Physics - Rotation Question with Solution | TestHub

The moment of inertia of a thin rod of mass $m$ and length $L$ about an axis passing through its center and perpendicular to its length is $I=\frac{m{L}^2}{12}$.

Given the length of the rod is $L=2l$, the moment of inertia is:

$I=\frac{m(2l{)}^2}{12}=\frac{4m{l}^2}{12}=\frac{m{l}^2}{3}$

The angular acceleration $\alpha$ is given by the rate of change of angular velocity:

$\alpha =\frac{\omega -0}{t}=\frac{\omega }{t}$

The torque $\tau$ acting on the rod is:

$\tau =I\alpha =\left(\frac{m{l}^2}{3}\right)\left(\frac{\omega }{t}\right)=\frac{m{l}^2\omega }{3t}$

Answer: $\frac{m{l}^2\omega }{3t}$