Question: 35: An energy of 484 J is spent in increasing the speed of a flywheel from 60 rpm to 360 rpm . The moment of inertia of the flywheel is :
(1) 0.7 \mathrm{~kg}-\mathrm{m}^{2}
(2) 3.22 \mathrm{~kg}-\mathrm{m}^{2}
(3) 30.8 \mathrm{~kg}-\mathrm{m}^{2}
(4) 0.07 \mathrm{~kg}-\mathrm{m}^{2}
Answer: Option (1)
Explanation:
The change in rotational kinetic energy of a flywheel is given by
\Delta K=\frac{1}{2}I\left(\omega_{2}^{2}-\omega_{1}^{2}\right).
Given energy spent \Delta K=484\,\mathrm{J}.
The initial speed is 60 rpm, which is 1\,\mathrm{rps}.
Hence, \omega_{1}=2\pi\,\mathrm{rad\,s^{-1}}.
The final speed is 360 rpm, which is 6\,\mathrm{rps}.
Hence, \omega_{2}=12\pi\,\mathrm{rad\,s^{-1}}.
Substituting in the energy equation, 484=\frac{1}{2}I\left[(12\pi)^{2}-(2\pi)^{2}\right].
Simplifying, 484=\frac{1}{2}I\left(144\pi^{2}-4\pi^{2}\right)=\frac{1}{2}I(140\pi^{2}).
Thus, I=\frac{2\times484}{140\pi^{2}}\approx0.7\,\mathrm{kg\,m^{2}}.
Therefore, the moment of inertia of the flywheel is 0.7\,\mathrm{kg\,m^{2}},
and the correct option is (1).