question_answer

A particle performs uniform circular motion with an angular momentum L. If the frequency of particle's motion is doubled and its K.E. is halved, what happens to its angular momentum?

Answer:

                \[L=\,\,m\,\upsilon \,r\]               and        \[\upsilon =r\omega =r\left( 2\pi n \right)\] \[r=\frac{\upsilon }{2\pi n}\]                       \[\therefore \]  \[L=m\upsilon \left( \frac{\upsilon }{2\pi n} \right)=\frac{m{{\upsilon }^{2}}}{2\pi n}\] As \[K.E.\,=\frac{1}{2}m{{\upsilon }^{2}}\], therefore, \[L=\frac{K.E.}{\pi n}\] When K.E. is halved and frequency (n) is doubled, \[L'=\frac{K.E.'}{\pi n'}=\frac{K.E./2}{\pi \left( 2n \right)}=\frac{K.E.}{4\pi n}=\frac{L}{4}\] i.e. angular momentum becomes one fourth.