In the instant shown in the figure, the board is moving vertically with constant velocity $(v) .$ The drum wind up at a constant rate $\omega=15 \mathrm{rad} / \mathrm{s}$. If the radius of the drum is $R=10 \mathrm{~cm}$ and the board always remains horizontal. Find the velocity of board at this moment( in $\mathrm{m} / \mathrm{s}$).

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Aicharya · Accepted Answer

Let us assume the lengths to be $z$ and $y$. $\therefore$ Total length $l=y+z$. Assuming the distance in the cower to be $\begin{aligned} l=y+\sqrt{y^{2}+l_{1}^{2}} \ \Rightarrow \frac{d l}{d t} =\frac{d y}{d t}+\frac{2 y}{x} \frac{d y}{\sqrt{y^{2}+l_{1}}} \frac{d y}{d t} \ \text { Here } \frac{d y}{d t}=v \ \omega R =v+\cos \theta \nu \ \omega_{R}=v(1+\cos \theta) \ v=\frac{\omega R}{1+\cos \theta} \end{aligned}$

In the instant shown in the figure, the board is moving vertically with constant velocity \((v) .\) The drum…

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