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MHT CET 2023 (14 May, Shift 1) · Rotational Kinetic Energy, Work & Power

Rotational Kinetic Energy, Work & Power question from MHT CET 2023 (14 May, Shift 1)

Q30MHT CET 2023 · 14 MAY · SHIFT 1Medium

Rotational Kinetic Energy, Work & Power

A thin uniform rod of mass 'mm' and length 'll' is suspended from one end which can oscillate in a vertical plane about the point of suspension. It is pulled to one side and then released. It passes through the equilibrium position with angular speed 'ω\omega'. The kinetic energy while passing through mean position is

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Answer: (c)

  1. At the mean position the rod rotates about the suspension point with I=ml23I = \frac{ml^2}{3}.
  2. KE=12Iω2=12⋅ml23ω2=ml2ω26KE = \frac{1}{2}I\omega^2 = \frac{1}{2}\cdot\frac{ml^2}{3}\omega^2 = \frac{ml^2\omega^2}{6}
  3. So the answer is (c) ml2ω26\frac{ml^2\omega^2}{6}.
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