Question

A uniform rod of mass $M$ and length $L$ lies radially on a disc of radius $2R$ 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. $Noneofthese$

Answer 1

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

Rotational KE of the rod of mass m is: $K{E}_{rot}=\frac{1}{2}I{\omega }^2$
But MI about the centre of the disc by applying parallel axis theorem is:
$I=\frac{m{L}^2}{12}+m{R}^2$

Hence, $K{E}_{rot}=\frac{1}{2}(\frac{m{L}^2}{12}+m{R}^2){\omega }^2=\frac{1}{2}m{\omega }^2(\frac{L^2}{12}+R^2)$