A rod length $AB$ is moving with ends remaining in contact with frictionless wall and floor. If at the instant shown, the velocity of end $B$ is 3 $ms^{- 1}$ towards negative $x$-direction, then magnitude of velocity of end $A$ will be [image:5_image46.emf]

Question

Dr. Abubacker · Accepted Answer

(b) From geometry, it is clear that $x = l\cos	heta$ and $y = l\sin	heta$
$	herefore\ v_{x} = \frac{dx}{dv} = - l\sin{	heta\frac{d	heta}{dt}}\ $ and $v_{y} = \frac{dy}{dt} = l\cos	heta\frac{d	heta}{dt}$
[image:5_image186.jpeg]
$	herefore\ \frac{v_{y}}{v_{x}} = - \cot	heta\ $ or $v_{y} = - v_{x}\cot	heta$
Since, $v_{x} = - 3\ ms^{- 1},$ hence
$v_{y} = - ( - 3)\cot{60{^\circ} = 3 	imes \frac{1}{\sqrt{3}} =}\sqrt{3}\ ms^{- 1}$

Q. A rod length $AB$ is moving with ends remaining in contact with frictionless wall and floor. If at the instant shown, the velocity of end $B$ is 3 $ms^{- 1}$ towards negative $x$-direction, then magnitude of velocity of end $A$ will be

Explanation

Reviewed by Dr. Abubacker

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