TITLE: Kinematics of a rolling disk on a static disk (variation of the Euler disk)
QUESTION [3 upvotes]: I'm puzzled by the following problem.  Consider a simple tilted disk $\mathcal{D}$ of radius 1 (in any unit) rolling without sliding on top of a static horizontal disk $\mathcal{S}$.  The normal $\mathbf{N}$ of $\mathcal{D}$ has a constant tilt $\alpha$, relative to the vertical $z$ axis.  The disk $\mathcal{S}$ has a radius of $\cos \alpha$, so the center $\mathcal{C}$ should stay fixed (no horizontal translation of the center $\mathcal{C}$).  See the picture below:

If the moving disk $\mathcal{D}$ was sliding on the static disk $\mathcal{S}$, the point $\mathcal{A}$ would simply rotate around the border of $\mathcal{S}$.  In this simple case, the center $\mathcal{C}$ remains static (vertically and horizontally).  The normal vector $\mathbf{N}$ and the radius $\mathbf{CA}$ would precess like this (I put the cartesian axes origin at the bottom, in the center of the static disk):
\begin{align}
\mathbf{N}(t) &= \sin \alpha \, \cos \omega t \, \mathbf{x} + \sin \alpha \, \sin \omega t \, \mathbf{y} + \cos \alpha \, \mathbf{z}, \tag{1} \\[2ex]
\mathbf{CA}(t) &= \cos \alpha \, \cos \omega t \, \mathbf{x} + \cos \alpha \, \sin \omega t \, \mathbf{y} - \sin \alpha \, \mathbf{z}, \tag{2} \\[2ex]
\mathbf{r}_{\mathcal{A}}(t) &= \cos \alpha \, \cos \omega t \, \mathbf{x} + \cos \alpha \, \sin \omega t \, \mathbf{y}. \tag{3}
\end{align}
I want to describe the motion of the material point $\mathcal{A}$ when $\mathcal{D}$ is rolling on $\mathcal{S}$, instead of simply sliding on it.  In this case, the material point $\mathcal{A}$ should get a vertical oscillation and wouldn't rotate around $\mathcal{S}$, so (2) and (3) aren't right.  How the material point $\mathcal{A}$ and vectors $\mathbf{N}$, $\mathbf{CA}$ should move?  My intuition tells me that $\mathbf{N}(t)$ should stay the same as (1) but I'm not sure this is right.  I'm expecting an oscillation of $\mathcal{A}$ around the fixed height $z_{\mathcal{C}} = \sin \alpha$, something like
$$\tag{4}
\mathbf{r}_{\mathcal{A}}(t) = \cos \alpha \, \cos \varphi_{\mathcal{A}} \, \mathbf{x} + \cos \alpha \, \sin \varphi_{\mathcal{A}} \, \mathbf{y} + \sin \alpha \, (1 - \cos(\varphi_{\mathcal{A}} - \omega t)) \, \mathbf{z},
$$
where $\varphi_{\mathcal{A}}$ is the static angular coordinate of the material point along the edge of the static disk.  Vector (4) is such that setting $\varphi_{\mathcal{A}} = \omega t$ (in the rotating reference frame) gives back vector (3).  But how can we prove that vector (4) describes a rolling disk without sliding?
The following video on YouTube shows the illusion of two rings rolling on top of each other:
https://www.youtube.com/watch?v=x7zZzPqfhlg
But it is clear that the rings are just rotating around the vertical axis, and aren't actually rolling on each other.

REPLY [0 votes]: I think that to obtain the rolling condition you have to allows additional
rotation of disc D
in this case the relative velocity at point A is:
$$v_r=\omega_d\rho_d-\omega_s\rho_s$$
the rolling condition is fulfilled if the velocity $~v_r~$  equal zero.
thus the angular velocity of disc D must be:
$$\omega_d=\frac{\rho_s}{\rho_d}\,\omega_s$$
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