Rotating Disk Energy

This solution is long overdue. I will lay out the steps without going into the derivation of each of the results.

Defining the orientation of an arbitrary object in 3D is not trivial. To do so, two coordinate frames are defined. One is the global coordinate frame which is fixed and is an inertial frame of reference. The other is a body fixed coordinate frame oriented such that the local Z axis coincides with the vector normal to the disk. All calculations are carried out in the body fixed frame.

The moment of inertia of the disk is defined as a matrix as follows:

$I = \left[\begin{matrix}I_{x}&0&0\\0&Iy&0\\0&0&I_z\end{matrix}\right]$

Which evaluates to:

$I = \left[\begin{matrix}\frac{1}{4}MR_d^2&0&0\\0&\frac{1}{4}MR_d^2&0\\0&0&\frac{1}{2}MR_d^2\end{matrix}\right]$

The orientation of the disk is a combination of successive rotations about the global Y axis and the local X axis. The vector normal to the plane of the disk can be rewritten as :

$\hat{n} = \left[\begin{matrix}\cos(\theta)&-\sin(\theta)&0\\ \sin(\theta)&\cos(\theta)&0\\0&0&1\end{matrix}\right] \left[\begin{matrix}\cos(\phi)&0&\sin(\phi)\\ 0&1&0\\-\sin(\phi)&0&\cos(\phi)\end{matrix}\right] \left[\begin{matrix}0\\0\\1\end{matrix}\right]$

The net rotation matrix is defined as :

$R = \left[\begin{matrix}\cos(\theta)&-\sin(\theta)&0\\ \sin(\theta)&\cos(\theta)&0\\0&0&1\end{matrix}\right] \left[\begin{matrix}\cos(\phi)&0&\sin(\phi)\\ 0&1&0\\-\sin(\phi)&0&\cos(\phi)\end{matrix}\right]$

Using this matrix, the angular velocity vector can be derived as follows:

$\tilde{\omega} = R^T\dot{R}$

Where

$\dot{R} = \frac{\partial R}{\partial \phi} \dot{\phi} + \frac{\partial R}{\partial \theta} \dot{\theta}$

The angular velocity vector is obtained by taking the elements of the matrix $\tilde{\omega}$ as follows:

$\vec{\omega} = (\tilde{\omega}(3,2),\tilde{\omega}(1,3),\tilde{\omega}(2,1))$

or

$\omega = \left[\begin{matrix}\tilde{\omega}(3,2)\\\tilde{\omega}(1,3)\\\tilde{\omega}(2,1)\end{matrix}\right]$

From here, the expression of kinetic energy is:

$KE = \frac{1}{2}\omega^TI\omega$

The expression comes out to be:

$KE = \frac{{\mathrm{\dot{\phi}}}^2}{8}-\frac{{\mathrm{\dot{\theta}}}^2\,{\sin\left(\phi \right)}^2}{8}+\frac{{\mathrm{\dot{\theta}}}^2}{4}$

Note that the expression is independent of the angle $\theta$ . I realise this approach may seem very abstract but this is the method I used to arrive at the answer. If necessary, I will update the solution to provide more clarity. I recommend referring to the solution provided by Steven Chase.

I will derive the kinetic energy from first principles, without making reference to moments of inertia. Derive two orthogonal unit vectors which lie in the plane of the disk. By inspection, the first is as follows:

$\vec{u}_{1} = \Big(\sin (\theta), -\cos (\theta), 0 \Big)$

The second is obtained by taking the cross product of $\vec{N}$ with $\vec{u}_1$ .

$\vec{u}_{2} = \Big(\cos (\theta) \, \cos (\phi), \sin (\theta) \, \cos (\phi), -\sin (\phi) \Big)$

The weights on these unit vectors are defined by the following polar coordinates:

$0 \leq r \leq R \\ 0 \leq \alpha \leq 2 \pi$

Expressions for disk coordinates:

$\vec{P } = r \, \cos (\alpha) \, \vec{u}_1 + r \, \sin (\alpha) \, \vec{u}_2 \\ x = r \, \cos (\alpha) \, \sin(\theta) + r \, \sin (\alpha) \, \cos (\theta) \, \cos (\phi) \\ y = -r \, \cos (\alpha) \, \cos(\theta) + r \, \sin (\alpha) \, \sin (\theta) \, \cos (\phi) \\ z = -r \, \sin (\alpha) \, \sin (\phi)$

Expressions for the velocities of an infinitesimal patch on the disk:

$\dot{x} = r \, \cos (\alpha) \, \cos(\theta) \, \dot{\theta}+ r \, \sin (\alpha) \, \Big(-cos (\theta) \, sin(\phi) \, \dot{\phi} - \sin(\theta) \, \cos(\phi) \, \dot{\theta} \Big) \\ \dot{y} = r \, \cos (\alpha) \, \sin(\theta) \, \dot{\theta} + r \, \sin (\alpha) \, \Big(-\sin(\theta) \, \sin(\phi) \, \dot{\phi} + \cos (\theta) \, \cos (\phi) \, \dot{\theta} \ \Big) \\ \dot{z} = -r \, \sin (\alpha) \, \cos (\phi) \, \dot{\phi}$

Infinitesimal disk patch energy and total kinetic energy (evaluated numerically):

$dm = M \frac{r \, dr \, d \alpha}{\pi R^2} \\ dE = \frac{1}{2 } \, dm \, (\dot{x}^2 + \dot{y}^2 + \dot{z}^2) \\ E = \int_0^{2 \pi} \int_0^R dE$