Question

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

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

Answer 1

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

$\therefore$ $K=\frac{1}{2}I{\omega }^2=\frac{1}{2}(I\times \omega )\times \omega$
$\Rightarrow K=\frac{1}{2}L\omega$
$\Rightarrow L=\frac{2K}{\omega }$
So, $\frac{L_2}{L_1}=\frac{K_2}{K_1}\times \frac{{\omega }_1}{{\omega }_2}=\frac{1}{2}\times \frac{1}{2}$

$\because 2{\omega }_1={\omega }_2$ and $2K_2=K_1$ (Given)
$\Rightarrow \frac{L_2}{L_1}=\frac{1}{4}$
$L_2=\frac{L}{4}$

Thus, Angular momentum becomes one-fourth of its initial value.