Question

A uniform rod of length $L$ and mass $M$ is pivoted freely at one end and placed in vertical position. What is the angular acceleration of the rod when it is at an angle $\theta$ with the vertical?

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

Answer 1

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


The torque due to the weight of the rod about $O$,
${\tau }_{mg}=\left(mg\sin \theta \right)\frac{L}{2}$

Also, using torque equation about $O$, we get
$\tau =I\alpha \Rightarrow \frac{mgL}{2}\sin \theta =\frac{m{L}^2}{3}\alpha$
Thus, $\alpha =\frac{3g\sin \theta }{2L}$