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 gravity) (a) (L²ω²)/(2g) (b) (Lω)/(6g) (c) (Lω)/(2g) (d) (L²ω²)/(6g)

Answer: (d). At the lowest point all energy is rotational KE; at the highest point it is all PE of the centre of mass. Moment of inertia about the end: I = ML²/3. 1/2²/3ω² = Mgh ⇒ h = (L²ω²)/(6g) So the answer is (d) (L²ω²)/(6g).

A thin uniform rod of length ' L ' and mass ' M ' is swinging freely… | MHT CET 2025 (25 Apr, Shift 2)

Q18MHT CET 2025 · 25 APR · SHIFT 2Medium

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 gravity)

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Answer: (d)

At the lowest point all energy is rotational KE; at the highest point it is all PE of the centre of mass.
Moment of inertia about the end: $I = \frac{ML^2}{3}$ .
$\frac{1}{2}\cdot\frac{ML^2}{3}\omega^2 = Mgh \Rightarrow h = \frac{L^2\omega^2}{6g}$
So the answer is (d) $\frac{L^2\omega^2}{6g}$ .

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Rotational Kinetic Energy, Work & Power question from MHT CET 2025 (25 Apr, Shift 2)

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