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(C) Let the position of the upper end be $(0, y)$ and the lower end be $(x, 0)$. Since the rod has a fixed length $L$, we have $x^2 + y^2 = L^2$.Diffe

Vedclass · Accepted Answer

The speed of the lower end gets smaller and smaller and vanishes when the upper end touches the ground.

Vedclass · Answer

(C) Let the position of the upper end be $(0, y)$ and the lower end be $(x, 0)$. Since the rod has a fixed length $L$, we have $x^2 + y^2 = L^2$.Differentiating with respect to time $t$, we get $2x(dx/dt) + 2y(dy/dt) = 0$.Let the velocity of the upper end be $v = -dy/dt$ (downward) and the velocity of the lower end be $v' = dx/dt$ (rightward).Substituting these, we get $x v' - y v = 0$, which implies $v' = v(y/x)$.Since $y/x = 	an 	heta$, where $	heta$ is the angle the rod makes with the floor, we have $v' = v 	an 	heta$.As the upper end moves downward, the angle $	heta$ decreases. As $	heta 	o 0$, $	an 	heta 	o 0$, and consequently $v' 	o 0$.Thus, the speed of the lower end decreases and vanishes when the upper end touches the ground.

$A$ rod of length $L$ leans against a smooth vertical wall while its other end is on a smooth floor. The end that leans against the wall moves uniformly vertically downward. Select the correct alternative.

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