1. Circular Motion; Gravitation
Chapter 5
Circular Motion; Gravitation

2. Objectives
Identify the force that is the cause of the centripetal acceleration and determine the direction of the acceleration vector.
Use Newton's laws of motion and the concept of centripetal acceleration to solve word problems.

3. 5-3 Highway Curves, Banked and Unbanked
When a car goes around a curve, there must be a net force towards the center of the circle of which the curve is an arc. If the road is flat, that force is supplied by friction.

4. 5-3 Highway Curves, Banked and Unbanked
If the frictional force is insufficient, the car will tend to move more nearly in a straight line, as the skid marks show.

5. 5-3 Highway Curves, Banked and Unbanked
As long as the tires do not slip, the friction is static. If the tires do start to slip, the friction is kinetic, which is bad in two ways:
The kinetic frictional force is smaller than the static.
The static frictional force can point towards the center of the circle, but the kinetic frictional force opposes the direction of motion, making it very difficult to regain control of the car and continue around the curve.

6. 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.

7. 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.

8. 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.

9. 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

10. 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.

11. Example: 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 = 21.9 m/s

12. Level Curves – Try this one.
A 1500 kg car moving on a flat, horizontal road negotiates a curve as shown. If the radius of the curve is 35.0 m and the coefficient of static friction between the tires and dry pavement is 0.523, find the maximum speed the car can have and still make the turn successfully.
April 24, 2017

13. Level Curves
The force of static friction directed toward the center of the curve keeps the car moving in a circular path.
April 24, 2017
Period 1 stopped here

14. 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

15. 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

16. 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

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

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

19. 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 = 10.40 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?

20. 5-3 Highway Curves, Banked and Unbanked
Banking the curve can help keep cars from skidding. In fact, for every banked curve, there is one speed where the entire centripetal force is supplied by the
horizontal component of the normal force, and no friction is required. This occurs when:

21. Banked Curves
A car moving at the designated speed can negotiate the curve. Such a ramp is usually banked, which means that the roadway is tilted toward the inside of the curve. Suppose the designated speed for the ramp is to be 13.4 m/s and the radius of the curve is 35.0 m. At what angle should the curve be banked?
April 24, 2017

22. Banked Curves
Show work for problem
April 24, 2017

23. A car of mass, m, is traveling at a constant speed, v, along a curve that is now banked and has a radius, R. What bank angle, q, makes reliance on friction unnecessary? N mg
frictionless q

24. homework
Questions p.129 # 8, 9
Problems 9, 10, and do example 5-7 on page 114 (cover up answer)

25. 5-4 Nonuniform Circular Motion
If an object is moving in a circular path but at varying speeds, it must have a tangential component to its acceleration as well as the radial one.

26. 5-4 Nonuniform Circular Motion
This concept can be used for an object moving along any curved path, as a small segment of the path will be approximately circular.

27. Tangential & Total Acceleration
An object may be changing its speed (speeding up or slowing down) as it moves in a circular path. In that case, there is a tangential acceleration as well as a centripetal acceleration.
The total acceleration atotal is the vector sum of the centripetal acceleration acp, which points toward the center of rotation, and the tangential acceleration at, which points in the direction of speed increase.

28. Tangential Acceleration
Tangential Acceleration – The instantaneous linear acceleration of an object directed along the tangent to the object’s circular path.
atan = Δv/Δt

29. Example 5-8

30. Example 5-8 Solution

31. The Centrifuge
The apparent weight of an object can be greatly increased using circular motion.
A centrifuge is a laboratory device used in chemistry, biology, and medicine for increasing the sedimentation rate and separation of a sample by subjecting it to a very high centripetal acceleration. Accelerations on the order of 10,000 g can be achieved. This can give a 12 g sample can be given an apparent weight of about 1130 N = 250 lb.

32. 5-5 Centrifugation
A centrifuge works by spinning very fast. This means there must be a very large centripetal force. The object at A would go in a straight line but for this force; as it is, it winds up at B.

33. Example: Big Gees
A centrifuge rotates at a rate such that the bottom of a test tube travels at a speed of 89.3 m/s. The bottom of the test tube is 8.50 cm from the axis of rotation.
What is the centripetal acceleration acp at the bottom of the test tube in m/s and in g (where 1 g = 9.81 m/s2)?

34. “Centrifugal Force”
“Centrifugal force” is a fictitious force - it is not an interaction between 2 objects, and therefore not a real force.
Nothing pulls an object away from the center of the circle.

35. “Centrifugal Force”
What is erroneously attributed to “centrifugal force” is actually the action of the object’s inertia - whatever velocity it has (speed + direction) it wants to keep.