Question

A disc of moment of inertia $\frac{9.8}{\pi^{2}}kgm^{2}$ is rotating at 600 rpm. If the frequency of rotation changes from 600 rpm to 300 rpm, then what is the work done:

• A. 1467 J
• B. 1452 J
• C. 1567 J
• D. 1632 J

Answer 1

Answer: (A)

1467 J

Answer 2

Work done = Change in rotational kinetic energy $= \frac{1}{2}I \times (\omega_{1}^{2} - \omega_{2}^{2}) = \frac{1}{2}I \times 4\pi^{2}(n_{1}^{2} - n_{2}^{2}) = \frac{1}{2} \times \frac{9.8}{\pi^{2}} \times 4\pi^{2}(10^{2} - 5^{2}) = 9.8 \times 2 \times 75 = 1470\mspace{6mu} J$