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 speed $v=3$ m/s and $2v=6$ m/s respectively perpendicular to the rod, find the Kinetic energy ( in J ) of the rod

Answer 1

The rod rotates about point about which the angular speed of both sides is same. Let the point be at a distance $x$ from end A.

Thus $\frac{v}{x}=\frac{2v}{L-x}$

$⟹x=\frac{L}{3}=\frac{1}{3}m$

The angular speed of rod about this point=$\frac{v}{x}=\frac{3m/s}{1/3}=9rad/s$

Moment of inertia about this axis=$I={\int }_{-1/3}^{2/3}\frac{m}{L}{x}^{2}dx=\frac{2}{9}kg{m}^2$

Thus kinetic energy of the rod=$\frac{1}{2}I{\omega }^2=9J$