question_answer

A particle performing uniform circular motion has angular momentum L. What will be the new angular momentum, if its angular frequency is doubled and its kinetic energy halved?

Answer:

                Rotational K.E., \[K=\frac{1}{2}I{{\omega }^{2}}\]             \[\therefore \]  \[I=\frac{2K}{{{\omega }^{2}}}\] Angular momentum, \[L=I\omega =\frac{2K}{{{\omega }^{2}}}.\omega =\frac{2K}{\omega }\] When angular frequency is doubled and kinetic energy is halved, the angular momentum becomes, \[L'=\frac{2(K/2)}{2\omega }=\frac{1}{4}.\frac{2K}{\omega }=\frac{\mathbf{L}}{\mathbf{4}}\]