A thin uniform rod of length $L$ is resting against a wall and the floor as shown in the figure. Its lower end $A$ is pulled towards left with a constant velocity $v$. Then the downward velocity $v^{\prime}$ of the other end $B$ when the rod makes an angle $\theta$ with the floor is

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Given, length of the uniform rod $=L$ the angle between rod and floor $=\theta$ the angle between wall and floor $=90^{\circ}$ Let the distance between point $O$ to $A$ is $x$ and the distance between point $O$ to $B$ is $y$. Now, By the velocity displacement relation w.r.t $t$

(- negative sign denoted that $y$ is decreasing.) Futher, $\quad x^2+y^2=L^2$ [From the figure] Differentiating above equation w.r.t., time $t$, $ \therefore \quad 2 x \frac{d x}{d t}+2 y \frac{d y}{d x}=0 $ Now, putting the values of $\frac{d x}{d t}$ and $\frac{d y}{d t}$ from Eqs. (i) and (ii), we get $ \therefore \quad x v=y v^{\prime} \text { or } v^{\prime}=\frac{x}{y} v $ or $v^{\prime}=v \cot \theta$ [From the figure, $\because \cot \theta=\frac{x}{y}$ ] So, the downward velocity of the other end $B$ when the rod makes an angle $\theta$ with the floor is $v^{\prime}=v \cot \theta$

A thin uniform rod of length $L$ is resting against a wall and the floor as shown in the figure. Its lower…

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