Question

A particle of mass m is describing a circular motion of radius r with uniform speed. If L is the angular momentum of the particle about the axis of the circle, find the kinetic energy of the particle.

Answer 1

Given: a particle of mass m is describing a circular motion of radius r with uniform speed. If L is the angular momentum of the particle about the axis of the circle

To find the kinetic energy of the particle.

Solution: Assume the particle has angular velocity $\omega$.

We define the angular momentum L of the particle about the point as

$ L = r p sin\theta$

where $r$ is the radius of the circular motion and $p$ is its momentum.

For a particle moving in a circular path $sin\theta =1$,

$L=rp=rmv=m{r}^2\omega =I\omega$

where v is the velocity of the particle.

And velocity of the particle can be written as $v=r\omega$

and the moment of inertial of the particle moving in circle is $m{r}^2$.

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

Then the kinetic energy of the particle moving in circle is

$E_k=\frac{1}{2}m{v}^2$

$E_k=\frac{1}{2}m(r\omega {)}^2$ (as $v=r\omega$)

$⟹E_k=\frac{1}{2}m{r}^2{\omega }^2$

But the moment of inertial of the particle moving in circle is $m{r}^2$.

$⟹E_k=\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(I\omega \right)\omega$

$⟹E_k=\frac{1}{2}L\omega$ (from eqn(i))

Hence this is the kinetic energy of the particle moving in circle.