In non-uniform circular motion with a constant magnitude of tangential acceleration, though the particle keeps completing the same circle its speed changes continuously and hence the circular motions each time are not identical. So it should not be periodic motion according to me. Am I right?


No, the situation you’ve described in the body of your question is not periodic. If there exists a $T\in \mathbb{R}$ such that the motion of a particle is described by

$$\mathbf{r}(t + T) = \mathbf{r}(t) $$


$$ \mathbf{p}(t + T) = \mathbf{p}(t)$$

for all $t$ in the domain of $\mathbf{r}$, then your motion may be considered to be periodic.

More generally, one can consider phase space $\{ (\mathbf{r, p}) \}$ of your system. Periodic motion corresponds to an orbit in phase space parametrized by time.

So while the position will come back on itself, the momentum never will. Hence the trajectory in phase space is not an orbital.

  • $\begingroup$ But the situation described in the question decidedly does meet the conditions, so the first would would be rather better as "No." $\endgroup$ – dmckee --- ex-moderator kitten Jan 4 '19 at 3:14
  • $\begingroup$ Ah yes, thank you. I was responding to the question posed in the title. $\endgroup$ – InertialObserver Jan 4 '19 at 3:21
  • $\begingroup$ @AaronStevens Yes. I did mean that. Good catch. But as InertialObserver has corrected the issue I will just delete all out comments. $\endgroup$ – dmckee --- ex-moderator kitten Jan 4 '19 at 3:24
  • $\begingroup$ @InertialObserver Thank you, can you give me reference to this, the way you presented this is very interesting, I would like to read more $\endgroup$ – sheshin Jan 4 '19 at 6:54
  • $\begingroup$ @sheshin Personally, I learned this from Goldstein, which may be a bit dense for you at this point. Perhaps John Taylor's classical mechanics section on phase space will be of use to you. $\endgroup$ – InertialObserver Jan 4 '19 at 7:03

A circle trajectory is described by this equation:

$x=r\,\cos(\varphi(t))\qquad (1)$

$y=r\,\sin(\varphi(t))\qquad (2)$

if the circle radius $r$ is constant we have periodic function

To calculate the so call phase diagram ($\dot{y}$ over $\dot{x}$) we take the time derivative (r=constant) of equations (1) and (2)



with $\dot{\varphi}=\frac{v(t)}{r}$ and $v(t)$ is the tangential circle velocity.

So only if the velocity $v(t)$ is constant we get again circle trajectory (periodic function)

enter image description here

  • $\begingroup$ I think circular motion with constant radius but varying speeds is possible. If I rotate a stone tied to a string in a circle and if I increase the speed of rotation, even then the stone will be in circular motion with constant radius (=the length of the string). I think the way you differentiated the equation for x and y might not be true. Thanks for the response. @InertialObserver has answered the question. $\endgroup$ – sheshin Jan 5 '19 at 6:39
  • $\begingroup$ I got the same answer that @InertialObserver gave? $\endgroup$ – Eli Jan 5 '19 at 9:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.