Rotation of a cone rolling on its side without slipping on a plane

[Wavefunction]

Homework Statement

A uniform right circular cone of height $h$, half angle $α$, and density $ρ$ rolls on its side without
slipping on a uniform horizontal plane in such a manner that it returns to its original position in
a time $\tau$. Find expressions for the kinetic energy and the components of the angular momentum
of the cone. Hint: If $\vec{v} = \vec{ω}\times\vec{r}$ in the inertial frame, points on the cone instantaneously at rest
in this frame will lie in the direction of $\vec{ω}$.

Homework Equations

$\mathcal{L}=T-U=T_{rot}$

$T_{rot}=\frac{1}{2}I_{ij}\omega_{i}\omega_{j} = \frac{1}{2}L_{j}\omega_{j}$

The Attempt at a Solution

Setup: I'll start off by defining an inertial coordinate system $\hat{x}'$ Also successive rotations of this coordinate system will be given by $\hat{x}'',\hat{x}''',...$ etc. Eventually I want to build to a body frame $\hat{x}$.

Rotation 1: about the $\hat{x_3}'$ axis by an angle $\theta$ given by the rotation matrix $\mathbf{A}$

$\begin{pmatrix}x''_1\\x''_2\\x''_3\end{pmatrix} = \begin{pmatrix}\cos\theta&\sin\theta&0\\-\sin\theta&\cos\theta&0\\0&0&1\end{pmatrix}\begin{pmatrix}x'_1\\x'_2\\x'_3\end{pmatrix}$

Rotation 2: about the $\hat{x_2}''$ axis by an angle $\alpha$ given by the rotation matrix $\mathbf{B}$

$\begin{pmatrix}x'''_1\\x'''_2\\x'''_3\end{pmatrix} = \begin{pmatrix}\cos\alpha&0&-\sin\alpha\\0&1&0\\\sin\alpha&0&\cos\alpha\end{pmatrix}\begin{pmatrix}x''_1\\x''_2\\x''_3\end{pmatrix}$

Rotation 3: about the $\hat{x_1}'''$ axis by an angle $ψ$ given by the rotation matrix $\mathbf{C}$

$\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1&0&0\\0&\cos ψ&\sin ψ\\0&-\sin ψ&\cos ψ\end{pmatrix}\begin{pmatrix}x'''_1\\x'''_2\\x'''_3\end{pmatrix}$

So now I have $\vec{x}=\mathbf{CBA}\vec{x}'$ also $[\mathbf{CBA}]^{T}\vec{x}=\vec{x}'$ which is a relationship between the inertial and body frames.

I can also get $\vec{\omega} = \dot{\theta}\hat{x_3}'+\dot{ψ}\hat{x_1}'''$

Putting $\vec{\omega}$ into the body frame: $\vec{\omega} = \begin{pmatrix}-\dot{\theta}\sin\alpha+\dot{ψ}\\\dot{\theta}\cos\alpha\sin ψ\\\dot{\theta}\cos\alpha\cos ψ \end{pmatrix}$

Okay now before I go any further I want to make sure what I have is correct. Also please if you find a mistake please explain in detail why it is wrong. I really want to get an understanding of rigid body rotations. I'm also attaching my drawing of the various transformations too, thanks in advance for your help.

 

[Wavefunction]

Homework Statement

A uniform right circular cone of height $h$, half angle $α$, and density $ρ$ rolls on its side without
slipping on a uniform horizontal plane in such a manner that it returns to its original position in
a time $\tau$. Find expressions for the kinetic energy and the components of the angular momentum
of the cone. Hint: If $\vec{v} = \vec{ω}\times\vec{r}$ in the inertial frame, points on the cone instantaneously at rest
in this frame will lie in the direction of $\vec{ω}$.

Homework Equations

$\mathcal{L}=T-U=T_{rot}$

$T_{rot}=\frac{1}{2}I_{ij}\omega_{i}\omega_{j} = \frac{1}{2}L_{j}\omega_{j}$

The Attempt at a Solution

Setup: I'll start off by defining an inertial coordinate system $\hat{x}'$ Also successive rotations of this coordinate system will be given by $\hat{x}'',\hat{x}''',...$ etc. Eventually I want to build to a body frame $\hat{x}$.

Rotation 1: about the $\hat{x_3}'$ axis by an angle $\theta$ given by the rotation matrix $\mathbf{A}$

$\begin{pmatrix}x''_1\\x''_2\\x''_3\end{pmatrix} = \begin{pmatrix}\cos\theta&\sin\theta&0\\-\sin\theta&\cos\theta&0\\0&0&1\end{pmatrix}\begin{pmatrix}x'_1\\x'_2\\x'_3\end{pmatrix}$

Rotation 2: about the $\hat{x_2}''$ axis by an angle $\alpha$ given by the rotation matrix $\mathbf{B}$

$\begin{pmatrix}x'''_1\\x'''_2\\x'''_3\end{pmatrix} = \begin{pmatrix}\cos\alpha&0&-\sin\alpha\\0&1&0\\\sin\alpha&0&\cos\alpha\end{pmatrix}\begin{pmatrix}x''_1\\x''_2\\x''_3\end{pmatrix}$

Rotation 3: about the $\hat{x_1}'''$ axis by an angle $ψ$ given by the rotation matrix $\mathbf{C}$

$\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1&0&0\\0&\cos ψ&\sin ψ\\0&-\sin ψ&\cos ψ\end{pmatrix}\begin{pmatrix}x'''_1\\x'''_2\\x'''_3\end{pmatrix}$

So now I have $\vec{x}=\mathbf{CBA}\vec{x}'$ also $[\mathbf{CBA}]^{T}\vec{x}=\vec{x}'$ which is a relationship between the inertial and body frames.

I can also get $\vec{\omega} = \dot{\theta}\hat{x_3}'+\dot{ψ}\hat{x_1}'''$

Putting $\vec{\omega}$ into the body frame: $\vec{\omega} = \begin{pmatrix}-\dot{\theta}\sin\alpha+\dot{ψ}\\\dot{\theta}\cos\alpha\sin ψ\\\dot{\theta}\cos\alpha\cos ψ \end{pmatrix}$

Okay now before I go any further I want to make sure what I have is correct. Also please if you find a mistake please explain in detail why it is wrong. I really want to get an understanding of rigid body rotations. I'm also attaching my drawing of the various transformations too, thanks in advance for your help.

The Inertia tensor for a rotatons about the apex of the cone is:

$\mathbf{J} = \begin{pmatrix}\frac{3MR^2}{10}&0&0\\0&\frac{3Mh^2}{5}+\frac{3MR^2}{20}&0\\0&0&\frac{3Mh^2}{5}+\frac{3MR^2}{20}\end{pmatrix}$

The the expression for energy is: $T_{rot} = \frac{1}{2}\begin{pmatrix}-\dot{\theta}\sin\alpha+\dot{ψ}&\dot{\theta}\cos\alpha\sin ψ &\dot{\theta}\cos\alpha\cos ψ \end{pmatrix} \begin{pmatrix}\frac{3MR^2}{10}&0&0\\0&\frac{3Mh^2}{5}+\frac{3MR^2}{20}&0\\0&0&\frac{3Mh^2}{5}+\frac{3MR^2}{20}\end{pmatrix} \begin{pmatrix}-\dot{\theta}\sin\alpha+\dot{ψ}\\\dot{\theta}\cos\alpha\sin ψ\\\dot{\theta}\cos\alpha\cos ψ \end{pmatrix}$

$= \frac{1}{2}[\frac{3MR^2}{10}(-\dot{\theta}\sin\alpha+\dot{\psi})^2+(\frac{3Mh^2}{5}+\frac{3MR^2}{20})(\dot{\theta})^2\cos^2\alpha]$

$= \frac{1}{2}[\frac{3MR^2}{10}[(\dot{\theta})^2\sin^2\alpha-2\dot{\theta}\dot{\psi}\sin\alpha+\dot{\psi}^2]+(\frac{3Mh^2}{5}+\frac{3MR^2}{20})(\dot{\theta})^2\cos^2\alpha]$

Also $\vec{L} = \begin{pmatrix}\frac{3MR^2}{10}&0&0\\0&\frac{3Mh^2}{5}+\frac{3MR^2}{20}&0\\0&0&\frac{3Mh^2}{5}+\frac{3MR^2}{20}\end{pmatrix} \begin{pmatrix}-\dot{\theta}\sin\alpha+\dot{ψ}\\\dot{\theta}\cos\alpha\sin ψ\\\dot{\theta}\cos\alpha\cos ψ \end{pmatrix}$

$= \begin{pmatrix}\frac{3MR^2}{10}(-\dot{\theta}\sin\alpha+\dot{\psi})\\\dot{\theta}(\frac{3Mh^2}{5}+\frac{3MR^2}{20})\cos\alpha \sin\psi \\\dot{\theta}(\frac{3Mh^2}{5}+\frac{3MR^2}{20})\cos\alpha \cos\psi \end{pmatrix}$