# Can non-uniform circular motion be called as periodic motion?

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

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

and

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

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.

• But the situation described in the question decidedly does meet the conditions, so the first would would be rather better as "No." Commented Jan 4, 2019 at 3:14
• Ah yes, thank you. I was responding to the question posed in the title. Commented Jan 4, 2019 at 3:21
• @AaronStevens Yes. I did mean that. Good catch. But as InertialObserver has corrected the issue I will just delete all out comments. Commented Jan 4, 2019 at 3:24
• @InertialObserver Thank you, can you give me reference to this, the way you presented this is very interesting, I would like to read more Commented Jan 4, 2019 at 6:54
• @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. Commented Jan 4, 2019 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)

$$\dot{x}=-r\,\sin(\varphi)\,\dot{\varphi}$$

$$\dot{y}=r\,\cos(\varphi)\,\dot{\varphi}$$

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)

• 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. Commented Jan 5, 2019 at 6:39
• I got the same answer that @InertialObserver gave?
– Eli
Commented Jan 5, 2019 at 9:11