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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:

enter image description here

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.

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  • $\begingroup$ Hmmm, drawing the curve described by (4) doesn't give a perfect circle. It's an ellipsis, so (4) isn't right! $\endgroup$
    – Cham
    Apr 7, 2022 at 17:36
  • $\begingroup$ The center of the disk does not stay in one place, but orbits around. The situation described above isn't realistic. You can see this as the sum of the horizontal forces acting on the disk are not zero, and hence the center of mass will accelerate. $\endgroup$
    – JAlex
    Apr 8, 2022 at 17:10
  • $\begingroup$ Great diagrams BTW! $\endgroup$
    – JAlex
    Apr 8, 2022 at 17:13

1 Answer 1

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enter image description here

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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