Question

In the given figure an L-shaped bar of mass $M$ is pivoted at one of its end, so that it can freely rotate in a vertical plane. Find the frequency of oscillation, if it is slightly disturbed from its equilibrium position.

• A. $\frac{1}{2\pi }\sqrt{\frac{g}{L}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{3g\sqrt{10}}{4L}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{3g}{4L}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{3g\sqrt{5}}{4L}}$

Answer 1

The correct option is B $\frac{1}{2\pi }\sqrt{\frac{3g\sqrt{10}}{4L}}$


We can consider AB and BC as individual rods of masses $M/2$ and length $L$ each.
Let $C_1$ and $C_2$ be the COM of AB and BC respectivley.
Then COM of L-shaped rod is
$x_{com}=\frac{\frac{M}{2}\frac{L}{2}}{M}=\frac{L}{4};Y_{com}=\frac{\frac{M}{2}\frac{L}{2}}{M}=\frac{L}{4}$



Hence, distance of COM from point of suspension is
$d=\sqrt{{\left(\frac{3L}{4}\right)}^2+{\left(\frac{L}{4}\right)}^2}=\frac{L}{4}\sqrt{10}$
Moment of inertia about $A$ is
$I_A=\left(\frac{M}{2}\right)\frac{L^2}{3}+\left(\frac{M}{2}\right)\frac{L^2}{12}+\frac{M}{2}[L^2+{\left(\frac{L}{2}\right)}^2]=\frac{M{L}^2}{3}$
Frequency of physical pendulum is
$f=\frac{1}{2\pi }\sqrt{\frac{Mgd}{I_A}}$
$f=\frac{1}{2\pi }\sqrt{\frac{MgL\sqrt{10}}{4\frac{M{L}^2}{3}}}=\frac{1}{2\pi }\sqrt{\frac{3g\sqrt{10}}{4L}}$
$f=\frac{1}{2\pi }{\left(\frac{3g\sqrt{10}}{4L}\right)}^{\frac{1}{2}}$