Question

The kinetic energy of a body rotating at 300 revolutions per minute is 62.8 J. Its angular momentum $(\text{in kg}\,{{m}^{2}}{{s}^{-1}})$ is approximately:

• A. 1
• B. 2
• C. 4
• D. 8

Answer 1

Answer: (C)

4

Answer 2

The correct answer is C:4
Kinetic energy of rotating body $K=\frac{1}{2}I{{\omega }^{2}}=62.8\,J$
Angular momentum, $L=I\omega$ $=\left( \frac{1}{2}I{{\omega }^{2}} \right)\times \frac{2}{\omega }$
$=\frac{62.8\times 2}{(2\times \pi \times 300/60)}$ $=4\,kg\,{{m}^{2}}{{s}^{-1}}$