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One end of a rod of length $L=1 \mathrm{~m}$ is fixed to a point on the circumference of a wheel of radius $\mathrm{R}=1 / \sqrt{3} \mathrm{~m}$. The other end is sliding freely along a straight channel passing through the center $\mathrm{O}$ of the wheel as shown in the figure below. The wheel is rotating with a constant angular velocity $\omega$ about $\mathrm{O}$.
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The correct answer is:
$\frac{2 \omega}{3}$
$\tan 60^{\circ}=\frac{\mathrm{L}}{\mathrm{R}}=\frac{1}{1 / \sqrt{3}}=\sqrt{3}=\tan \theta$
$\therefore \quad \mathrm{x}=\sqrt{1+\frac{1}{3}}=\frac{2}{\sqrt{3}}$
$\cos \theta=\frac{\mathrm{R}^{2}+\mathrm{x}^{2}-\mathrm{L}^{2}}{2 \mathrm{Rx}}$
$\Rightarrow \mathrm{R}^{2}+\mathrm{x}^{2}-\mathrm{L}^{2}=2 \mathrm{Rx} \cos \theta$
$\begin{aligned} \Rightarrow 2 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{R}\left[\mathrm{x}(-\sin \theta)+\cos \theta \frac{\mathrm{dx}}{\mathrm{dt}}\right] \\ & \frac{\mathrm{dx}}{\mathrm{dt}}[\mathrm{x}-\mathrm{R} \cos \theta]=-\mathrm{Rx} \sin \theta \frac{\mathrm{d} \theta}{\mathrm{dt}} \\ \therefore \quad-&{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v} \& \frac{\mathrm{d} \theta}{\mathrm{dt}}=\omega\end{aligned}$
$\Rightarrow \mathrm{v}=\frac{\mathrm{R} \mathrm{x} \sin \theta \omega}{\mathrm{x}-\mathrm{R} \cos \theta}$
$\begin{aligned} &=\frac{\frac{1}{\sqrt{3}} \frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} \cdot \omega}{\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{3}} \cdot \frac{1}{2}}=\frac{\frac{\omega}{\sqrt{3}}}{\frac{1}{\sqrt{3}} \frac{3}{2}}=\frac{2 \omega}{3} \\ \mathrm{v}=& \frac{2 \omega}{3} \end{aligned}$
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