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MHT CET 2023 (13 May, Shift 2) · Rotational Kinetic Energy, Work & Power

Rotational Kinetic Energy, Work & Power question from MHT CET 2023 (13 May, Shift 2)

Q31MHT CET 2023 · 13 MAY · SHIFT 2Medium

Rotational Kinetic Energy, Work & Power

A thin uniform rod AB\text{rod } AB of mass mm and length ll is hinged at one end AA to the ground level. Initially the rod stands vertically and is allowed to fall freely to the ground in the vertical plane. The angular velocity of the rod when its end BB strikes the ground is ( g=g = acceleration due to gravity)

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Answer: (c)

  1. Centre of mass falls from height l2\frac{l}{2} to ground level; this PE becomes rotational KE about AA.
  2. IA=ml23I_A = \frac{ml^2}{3}
  3. mgl2=12⋅ml23ω2⇒ω2=3glmg\frac{l}{2} = \frac{1}{2}\cdot\frac{ml^2}{3}\omega^2 \Rightarrow \omega^2 = \frac{3g}{l}
  4. So the answer is (c) 3gl\sqrt{\frac{3g}{l}}.
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