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Unformatted text preview: Dynamics III: Motion in a Circle 71 7.32. Model: We will use the particle model for the car which is in uniform circular motion. Visualize: Solve: The centripetal acceleration of the car is ( ) 2 2 2 15 m/s 4.5 m/s 50 m r v a r = = = The acceleration is due to the force of static friction. The force of friction is ( ) ( ) 2 s 1500 kg 4.5m s 6750 N r f ma = = = . Assess: The model of static friction is s max s s 15,000 N f n mg mg μ μ = = ≈ ≈ since s 1 μ ≈ for a dry road surface. We see that s s max , f f < which is reasonable. Dynamics III: Motion in a Circle 72 7.44. Model: Use the particle model for the car which is undergoing circular motion. Visualize: Solve: The car is in circular motion with the center of the circle below the car. Newton’s second law at the top of the hill is 2 r r r r mv F w n mg n ma r = = = = ∑ 2 n v r g m ⇒ = Maximum speed is reached when n = 0 and the car is beginning to lose contact with the road. ( ) ( ) 2 max 50 m 9.8 m/s 22.1 m/s v rg = = = Assess: A speed of 22.1 m/s is equivalent to 49.5 mph, which seems like a reasonable value. 7.50. Model: Model the person as a particle in uniform circular motion. Dynamics III: Motion in a Circle 73 Visualize: Solve: The only force acting on the passengers is the normal force of the wall. Newton’s second law along the raxis is: r F = ∑ 2 n mr ϖ = To create “normal” gravity, the normal force by the inside surface of the space station equals mg . Therefore, 2 2 2 500 m 2 2 45.0 s 9.8 m/s g r mg mr T T r g π ϖ ϖ π π = ⇒ = = ⇒ = = = Assess: This is a fast rotation. The tangential speed is This is a fast rotation....
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 Spring '08
 Nandi
 Physics, Acceleration, Circular Motion, Force, Friction, Mass

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