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A thin rod $\mathrm{MN}$, free to rotate in the vertical plane about the fixed end $\mathrm{N}$, is held horizontal. When the end $\mathrm{M}$ is released the speed of this end, when the rod makes an angle $\alpha$ with the horizontal, will be proportional to: (see figure)
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Verified Answer
The correct answer is:
$\sqrt{\cos \alpha}$
$\sqrt{\cos \alpha}$
When the rod makes an angle $\alpha$
$$
\begin{aligned}
&\text { Displacement of centre of mass }=\frac{l}{2} \cos \alpha \\
&m g \frac{l}{2} \cos \alpha=\frac{l}{2} I \omega^2 \\
&m g \frac{l}{2} \cos \alpha=\frac{m l^2}{6} \omega^2 \quad(\because \quad \text { M.I. of thin }
\end{aligned}
$$
uniform rod about an axis passing through its centre of mass and perpendicular to the
$$
\begin{aligned}
&\left.\operatorname{rod} I=\frac{m l^2}{12}\right) \\
&\Rightarrow \omega=\sqrt{\frac{3 g \cos \alpha}{l}}
\end{aligned}
$$
Speed of end $=\omega \times l=\sqrt{3 g \cos \alpha l}$
i.e., Speed of end, $\omega \propto \sqrt{\cos \alpha}$
$$
\begin{aligned}
&\text { Displacement of centre of mass }=\frac{l}{2} \cos \alpha \\
&m g \frac{l}{2} \cos \alpha=\frac{l}{2} I \omega^2 \\
&m g \frac{l}{2} \cos \alpha=\frac{m l^2}{6} \omega^2 \quad(\because \quad \text { M.I. of thin }
\end{aligned}
$$
uniform rod about an axis passing through its centre of mass and perpendicular to the
$$
\begin{aligned}
&\left.\operatorname{rod} I=\frac{m l^2}{12}\right) \\
&\Rightarrow \omega=\sqrt{\frac{3 g \cos \alpha}{l}}
\end{aligned}
$$
Speed of end $=\omega \times l=\sqrt{3 g \cos \alpha l}$
i.e., Speed of end, $\omega \propto \sqrt{\cos \alpha}$
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