An encounter in u.c.m.

Data

R = 30 m
T₁ = 20 s
T₂ = 60 s
${\phi }_{01}={\phi }_{02}= 0 rad$
 

Since their trajectory is a circumference and their angular velocities are constant we face a problem of uniform circular motion or u.c.m.

Question a)

Since we know the period of each one, we can calculate the angular velocities by means of the following equation:

$$w=\frac{2\pi }{T}$$

Therefore:

$$\begin{gathered} w_1=\frac{2\pi }{T_1}=\frac{2\pi }{20}=\frac{\pi }{10}rad \\ {w}_2=\frac{2\pi }{T_2}=\frac{2\pi }{60}=\frac{\pi }{30}rad \end{gathered}$$

Both bodies will meet each other before completing a rotation. In particular, when they meet the sum of their angular positions will be exactly 2π radians.

$${\phi }_1+{\phi }_2=2\pi$$

Therefore:

$${\phi }_1=2\pi -{\phi }_2$$

If we use the equation of position of ${\phi }_1$ , and substitute, we get:

$$\begin{gathered} {\phi }_1={\phi }_{01}+{\omega }_1·t \Rightarrow \\ 2\pi -{\phi }_2=0+{\omega }_1·t \end{gathered}$$

Knowing that similarly ${\phi }_2={\omega }_2·t$, then:

$$\begin{gathered} 2\pi -{\phi }_2={\omega }_1·t \Rightarrow \\ 2\pi -{\omega }_2·t={\omega }_1·t \Rightarrow \\ ({\omega }_1+{\omega }_2)·t=2\pi \Rightarrow \\ \left(\frac{\pi }{10}+\frac{\pi }{30}\right)·t=2\pi \Rightarrow \\ \boxed{t=15 s} \end{gathered}$$

Question b)

If we replace the time at which they meet (t = 15 s) and the angular velocity in the equation of position of ${\phi }_1$, we get that the angle traveled by c1, which is:

$$\begin{gathered} {\phi }_1=\frac{\pi }{10}·15 \Rightarrow \\ \boxed{{\phi }_1=1.5\pi rad} \end{gathered}$$

And that the angle traveled by c2 is:

$$\begin{gathered} {\phi }_2=2\pi -{\phi }_1\Rightarrow \\ {\phi }_2=2\pi -1.5\pi \Rightarrow \\ \boxed{{\phi }_2=0.5\pi rad} \end{gathered}$$

To calculate now distance traveled (s) for each of the two bodies, simply apply the equation $s=\phi ·R$:

$$\begin{gathered} s_1={\phi }_1·R=1.5\pi ·30 = 45\pi m \\ {s}_2={\phi }_2·R=0.5\pi ·30 = 15\pi m \end{gathered}$$