Question

A uniform pole of length $L$ and mass $M$ is pivoted on the ground with a frictionless hinge. The pole makes an angle $\theta$ with the horizontal. The moment of inertia of the pole about one end is $\frac{1}{3}M{L}^2$. If it starts falling from the position shown in the accompanying figure, the linear acceleration of the free end of the pole immediately after release would be :

• A. $\frac{2}{3}gcos\theta$
• B. $\frac{2}{3}g$
• C. $g$
• D. $\frac{3}{2}gcos\theta$

Answer 1

The correct option is D $\frac{3}{2}gcos\theta$

We write the torque $\tau$ as
$Mgcos\theta \times \frac{L}{2}=I\alpha$
or
$Mgcos\theta \times \frac{L}{2}=\frac{M{L}^2}{3}\alpha$
or
$\alpha =\frac{3}{2}\frac{gcos\theta }{L}$
or
$a=\alpha L=\frac{3}{2}gcos\theta$