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

Question

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?

Answer 1

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.

Answer 2

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)