Answer to Question #124709 in Atomic and Nuclear Physics for Hivi

Question

Show that the simple pendulum is a special case of compound pendulum when the moment of inertia is mLsqure when the center of mass distance of compound pendulum equal to length of simple pendulum

Answer 1

since, equation which compound pendulum is governed is

$$\ddot{\theta}+\omega^2\theta=0$$

Where,

$$\omega^2=\frac{MgD}{I}$$

Now, we have given $I=ML^2,D=L$ ,thus $\omega^2=\frac{g}{L}$ but equation of simple pendulum is given by

$$\ddot{\theta}+\omega_1^2\theta=0$$

where,

$$\omega_1^2=\frac{g}{L}$$

Thus,

$$\omega=\omega_1$$

Therefore, compound pendulum becomes simple pendulum.