = 5 m; v = 8 m/s
F = (3 kg)(12.8 m/s2)
Fc =

2. Car Negotiating a Flat Turnv Fc
What is the direction of the force ON the car?
Ans. Toward Center
This central force is exerted BY the road ON the car.

3. Car Negotiating a Flat Turnv Fc
Is there also an outward force acting ON the car?
Ans. No, but the car does exert a outward reaction force ON the road.

4. Car Negotiating a Flat Turn
The centripetal force Fc is that of static friction fs: R v m Fc n
Fc = fs fs R mg
The central force FC and the friction force fs are not two different forces that are equal. There is just one force on the car. The nature of this central force is static friction.

5. Finding the maximum speed for negotiating a turn without slipping.
Fc = fs n mg fs R R v m Fc
The car is on the verge of slipping when FC is equal to the maximum force of static friction fs.
Fc =
mv2 R
fs = msmg
Fc = fs

6. Maximum speed without slipping (Cont.)
Fc = fs n mg fs R v m Fc
mv2 R
= msmg
v = msgR
Velocity v is maximum speed for no slipping.

7. Example 4: A car negotiates a turn of radius 70 m when the coefficient of static friction is 0.7. What is the maximum speed to avoid slipping? R v m Fc
ms = 0.7
Fc =
mv2 R
fs = msmg
From which:
v = msgR
g = 9.8 m/s2; R = 70 m
v =

8. Optimum Banking Angle n n n m
By banking a curve at the optimum angle, the normal force n can provide the necessary centripetal force without the need for a friction force. R v m Fc n
fs = 0 n fs q
slow speed q n fs q w w w
fast speed
optimum

9. Free-body Diagram n cos q n sin q n n n x mg + ac mg mg
Acceleration a is toward the center. Set x axis along the direction of ac , i. e., horizontal (left to right). n mg q x
n cos q n mg q n q
+ ac
n sin q q mg

10. Optimum Banking Angle (Cont.)mg q
n sin q
n cos q q n mg
mv2 R
n sin q =
Apply Newton’s 2nd Law to x and y axes.
SFx = mac
n cos q = mg
SFy = 0

11. Optimum Banking Angle (Cont.)mg q
n sin q
n cos q q n mg
mv2 R
n sin q =
n cos q = mg

12. Optimum Banking Angle (Cont.)mg q
n sin q
n cos q
Optimum Banking Angle q

13. How might you find the centripetal force on the car, knowing its mass?
Example 5: A car negotiates a turn of radius 80 m. What is the optimum banking angle for this curve if the speed is to be equal to 12 m/s? q n mg
tan q = = v2 gR
(12 m/s)2
(9.8 m/s2)(80 m)
q = n mg q
n sin q
n cos q
tan q = 0.184
How might you find the centripetal force on the car, knowing its mass?

14. The Conical Pendulum
A conical pendulum consists of a mass m revolving in a horizontal circle of radius R at the end of a cord of length L.
T cos q q h T L R T q mg
T sin q
Note: The inward component of tension T sin q gives the needed central force.

15. Solve two equations to find angle q
Angle q and velocity v: q h T L R mg
T sin q
T cos q
Solve two equations to find angle q
mv2 R
T sin q =
tan q = v2 gR
T cos q = mg

16. 2. Recall formula for pendulum.
Example 6: A 2-kg mass swings in a horizontal circle at the end of a cord of length 10 m. What is the constant speed of the mass if the rope makes an angle of 300 with the vertical?
1. Draw & label sketch.
q = 300 q h T L R
2. Recall formula for pendulum.
Find: v = ?
3. To use this formula, we need to find R = ?
R = L sin 300 = (10 m)(0.5)
R =

17. Example 6(Cont.): Find v for q = 300h T L R
q = 300
R = 5 m
4. Use given info to find the velocity at 300.
R = 5 m
g = 9.8 m/s2
Solve for v = ?
v =

18. Swinging Seats at the Fair
This problem is identical to the other examples except for finding R. q h T L R b d
R = d + b
R = L sin q + b
tan q = v2 gR
and
v = gR tan q

19. Example 9. If b = 5 m and L = 10 m, what will be the speed if the angle q = 260?
tan q = v2 gR q T L R d b
R = d + b
d = (10 m) sin 260 = 4.38 m
R = 4.38 m + 5 m = 9.38 m
v =

20. Motion in a Vertical Circle+ T mg v
Top Right
Weight causes small decrease in tension T
Consider the forces on a ball attached to a string as it moves in a vertical loop. v T mg
Top of Path
Tension is minimum as weight helps Fc force + + v T mg
Top Right
Weight has no effect on T v T mg +
Left Side
Weight has no effect on T + T mg v
Bottom
Maximum tension T, W opposes Fc + T mg v
Bottom
Note also that the positive direction is always along acceleration, i.e., toward the center of the circle.
Note changes as you click the mouse to show new positions.

21. The tension T must adjust so that central resultant is 40 N.
As an exercise, assume that a central force of Fc = 40 N is required to maintain circular motion of a ball and W = 10 N. R v T
10 N + + T
10 N
The tension T must adjust so that central resultant is 40 N.
At top: 10 N + T = 40 N
T =
Bottom: T – 10 N = 40 N
T = __?___

22. Motion in a Vertical Circle
Fc =
mv2 R
Resultant force toward center T mg
Consider TOP of circle:
mg + T =
mv2 R
AT TOP: + T mg
T = mg
mv2 R

23. Vertical Circle; Mass at bottom
Fc =
mv2 R
Resultant force toward center T mg
Consider bottom of circle:
T - mg =
mv2 R
AT Bottom: T mg +
T = mg
mv2 R

24. Centripetal acceleration:
Summary
Centripetal acceleration:
tan q = v2 gR
v = msgR
v = gR tan q
Conical pendulum:

25. Summary: Motion in Circlev R
AT TOP: T mg +
T = mg
mv2
AT BOTTOM:
T = mg

26. Summary: Ferris Wheel n n = mg - n = + mg AT TOP: mg + AT BOTTOM:
mv2 R
n = mg v
n = mg -

27. CONCLUSION: Chapter 10 Uniform Circular Motion