Question

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

Answer 1

Mark the intersection of the horizontal and vertical wall as origin, O.

Now,

$OA=y$

$OB=x$

In triangle $AOB$

$y=lsin\theta$

$x=lcos\theta$

Velocity of end A $u$ will be given by the rate of change of $y$ with respect to time $t$.

Thus,

$u=y^{\prime }=lcos\theta {\theta }^{\prime }$

$v=x^{\prime }=-lsin\theta {\theta }^{\prime }$

As

$l{\theta }^{\prime }=\frac{u}{cos\theta }$

Thus,

$v=-\frac{u}{cos\theta }\times sin\theta$

$v=-utan\theta$

Thus the velocity of the point B will be given by $-utan\theta$

Let the Position of Centre of Mass of the rod be $(x_g,y_g)$

The CM of the rod will lie at a distance $\frac{l}{2}$ of the rod.

Thus,

$x_g=\frac{l}{2}cos\theta =\frac{x}{2}$

$y_g=\frac{l}{2}sin\theta =\frac{y}{2}$

Trajectory of Center of rod will be given by:

$x_g\hat{i}+y_g\hat{j}=\frac{1}{2}(x\hat{i}+y\hat{j})$

Therefore, Velocity of centre of rod is given by:

$\frac{d(x_g\hat{i}+y_g\hat{j})}{dt}=\frac{1}{2}\frac{d(x\hat{i}+y\hat{j})}{dt}$

${\overrightarrow{v}}_g=\frac{1}{2}(\overrightarrow{v}+\overrightarrow{u})$

$v_g=\frac{1}{2}\sqrt{(lcos\theta {\theta }^{\prime }{)}^2+(-lsin\theta {\theta }^{\prime }{)}^2}$

$v_g=\frac{l}{2}\frac{d\theta }{dt}$

or

$v_g=\frac{u}{2cos\theta }$

Therefore, the velocity of centre of rod will be given by; $v_g=\frac{u}{2cos\theta }$