enter image description hereSay I have a ring with the center of mass G. Also I rotate the ring around a point P on the ring, and I have a "anything" moving along the ring at the same time with velocity v.

How would one explain the motion of the "anything"? Would one say that it's absolute velocity is the velocity of G + the relative velocity (velocity measured from P) + the rotational velocity of the entire system? Or is some of these terms zero?

Thank you.

  • $\begingroup$ This seems like an interesting question, but could you clarify a bit? Does the main ring spin in its own plane, like spinning a hoola hoop around your waist, or does it spin about an axis in its own plane, like spinning a coin on its side on a table? Is there perhaps an existing object that behaves like this that you can give as an example? $\endgroup$ – ChP Apr 5 '18 at 11:03
  • $\begingroup$ One way to describe it would be to define two separate rotation matrices for the "ring" and the "thing" and crossing the two and apply it to the "thing" $\endgroup$ – ChP Apr 5 '18 at 11:08
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    $\begingroup$ A sketch or diagram would help clarify the question. $\endgroup$ – Daniel K Apr 5 '18 at 11:42
  • $\begingroup$ I have added a picture. $\endgroup$ – 123 Apr 5 '18 at 14:48

The Anything path is a circle with G at its center. So in that reference frame

$$x= \sqrt{1-y^2} $$ But G is turning with its $$ X_p = \sqrt{1-Y^2_p} $$

you have to transfer the X1 and Y1 to new frame by substituting X1, with (X1-Xp) , and same for y, Y1 substitute with Y1-Yp.

The final path will look like a Lotus pattern of a round flower with oval petals. The radius of this pattern will be twice the radius of your ring and the number of petals depends on the ratio of the to rotations.

  • $\begingroup$ Thank you, but also do you know if there is a centripetal force on the thing for rotating around G? $\endgroup$ – 123 Apr 7 '18 at 4:47
  • $\begingroup$ Not sure centripetal, but a mix. It is constantly changing the acceleration vector as Asub x= dx^2/ dt^2 and Asub y = dy^2/dt^2. $\endgroup$ – kamran Apr 7 '18 at 4:55

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