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 - Page 298: 9.60

Answer

The thickness of the wheel should be 5.26 cm.

Work Step by Step

We can find the angular speed at $t = 8.00~s$: $\omega = \alpha ~t$ $\omega = (3.00~rad/s^2)(8.00~s)$ $\omega = 24.0~rad/s$ The kinetic energy is required to be 800 J. We can find the required mass $M$. $KE = 800~J$ $\frac{1}{2}I\omega^2 = 800~J$ $\frac{1}{2}(\frac{1}{2}MR^2)\omega^2 = 800~J$ $M = \frac{(4)(800~J)}{R^2~\omega^2}$ $M = \frac{3200~J}{(0.250~m)^2~(24.0~rad/s)^2}$ $M = 88.89~kg$ We can find the thickness $x$ of the wheel. $\rho \times volume = M$ $\rho\times x\times Area = M$ $x = \frac{M}{\rho ~\pi ~R^2}$ $x = \frac{88.89~kg}{(8600~kg/m^3)(\pi)(0.250~m)^2}$ $x = 0.0526~m = 5.26~cm$ The thickness of the wheel should be 5.26 cm.
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