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A thin rod $MN$, free to rotate in the vertical plane about the fixed end $N$, is held horizontal . When the end $M$ is released the speed of this end, when the rod makes an angle $\alpha $ with the horizontal, will be proportional to ( see figure)
$\sqrt {\cos \alpha } $
$\cos \alpha $
$\sin \alpha $
$\sqrt {\sin \alpha } $
Solution
When the rod makes an angle $\alpha $
Displacement of center of mass $ = \frac{1}{2}\cos \,\alpha $
$mg\frac{1}{2}\cos \alpha = \frac{l}{2}I{\omega ^2}\,$
$mg\frac{1}{2}\cos \alpha = \frac{{m{l^2}}}{6}{\omega ^2}$ (M.I. of thin uniform rod about an axis passing through its center of mass and perpendicular to the rod $I = \frac{{m{l^2}}}{{12}}$)
$ \Rightarrow \omega = \sqrt {\frac{{3g\cos \alpha }}{l}} $
speed of end $ = \omega \times l = \sqrt {3g\cos \alpha l} $
i.e., Speed of end, $\omega \, \propto \,\sqrt {\cos \alpha } $
