1. Circular Motion For a car going around a curve at constant speed, the free-body diagram is: where F w is the weight of the car, F N is the normal (perpendicular) force, and F cp is the centripetal force. FwFw FNFN F cp

2. ΣF y = 0 ΣF X = F cp F N = F w  The vector sum of the forces in a vertical direction equals zero which makes sense because there is no vertical acceleration.  The vector sum of the forces in a horizontal direction is greater than zero which makes sense because the car is accelerating in a horizontal plane.

3.  F cp is the net force which by Newton’s 2 nd law causes the car to accelerate toward the center of the circle. The centripetal force is given by: F cp = mv 2 /r where m is the mass in kg, v is the velocity in m/s, and r is the radius of the circle in m.

4. An object undergoing circular motion is constantly accelerating because its instantaneous velocity is always changing direction.  The instantaneous velocity is always tangent to the circle.

5. The centripetal acceleration is always directed toward the center of the circle is given by:  a c = v 2 /r As is the case with projectile motion, Newton’s three laws can be found in circular motion. acac

6. Newton’s 1 st law is found when the object wants to continue its uniform motion in a straight line.  An object does want to accelerate but can be made to do so with a net force. Newton’s 2 nd law is found by the net force and the acceleration acting in the same direction, toward the center of the circle. Newton’s 3rd law is also found but this is where a colossal misinterpretation is made.

7. Remember that Newton’s 3rd law requires two different objects.  When you go around a corner in a car, who or what is accelerating?  That’s right, you and the car are accelerating by a net force which is the centripetal force.  Both you and the car have mass (hopefully not the same mass) and because of inertia both of you want a state of zero acceleration.

8.  The centripetal force is the force that the road exerts on the car (and you) via friction.  The reaction to this force is the car exerting a force on the road which is called the centrifugal force.  When the driver takes an especially sharp turn while you are the passenger, many people say that centrifugal force pushes them into the door.

9.  This is where your “physical” intuition can fool you.  The centrifugal force acts on the road, not you!  In actuality, it is nothing more than Newton’s 1 st law kicking into action.

10. Acceleration and Velocity Any net force produces an acceleration which is in the same direction as the net force.  When the acceleration is parallel to the velocity, the speed will increase.  When the acceleration is in the opposite direction of the velocity, the speed will decrease.

11.  If the acceleration is perpendicular to the velocity, the speed remains the same but the velocity changes in direction.

12. Circular Motion Problem A 0.017 kg rubber stopper is attached to a 0.89 m length of string. The stopper is swung in a horizontal circle making one revolution in 1.2 s. (a)Draw a free body diagram for the rubber stopper and identify the origin of each force.

13. F cp is the centripetal force and represents the force the string exerts on the rubber stopper. F w is the weight of the stopper which represents the earth pulling down on the stopper. Remember that in a force diagram you only include the forces that act on the object! FwFw F cp

14. (b) What is the speed of the stopper? v ave = Δx/Δt = C/Δt = 2πr/Δt v ave = 2 × 3.14 × 0.89 m/1.2 s = 4.7 m/s (c) What is the instantaneous velocity? v = 4.7 m/s tangent to the circle.v

15. (d) What is the instantaneous acceleration? a = v 2 /r = (4.7 m/s) 2 /0.89 m = 24 m/s 2 a = 24 m/s 2 toward the center of the circle. (e) Determine both centripetal and centrifugal force. F cp = ma = 0.017 kg × 24 m/s 2 = 0.41 N F cf = F cp = 0.41 N

16. (f) What happens if the string breaks? When the string breaks, F cp = 0, the stopper will be in free fall and will attempt to travel in a straight line with a velocity of 4.7 m/s while accelerating vertically.

17. Vertical Circular Motion When an object moves in a vertical circle, the weight of an object can not be ignored. Free-body diagram at the top of a circle: Fw T

18. F net = F c F net = F w + T F c = F w + T Free-body diagram at the bottom of a circle: Fw T

19. Vertical Circle Problem A 0.147 kg ball at the end of a 1.20 m long string is swung in a vertical circle. If the mass of the string is neglected: (a)What is the minimum speed the ball must have at the top of the circle so that it continues to uniformly accelerate towards the center?

20. m = 0.147 kgr = 1.20 m Fw At the top, the minimum speed occurs when T = 0, that is when the weight of the object supplies the centripetal force.

21. F net = F c = F w F c = F w mv 2 r = mg v = (rg) 1/2 = (1.20 m × 9.80 m/s 2 ) 1/2 =3.43 m/s

22. (b) What is the tension in the cord at the bottom of its path if the ball is moving at three times the speed at the top? FwFw T

23. At the bottom, the tension is a maximum because it must support the weight of the object and also provide the centripetal force. F net = F c = T – F w F c = T- F w = mv 2 /r – mg T = F c + F w T = 0.147 kg × (10.3 m/s) 2 + 0.147 kg × 9.80 m/s 2 T = 17.0 N

24. Another Circular Motion Problem A boy weighing 294 N sits 5.0 m from the axis of rotation on a merry-go-round. The merry-go-round spins at 4.7 rev/min. (a)What is the frictional force required to keep the boy on the merry-go-round. FwFw FNFN FcFc

25. F w = 294 Nr = 5.0 mv = 4.7 rev/min FfFf =FcFc = mv 2 r FfFf = 294 N/9.80 m/s 2 × v= rev 4.7 × 1 min 2 × π × 5.0 m 1 rev × 1 min 60 s v= 2.5 m/s (2.5 m/s) 2 5.0 m = 38 N

26. (b) Determine the coefficient of friction. µ= FfFf FNFN = 38 N 294 N = 0.13

27. More Circular Motion Thoughts In the case of a merry-go-round, friction must provide the centripetal force. You must always pay attention to units! Because 1 N = kgm/s 2, all the units must be compatible to cancel out. One complete revolution is equal to the circumference of the circle, 2πr.

28. An interesting tidbit is that the coefficient of friction (µ) is independent of mass in this problem. How do you know? I’m glad you asked! Aren’t you glad you asked? µ = FfFf FNFN = FcFc FwFw mv 2 /r mg == v2v2 rg

29. Wrap Up Questions If the average velocity of an object is zero during a time interval, what can be said for the displacement of that object for the same time interval? The displacement equals zero because displacement is used to define velocity.

30. If the displacements of a particle are known at two distinct points and the time to travel that displacement is known, can you determine the instantaneous velocity? The instantaneous velocity can not be determined unless you are told that the object is traveling at a constant velocity, in which case the two velocities would be the same. You could determine the instantaneous velocity from a Displacement vs Time graph.

31. If the speed of an object is constant, can it be accelerating? It could accelerate by following a curved path, either going around in circular motion or turning a corner in a car. If the velocity of an object is constant, can it be accelerating? No, because acceleration is defined as the rate at which the velocity changes.