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Centripetal Force

Centripetal Force. Second Law Review. According to Newton’s second law if there is acceleration there must be a net force acting on the object in the direction of the acceleration. Centripetal Force. It is the net force acting towards the center

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Centripetal Force

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  1. Centripetal Force

  2. Second Law Review • According to Newton’s second law if there is acceleration there must be a net force acting on the object in the direction of the acceleration.

  3. Centripetal Force • It is the net force acting towards the center • The inward force causing the inward acceleration of an object moving in a circular motion. • Means “ center seeking”

  4. This is not a new force it is just describing the direction of the net force acting upon the object moving in the circle.

  5. As a car makes a turn, the force of friction acting upon the turned wheels of the car provide the centripetal force required for circular motion Examples of Centripetal Force

  6. As a bucket of water is tied to a string and spun in a circle the force of tension acting upon the bucket provides the centripetal force

  7. As the moon orbits the Earth, the force of gravity acting upon the moon provides the centripetal force

  8. Why does an object need Centripetal Force? • All objects have a natural tendency to continue moving in the same direction unless some net force acts upon the object • Centripetal Force pulls the object toward the center of the circle to force the object to follow the circular path.

  9. The movement of a person away from the center of the circle is due to inertia. • Your body is resisting the change in velocity • If there were no force pulling you toward the center you would continue to move tangent to the circle

  10. There is no outward force pushing on an object

  11. Calculating Fc • Fc = mAc = m (vt)2/r = mrw2

  12. A pilot is flying a small plane at 30.0 m/s in a circular path with a radius of 100.0m. The mass of pilot and plane is 70.6 kg. What is the centripetal force needed to keep the pilot in circular motion?

  13. Vt = 30.0 m/s r= 100.0m m= 70.6 kg • Fc = m (vt)2/r 70.6 (30.0)2/100.0 Fc = 635 N

  14. A dog sits 1.50m from the center of a merry-go-round with an angular speed of 1.20 rad/s. If the dog’s mass is 18.5 kg, what is the magnitude of the force to keep the dog in circular motion?

  15. r = 1.50m w = 1.20 rad/s m = 18.5 kg • Fc = mrw2 (18.5)(1.50)(1.20) 2 Fc = 40.0N

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