Question

A uniform rod of length $L$ and mass $M$ is pivoted freely at one end. The angular acceleration of the rod when it is at angle $\theta$ to the vertical is:

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

Answer 1

The correct option is C $\frac{3g\sin \theta }{2L}$

Torque, $T=I\alpha$

where $I$ is moment of inertia, and $\alpha$ is angular acceleration.

Also, $T=rFsin\theta$

$\therefore mgsin\theta \left(\frac{l}{2}\right)=\frac{{ml}^2\alpha }{3}$

$\alpha =\frac{3gsin\theta }{2l}$