Week8 3 CH6 5 - Lecture notes Ch. 6_5 PDF

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Covers. a portion of Ch.6...

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6.5 DEFINITION OF UNIFORM CIRCULAR MOTION Uniform circular motion is the motion of an object traveling at a constant speed on a circular path. Velocity = distance/ elapsed time = circumference / (time for complete circular motion) =circumference / period of the circular motion

2 r v T Let T be the time it takes for the object to travel once around the circle.

radius

period 36

6-5 Circular Motion • An object moving in a circle must have a force acting on it; otherwise it would move in a straight line. • The direction of the force is towards the center of the circle. A force is needed to change the direction of the motion.

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6-5 Circular Motion •

Some algebra gives us the magnitude of the acceleration, and therefore the force, required to keep an object of mass m moving in a circle of radius r.

•

The magnitude of the force is given by:

(6-16) •

The direction of the centripetal force always points toward the center of the circle and continually changes direction as the object moves.

𝑓  centripetal force

𝑎  centripetal acceleration 38

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6-5 Circular Motion •

This force may be provided by the tension in a string, the normal force, or friction, among others.

Normal force Static friction Tension 𝑓  centripetal force should come from one of the sources we talked about before. 39 © 2017 Pearson Education, Inc.

Example 7: Tension force The model airplane has a mass of 0.90 kg and moves at constant speed on a circle that is parallel to the ground. The path of the airplane and the guideline lie in the same horizontal plane because the weight of the plane is balanced by the lift generated by its wings. Find the tension in the 17 m guideline for a speed of 19 m/s. Towards the center, Fnet = mapc

  2  19 m s  T  0.90 kg 

17 m

 19 N 40

On an unbanked curve, the static frictional force provides the centripetal force. Unbanked curves, the static friction force, 𝑓 , provides the necessary force directing a vehicle towards the center of the curve. The static friction force is when there is no sliding between the rod and the tires of the vehicle. If there is s sliding, that would be away from the center. So 𝑓 is towards the center.

Flat roadbed.

Without static friction the vehicle can 



not safely navigate a unbanked curve.

 41

Example: Friction force A car enters a horizontal, curved roadbed of radius 50 m. The coefficient of static friction between the tires and the roadbed is 0.20. What is the maximum speed with which the car can safely negotiate the unbanked curve? y

∑𝐹  𝑓  𝑚 𝑎𝑐𝑝    

𝑐𝑝

x

, 𝑠

 

 

42

To find the normal force. ∑𝐹  0 𝑁  𝑚𝑔  0 𝑁  𝑚𝑔

To find the maximum speed.  



𝑣  𝜇 𝑔𝑟 

 

0.2  9.8

  

50 m  9.9

  43...