Physics - Rotation Question with Solution | TestHub

The moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis passing through its center and perpendicular to its length is $I=\frac{1}{12}M{L}^2$.

The force $F$ is applied perpendicularly at one end, so the distance from the pivot (center) to the point of application of force is $r=\frac{L}{2}$. The torque produced by the force is $\tau =F\times r=F\times \frac{L}{2}$. Using Newton's second law for rotation, $\tau =I\alpha$, where $\alpha$ is the angular acceleration. So, $$F\frac{L}{2}=\left(\frac{1}{12}M{L}^2\right)\alpha$$. Solving for $\alpha$: $$\alpha =\frac{FL/2}{M{L}^2/12}=\frac{FL}{2}\times \frac{12}{M{L}^2}=\frac{6F}{ML}$$.