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: 39

Answer

The marble will spin at a constant height.

Work Step by Step

We first find the angular velocity in rad/s: $\omega = (150~rpm(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s})$ $\omega = 15.7~rad/s$ The normal force $F_N$ from the wall provides the centripetal force to keep marble moving in a circle. $F_N = m\omega^2~r$ We can find the maximum possible force of static friction (directed upward) which could prevent the marble from sliding down the wall. We can assume that the coefficient of static friction is 0.70. Therefore; $F_f = F_N~\mu_s$ $F_f = m\omega^2~r~\mu_s$ $F_f = (0.010~kg)(15.7~rad/s)^2(0.060~m)(0.70)$ $F_f = 0.104~N$ We can find the weight of the marble as: $weight = mg$ $weight = (0.010~kg)(9.80~m/s^2)$ $weight = 0.098~N$ Since the maximum possible force of static friction (directed upward) on the marble is greater than the marble's weight, the marble will spin at a constant height.
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