Question
A pendulum consists of a heavy but very small bob of mass M suspended at the end of a rigid rod of mass m and length L. The time period of small oscillations in the vertical plane, about a horizontal axis through the upper end of the rod is :
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Solution
[Key Concept: A rod plus end bob is a physical pendulum: T = 2π√(I/(Mtotal g d)), with I and the gravity torque taken about the pivot.]
$$\begin{matrix} & \tau = I\alpha \\ & \ - \left\{ (mgsin\theta)\frac{L}{2} + (Mgsin\theta)L \right\} = \left( {ML}^{2} + \frac{{mL}^{2}}{3} \right)\alpha \end{matrix}$$
$\theta$ is very small
So, $sin\theta \approx \theta$
$$\begin{matrix} & \ - \left( \frac{m}{2} + M \right)g\theta = \left( \frac{3M + m}{3} \right)L\alpha \\ & \ - \left( \frac{m + 2M}{2} \right)g\theta = \left( \frac{3M + m}{3} \right)L\alpha \\ & \alpha = - \left( \frac{m + 2M}{\text{ }m + 3M} \right)\frac{3\text{ }g}{2\text{ }L}\theta \end{matrix}$$
By comparing standard equation
$$\begin{matrix} & \alpha = - \omega^{2}\theta \\ & \omega = \sqrt{\left( \frac{m + 2M}{m + 3M} \right)\frac{3g}{2L}} \\ & T = \frac{2\pi}{\omega} \\ & T = 2\pi\sqrt{\left( \frac{m + 3M}{m + 2M} \right)\frac{2L}{3g}} \end{matrix}$$
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