Question

A slender uniform rod of mass $M$ and length $l$ is provided at one end so that it can rotate in a vertical plane$.$ There is negligible friction at the pivot$.$ The free end is held vertically above the pivot and then released$.$ The angular acceleration of the rod when it makes an angle $\theta$ with me

• A. $\frac{3g}{2l}\cos \theta$
• B. $\frac{3g}{2l}\sin \theta$
• C. $\frac{2g}{3l}\sin \theta$
• D. $\frac{2g}{3l}\cos \theta$

Answer 1

The correct option is B $\frac{3g}{2l}\sin \theta$

Take torque about pivot $T=l\alpha$

$mg\sin \theta \frac{l}{2}=\frac{m{l}^2}{3}\alpha$

$\alpha =\frac{3g}{2l}\sin \theta$

Hence,

option $\left(B\right)$ is correct answer.