Moment of Inertia

[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?

 

[adjacent]

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

 

[GeneralOJB]

Ah, didn't know we had LaTeX.

 

[A.T.]

Shouldn't the momentum just be the sum of the momentum of all the particles.

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.