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QJEEM17Phe9b6: JEE Main 2017 Physics - Class 11 JEEM17Phe9b6 Solution: A slender uniform rod of mass M and length is pivoted at one…

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

JEE Main 2017 (Offline) Physics - Rotational Motion Question 192 English

Multiple Choice Options

  • A 2g3lcos⁡θ{{2g} \over {3l}}\cos \theta
  • B 3g2lsin⁡θ{{3g} \over {2l}}\sin \theta
  • C 2g3lsin⁡θ{{2g} \over {3l}}\sin \theta
  • D 3g3lsin⁡θ{{3g} \over {3l}}\sin \theta

Step-by-Step Verified Solution

Forces acting on the rod are shown below. JEE Main 2017 (Offline) Physics - Rotational Motion Question 192 English Explanation Torque about pivot point O due to force Nx and Ny are zero.

Mgcos θ\theta is passing through O, so torque will be zero due to this force.

So torque about the point O is

τ=Mgsin⁡θ×l2\tau = Mg\sin \theta \times {l \over 2}

We know, τ=Iα\tau = I\alpha

∴\therefore Iα=Mgsin⁡θ×l2I\alpha = Mg\sin \theta \times {l \over 2}

Moment of inertia of rod about point O, II = Ml23{{M{l^2}} \over 3}

∴\therefore Ml23×α=Mgsin⁡θ×l2{{M{l^2}} \over 3} \times \alpha = Mg\sin \theta \times {l \over 2}

⇒α=32glsin⁡θ\Rightarrow \alpha = {3 \over 2}{g \over l}\sin \theta

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