The end B of the rod AB which makes an angle | vedclass

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Question

(A) Let the coordinates of $A$ be $(0, y)$ and $B$ be $(x, 0)$. The length of the rod is $l = \sqrt{x^2 + y^2}$.Since $x = l \cos 	heta$ and $y = l \

Vedclass · Accepted Answer

velocity of end $A$ is $\frac{4}{3} v_0$ downwards

Vedclass · Answer

(A) Let the coordinates of $A$ be $(0, y)$ and $B$ be $(x, 0)$. The length of the rod is $l = \sqrt{x^2 + y^2}$.Since $x = l \cos 	heta$ and $y = l \sin 	heta$,we have $v_0 = \frac{dx}{dt} = -l \sin 	heta \frac{d	heta}{dt}$.Thus,the angular velocity $\omega = |\frac{d	heta}{dt}| = \frac{v_0}{l \sin 	heta}$.For end $A$,$v_A = |\frac{dy}{dt}| = |l \cos 	heta \frac{d	heta}{dt}| = l \cos 	heta (\frac{v_0}{l \sin 	heta}) = v_0 \cot 	heta$.At $	heta = 37^o$,$\cot 37^o = \frac{4}{3}$,so $v_A = \frac{4}{3} v_0$.Also,$\omega = \frac{v_0}{l \sin 37^o} = \frac{v_0}{l (3/5)} = \frac{5}{3} \frac{v_0}{l}$.Since $v_A$ and $\omega$ depend on $	heta$,they are not constant. Thus,option $A$ is correct.

The end $B$ of the rod $AB$ which makes an angle $\theta$ with the floor is being pulled with a constant velocity $v_0$ as shown. The length of the rod is $l$. At the instant when $\theta = 37^o$,then:

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