Simple Harmonic Motion
12.0 Physical Pendulum
12.0 Physical Pendulum
Physical pendulum is a rigid body suspended from a fixed support constitutes a physical pendulum.
This pendulum oscillates about the horizontal axis passing through it.
Consider the situation when the body is displaced through small angle, $\theta$.
Torque on the body about point $O$ is given by
$$\tau = - mglsin\theta \quad ..\left( i \right)$$
Note: -ve sign indicates that torque is acting in opposite direction to that of the displacement.
Where,
$l=$ distance between point of suspension ($O$) and the center of mass of the body ($G$)
If $I$, be the moment of inertia of the body about point $O$ then, $$\tau = I\alpha \quad ..\left( {ii} \right)$$
From equation $(i)$ & $(ii)$ we get,
$$\begin{equation} \begin{aligned} I\alpha = - mglsin\theta \\ \alpha = \frac{{ - mgl\theta }}{I}\left( {sin \approx \theta\ \ {\text{for small oscillation}}} \right)\quad ..\left( {iii} \right) \\ \alpha = - {\omega ^2}\theta \quad ..\left( {iv} \right) \\\end{aligned} \end{equation} $$
From equation $(iii)$ & $(iv)$ we get,
$$\begin{equation} \begin{aligned} \omega = \sqrt {\frac{{mgl}}{I}} \\ \therefore T = 2\pi \sqrt {\frac{I}{{mgl}}} \\\end{aligned} \end{equation} $$
Example 8. A disc is suspended at a point $\frac{R}{2}$ above its center. Find its period of oscillations.
Solution: First we will calculate the moment of inertia about point $O$ i.e., $I_O$.
We know moment of inertia about point $C$, i.e., $I_C$ $${I_C} = \frac{1}{2}m{R^2}$$
By using parallel axis theorem, we get
$$\begin{equation} \begin{aligned} {I_O} = {I_C} + m{l^2} \\ {I_O} = \frac{1}{2}m{R^2} + m{\left( {\frac{R}{2}} \right)^2} \\ {I_O} = \frac{1}{2}m{R^2} + \frac{1}{4}m{R^2} \\ {I_O} = \frac{3}{4}m{R^2}\quad ..\left( i \right) \\\end{aligned} \end{equation} $$
After displacing the disc by a small angle $\theta$, the restoring toque act is,
$$\tau = - mglsin\theta $$
(-ve sign indicated that torque is restoring).
$$\begin{equation} \begin{aligned} \tau = - mgl\theta \left( {{\text{for small oscillation }}\ sin\theta \approx \theta } \right) \\ {I_O}\alpha = - mgl\theta \left( {{\text{As }}\tau = {I_O}\alpha } \right) \\ \alpha = \frac{{ - mgl\theta }}{{{I_O}}}\quad ..\left( {ii} \right) \\\end{aligned} \end{equation} $$
We know, $$\alpha = - {\omega ^2}\theta \quad ..\left( {iii} \right)$$
$$\omega = \sqrt {\frac{{mgl}}{{{I_O}}}} \quad ..\left( {iv} \right)$$
As we know, $$T = \frac{{2\pi }}{\omega } = 2\pi \sqrt {\frac{{{I_O}}}{{mgl}}} \quad ..\left( v \right)$$
From equation $(i)$ & $(v)$ we get,
$$\begin{equation} \begin{aligned} T = 2\pi \sqrt {\frac{{\frac{3}{4}m{R^2}}}{{mg\frac{R}{2}}}} \\ T = 2\pi \sqrt {\frac{{3R}}{{2g}}} \\\end{aligned} \end{equation} $$
