The velocity of end A of a rigid rod placed b | vedclass

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(B) Let the rod have length $L$. The coordinates of ends $A$ and $B$ are $(0, y)$ and $(x, 0)$ respectively.Since the rod is rigid,$x^2 + y^2 = L^2$.D

Vedclass · Accepted Answer

$u 	an 	heta$

Vedclass · Answer

(B) Let the rod have length $L$. The coordinates of ends $A$ and $B$ are $(0, y)$ and $(x, 0)$ respectively.Since the rod is rigid,$x^2 + y^2 = L^2$.Differentiating with respect to time $t$,we get $2x(dx/dt) + 2y(dy/dt) = 0$.Here,$dy/dt = -u$ (as $y$ is decreasing) and $dx/dt = v$ (velocity of end $B$).So,$xv + y(-u) = 0$,which gives $xv = yu$.From the geometry,$x = L \cos 	heta$ and $y = L \sin 	heta$.Substituting these,$(L \cos 	heta)v = (L \sin 	heta)u$.Therefore,$v = u \frac{\sin 	heta}{\cos 	heta} = u 	an 	heta$.

The velocity of end $A$ of a rigid rod placed between two smooth perpendicular walls moves with velocity $u$ along the vertical direction. Find the velocity of end $B$ of the rod,given that the rod always remains in contact with the walls.

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