Q. A flywheel is in the form of solid circular wheel of mass $72\ kg$ and radius of $0.5m$ and it takes $70\ r.p.m$, then the energy of revolution is
A) $24\ J$
B) $240\ J$
C) $2.4\ J$
D) $2400\ J$
Explanation
(b) $K_{R} = \frac{1}{2}I\omega^{2} = \frac{1}{2}\left( \frac{1}{2}MR^{2} \right)(2\pi n)^{2}$
$= \frac{1}{2}\left( \frac{1}{2} \times 72 \times (0.5)^{2} \right) \times 4\pi^{2} \times \left( \frac{70}{60} \right)^{2} = 240\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.