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# Chapter 10 CIRCULAR MOTION

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### Chapter 10 CIRCULAR MOTION

1. 1. Circular Motion Centripetal forces keep these children moving in a circular path.
2. 2. Circular Motion Constant force is toward center. velocity is tangent to path. v Fc Circular Motion is motion along a circular path in which the direction of its velocity is always changing. Thus, it is accelerating.
3. 3. Initial quantities involved in circular motion: π‘ β period for N revolutions π β frequency of N revolutions π£ = 2ππ π‘ = 2πππ π = 1 π‘
4. 4. Deriving Central Acceleration Consider initial velocity at A and final velocity at B: R vo v v -vo A B R vo Dv Ξs R
5. 5. Deriving Acceleration v -vo R vo Dv Ξs Definition: Similar Triangles mass m Centripetal acceleration: π = βπ£ βπ‘ βπ£ π£ = βπ  π π π = π£ π βπ  βπ‘ = π£ π π£ π π = π£2 π R
6. 6. Dynamics of Circular Motion The question of an outward force can be resolved by asking what happens when the string breaks! When central force is removed, ball continues in straight line. v Ball moves tangent to path, NOT outward as might be expected. Centripetal force is needed to change direction.
7. 7. Car Example There is an outward force, but it does not act ON you. It is the reaction force exerted BY you ON the door. It affects only the door. The centripetal force is exerted BY the door ON you. (Centrally) Fc Fβ Reaction
8. 8. From Newtonβs Second Law: πΉπππ‘ = ππ πΉπππ‘ = πΉπ In circular motion, the net force is the centripetal force! πΉπππ‘ = πΉπ = π π£2 π
9. 9. 1. Horizontal circular motion or Uniform Circular Motion - motion along a circular path in which there is no change in speed, only a change in direction. 2. Vertical circular motion β non-uniform circular motion Kinds of Circular Motion
10. 10. Car Negotiating a Flat Turn R v Is there also an outward force acting ON the car? Ans. No, but the car does exert an outward reaction force ON the road. Fc Applications: Circular Motion
11. 11. Car Negotiating a Flat Turn The centripetal force Fc is that of static friction fs: The central force FC and the friction force fs are not two different forces that are equal. The nature of this central force is static friction. Fc = fsR v m Fc N mg fs R
12. 12. Finding the maximum speed for negotiating a turn without slipping. The car is on the verge of slipping when FC is equal to the maximum force of static friction fs. R v m Fc Fc = fs n mg fs R
13. 13. q Optimum Banking Angle 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. optimum N fs = 0 w R v m Fc
14. 14. Free-body Diagram N mg q q Acceleration a is toward the center. Set x axis along the direction of ac , i. e., horizontal (left to right). N mg q N sin q N cos q + ac q N mg x
15. 15. Motion in a Vertical Circle Consider the forces on a ball attached to a string as it moves in a vertical loop. Note also that the positive direction is always along acceleration, i.e., toward the center of the circle. Note changes in positions. + T mg v Bottom Maximum tension T, W opposes Fc + v T mg Top Right Weight has no effect on T + T mg v Top Right Weight causes small decrease in tension T v T mg + Left Side Weight has no effect on T + T mg v Bottom v T mg Top of Path Tension is minimum as weight helps Fc force +
16. 16. R v v 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. The tension T must adjust so that central resultant is 40 N. At top: 10 N + T = 40 N Bottom: T β 10 N = 40 N T = __?___T = 50 N T = 30 NT = _?_ T 10 N + + T 10 N
17. 17. Homework: 1. In the banking of curves at optimum angle π, an object must move with a certain speed so that it can move in the curve without friction. Prove that this speed is π£ = ππ sin π. 2. A 3.0-kg rock swings in a horizontal circle of radius 2.0 m. If it takes 2.0 s to complete one revolution, what is its tangential speed? What is its centripetal acceleration? What is the tension of the string used? Draw the FBD of the rock. 3. In the frictionless loop-the-loop apparatus, derive the equation of the normal force at the top of the loop in terms of the mass of the cart m, the radius of the loop R, the tangential speed v, and the acceleration of gravity g. Note: Draw FBD. R v v