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

Answer 1

Answer: (A)

$\frac{3g}{\sqrt{10}\mathcal{l}}$

Answer 2

Moment of inertia about axis AA¢ is

$I_{AA'}$ = $\left\{ \left( \frac{M}{2} \right)\frac{\mathcal{l}^{2}}{12} + \left( \frac{M}{2} \right).\frac{5\mathcal{l}^{2}}{4} + \left( \frac{M}{2} \right)\frac{\mathcal{l}^{2}}{3} \right\}$

\ OD = $\frac{\sqrt{5}\mathcal{l}}{2}$

OC = $\frac{\sqrt{10}\mathcal{l}}{4}$ Torque of Mg will maximum when ‘OC’ is horizontal

\ a = $\frac{Mg.\sqrt{10}\mathcal{l}/4}{\frac{5M\mathcal{l}^{2}}{6}}$=$\frac{3g}{\sqrt{10}\mathcal{l}}$