Question

A rod $PQ$ of mass $m$ and length $L$ is hinged at its one end $P$. The rod is kept horizontal by a massless string tied to end $Q$. When the string is cut, initial angular acceleration of the rod is

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

Answer 1

The correct option is A $\frac{3g}{2L}$

When string is cut, the rod will rotate about hinge point $P$.
Let the initial angular acceleration be $\alpha$.


Let $N_x\&N_y$ be the hinge forces but they pass through point $P$, hence their torques about $P$ vanish.

Applying the equation of torque about $P$,
$\sum \tau =I\alpha$
$\Rightarrow {\tau }_{mg}=I\alpha ...\left(i\right)$
From parallel axis theorem, the $\text{MOI}$ of the rod about $P$,
$I=\frac{m{L}^2}{12}+\frac{m{L}^2}{4}$
Substituting in Eq.$\left(i\right)$,
$\frac{L}{2}\times mg=(\frac{m{L}^2}{12}+\frac{m{L}^2}{4})\alpha$
$\Rightarrow \frac{mgL}{2}=\frac{m{L}^2}{3}\alpha$
$\therefore \alpha =\frac{3g}{2L}$
Initial angular acceleration of the rod about $P$ is $\frac{3g}{2L}$