Question

A fan of moment of inertia $0.6kg{m}^2$ is to run upto a working speed of $0.5$ revolution per second. Indicate the correct value of the angular momentum of the fan

• A. $0.6\pi kg\times \frac{metr{e}^2}{sec}$
• B. $6\pi kg\times \frac{metr{e}^2}{sec}$
• C. $3\pi kg\times \frac{metr{e}^2}{sec}$
• D. $\frac{\pi }{6}kg\times \frac{metr{e}^2}{sec}$

Answer 1

The correct option is A $0.6\pi kg\times \frac{metr{e}^2}{sec}$

Angular momentum=I$\omega =0.6\times 0.5\times 2\pi =0.6\pi kg\frac{metr{e}^2}{sec}$