# Moment of Inertia

1. Feb 14, 2014

### GeneralOJB

I read that for a rotating body the kinetic energy $E_k = \sum \frac{1}{2}mv^2 = \frac{1}{2}{\omega}^2∑mr^2 = \frac{1}{2}I{\omega}^2$ where $I$ is the moment of inertia.

If we did the same thing for momentum then $P = ∑mv = \omega\sum mr$

So why is angular momentum $I\omega=\omega\sum mr^2$? Shouldn't the momentum just be the sum of the momentum of all the particles, like we did with kinetic energy?

Also why should I believe that this quantity $I\omega$ is conserved?

Last edited: Feb 14, 2014
2. Feb 14, 2014

Use LaTeX
$E_k = \sum\frac{1}{2}mv^2 = \frac{1}{2}\sum m(\omega r)^2 = \frac{1}{2}\omega^2\sum mr^2 = \frac{1}{2}I\omega^2$

3. Feb 14, 2014

### GeneralOJB

Ah, didn't know we had LaTeX.

4. Feb 14, 2014

### A.T.

That would be the total linear momentum, not the total angular momentum.

Note that linear and rotational kinetic energy are both of the same physical scalar quantity. While linear and angular momentum are two different vector quantities. You should look at the vector formulas for momentum to understand it better.