In my physics class we saw this problem:

> A disc of mass $M$ and radius $r$ is standing vertically and can rotate freely through an axis thats goes through its center of mass. A small particle of mass $m$ is attached to the top border of the disc. A small perturbance makes the disc rotate and so, the particle goes down.

>Determine the angular velocity of the disc when the particle is at the lowest part.

And my professor said this:

>As the energy is conserved in this case, we have that the initial energy: $2r\,m\,g$ is equal to the final energy $\frac m 2 v^2+\frac {I}{2}\omega^2$...

I don't understand why we have to put the $\frac I2\omega^2$ part, when I was trying to solve the exercise I just put the (regular) kinetic energy $\frac {m}{2}v^2$ and I was told this was wrong, but I was not explained why.

By putting the two kinetic energies, it feels like I'm counting the same thing twice, as the particle is just rotating!

Could someone clear my confusion?