Question

A rod of length $l=1m$ leaning against a vertical wall is pulled at its lowest point $A$ with a constant velocity $v=4m{s}^{-1}$. In consequence, the rod rotates in the vertical plane. When the rod makes an angle $\theta ={37}^0$ with the vertical, find the angular velocity of the rod $\left(inrad{s}^{-1}\right)$

Answer 1

Let the horizontal be X-axis and Y-axis be vertical.

And distance between contact with horizontal and vertical wall be'$x$', and distance between contact with vertical and horizontal be'$y$'.

Since the length of ROD is constant.

BY hypotenus theorm ,

$x^2+y^2=l^2$

by differtiating the above equation with '$t$'.

$x\frac{dx}{dt}+y\frac{dy}{dt}=0$ (since , l is constant)

Since,$\frac{dx}{dt}=v_x$ and $\frac{dy}{dt}=v_y$

therefore,

$v_y=v_x\frac{x}{y}$

Given, at positon when the rod makes an angle $={37}^0$ with vertical,

$\Rightarrow tan\left(\theta \right)=tan\left(37\right)=0.75$

from figure ,$tan\left(\theta \right)=\frac{x}{y}$

and given $v_x=4$ .

by substituting the above results . we get,

$v_y=3$.

angular velocity$=|\frac{v_x-v_y}{l}|=|\frac{4i-(-3j)}{1}|=|4i+3j|=5$