Deriving Rotational Energy Equation for Rigid Bodies

[Crosshash]

Hello, I think I've got the right idea on how to perform this question but I just need a little bit of help.

Homework Statement

Show that for a rigid body rotating with angular velocity $\omega$ the energy of rotation may be written as:
$E = \dfrac {1}{2}I\omega^{2}$

where the moment of inertia of the body about the axis of rotation is given by:
$I = \int dV \rho r^{2}$

where $r$ is the distance from the rotation axis to the volume element $dV$ and $\rho$ is the density of the object in that region

Homework Equations

$E = \dfrac {1}{2} mv^{2}$
$v = \omega r$
$I = \int r^{2} dm$

The Attempt at a Solution

I can firstly identify that

$E = \dfrac {1}{2} mv^{2}$ which looks similar to the rotation energy equation.

I know that
$v = \omega r$

But what's confusing me is that usually, the moment of Inertia is represented as
$I = \int r^{2} dm$ and I don't really know how to link the two.

Substiting angular velocity into the energy equation

$E = \dfrac {1}{2} m(r\omega)^{2}$

but where do I go now?

Thanks

 

[CompuChip]

So if you have a little volume dV at radius r, its mass is rho*dV, right? What is the kinetic energy of this little piece? For the total energy you should then sum (read: integrate) all these pieces.

 

[Crosshash]

Thanks for the reply CompuChip, sorry for this late reply. I took onboard what you said, perhaps you could verify my answer.

I have $v = \omega r$

$I = \int dV \rho r^{2}$

$dm = \rho dV$

$I = \int \dfrac {dm}{\rho} \rho r^{2}$

$I = \int dm r^{2}$

$I = r^{2} \int dm$

$I = r^{2}m$

$m = \dfrac {I}{r^{2}}$

$E = \dfrac {1}{2} m v^{2}$

$E = \dfrac {1}{2} m \omega^{2} r^{2}$

$E = \dfrac {1}{2} \dfrac {I}{r^{2}} \omega^{2} r^{2}$

$E = \dfrac {1}{2} I \omega^{2}$

Is that the best way?