Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems: 47

Answer

The magnitude of the net force that the track exerts on the ball is 0.38 N

Work Step by Step

We can express the angular velocity in units of rad/s $\omega = (60~rpm)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s})$ $\omega = 2\pi~rad/s$ The horizontal force $F_x$ from the track on the ball provides the centripetal force to keep the ball moving in a circle. $F_x = m\omega^2~r$ $F_x = (0.030~kg)(2\pi)^2(0.20~m)$ $F_x = 0.237~N$ The vertical force $F_y$ from the track is equal to the ball's weight. $F_y = mg$ $F_y = (0.030~kg)(9.80~m/s^2)$ $F_y = 0.294~N$ We can find the magnitude of the net force that the track exerts on the ball. $F = \sqrt{(F_x)^2+(F_y)^2}$ $F = \sqrt{(0.237~N)^2+(0.294~N)^2}$ $F = 0.38~N$ The magnitude of the net force that the track exerts on the ball is 0.38 N
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