Question

A rigid body is made of three identical thin rods, each of length $L$ fastened together in the form of letter $H$. The body is free to rotate about a horizontal axis that runs along the length of one of the legs of the $H$. The body is allowed to fall from rest from a position in which the plane of $H$ is horizontal. What is the angular speed of the body when the plane of $H$ is vertical?

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

Answer 1

The correct option is C $\frac{3}{2}\sqrt{\frac{g}{L}}$

Let length of each rod be $L$ and mass of each rod is $m$

CM of the middle rod is at $\frac{L}{2}$

CM of the end arm rod of the $H$ is at $\frac{L}{2}$

CM of the system containing end and middle rod $=\frac{m\frac{l}{2}+ml}{2m}=\frac{3}{4}L$

MI about the axis for the middle rod

$=\frac{m{l}^2}{3}$

MI of the end arm with entire mass concentrated at the end of the middle rod

$m{l}^2$

Total MI

$=\frac{m{l}^2}{3}+m{l}^2=\frac{4}{3}{ml}^2$

Total PE at the time of its fall$=$PE$=$Total mass $\times$g$\times$height of CM

$PE=2mg\times \frac{3}{4}l$

KE of the rotating rod system is given by

$=\frac{1}{2}I{w}^2$

$=\frac{1}{2}\frac{4}{3}m{l}^2{w}^2$

At the vertical position all PE is converted to KE. So,

$=\frac{1}{2}\times \frac{4}{3}m{l}^2{w}^2=2mg\times \frac{3}{4}lw=\frac{3}{2}\sqrt{\frac{g}{l}}$