Question

A rigid body is made up of three identical thin rods $A,B$ and $C$, 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 ($A$) of the body $H$. 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 $H$ becomes 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 $\frac{3}{2}\sqrt{\frac{g}{L}}$

Moment of inertia of the system about the given axis,
$I=I_A+I_B+I_C...\left(i\right)$
The moment of inertia of rod $A$ about axis passing through $A$ will be zero, since the $\perp$ distance of particles of rod from the axis of rotation will be zero (considering rod is thin).
$\therefore I_A=0$
For the axis of rotation passing through rod $A$, rod $B$ is rotating about its one end.
$\Rightarrow I_B=\frac{M{L}^2}{3}$
For rod $C$ all the points on rod are at a $\perp$ distance $L$ from the axis of rotation, so
$\Rightarrow I_C=\sum m_i{L}^2$
$\Rightarrow I_C=L^2\sum m_i=M{L}^2$
Where $m_i$ is mass of each particle.
Substituting the values in Eq.$\left(i\right)$,
$\Rightarrow I=0+\frac{M{L}^2}{3}+M{L}^2=\frac{4}{3}M{L}^2$

let $\omega$ be the angular speed of body$\left(H\right)$ when it is rotated from horizontal to vertical position.

Gain in rotational kinetic energy of body is,
$\Delta {\text{K.E}}_{\text{Rot}}=\frac{1}{2}I{\omega }^2=\frac{1}{2}\left[\frac{4}{3}M{L}^2\right]{\omega }^2$
$\Rightarrow$When body gets oriented to vertical position, the loss in gravitational $\text{P.E}$ of individual rods is,
$\Delta {\text{P.E}}_A=0$, because position of centre of mass of rod remains same.
$\Delta {\text{P.E}}_B=\frac{MgL}{2}$, because centre of mass of rod $B$ gets lowered through a distance $\frac{L}{2}$
$\Delta {\text{P.E}}_C=MgL$, centre of mass of rod $C$ gets lowered through a distance $L$
$\Rightarrow$Loss in potential energy of the system during this process of rotation will be,
$\Delta {\text{P.E}}_{\text{system}}=0+Mg\frac{L}{2}+MgL$
$\Rightarrow \Delta {\text{P.E}}_{\text{system}}=\frac{3}{2}MgL$
Applying conservation of mechanical energy on the body:
(Gain in Rotational $\text{K.E}$ equals the loss in gravitational $\text{P.E}$ of system)
$\Rightarrow \Delta {\text{K.E}}_{\text{Rot}}=\Delta {\text{P.E}}_{\text{system}}$
$\Rightarrow \frac{2}{3}M{L}^2{\omega }^2=\frac{3}{2}MgL$
$\therefore \omega =\frac{3}{2}\sqrt{\frac{g}{L}}$