Question

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

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

Answer 1

Answer: (A)

Moment of inertia about axis $AA'$ is

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