Question

A rod of length 'I' is inclined at an angle theta with the floor against a smooth vertical wall. If the end A moves instantaneously with velocity v1, what is the velocity of end B at the instant when rod makes theta angle with the horizontal​

Answer 1

AnSwer :

Let at any instant, end B and A are at a distance x and y respectively from the point 'O'.

|| refer the attachment ||

thus, we can say, x² + y² = l² ............(1)

here l is the length of the rod, which is constant.

now, let's differentiate eq (1) with respect to time, we get

${ : \implies {\sf {\dfrac{d}{dt} ( {x}^{2} + {y}^{2} ) = \dfrac{d}{dt} ( {l}^2)}}}$

${ : \implies {\sf { 2x \dfrac{dx}{dt} + 2y \dfrac{dy}{dt} = 0}}}$

${\sf {lf \: \dfrac{dx}{dt} = v_{2} \: \: \: \: and \: \: \: \: \dfrac{dy}{dt} = - v_{1}}}$

${ : \implies {\sf {x( v_{2}) + y( - v_{1}) = 0}}}$

${ : \implies{ \sf { v_{2} = \bigg( \dfrac{y}{x} \bigg) v_{1}}}}$

${ : \implies {\sf { v_{2} = v_{1}tan \theta \: \: \: \: \: \bigg[ \because \: \dfrac{y}{x} = tan \theta \bigg] }}}$

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#sanvi....