Question

The velocity of the end 'A' of a rigid rod placed between two smooth vertical walls is ${}^{\prime }{u}^{\prime }$ along the vertical direction. Find out the velocity of end 'B' of the rod if rod always remains in contact with the vertical wall and also find the velocity of the centre of the rod. Also, find the equation of path of the centre of the rod.

Answer 1

From diagram, let length of rod be $l$.

At any instant,

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

$\Rightarrow \frac{2xdx}{dt}+\frac{2ydy}{dt}=0$

$\Rightarrow x.\frac{dx}{dt}=-y\frac{dy}{dt}$

$\Rightarrow l\cos \theta .\frac{dx}{dt}=l\sin \theta .u$

$\Rightarrow \frac{dx}{dt}=u\tan \theta$,

So, velocity of end $B$ is $u\tan \theta$

similarly for centre,

$V_x=\frac{d\left(\frac{x}{2}\right)}{dt}$; $V_y=\frac{d\left(\frac{y}{2}\right)}{dt}$

$\Rightarrow V_x=\frac{1}{2}\frac{dx}{dt}=\frac{u\tan \theta }{2}$; $V_y=\frac{1}{2}\frac{dy}{dt}=\frac{-u}{2}$

so, velocity of $C$ is $(\frac{u\tan \theta }{2}\hat{i},-\frac{u}{2}\hat{i})$.

$x$ coordinate of $C$ is always $\frac{1}{2}\cos \theta$; when $\theta$ is parameter.

$y$ coordinate of $C$ is always $\frac{1}{2}\sin \theta$.

Eliminating $\theta$ from both we get.

$x^2+y^2={\left(\frac{l}{2}\right)}^2({\sin }^2\theta +{\cos }^2\theta )$

$\Rightarrow x^2+y^2={\left(\frac{l}{2}\right)}^2$