Question 25 of 126

A $2kg$ steel rod of length $0.6m$ is clamped on a table vertically at its lower end and is free to rotate in the 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 the rod when it passes through its lowest position is _______m{s}^{-1}.
(Take $g=10{ms}^{-2})$

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Answer: 6

In the above case the torque due to weight of the rod, here use the concept of translation work done and rotational kinetic energy,
$\Rightarrow W=K.E.\Rightarrow mgl=\dfrac{1}{2}I{\omega }^{2}$, now the moment of inertia of the rod when the axis of rotation passes through one end of the rod, $I=\dfrac{1\times m{l}^{2}}{3}$,
$\Rightarrow mgl=\dfrac{1}{2}\times \dfrac{m{l}^{2}}{3}{\omega }^{2}$
$\Rightarrow \omega =\sqrt{\dfrac{6g}{l}}$, now use the relation between linear velocity and angular velocity,
$\Rightarrow v=l\omega =\sqrt{3.6g}=6m{s}^{-1}$