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Suppose a circumference of radius r and center $\Omega$ rotates with constant angular velocity $\omega_D=\dot\phi e_3$ (D stands for disk) around an axis parallel to $e_3$ through $\Omega$. Let $\Omega$ rotate on the circumference of radius $R+r$ with constant angular velocity $\omega_C=\dot\psi e_3$ (C stands for center). In other words, the circumference of radius $r$ rolls on the circumference of radius $R$. Since we can assume that the large circumference is not moving, it is pretty straightforward to check that the condition for the small circumference to roll without slipping is $(R+r)\dot\psi=r\dot\phi$, which basically tells me that the modulus of the velocity of $\Omega$ should be equal to the modulus of the velocity of any point on the small circumference if it only rotated, which does not make any sense to me. How can I interpret this result? Is there a nice way to look at this? Thanks in advance. enter image description here

Please ignore the green lines.

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  • $\begingroup$ What do you mean by "interpret"? You've already given one interpretation of the results as far as I understand the word "interpret". $\endgroup$
    – David K
    Commented Jul 8 at 19:57
  • $\begingroup$ I mean: what's the sense of this explanation? Or, is there a better way to look at it which will make me appreciate and understand the result more deeply? $\endgroup$
    – Davide Masi
    Commented Jul 8 at 20:04

2 Answers 2

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The circle with center $O$ is not moving, so the velocity of the point of contact must be zero.

Let $P$ be the point on the circle with center $\Omega$ that is initially in contact with circle $O$. The point $P$ has a velocity of magnitude $r\dot\phi$ relative to $\Omega$. That velocity is added to the velocity of $\Omega$, which has magnitude $(R+r)\dot\psi$, to obtain the total velocity of $P$.

The only way for two vectors to sum to zero is for them to be of equal magnitude and in opposite directions. Setting the magnitudes equal,

$$ (R+r)\dot\psi = r\dot\phi.$$

Another way to consider this is from the point of view of an observer who sits on the line $O\Omega$ and rotates along with that line. That observer sees the circle with center $O$ rotating at angular speed $\lvert\dot\psi\rvert$ and the circle with center $\Omega$ rotating at angular speed $\lvert\dot\phi - \dot\psi\rvert$. (It may take some time to visualize this correctly. Recognize that the observer is rotating in the same direction as the circle with center $\Omega$, only at a different speed.)

The observer sitting on the line $O\Omega$ then sees two circles with stationary centers rolling against each other, such that the velocity at the point of contact is $R\lvert\dot\psi\rvert$ for one circle $r\lvert\dot\phi - \dot\psi\rvert$ for the other circle.

Careful examination of the figure will show that $R\dot\psi = r(\dot\phi - \dot\psi)$, that is, you can remove the absolute values, after which your result immediately follows.

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  • $\begingroup$ This does not reply to my question though :/ $\endgroup$
    – Davide Masi
    Commented Jul 8 at 20:18
  • $\begingroup$ Then you have not made your question clear. What are you looking for? $\endgroup$
    – David K
    Commented Jul 8 at 20:30
  • $\begingroup$ Basically, I know $(R+r)\dot\psi=r\dot\phi$ holds. I believe every result has a meaning, so I want to understand what this equality really tells me, since as I said the left-hand side is the magnitude of the velocity of $\Omega$ and the right-hand side is the magnitude of the velocity which any point on the small circumference would have if the small circumference was just rotating (not rolling around the large circumference). However, I don't see any connection between these two velocities. Hope I was clear enough $\endgroup$
    – Davide Masi
    Commented Jul 8 at 20:44
  • $\begingroup$ Oh, I have just seen your edit. I hadn't considered this way to see the problem, which leads, as you pointed out, to $R\dot\psi = r(\dot\phi - \dot\psi)$ which is equivalent to the result I knew. This is exactly what I was looking for. Thank you very much! $\endgroup$
    – Davide Masi
    Commented Jul 8 at 20:57
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That is exactly what "without slipping" means. Points in contact have no relative motion. If the two centers are fixed the angular velocity of the small circle will be $\frac Rr$ times the angular velocity of the large circle so that the linear velocities at the point of contact match.

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  • $\begingroup$ I thought this could be a possible explanation, but in this case I guess it should be $R\dot\psi=r\dot\phi$ and not $(R+r)\dot\psi=r\dot\phi$ $\endgroup$
    – Davide Masi
    Commented Jul 8 at 20:07
  • $\begingroup$ It depends on whether and at what rate the large circle is rotating. You did not specify it. $R\dot \psi = r\dot \phi$ is correct if both centers are stationary. If could be that the large circle is fixed and the small one is rolling around it. Then you get a relationship between the rotation rate of the small circle and the rotation of the center of the small circle around the large one. You did not specify the situation completely. $\endgroup$
    – Ross Millikan
    Commented Jul 8 at 20:31
  • $\begingroup$ I said we can assume the large circle is not moving, so not even rotating. The only motions are the rotation of the small circle around itself and the small circle rolling without slipping on the large one, in other words the center of the small circle moving on the circumference of radius $R+r$ $\endgroup$
    – Davide Masi
    Commented Jul 8 at 20:36
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    $\begingroup$ So then you get a relation between the angular velocity of the small circle and the angular velocity of the center of the small circle as David K shows. $\endgroup$
    – Ross Millikan
    Commented Jul 8 at 21:33

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