Question

A body of mass 20 kg is moving on a circular path of diameter 0.20 m at the rate of 100 revolutions in 3 s. Find:

(i) rotational energy of the body

(ii) angular momentum of the body, π² = 9.86.

Answer 1

\(m =20\ kg;R = \frac{D}{2} = \frac{0.20}{2} = 0.10\ m;\) frequency \(n = \frac{100}{3}\ c.s^{-1}\)

\(\omega = 2\pi n = \frac{2\pi\times100}{3} = \frac{200}{3}\pi\ rad.s^{-1}\)

∴ I = MR² = 20 x 0.1 x 0.1 = 0.2 kg.m²

(i) Rotational energy,