Question

A particle performs uniform circular motion with an angular momentum L. If the frequency of particle's motion is doubled and its kinetic energy halved, the angular momentum becomes:

• A. $2L$
• B. $4L$
• C. $L/2$
• D. $L/4$

Answer 1

Answer: (D)

$L/4$

$\begin{array}{c}\text{we have}L=mvr\text{and}kE=\frac{1}{2}m{v}^2\\ \text{frequency}D=\frac{v}{2\pi r}\\ \text{From the above relations we get}\\ L=\frac{kE}{\pi 0}\end{array}$

$\begin{array}{c}\text{If kineticenergy is halved and frequency is doubled}\\ L_f=\frac{kE}{4\pi 0}=\frac{L}{4}L\to \frac{L}{4}\end{array}$