JEE MAIN - Physics (2017 (Offline) - No. 20)
A slender uniform rod of mass M and length $$l$$ 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 $$\theta$$ with the vertical is
_en_20_1.png)
$${{2g} \over {3l}}\cos \theta $$
$${{3g} \over {2l}}\sin \theta $$
$${{2g} \over {3l}}\sin \theta $$
$${{3g} \over {3l}}\sin \theta $$
Explanation
Forces acting on the rod are shown below.
_en_20_2.png)
Torque about pivot point O due to force Nx and Ny are zero.
Mgcos$$\theta $$ is passing through O, so torque will be zero due to this force.
So torque about the point O is
$$\tau = Mg\sin \theta \times {l \over 2}$$
We know, $$\tau = I\alpha $$
$$\therefore$$ $$I\alpha = Mg\sin \theta \times {l \over 2}$$
Moment of inertia of rod about point O, $$I$$ = $${{M{l^2}} \over 3}$$
$$\therefore$$ $${{M{l^2}} \over 3} \times \alpha = Mg\sin \theta \times {l \over 2}$$
$$ \Rightarrow \alpha = {3 \over 2}{g \over l}\sin \theta $$
Torque about pivot point O due to force Nx and Ny are zero.
Mgcos$$\theta $$ is passing through O, so torque will be zero due to this force.
So torque about the point O is
$$\tau = Mg\sin \theta \times {l \over 2}$$
We know, $$\tau = I\alpha $$
$$\therefore$$ $$I\alpha = Mg\sin \theta \times {l \over 2}$$
Moment of inertia of rod about point O, $$I$$ = $${{M{l^2}} \over 3}$$
$$\therefore$$ $${{M{l^2}} \over 3} \times \alpha = Mg\sin \theta \times {l \over 2}$$
$$ \Rightarrow \alpha = {3 \over 2}{g \over l}\sin \theta $$
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