One end of a rod of length L=1 ,m is fixed to | vedclass

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Question

$(a)$ Let rod $O A$ makes angle $	heta$ at some instant $t$ with $O P$ as shown below in the figure.

'Then, by applying cosine formula to $	r

Vedclass · Accepted Answer

$\frac{2 \omega}{3}$

Vedclass · Answer

$(a)$ Let rod $O A$ makes angle $	heta$ at some instant $t$ with $O P$ as shown below in the figure.

'Then, by applying cosine formula to $	riangle O A P$, we get

$\cos 	heta =\frac{R^{2}+x^{2}-L^{2}}{2 R x}$

$=R^{2}+x^{2}-L^{2}-2 R x \cos 	heta$

Differentiating above equation with respect to time $t$, we get

$\Rightarrow \frac{d x^{2}}{d t}=2 R\left(x \cdot \frac{ d }{d t} \cos 	heta+\cos 	heta \frac{d x}{d t}ight)$

$\Rightarrow 2 x \frac{d x}{d t}=2 R\left(-x \sin 	heta \frac{d 	heta}{d t}+\cos 	heta \frac{d x}{d t}ight)$

$\Rightarrow \quad x v=-x R \omega \sin 	heta+R v \cos 	heta$

$	herefore \quad \frac{d 	heta}{d t}=\omega$ and $\frac{d x}{d t}=v$

$\Rightarrow \quad v=\frac{R x \omega \cdot \sin 	heta}{x-R \cos 	heta} \quad \dots(i)$

Now, when angle $	heta=60^{\circ}$, then

In $	riangle O A P$, we have $\angle A=90^{\circ}, L=1 \,m$ and $R=\frac{1}{\sqrt{3}} \,m$ So, $x=\sqrt{L^{2}+R^{2}}=\sqrt{1+\frac{1}{3}}=\frac{2}{\sqrt{3}} m$

Substituting values in Eq. $(i)$, we get

$v=\frac{\left(\frac{1}{\sqrt{3}} 	imes \frac{2}{\sqrt{3}} 	imes \sin 60^{\circ}ight) \cdot \omega}{\left(\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{3}} \cos 60^{\circ}ight)}$

$\Rightarrow$ $v=\left(\frac{\frac{2}{3} 	imes \frac{\sqrt{3}}{2}}{\frac{2-1}{\sqrt{3}-2 \sqrt{3}}}ight) \omega \Rightarrow v=\frac{2}{3} \omega$

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