Question

A uniform rod of mass $M$ and length $L$ is placed in a horizontal plane with one end hinged about the vertical axis. A horizontal force $F=\frac{Mg}{2}$ is applied at a distance $\frac{5L}{6}$ from the hinged end. The angular acceleration of the rod will be:

• A. $F=\frac{4g}{5L}$
• B. $F=\frac{5g}{4L}$
• C. $F=\frac{3g}{4L}$
• D. $F=\frac{4g}{3L}$

Answer 1

The correct option is B $F=\frac{5g}{4L}$

Torque applied $\tau =Force\times perpendiculardistance=\frac{Mg}{2}\frac{5L}{6}=\frac{5MgL}{12}=I\alpha$ .....(1) where $I$ is the moment of inertia about the hinge and $\alpha$ is the angular acceleration.
Moment of inertia of the rod about an axis passing through an end of the rod $I=\frac{M{L}^2}{3}$
Substituting $I$ in eqn(1) we get $\frac{5MgL}{12}=\frac{M{L}^2}{3}\alpha$
$\Rightarrow \alpha =\frac{5g}{4L}$