Q. A ring of mass $10\ kg$ and diameter $0.4\ m$ is rotated about its axis. If it makes 2100 revolutions per minute, then its angular momentum will be
A) $44\ kg\ \times m^{2}/s$
B) $88\ kg\ \times m^{2}/s$
C) $4.4\ kg\ \times m^{2}/s$
D) $0.4\ kg\ \times m^{2}/s$
Explanation
(b) $I = mr^{2} = 10 \times (0.2)^{2} = 0.4kg - m^{2}$
$\omega = 2\pi\ n = 2\pi \times \frac{2100}{60}rad/s$
$\therefore L = I\omega = \frac{0.4 \times 2\pi \times 210}{6} = 88kg - m^{2}/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.