University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 9 - Rotation of Rigid Bodies - Problems - Exercises: 9.40

Answer

(a) $I = 0.0225~kg~m^2$ (b) The mass of the turntable should be 0.50 kg.

Work Step by Step

(a) We can convert the angular speed to units of rad/s. $\omega = (45~rpm)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s})$ $\omega = 4.71~rad/s$ We can find the moment of inertia of the turntable. $\frac{1}{2}I~\omega^2 = K$ $I = \frac{2~K}{\omega^2}$ $I = \frac{(2)(0.250~J)}{(4.71~rad/s)^2}$ $I = 0.0225~kg~m^2$ (b) We can find the mass of the turntable. $\frac{1}{2}MR^2 = I$ $M = \frac{2I}{R^2}$ $M = \frac{(2)(0.0225~kg~m^2)}{(0.30~m)^2}$ $M = 0.50~kg$ The mass of the turntable should be 0.50 kg.
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