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.
• B.
• C.
• D.

Answer 1

The correct option is B

$\tau =l\alpha$

$\alpha =\frac{\tau }{I}=\frac{\tau F}{I}=\frac{\left(\frac{5L}{6}\right)\left(\frac{Mg}{2}\right)}{\frac{M{L}^2}{3}}$

$\Rightarrow \alpha =\frac{5g}{4L}$