Question

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

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

Answer 1

Answer: (D)

$\frac{L}{4}$

$L=m\nu R=m\omega R^2$ -----(1)

now angular frequency is doubled

i.e.$\omega \Rightarrow 2\omega$

also KE is halved $\frac{1}{2}m{\omega }^2{R}^2\to \frac{1}{2}\left(\frac{m}{2}{\omega }^2{R}^2\right)$

${\omega }_1=\omega ,{\omega }_2=2\omega$

$\frac{1}{2}\left[\frac{m}{2}{\omega }^2{R}_1^2\right]=\left[\frac{m}{2}\right(2\omega {)}^2\left(R_2{)}^2\right]$

$\frac{R_1^2}{2}=4R_2^2.$

$R_2=\frac{R_1}{2\sqrt{2}}.$

put it in ---(1)

$L^{\prime }=m\left(2\omega \right){\left(\frac{R}{2\sqrt{2}}\right)}^2$

$=\frac{L}{4}$