Question

A particle performs uniform circular motion with an angular momentum $L$. If the angular frequency 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

The correct option is D $\frac{L}{4}$

$L=mvr$

$\alpha =0$

$v=\omega r$

$L=m\omega r^2$

$\frac{1}{2}m{v}^2=k$

$L=I\omega$

$k=\frac{1}{2}I{\omega }^2$

${\omega }^1=2\omega$

$k^1=1/2k$

$\therefore I^1=\frac{1}{8}I$

$L^1=\frac{1}{8}I\times 2\omega$

$=\frac{1}{4}I\omega$

$=L/4$