# Energy of a rotating disc around different moments of inertia

Lets take a disc that is rolling without slipping which has moment of inertia $I=kmR^2$. It will have total kinetic energy $E=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2=\frac{1}{2}mv^2(1+k)$.

Lets now use the parallel axis theorem to measure the moment of inertia around a different axis, lets say the point of contact with the ground. So $I=kmR^2+mR^2$ (since the distance from the center of mass is R). To my understanding, in this instance, we need to add another term to the equation, $m \vec{v}\cdot(\vec{\omega} \times \vec{R})=-m |\vec{\omega}|^2R^2$ where $\vec{R}$ is the distance from the point of calculating moment of inertial to the axis of rotation.

So \begin{align*}E &= \frac{1}{2}mv^2 + \frac{1}{2}(kmR^2+mR^2)\omega^2 + (-m |\vec{\omega}|^2R^2)\\ &= \frac{1}{2}mv^2 + \frac{1}{2}(k+1)mv^2 + (-m v^2)\\ &= \frac{1}{2}mv^2(1+k) - \frac{1}{2}mv^2\\ &= \frac{1}{2}kmv^2\end{align*} which does not seem to equal $\frac{1}{2}mv^2(1+k)$ at all.

Is there something I did wrong?

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Is it just me, or did you lose a $\frac{1}{2}mv^2$ term in the last calculation? It seems like just a little arithmetic error... –  David Z Sep 19 '11 at 3:21
I don't believe there is an error... The $\frac{1}{2}mv^2$ just cancels out with the $-\frac{1}{2}mv^2$. –  Roger Hampton Sep 19 '11 at 3:46
I cleaned up the question a bit to incorporate your edit. –  David Z Sep 19 '11 at 4:50

$T = \frac{1}{2}m\vec{a}\cdot\vec{a}+\frac{1}{2}\vec{\omega}\cdot\vec{\omega}I+m\vec{a}\cdot(\vec{\omega}\times\vec{R})$, where $\vec{a}$ is the linear velocity of the point you are calculating I around.
So for the point of contact of a rotating wheel, you have $\vec{a}=0$, as the instantaneous linear velocity of the point of contact is zero,
$T= \frac{1}{2}m\vec{a}\cdot\vec{a}+\frac{1}{2}\vec{\omega}\cdot\vec{\omega}I+m\vec{a}\cdot(\vec{\omega}\times\vec{R})= 0 + \frac{1}{2}\vec{\omega}\cdot\vec{\omega}I + 0 = \frac{1}{2}\omega^2I$