Question

The angular momentum of a particle performing uniform circular motion is $L$. If the kinetic energy of the particle is doubled and frequency is halved, then angular momentum becomes

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

Answer 1

The correct option is D $4L$

Angular momentum, $L=I\omega ...........\left(i\right)$

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))

$L=\frac{2K}{\omega }.......ii)$

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,

$L^{\prime }=\frac{2K^{\prime }}{{\omega }^{\prime }}⟹L^{\prime }=\frac{2\left(2K\right)}{\frac{\omega }{2}}⟹L^{\prime }=4\frac{2K}{\omega }⟹L^{\prime }=4L$

So, angular momentum becomes four times its original value.