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 14 - Oscillations - Exercises and Problems - Page 405: 56

Answer

$T=\pi\;\sqrt{\dfrac{7L }{ 3g }}$

Work Step by Step

First, we need to find the rod's moment of inertia around the pivot point. So we need to use the parallel-axis theorem. $$I_{pivot}=I_{cm}+Md^2$$ The pivot point is at a distance of $L/4$ from one end of the rod, and the moment of inertia of a uniform rod around its center of mass is $ML^2/12$. Thus, $$I_{pivot}=\frac{1}{12}ML^2+M\left[\dfrac{L}{4}\right]^2$$ $$I_{pivot}=\frac{7}{48}ML^2\tag 1$$ Now we need to find the period of our physical pendulum and we know that the angular frequency of such one is given by $$\omega=2\pi f=\sqrt{\dfrac{mgl}{I}}$$ where m is the mass of the system, and l is the distance from the pivot point to the center of mass of the system. So, in our case, $$2\pi f=\sqrt{\dfrac{Mg\frac{L}{4}}{I}}$$ Plugging from (1); $$2\pi f=\sqrt{\dfrac{ \color{red}{\bf\not} Mg \color{red}{\bf\not} L}{4(\frac{7}{48} \color{red}{\bf\not} ML^{ \color{red}{\bf\not} 2})}}$$ $$2\pi f=\sqrt{\dfrac{ g }{ \frac{7}{12} L }}$$ Recalling that the period is given by $T=1/f$, so $$2\pi \dfrac{1}{T}=\sqrt{\dfrac{ g }{ \frac{7}{12} L }}$$ Solving for $T$; $$T=2\pi\;\sqrt{\dfrac{7L }{ 12g }}$$ $$\boxed{T= \pi\;\sqrt{\dfrac{7L }{ 3g }} }$$
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