JEE MAIN - Physics (2009 - No. 27)
A thin uniform rod of length $$l$$ and mass $$m$$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $$\omega $$. Its center of mass rises to a maximum height of:
$${1 \over 6}\,\,{{l\omega } \over g}$$
$${1 \over 2}\,\,{{{l^2}{\omega ^2}} \over g}$$
$${1 \over 6}\,\,{{{l^2}{\omega ^2}} \over g}$$
$${1 \over 3}\,\,{{{l^2}{\omega ^2}} \over g}$$
Explanation
The moment of inertia of the rod about $$O$$ is $${1 \over 2}m{\ell ^2}.$$ The maximum angular speed of the rod is when the rod is instantaneously vertical. The energy of the rod in this conditions is $${1 \over 2}I{\omega ^2}$$ where $$I$$ is the moment of inertia of the rod about $$O.$$ when the rod is in its extreme portion, its angular velocity is zero momentarily. In this case, the center of mass is raised through $$h$$, so the increase in potential energy is $$mgh$$. This is equal to kinetic energy $${1 \over 2}I{\omega ^2}$$.
$$\therefore$$ $$mgh = {1 \over 2}I{\omega ^2} = {1 \over 2}\left( {{1 \over 3}m{l^2}} \right){\omega ^2}$$.
$$ \Rightarrow h = {{{\ell ^2}{\omega ^2}} \over {6g}}$$
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