Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 13 - Exercises and Problems - Page 240: 75


$T=2\pi \sqrt{\frac{7}{10ga}}$

Work Step by Step

We first find the kinetic energy: $ E_k = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 $ $ E_k = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)\frac{v^2}{r^2} $ $ E = \frac{7}{10}mv^2 $ $E = \frac{7}{10}m(\frac{dx}{dt})v^2 $ Thus, we can find omega. Using the equation for omega, we obtain: $\omega = \frac{x}{\frac{d^2x}{dt^2}}$ $\omega= \sqrt{\frac{7}{10ga}}$ The period is equal to the angular velocity times 2 pi, so we find: $T=2\pi \sqrt{\frac{7}{10ga}}$
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