Question

Angular momentum of a particle executing uniform circular motion is L If its angular frequency is doubled and kinetic energy is halved, then its new angular momentum will be:

(a) \(\frac{L}{4}\)

(b) 2L

(c) 4L

(d) \(\frac{L}{2}\)

Answer 1

Angular momentum L = Iω and kinetic energy K = \(\frac{1}{2}\) Iω²

When angular velocity is doubled and kinetic energy is halved, then angular momentum,

\(L' = \frac{2\left(\frac{K}{2}\right)}{2\omega} = \frac{K}{2\omega} = \frac{2K}{4\omega} = \frac{L}{4}\ or\ L' = \frac{L}{4}\)