Physics - Rotation Question with Solution | TestHub

The torque acting on the rod about the pivot is due to the gravitational force acting at its center of mass.

The center of mass of a uniform rod of length $L$ is at a distance of $\frac{L}{2}$ from the pivot.

When the rod makes an angle $\theta$ with the vertical, the perpendicular distance from the pivot to the line of action of the gravitational force is $\frac{L}{2}\sin \theta$.

Therefore, the torque $\tau$ is given by:

$\tau =mg\left(\frac{L}{2}\sin \theta \right)$

The moment of inertia $I$ of the rod about an axis passing through its end is:

$I=\frac{m{L}^2}{3}$

Using Newton's second law for rotational motion, $\tau =I\alpha$, where $\alpha$ is the angular acceleration:

$mg\left(\frac{L}{2}\sin \theta \right)=\left(\frac{m{L}^2}{3}\right)\alpha$

Solving for $\alpha$, we get:

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

Answer: $\frac{3\mathrm{g}\sin \theta }{2\mathrm{L}}$