Question

A particle is executing uniform circular motion with angular momentum $J$. If its kinetic energy is reduced to half and its angular frequency is doubled then its angular momentum becomes:

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

Answer 1

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

We have $KE=\frac{1}{2}I{\omega }^2$
or
$KE=\frac{1}{2}J\omega$ (as $J=I\omega$)
Thus we get
$J=\frac{2\times KE}{\omega }$
$J^{\prime }=\frac{2\times (KE{)}^{\prime }}{{\omega }^{\prime }}$
$\frac{J^{\prime }}{J}=\frac{(KE{)}^{\prime }}{\left(KE\right)}\times \frac{\omega }{{\omega }^{\prime }}=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$