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JEE Main 2009 (AIEEE) · Rotational Kinetic Energy, Work & Power

Rotational Kinetic Energy, Work & Power question from JEE Main 2009 (AIEEE)

Q51JEE Main 2009 · AIEEEMedium

Rotational Kinetic Energy, Work & Power

A thin uniform rod of length $l$ and mass $m$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega$ . Its center of mass rises to a maximum height of:

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Answer: (c)

Rotational KE at the lowest point equals the gain in PE of the centre of mass.
$\frac{1}{2}\cdot\frac{ml^2}{3}\omega^2 = mgh \Rightarrow h = \frac{l^2\omega^2}{6g}$
So the answer is (c) $\frac{1}{6}\frac{l^2\omega^2}{g}$ .

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