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JEE Main 2017 · Rotational Kinetic Energy, Work & Power

Rotational Kinetic Energy, Work & Power question from JEE Main 2017

Q48JEE Main 2017Medium

Rotational Kinetic Energy, Work & Power

A slender uniform rod of mass M and length ll is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle θ\theta with the vertical is

Figure

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Answer: (b)

  1. Torque of weight about the pivot: τ=Mgl2sin⁡θ\tau = Mg\frac{l}{2}\sin\theta.
  2. I=Ml23I = \frac{Ml^2}{3} about the end.
  3. α=τI=3g2lsin⁡θ\alpha = \frac{\tau}{I} = \frac{3g}{2l}\sin\theta
  4. So the answer is (b) 3g2lsin⁡θ\frac{3g}{2l}\sin\theta.
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