Q. A disc of moment of inertia $\frac{9.8}{\pi^{2}}kg\ m^{2}$ is rotating of $600\ rpm$. If the frequency of rotation changes from $600\ rpm$ to $300\ rpm$, then what is the work done
A) $1467\ J$
B) $1452\ J$
C) $1567\ J$
D) $1632\ J$
Explanation
(a) Work done = Change in rotational kinetic energy
$= \frac{1}{2}I \times \left( \omega_{1}^{2} - \omega_{2}^{2} \right) = \frac{1}{2}I \times 4\pi^{2}(n_{1}^{2} - n_{2}^{2})$
$= \frac{1}{2} \times \frac{9.8}{\pi^{2}} \times 4\pi^{2}\left( 10^{2} - 5^{2} \right) = 9.8 \times 2 \times 75 = 1470\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.