Finding coordinates of a point on a rolling circle

Question

Ok, here's the situation:

I'm imagining a circle of radius A, and a circle of (a smaller) radius B. Circle A is centred at the point (0,0) and circle B starts centred at the point (b, 0). On circle B is a point P, which starts at (a+b, 0). The centre of circle B then travels along the circumference of circle A at a given rate (the rate itself is unimportant); simultaneously, P travels along the circumference of circle B, again at some arbitrary rate.

Now for the problem: given A, B, their respective rates of rotation, and the time elapsed, how do I calculate the new x and y coordinates for P? I've tried drawing countless diagrams, yet I just can't crack it. I suspect there's something missing in my intuition, or else I'm just overthinking everything. In any case, any help would be welcome.

Here's a rough diagram of what I'm trying to do:

enter image description here

You can ignore the bottom circle, as that was just me trying to figure something else out.

Answer 1

You wrote that at $t=0$ the center of $B$ is at $(b,0)$ and the point $P$ at $(a+b,0)$. I'm assuming both of these indicate the rightmost possible position, so circle $A$ would have radius $b$ and circle $B$ would have radius $a$. Confusing. May I switch names? Let $a$ be the radius of circle $A$ and $b$ be the radius of circle $B$. Also let $\alpha$ be the rotational speed (in angle per time) along the circumference of $A$, and $\beta$ be the speed around $B$. Then the position of $P$ is

$$P(t)=\begin{pmatrix}a\cos(\alpha t)+b\cos(\beta t)\\a\sin(\alpha t)+b\sin(\beta t)\end{pmatrix}$$

because you simply have

$$\begin{pmatrix}a\cos(\alpha t)\\a\sin(\alpha t)\end{pmatrix}$$

as the position of the center of $B$ on $A$, and you have

$$\begin{pmatrix}b\cos(\beta t)\\b\sin(\beta t)\end{pmatrix}$$

as the offset of $P$ relative to the center of $B$. These two add up using regular vector addition. Done.

Answer 2

By vector addition,

$$x=a\cos\omega_at+b\cos\omega_bt,\\y=a\sin\omega_at+b\sin\omega_bt.$$

Note that $\omega_b$ is the angular speed wrt the fixed frame. If you want to express it wrt the first rotating frame, the relative speed is $\omega_b'=\omega_b-\omega_a$.

If the circle $b$ is rolling on $a$ without gliding, when the contact point moves by $a\omega_a\Delta t$ units, the rolling circle rotates so that the contact point moves by $b\omega_b'\Delta t$ units in the revolving frame. Hence,

$$\frac{b\omega_b'}{a\omega_a}=\frac{a}{b}.$$