A 2 kg steel rod of length 0.6 m is clamped on a table vertically at its lower end and is free to rotate in vertical plane. The upper end is pushed so that the rod falls under gravity. Ignoring the friction due to clamping at its lower end, the speed of the free end of rod when it passes through its lowest position is _ _ _ _ $m s^{-1}$.

*Solution*

So, $mgl = \frac {1}{2} I \omega ^2 = \frac {1}{2} .\frac {1}{3} ml^2 \omega ^2 $

$\Rightarrow 6g = l \omega ^2 $

Using $v = l \omega $, we have

$6 g = l. \frac {v^2}{l^2}$

$\Rightarrow v^2 = 6gl $

$\Rightarrow v = \sqrt {6gl} = \sqrt {6\times 10 \times 0.6 } = 6ms^{-1} $

Answer: 6