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?

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?

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? Thanks a lot.

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?