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

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(A) Let the rod $OA$ make an angle $	heta$ with the line $OP$ at any instant $t$. Let $x$ be the distance $OP$. By applying the law of cosines in $

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the rod $OA$ make an angle $	heta$ with the line $OP$ at any instant $t$. Let $x$ be the distance $OP$. By applying the law of cosines in $	riangle OAP$:$L^2 = R^2 + x^2 - 2Rx \cos 	heta$$x^2 - (2R \cos 	heta)x + (R^2 - L^2) = 0$Differentiating with respect to time $t$:$2x \frac{dx}{dt} - 2R \cos 	heta \frac{dx}{dt} + 2Rx \sin 	heta \frac{d	heta}{dt} = 0$Given $\frac{d	heta}{dt} = \omega$ and $\frac{dx}{dt} = v$:$v(x - R \cos 	heta) = -Rx \omega \sin 	heta$$v = \frac{Rx \omega \sin 	heta}{R \cos 	heta - x}$At $	heta = 60^{\circ}$,$R = 1/\sqrt{3}$,$L = 1$. From the geometry,$x = R \cos 	heta + \sqrt{L^2 - R^2 \sin^2 	heta} = \frac{1}{\sqrt{3}} \cdot \frac{1}{2} + \sqrt{1 - \frac{1}{3} \cdot \frac{3}{4}} = \frac{1}{2\sqrt{3}} + \sqrt{\frac{3}{4}} = \frac{1}{2\sqrt{3}} + \frac{\sqrt{3}}{2} = \frac{1+3}{2\sqrt{3}} = \frac{2}{\sqrt{3}} \,m$.Substituting values into the expression for $v$:$v = \frac{(1/\sqrt{3}) \cdot (2/\sqrt{3}) \cdot \omega \cdot \sin 60^{\circ}}{(1/\sqrt{3}) \cdot \cos 60^{\circ} - (2/\sqrt{3})}$$v = \frac{(2/3) \cdot \omega \cdot (\sqrt{3}/2)}{(1/2\sqrt{3}) - (2/\sqrt{3})} = \frac{\omega / \sqrt{3}}{-3 / 2\sqrt{3}} = -\frac{2}{3} \omega$.The speed is the magnitude,so $|v| = \frac{2}{3} \omega$.

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