Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

As we know that, angular momentum, l=mvr
$\Rightarrow L=mv\left(\frac{v}{\omega }\right)[\because v=r\omega ]$
$\Rightarrow L=\frac{m{v}^2}{\omega }\Rightarrow L=\frac{2}{\omega }\left(\frac{1}{2}m{v}^2\right)$
$\Rightarrow L=\frac{2}{\omega }(K.E)\Rightarrow L\propto \frac{K.E}{\omega }$
$\Rightarrow \frac{L_2}{L_1}=\frac{K{E}_2}{\left(K{E}_1\right)}\frac{{\omega }_1}{{\omega }_2}=\left(\frac{K{E}_1}{2K{E}_1}\right)\left(\frac{{\omega }_1}{2{\omega }_1}\right)$
$L_2=\frac{L_1}{4}$