Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 13 - Exercises and Problems - Page 247: 55

Answer

$\frac{\pi}{3}\sqrt {\frac{58D}{g}}$

Work Step by Step

For a physical pendulum $T=2\pi \sqrt {\frac{I}{mgL}}$ , where T - oscillation period, I - Rotational inertia of the system through the pivot, L - distance from the pivot to center of mass of the ball, g - gravitational acceleration ($9.8\space m/s^{2}$) $T=2\pi \sqrt {\frac{I}{mgL}}-(1)$ According to the parallel axis theorem we can write, $I=I_{cm}+mL^{2}$ (Please see the attached image) $I=\frac{2}{3}m(\frac{D}{2})^{2}+m(D+\frac{D}{2})^{2}= \frac{mD^{2}}{6}+\frac{9mD^{2}}{4}$ $I=\frac{2mD^{2}+27mD^{2}}{12}=\frac{29mD^{2}}{12}-(2)$ (2)=>(1) $T=2\pi\sqrt {\frac{29mD^{2}/12}{mg\times\frac{3D}{2}}}=2\pi \sqrt {\frac{29D}{18g}}=2\pi \sqrt {\frac{58D}{36g}}$ $T=\frac{2\pi}{6}\sqrt {\frac{58D}{g}}=\frac{\pi}{3}\sqrt {\frac{58D}{g}}$
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