Question

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

• A. $2L$
• B. $4L$
• C. $\frac{L}{2}$
• D. $\frac{L}{4}$

Answer 1

Answer: (D)

$\frac{L}{4}$

$\begin{array}{c}\text{Given -}\ast \text{Angular momentum is}\left(L\right)\\ \text{* Frequency of particle's motion (0)}\\ \text{is doubled}⟶\left(20\right)\\ \text{and kinetic energy}kE\to \frac{kE}{2}\\ \text{(halved)}\end{array}$

$\begin{array}{c}\text{we have}L=mvr\text{and}kE=\frac{1}{2}m{v}^2\\ \text{foequency}v=\frac{v}{2\pi }v\\ \text{From the above relations we get}\\ L=\frac{kE}{\pi 0}\\ \text{H kineticenergy is halved and frequency is doubled}\\ L_f=\frac{kE}{4\pi 0}=\frac{L}{4}L\to \frac{L}{4}\\ \text{Henu, option}\left(d\right)\text{is correct.}\end{array}$