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 ω. Its centre of mass rises to a maximum height of ( g = acceleration due to gravity) (a) (L²ω²)/(2g) (b) (Lω)/(6g) (c) (Lω)/(2g) (d) (L²ω²)/(6g)

Answer: (d). At the lowest point KE is maximum: 1/2²/3ω² All of it becomes PE at the highest point: mgh = (mL²ω²)/(6) h = (L²ω²)/(6g) So the answer is (d) (L²ω²)/(6g).

A thin uniform rod of length L and mass m is swinging freely about a… | MHT CET 2026 (15 May, Shift 1)

Q10MHT CET 2026 · 15 MAY · SHIFT 1Medium

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 centre of mass rises to a maximum height of ( $g$ = acceleration due to gravity)

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Answer: (d)

At the lowest point KE is maximum: $\frac{1}{2}\cdot\frac{mL^2}{3}\omega^2$
All of it becomes PE at the highest point: $mgh = \frac{mL^2\omega^2}{6}$
$h = \frac{L^2\omega^2}{6g}$
So the answer is (d) $\frac{L^2\omega^2}{6g}$ .

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Rotational Kinetic Energy, Work & Power question from MHT CET 2026 (15 May, Shift 1)

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