# 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. 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. Supposing I want to use the inertial frame, I will need$$T ~~=~~ T_{\text{trans}} + T_{\text{rot}} ~~=~~ \frac{1}{2}Mv_{cm}^2+\frac{1}{2}I_1 \left(\omega_1^2+\omega_2^2 \right)+\frac{1}{2}I_3\omega_3^2$$ and also, my components of $\bf{L}$ would be $L_i = I_i \omega_i$

but now I need expressions for the components of $\omega$. I call $\omega$ the angular velocity of the cone in the body about it's $z'$ axis. This must be expressed in Euler angles, OK well fortunately these are readily found in the book as \begin{alignat}{5} \omega_x & ~=~ & \dot{\phi} \, & \sin(\theta) \sin(\psi) & + & \dot{\theta} \cos(\psi) \\ \omega_y & ~=~ & \dot{\phi} \, & \sin(\theta) \cos(\psi) & + & \dot{\theta} \sin(\psi) \\ \omega_z & ~=~ & \dot{\phi} \, & \cos(\theta) & + & \dot{\psi} \end{alignat} however if I have rotated the cone, and I knew where $\omega$ pointed (exclusively in the $z$ before rotation) can't I just rotate $\vec{\omega}= \left[0;0;\omega \right]$ by using the Euler rotation matrix? this would give me $$A \begin{bmatrix} 0 \\ 0 \\ \omega \end{bmatrix} = \omega \begin{bmatrix} \sin(\psi) \sin(\theta) \\ \cos(\psi) \sin(\theta) \\ \cos(\theta) \end{bmatrix}$$

If I use the book's expressions, from the geometry of the problem we see that $\dot{\theta}$ is zero so these terms drop and I am left with

\begin{alignat}{4} \omega_x &~=~ & \dot{\phi} \, & \sin(\theta) \sin(\psi) \\ \omega_y &~=~ & \dot{\phi} \, & \sin(\theta) \cos(\psi) \\ \omega_z &~=~ & \dot{\phi} \, & \cos(\theta) + \dot{\psi} \end{alignat} _{\Large{,}} which is close to the other having just rotated $\vec{\omega}$ but missing that $\dot{\psi}$ which I'm not sure if this is zero or $\omega$. I also see that since $\theta$ is fixed, we can note that $\theta + \alpha = \frac{\pi}{2}$ and so $\sin(\theta) = \cos(\alpha)$.

Here's where I stall out. I'm not sure, but I think $\dot{\psi}$ is what I've defined as $\omega$. I see that the instantaneous axis of rotation is in the $\left(x,\, y\right)$-plane, but I'm not sure how this is relevant or how to apply this knowledge. It seems there are different ways to do this problem as previously mentioned, and I'm not seeing which information is required for each method. If the instantaneous axis of rotation is in the $\left(x,\, y\right)$-plane, how does this impact my expression for $\bf{I}$? Can the non-inertial body frame be used at all? Does it HAVE to be used? And how does this change the expressions for $\bf{I}$ and $T$?

I feel like I'm very close to being able to solve this in multiple ways but I'm missing a concept that ties it all together and so I'm just getting lost at the end. I guess the main issue I'm having is identifying the components of $\omega$ and how the frames impact the equations. Thanks to anyone that can help explain

Edit: as it was brought up in a comment, there are some physics instructors that have their versions of this solution online. Upon evaluation, at least one uses the instantaneous axis of rotation but one of them omits the contribution of $\omega_2$ from the kinetic energy, that solution is found here (pages 6-8). To me, this seems incorrect, but I'm the one asking the question so obviously I don't know for sure. Then, another solution available here (pages 18-23) arrives at a slightly different expression for $T$, however they omit the contribution of the center of mass translation. Also seems incorrect but again... I'm the one trying to tie this all together. Even more contradictory solutions from another instructor found here which arrives at the same solution as one of the others however from an approach that makes most sense to me. Solutions are not what is sought here, I would like to be able to argue the validity of the approach. I think that if we consider the Euler angles to rotate the cone from the "original" frame from which the inertia moments are found and where $\omega$ has only a $z$ component, then the $\omega$' should be found by a simple rotation by Euler angles, then use $\bf{I}_{CM}$ for the rotational component of T. I still don't understand the necessity (or if there really is one) of identifying the instantaneous rotation axis.