Question

A uniform rod of length $\ell$ and mass $m$ is free to rotate in a vertical plane about $A$. The rod initially in horizontal position is released. The initial angular acceleration of the rod is (Moment of inertia of rod about $A$ is $\frac{m \ell^{2}}{3}$ ):

• A. $\frac{3 g}{2 \ell}$
• B. $\frac{2 \ell}{3 g }$
• C. $\frac{3 g }{2 \ell^{2}}$
• D. $mg \frac{\ell}{2}$

Answer 1

Answer: (A)

Here $\tau=\operatorname{I} \alpha$
$\Rightarrow ( mg )\left(\frac{\ell}{2}\right)=\left(\frac{ m \ell^{2}}{3}\right)(\alpha) $
$\Rightarrow \alpha=\frac{3 g }{2 \ell}$