Question

A uniform rod of mass m and length L lies radially on a disc rotating with angular speed $\omega$ in a horizontal plane about its axis The rod does not slip on the disc and the centre of the rod is at a distance R from the centre of the disc. Then the kinetic energy of the rod is

• A. $\frac{1}{2}m{\omega }^2[R^2+\frac{L^2}{12}]$
• B. $\frac{1}{2}m{\omega }^2{R}^2$
• C. $\frac{1}{24}m{\omega }^2{L}^2$
• D. $\frac{1}{12}m{\omega }^2{L}^2$

Answer 1

The correct option is A $\frac{1}{2}m{\omega }^2[R^2+\frac{L^2}{12}]$

Moment of inertia of the rod w.r.t. the axis through centre of the disc is (by parallel axis theorem)
$I=\frac{m{L}^2}{12}+m{R}^2$ and K.E. of rod w.r.t. disc
=$\frac{1}{2}I{\omega }^2=\frac{1}{2}m{\omega }^2[R^2+\frac{L^2}{12}]$