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(D) Let the rod be of length $L$. The coordinates of end $A$ are $(x, 0)$ and end $B$ are $(0, y)$.From the geometry,$x = L \cos 	heta$ and $y = L \s

Vedclass · Accepted Answer

$V_A \cot 	heta$

Vedclass · Answer

(D) Let the rod be of length $L$. The coordinates of end $A$ are $(x, 0)$ and end $B$ are $(0, y)$.From the geometry,$x = L \cos 	heta$ and $y = L \sin 	heta$.The velocity of end $A$ is $V_A = \frac{dx}{dt} = -L \sin 	heta \frac{d	heta}{dt}$. (Taking magnitude,$V_A = L \sin 	heta \frac{d	heta}{dt}$)The velocity of end $B$ is $V_B = \frac{dy}{dt} = L \cos 	heta \frac{d	heta}{dt}$.Dividing the two expressions: $\frac{V_B}{V_A} = \frac{L \cos 	heta \frac{d	heta}{dt}}{L \sin 	heta \frac{d	heta}{dt}} = \cot 	heta$.Therefore,$V_B = V_A \cot 	heta$.

$A$ rod is constrained to move along a vertical wall and a horizontal floor. If the velocity of end $A$ is $V_A$ along the horizontal floor,find the velocity $V_B$ of end $B$ along the vertical wall in terms of $V_A$ and $\theta$.

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