Question

The angular momentum of a particle describing uniform circular motion is $L$. If its kinetic energy is halved and angular velocity doubled, its new angular momentum is

• A. $4L$
• B. $\frac{L}{4}$
• C. $\frac{L}{2}$
• D. $2L$
• E. $\frac{L}{8}$

Answer 1

Answer: (B)

Angular momentum, $L=I\omega \dots \dots \dots ..$ (i) Where, I is the moment of inertia Kinetic energy, $K=\frac{1}{2}I{\omega }^2$
$⟹K=\frac{1}{2}(I\times \omega )\omega$
$⟹K=\frac{1}{2}L\omega$
$($ from eqn(i)) $L=\frac{2K}{\omega }\dots \dots .ii)$
Now, given angular frequency is halved, so, let ${\omega }^{\prime }=\frac{\omega }{2}$ And the kinetic energy is doubled, so, $K^{\prime }=2K$
If $L^{\prime }$ is the new angular momentum, then using (ii) we can write, $L^{\prime }=\frac{2K^{\prime }}{{\omega }^{\prime }}$
$⟹L^{\prime }=4\frac{2K}{\omega }$
$⟹L^{\prime }=4L$
So, angular momentum becomes four times its original value.