# Regarding (regular) kinetic energy and rotational kinetic energy

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?

• You don't count the same thing twice : The 2nd part $\frac {I}{2}\omega^2$ concerns the kinetic energy of the rotating disc, not of the small particle. Here $\:I = \textit{moment of inertia of the disc around the axis of rotation}\:$. The 1rst part $\:\frac m 2 v^2\:$ is the kinetic energy of the rotating particle, expressed also as $\:\frac m 2 r^2 \omega^2\:$, where $\:m r^2\:$ is the "moment of inertia" of the small particle. May 16 '16 at 21:08
• @Frobenius Thanks! After a while I understood that we were counting the energy of the system and not just that of the particle. It's all clear now. May 16 '16 at 22:17

The $\frac m 2 v^2$ term is the kinetic energy of the "small particle". The $\frac {I}{2}\omega^2$ is rotational kinetic energy of the disc of mass $M$. You are just counting the kinetic energy of each mass once.

The PE of the particle is converted into the KE of the particle $\frac12mv^2$ plus the KE of the disc $\frac12I\omega^2$ which is also moving.