A thin uniform rod of length ' L ' and mass ' M ' is swinging freely along a horizontal axis passing through its centre. Its maximum angular speed is ' ω '. Its centre of mass rises to a maximum height of [ g = gravitational acceleration] (a) (ω²L²)/(12g²) (b) (ω²L²g)/(6) (c) (ω²g)/(12L²) (d) (ω²L²)/(24g)

Answer: (d). Intended method: equate rotational KE to rise in PE, with I about the centre = (ML²)/(12). 1/2·(ML²)/(12)ω² = Mgh h = (ω²L²)/(24g) So the answer is (d) (ω²L²)/(24g).

A thin uniform rod of length ' L ' and mass ' M ' is swinging freely… | MHT CET 2024 (9 May, Shift 1)

Q26MHT CET 2024 · 9 MAY · SHIFT 1Medium

Rotational Kinetic Energy, Work & Power

A thin uniform rod of length ' $L$ ' and mass ' $M$ ' is swinging freely along a horizontal axis passing through its centre. Its maximum angular speed is ' $\omega$ '. Its centre of mass rises to a maximum height of [ $g =$ gravitational acceleration]

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Answer: (d)

Intended method: equate rotational KE to rise in PE, with $I$ about the centre $= \frac{ML^2}{12}$ .
$\frac{1}{2}\cdot\frac{ML^2}{12}\omega^2 = Mgh$
$h = \frac{\omega^2L^2}{24g}$
So the answer is (d) $\frac{\omega^2L^2}{24g}$ .

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Rotational Kinetic Energy, Work & Power question from MHT CET 2024 (9 May, Shift 1)

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