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?