JEE Main 2022 · Physics · Rotational Motion · medium

A uniform thin rod of length 1.2 m is pivoted at one end so that it can rotate freely in a vertical plane. It is released from the horizontal…

Numerical answer

A uniform thin rod of length

1.2\ \mathrm{m}

is pivoted at one end so that it can rotate freely in a vertical plane. It is released from the horizontal position. Taking

g=10\ \mathrm{m\,s^{-2}}

, the speed (in

\mathrm{m\,s^{-1}}

) of the free end when the rod passes through the vertical position is

Show answer and solution

Answer: 6

Energy conservation: the centre of mass falls by

\dfrac L2

, so

mg\dfrac L2=\dfrac12I\omega^{2}

with

I=\dfrac{mL^{2}}{3}

. This gives

\omega^{2}=\dfrac{3g}{L}=\dfrac{30}{1.2}=25

, so

\omega=5\ \mathrm{rad\,s^{-1}}

. The free end has speed

v=\omega L=5\times1.2=6\ \mathrm{m\,s^{-1}}

. Treating the rod as a point mass at its end (

v=\sqrt{2gL}

) would overestimate the speed.

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