Question

A light rigid rod of length L has a bob of mass M attached to one of its end just like a simple pendulum. Speed at the lowest point when it is inverted and released is

• A. $\sqrt{gL}$
• B. $\sqrt{2gL}$
• C. $2\sqrt{gL}$
• D. $\sqrt{5gL}$

Answer 1

The correct option is C $2\sqrt{gL}$

Taking the initial lowest point as the reference for Potential energy.
So PE of the bob at the inverted position is
$PE=mg\left(2L\right)$
Conserving energy between the top most point and the lowest point
$P{E}_t=K{E}_l$
$\Rightarrow 2mgL=\frac{1}{2}m{v}^2$
$\Rightarrow v=2\sqrt{gL}$
Option C.