Circle rotating within a circle I saw this video and I was wondering how to write an equivalence between the motion of oscillating dots and the rolling circle within the bigger circle, assuming that as the number of oscillating dots approaches infinity they become geometrically equivalent to a smaller circle rotating within the bigger one.
Therefore I'd like to know how we can relate the angular momentum of the rotation of the smaller circle to the oscillating dots. Basically I want to know how to find the angular momentum of the smaller circle if I had the frequency of oscillation of the independent dots. Is that possible?
PS: I apologize in advance if my question is ill-formed. I'm not a professional and I'm sorry for the badly written question. Thank you for your time.
 A: Let's consider circle of radius $r$ rotating inside circle of radius $R$ without slippage. The center of small circle C is moving at the circular line of radius $R-r$.
Using coordinate system with origin in the center of large circle: $x_C=(R-r)\cos \omega t$, $y_C=(R-r)\sin \omega t$.
During samll time $\Delta t$ the center of small circle goes the way $\omega (R-r) \Delta t$. Without slippage this way can be represented as result of rotation around touching point with angular velocity $\omega_1=\omega\frac{R-r}{r}$ in direction opposite to $\omega$.
Consider point A of small circle. $x_A=x_C+r \cos (\phi-\omega_1 t)$, $y_A=y_C+r\sin (\phi-\omega_1 t)$.
If $R=2r$ then $\omega_1=\omega$ and $x_A=r\cos\omega t+r\cos(\phi-\omega t)=r(1+\cos\phi)\cos\omega t+r\sin\phi \sin \omega t$, $y_A=r\sin\omega t+r\sin(\phi-\omega t)=r(1-\cos\phi)\sin\omega t+r \sin\phi \cos\omega t$.
Consider expression $x_A \sin \phi-y_A(1+\cos\phi)=$ $r(1+\cos\phi)\sin\phi\cos\omega t+r\sin^2\phi\sin\omega t-r(1-\cos^2\phi)\sin\omega t-r(1+\cos\phi)\sin\phi\cos\omega t=0$, that means $mx_A-ny_A=0$ with constants $m$, $n$ depending on $\phi$. This is equation of straight line going through the origin, except the case $\phi=\pi$, when $x_A=0$, which is also straight line.
A: I refer to the setting shown here and here: The outer circle has radius $R$, the inner circle has radius $R/2$. We can label each point of the inner circle with a parameter $0\leq a<2 \pi$. The motion of each point in the inner circle is given by the simultaneous motion of the white arrow
$$
w(t) = \frac{R}{2}(\cos (\omega t), \sin(\omega t) )
$$
and of the blue arrow relative to the point labelled by $a$ (note that the two arrows must move with the same angular velocity $\omega$ but in opposite directions):
$$
b_a(t) = \frac{R}{2}(\cos(-\omega t+a), \sin (-\omega t+a) )
$$

Therefore, for each point $a$, its trajectory is given by
$$
r_a(t) = b_a(t) + w(t) = \frac{R}{2}\big( \cos (\omega t)+\cos(a+\omega t) \, , \,  \sin(\omega t) +\sin (a-\omega t) \big)
$$
The velocity of the points in the inner circle is given by the derivative in $t$:
$$
r_a'(t) = \frac{R \, \omega }{2}\big(   -\sin(\omega t) + \sin(a-\omega t) \, , \,  \cos(\omega t ) - \cos(a - t \omega) \big)
$$
Finally, the angular momentum of the point labelled by $a$ with respect to the origin is $l_a(t) = r_a(t) \times r_a'(t)$, namely (using the somewhat standard notation $r_a = (x_a,y_a)$ and $r_a' = (x_a',y_a')$)
$$
l_a(t) = x_a(t) y_a'(t) - y_a(t) x_a'(t)
$$
Note: we are extending the cross product to 2D, the result it's a scalar.
You can check by direct calculation that
$$l_a(t)=0 \qquad \forall a$$
This is not surprising, since each point identified by $a$ moves on a straight line that goes through the origin. It follows that also the angular momentum of the whole inner circle is zero (as it is a sum, or better, an integral, of zero contributions). Assume that the inner circle has a total mass $M$, uniformly distributed (the mass density per unit length is $M/(2 \pi (R/2))= M/( \pi R)$ and the line element is $(R/2) da$)
$$
l_{tot} = \frac{M}{\pi R} \int_{0}^{2 \pi} da \, \frac{R}{2}  \, l_a(t) = \frac{M}{2\pi } \int_{0}^{2 \pi} da  \, l_a(t)  = 0
$$
Why is the angular momentum zero? because the inner circle undergoes two rotations in opposite directions at the same time. This result is consistent with the König's theorem.
