Consider two circles of radii $4\;cm$ and $8\;cm$, respectively, both circles have the same center $C$ and is two bodies $A$ and $B$, so that $A$ is smaller circumference of the trajectory at a constant speed of $5\; km/h$, while $B$ is also the greatest circumference at a constant speed of $10\;km/h$.
Knowing that the bodies initially form an angle of $90^o$ with respect to the center $C$, where the time needed and how to calculate so that the points $A$, $B$ and $C$ are aligned?
And how to calculate the next time they met aligned?
The angular velocity of bodies A and B are : $$\omega_A=\frac{v_A}{r_A}=\frac{5km/h}{4cm}=\frac{1250}{36}rad/sec$$ $$\omega_B=\frac{v_B}{r_B}=\frac{10km/h}{8cm}=\frac{1250}{36}rad/sec$$
Since both bodies have same angular speed, the angle between two bodies (with respect to the center C) does not change with time, and considering the fact that at the initial time this angle is $90^0$, the points A, B and C can not be aligned in any time.
Proof:
$$\theta_A(t)=\omega_A t$$ $$\theta_B(t)=\omega_B t\pm\frac{\pi}{2}$$
(In the latter equation, choose $\frac{\pi}{2}$ if at $t=0$, $\theta_B$ is larger than $\theta_A$, or choose $-\frac{\pi}{2}$ if at $t=0$, $\theta_A$ is larger than $\theta_B$) Therefore,
$$\theta_B(t)-\theta_A(t)=(\omega_B -\omega_A)t\pm\frac{\pi}{2}=\pm\frac{\pi}{2} \neq 0, \pm \pi, \pm 2\pi , ...$$