Question

A smooth uniform rod $AB$ of mass $M$ and length $l$ rotates freely with an angular velocity ${\omega }_0$, in a horizontal plane about a stationary vertical axis passing through its end $A$. A small sleeve of mass $m$ starts sliding along the rod from the point $A$. The velocity $v^{\prime }$ of the sleeve relative to the rod at the moment it reaches its other end $B$ is $\frac{{\omega }_{0}l}{\sqrt{1+\frac{xm}{M}}}$. Find the value of $x$.

Answer 1

In the process of motion of the given system the kinetic energy and the angular momentum relative to rotation axis do not vary.

Hence, it follows that:

$\frac{1}{2}\frac{M{l}^2}{3}{\omega }_0^2=\frac{1}{2}m({\omega }^2{l}^2+{v^{\prime }}^2)+\frac{1}{2}\frac{M{l}^2}{3}{\omega }^2$ ($\omega$ is the final angular velocity of the rod)

$\frac{M{l}^2}{3}{\omega }_0=\frac{M{l}^2}{3}\omega +m{l}^2\omega$

From these equations we obtain:
$\omega =\frac{{\omega }_0}{(1+\frac{3M}{M})}$

$v^{\prime }=\frac{{\omega }_{0}l}{\sqrt{1+\frac{3m}{M}}}$