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Note: Not a duplicate of Tracing the path of the point as I have my own method of arriving at the answer which seems wrong

I saw this video in which they show that the trajectory of a point on a smaller circle sliding inside along the circumference of the larger circle is a straight line. I also remember someone saying that it is true only if the inner circle is half the radius of the bigger circle.

I just wanted to prove it mathematically. Here is my failed try:

Let the radius of the bigger circle be R and so the radius of the smaller circle be R/2. Let the center of the bigger circle be the origin and we want to prove that the trajectory of the point x is a straight line. Here I am using a complex plane for positioning the point and circles. (x,y) <==> x+iy. Let $\theta$ be the angle of rotation of the inner circle about an imaginary vertical axis passing via OP.

Initially, x is at the origin.

enter image description here

1) The position of point P w.r.t O is $\frac{R }{2}{e}^{-i(\Theta + \frac{3\pi }{2}))}$

2) As the radius of the bigger circle is twice the radius of the smaller circle, it takes two complete rotations for the inner circle to complete one round around the bigger circle

3) The position of point x w.r.t point P is $\frac{R }{2}{e}^{-i(2\Theta + \frac{\pi }{2}))}$

Angle is $2\Theta$ because of the reason mentioned in point 2

4) By change of coordinates, the position of point x w.r. the origin O is ${e}^{-i(2\Theta + \frac{\pi }{2}))} - i\frac{R }{2}$

5) So, the position of the point x w.r.t O as the circle rotates is given by the result of 1 + result 4 which is

$\frac{R }{2}{e}^{-i(\Theta + \frac{3\pi }{2}))} + {e}^{-i(2\Theta + \frac{\pi }{2}))} - i\frac{R }{2}$

This is clearly not a straight line equation. What am I missing?

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You have made two mistakes:

  1. If the centre of the smaller circle is rotating at a rate $\omega$ radians per second relative to the centre of the larger circle, then the rate of rotation of the smaller circle relative to its own centre must be $-\omega$ in order for the point of contact between the two circles not to slip. This means that the angle in your steps 3 and 4 should $-\Theta$ instead of $2\Theta$.

  2. You have missed out a factor of $\frac R2$ in moving from step 3 to step 4.

With these two corrections you should find that the rate of rotation of a point on the smaller circle relative to the centre of the larger circle is $\omega - \omega = 0$. This means that each point on the circumference of the smaller circle moves along a straight line which is a diameter of the larger circle. If you show the time dependence explicitly by replacing $\Theta$ by $\omega t$ then you will see this motion is simple harmonic motion.

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  • $\begingroup$ Thanks for the answer. I understand the negative sign, but if P covers 2*pi radians w.r.t O in time T, x is covering 4*pi radians in the same time T, right? So I thinking like it should be -2w (read it as negative 2 omega). Can you please explain to me why it is not twice multiplied? $\endgroup$ – user3219492 Jan 4 '19 at 1:26
  • $\begingroup$ The point of contact between the small circle and the big circle goes twice around the small circle each time the small circle goes once around the big circle. But this point of contact is not a fixed point on the circumference of the small circle so this is not related to the angular speed of the small circle. Each fixed point on the circumference of the small circle has an angular speed of $-\omega$ relative to the centre of the small circle and an angular speed of $0$ relative to the centre of the big circle. $\endgroup$ – gandalf61 Jan 5 '19 at 10:56

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