Question

A fan of moment of inertia $0.6$ kg $\times$ $metr{e}^2$ 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, J= I$\omega$

J= (0.6$\times 2\pi r$) Kg.$m^2{s}^{-1}$

= 0.6$\times \pi (2\times 0.5)$ kg.$m^2{s}^{-1}$

= 0.6$\pi$ kg.$m^2{s}^{-1}$