Tracing the path of the point

Question

A wheel of radius $R$ is rolling inside a fixed circular cylinder of radius $2R$ as shown. What is the trajectory followed by a point on the rim of the wheel?

By observation, the only two points that seem to move in a straight line, are one at the centre of the cylinder and one at the common point of wheel and cylinder. I proved this too, using vector calculations. But I am not able to prove it for a general point. The answer given states that any general point on the rim of the wheel moves in a straight line. How to prove this?

Answer 1

If you rotate the circle sufficiently, it should be possible to show that every point on the smaller circle is, after some amount of rotation, the center of the larger circle. (Or, equally useful, on the edge of the larger circle.)

And then your results for the points at the center (or edge) will apply, yielding the desired result.

So what I'd try, then, is to prove that after rotating the smaller circle along the edge of the outer circle, every point on it is at some point either the center of the large circle, or on the edge of the large circle.

I'm not sure how easy that will be, it's just the immediate solution that screams to me.

Edit: A possible way seems to be using the fact that the rotation is continuous, and that the inner circle rotates through a whole $360$ degrees at least one time as it goes around the outer one. You could probably model this mathematically some way and use the intermediate value theorem or certain results about monotonic functions - it depends on how you'd model it. Sadly, without some sort of function or model, I won't be a ton of help and I don't trust myself particularly well to generate a good one. But if nothing else I feel like this might be a nudge in the right direction, so hopefully it was some help.

Answer 2

For the sake of simplicity, consider trajectory of point $A$ on the rim of the wheel. The initial position of this point is denoted with $A_0$.

$$x_A=R\sin\alpha-R\sin(\beta-\alpha)\tag{1}$$

$$y_A=R\cos\alpha+R\cos(\beta-\alpha)\tag{2}$$

One more condition has to be added. There is no slipping between the two cylinders so lenghts of arcs $AB$ and $A_0B$ must be equal:

$$R\beta=2R\alpha$$

$$\beta=2\alpha$$

Replace this into (1) and (2) and you get:

$$x_A=0$$

$$y_A=2R\cos\alpha$$

So for any given angle $\alpha$ the $x$ coordinate of point $A$ is equal to zero and therefore its trajectory must be a straight line (it lies on $y$-axis).

Trajectory is a straight line only in this very simple case. In a general case, the trajectory of point $A$ is an epicycloid.

Answer 3

This orbit is called hypotrochoid and has the parametric equations

$$ x(\theta) = (R-r)\cos\theta+r\cos\left(\frac{(R-r)\theta}{r}\right)\\ y(\theta) = (R-r)\sin\theta-r\sin\left(\frac{(R-r)\theta}{r}\right)\\ $$

in the present case we have $R=2R_0$ and $r = R_0$ so we have

$$ x(\theta) = R_0(\cos\theta+\cos\theta)\\ y(\theta) = R_0(\sin\theta-\sin\theta)\\ $$

so the parametric equations are

$$ x(\theta) = 2R_0\cos\theta\\ y(\theta) = 0 $$

In red the sought path inside the external circle (blue)