Question

A flywheel of mass $$60\ kg$$, radius $$40\ cm$$ is revolving $$300$$ revolutions per min. Its kinetic energy will be

Answer 1

Correct option is A. $$480\pi^{2} J$$
$$n = \dfrac {300}{60} = 5\ rev/s$$
$$\omega = 2\pi n = 10\pi$$
$$KE = \dfrac {1}{2} mv^{2} = \dfrac {1}{2} mr^{2}\omega^{2}$$
$$= \dfrac {1}{2}\times 60\times (0.4)^{2} \times (10\pi)^{2}$$
$$= \dfrac {1}{2} \times 60\times 0.16 \times 100\pi^{2} = 480 \pi^{2} J$$