A uniform rod of length $60 \mathrm{~cm}$ is placed with one end in contact with the horizontal table and is then inclined at an angle $30^{\circ}$ to the horizontal and allowed to fall. The angular velocity of the rod when it becomes horizontal is (acceleration due to gravity $=10…

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A uniform rod of length $60 \mathrm{~cm}$ is placed with one end in contact with the horizontal table and is then inclined at an angle $30^{\circ}$ to the horizontal and allowed to fall. The angular velocity of the rod when it becomes horizontal is (acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

Aicharya · Accepted Answer

( As rod falls, its loss of gravitational potential energy appears as gain of angular rotational energy.

If rod is rotating at angular speed $\omega$ when it is horizontal then, Rotational K E = Loss of gravitational PE $ \Rightarrow \quad \frac{1}{2} I \omega^2=m g h $ Now for rod, Moment of inertia of a solid rod about its one end, $ I=\frac{M L^2}{3} $ and $h=\frac{l}{2} \sin \theta$ (see figure $\theta=30^{\circ}$ ) $ =\frac{l}{2} \times \frac{1}{2}=\frac{l}{4} $ So from eq. (i), we have $ \frac{1}{2} M \frac{l^2}{3} \times \omega^2=M g \frac{l}{4} $ $ \Rightarrow $ $ \omega^2=\frac{3}{2} \times \frac{g}{l} $ Here, $g=10 \mathrm{~m} / \mathrm{s}^2$ and $l=0.6 \mathrm{~m}$ So, $\omega^2=\frac{3}{2} \times \frac{10}{0.6}=25$ or $\quad \omega=5 \mathrm{rad} / \mathrm{s}$

A uniform rod of length $60 \mathrm{~cm}$ is placed with one end in contact with the horizontal table and is…

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