Question

A uniform rod AB of mass $m=2$ kg and length $l=1$ m is kept on a smooth horizontal plane. If the ends A and B of the rod move with speeds $v=3$ m/s and $2v=6$ m/s respectively perpendicular to the rod, find the kinetic energy(in J) of the rod.

Answer 1

$v_{cm}=\frac{6+3}{2}=9/2$

$\omega =\frac{6-3}{l}=3$ ,moment of inertia =$\frac{m{l}^2}{12}$

so $K.E=\frac{m{v}_{cm}^2}{2}+\frac{I{\omega }^2}{2}=21J$