A thin uniform metal rod of mass ' M ' and length ' L ' is swinging about a horizontal axis passing through its end. Its maximum angular velocity is ' ω '. Its centre of mass rises to a maximum height of ( g = Acceleration due to gravity) (a) (L²ω²)/(3g) (b) (L²ω²)/(2g) (c) (L²ω²)/(6g) (d) (L²ω²)/(4g)

Answer: (c). Rotational KE at the lowest point becomes PE of the centre of mass at the highest point. I = ML²/3 about the end. 1/2²/3ω² = Mgh ⇒ h = (L²ω²)/(6g) So the answer is (c) (L²ω²)/(6g).

A thin uniform metal rod of mass ' M ' and length ' L ' is swinging… | MHT CET 2024 (15 May, Shift 1)

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Rotational Kinetic Energy, Work & Power

A thin uniform metal rod of mass ' $M$ ' and length ' $L$ ' is swinging about a horizontal axis passing through its end. Its maximum angular velocity is ' $\omega$ '. Its centre of mass rises to a maximum height of ( $g =$ Acceleration due to gravity)

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Answer: (c)

Rotational KE at the lowest point becomes PE of the centre of mass at the highest point.
$I = \frac{ML^2}{3}$ about the end.
$\frac{1}{2}\cdot\frac{ML^2}{3}\omega^2 = Mgh \Rightarrow h = \frac{L^2\omega^2}{6g}$
So the answer is (c) $\frac{L^2\omega^2}{6g}$ .

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Rotational Kinetic Energy, Work & Power question from MHT CET 2024 (15 May, Shift 1)

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