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}