Question

A $L$ shaped rod of mass $M$ is free to rotate in a vertical plane about axis $A{A}^{\prime }$ as shown in figure. Maximum angular acceleration of rod is

• A. $\frac{3g}{\sqrt{10}l}$
• B. $\frac{9g}{10l}$
• C. $\frac{9g}{5l}$
• D. $\frac{6g}{5\sqrt{2}l}$

Answer 1

Answer: (A)

Moment of inertia about axis $A{A}^{\prime }$ is

$I_{A{A}^{\prime }}=\left\{\left(\frac{M}{2}\right)\frac{l^2}{12}+\left(\frac{M}{2}\right)\frac{5l^2}{4}+\left(\frac{M}{2}\right)\frac{l^2}{3}\right\}$
$\therefore OD=\frac{\sqrt{5}l}{2}$
$OC=\frac{\sqrt{10}l}{4}$ ( point $C$ is centre of mass of the system )
Torque of $Mg$ will be maximum when $OC$ is horizontal
$\therefore \alpha =\frac{\tau }{I_{A{A}^{\prime }}}=\frac{\frac{Mg\sqrt{10}l}{4}}{\frac{5M{l}^2}{6}}$
$\alpha =\frac{3g}{\sqrt{10}l}$