Goldstein Classical Mechanics exercise 5.17 (3ed) conceptual
I am having difficulty understanding a concept in the "Heavy symmetric top" type of problems. I will include all of my efforts as to hopefully have someone easily point out what it is that I'm missing. The question reads:
A uniform right circular cone of height $h$, half-angle $\alpha$ and density $\rho$ rolls on it's side without slipping on a uniform horizontal plane in such a manner that it returns to it's original position in a time $\tau$ . Find expressions for the kinetic energy and the components of the angular momentum of the cone.
I begin by finding the volume (since I memorize no such things) and decide on the obvious choice of cylindrical coordinates. 15
My mass will then be $$\rho\int_0^h \int_0^{2\pi} \int_0^{z \tan{\left(\alpha\right)}}r \, \mathrm{d}r \, \mathrm{d}\theta \, \mathrm{d}z = \frac{\pi \rho R^2h}{3} $$ then while I'm at it I likewise find the center of mass (in the $z$) $$\frac{1}{V}\int_0^h \int_0^{2\pi} \int_0^{z \tan{\left(\alpha\right)}}zr \, \mathrm{d}r \, \mathrm{d}\theta \, \mathrm{d}z = \frac{3h}{4} $$ moving onto the moments for the inertia tensor, I use$$ I_{ij} = \int_V\rho \left(\delta_{ij} \Sigma_k x_k - x_i x_j\right) \, \mathrm{d}V $$ and find that $\bf{I}$ is already diagonal, I get elements $I_1 = I_2 = \frac{3M}{20}\left(R^2+4h^2\right)$; and $I_3 =\frac{3M}{20}\left(2R^2\right)$ so that the inertia tensor about the apex at the origin of the coordinate system is given by $$ \bf{I} = \begin{bmatrix} \frac{3M}{20}\left(R^2+4h^2\right) & 0 & 0 \ 0 & \frac{3M}{20}\left(R^2+4h^2\right) & 0 \ 0 & 0 & \frac{3M}{20}\left(2R^2\right)\end{bmatrix} $$
now in the event I should need $\bf{I}{CM}$ I also get that out of the way and find that using the parallel axis theorem, $ I{1_{CM}} = I_{2_{CM}} = \frac{3M}{20}(R^2+\frac{h^2}{4}) = \frac{3M}{80}\left(4R^2+h^2\right)$; and $ I_{3_{CM}} =\frac{3M}{80}\left(8R^2\right)$ so that $$ \bf{I}_{CM} = \begin{bmatrix} \frac{3M}{80}\left(4R^2+h^2\right) & 0 & 0 \ 0 & \frac{3M}{80}\left(4R^2+h^2\right) & 0 \ 0 & 0 & \frac{3M}{80}\left(8R^2\right)\end{bmatrix} $$
great. So it seems that I may use either the inertial frame OR the body frame, however this is where I'm getting confused. As I understand it, if I change my coordinates to the center of mass body coordinates, I will not need to consider the translational kinetic energy of the center of mass. But can I even do this here? In any event I find $v_{CM} = \frac{3h}{4} \cos(\alpha)\dot{\phi}$ where $\phi$ is the azimuthal angle about the fixed inertial $z$-axis, and I can find various relations for this based on the no-slipping condition though its not clear which I should want. Now that I have gotten most of the legwork done, I want to write my expression for the kinetic energy. 402