Physics, asked by jgghvfdxgu, 4 months ago

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​

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Answered by BrainlyTwinklingstar
7

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....

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