Question

Consider a physical pendulum consisting of an inverted $T$ shaped rigid body made out of two identical, thin, uniform rods of mass $M$ and length $L$ each joined together and hinged at the top point as shown, free to rotate in a vertical plane about an axis perpendicular to the plane of the figure, the time period for $small$ angular oscillations is $T=2\pi \sqrt{\frac{\alpha L}{\beta g}}$. Find $|\alpha -\beta |$.

Answer 1

As we know,

$T=2\pi \sqrt{\frac{\alpha L}{\beta g}}$

where, $I$ is the moment of inertia about the axis of rotation at the hinge,

$M$ the total mass

$y_{CM}$ the position of the centre of mass w.r.t the axis of rotation. The moment of inertia of system,

$I=\left[\frac{M{L}^2}{3}\right]+[\frac{M{L}^2}{12}+M{L}^2]=\frac{17M{L}^2}{12}$

and total mass $=2M$

$y_{CM}=3L/4$

Time period of $small$ angular oscillations,

$T=2\pi \sqrt{\frac{I}{Mg{y}_{CM}}}$

Substitute the values, we get

$T=2\pi \sqrt{\frac{\frac{17M{L}^2}{12}}{2M\times 3L/4\times g}}$

Compare with,

$T=2\pi \sqrt{\frac{\alpha L}{\beta g}}$

we get,

$|\alpha -\beta |=1$