## Statement

Two bodies, c_{1} and c_{2}, begin to move from the same point at constant angular velocity, but in opposite directions, along a 30 m radius circumference. If the first one takes 20 seconds to complete a rotation and the second takes 60 seconds, calculate:

a) The time that they take to meet.

b) The angle and distance traveled by each one.

## Solution

**Data**

R = 30 m

T_{1} = 20 s

T_{2} = 60 s

${\phi}_{01}={\phi}_{02}=0rad$

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:

$${w}_{1}=\frac{2\pi}{{T}_{1}}=\frac{2\pi}{20}=\frac{\pi}{10}rad\phantom{\rule{0ex}{0ex}}{w}_{2}=\frac{2\pi}{{T}_{2}}=\frac{2\pi}{60}=\frac{\pi}{30}rad$$

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:

$${\phi}_{1}={\phi}_{01}+{\omega}_{1}\xb7t\Rightarrow \phantom{\rule{0ex}{0ex}}2\pi -{\phi}_{2}=0+{\omega}_{1}\xb7t$$

Knowing that similarly ${\phi}_{2}={\omega}_{2}\xb7t$, then:

$$2\pi -{\phi}_{2}={\omega}_{1}\xb7t\Rightarrow \phantom{\rule{0ex}{0ex}}2\pi -{\omega}_{2}\xb7t={\omega}_{1}\xb7t\Rightarrow \phantom{\rule{0ex}{0ex}}({\omega}_{1}+{\omega}_{2})\xb7t=2\pi \Rightarrow \phantom{\rule{0ex}{0ex}}\left(\frac{\pi}{10}+\frac{\pi}{30}\right)\xb7t=2\pi \Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)t=15s}$$

**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:

$${\phi}_{1}=\frac{\pi}{10}\xb715\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{){\phi}_{1}=1.5\pi rad}$$

And that the angle traveled by c2 is:

$${\phi}_{2}=2\pi -{\phi}_{1}\Rightarrow \phantom{\rule{0ex}{0ex}}{\phi}_{2}=2\pi -1.5\pi \Rightarrow \phantom{\rule{0ex}{0ex}}\overline{){\phi}_{2}=0.5\pi rad}$$

To calculate now distance traveled (s) for each of the two bodies, simply apply the equation $s=\phi \xb7R$:

$${s}_{1}={\phi}_{1}\xb7R=1.5\pi \xb730=45\pi m\phantom{\rule{0ex}{0ex}}{s}_{2}={\phi}_{2}\xb7R=0.5\pi \xb730=15\pi m$$