Ratio of masses of double star problem I am currently studying Classical Mechanics, fifth edition, by Kibble and Berkshire. Problem 2 of chapter 1 is as follows:

The two components of a double star are observed to move in circles of radii $r_1$ and $r_2$. What is the ratio of their masses? (Hint: Write down their accelerations in terms of the angular velocity of rotation, $\omega$.)

The only relevant information provided by the chapter is as follows:

If we isolate the two bodies from all other matter, and compare their mutually induced accelerations, then according to (1.1) and (1.3),


$$m_1 \mathbf{a}_1 = - m_2 \mathbf{a}_2 \tag{1.7}$$

Since the textbook chapter does not provide enough information to complete this problem, I referred to the Wikipedia article for angular velocity. Writing linear velocity as $v = \omega r$, we get
$$m_1 \mathbf{a}_1 = -m_2 \mathbf{a}_2$$
$$\therefore m_1 \left( r_1 \dfrac{d \omega_1}{dt} \right) = -m_2 \left( r_2 \dfrac{d \omega_2}{dt} \right)$$
$$\Rightarrow \dfrac{m_1}{m_2} = \dfrac{\left( -r_2 \dfrac{d \omega_2}{dt} \right)}{\left( r_1 \dfrac{d \omega_1}{dt} \right)}$$
The only way that I can see to proceed would be to assume that the angular velocities $\omega_1$ and $\omega_2$ are the same (I have no idea if this is implied by the physics of a "double star"):
$$\therefore \dfrac{m_1}{m_2} = - \dfrac{r_2}{r_1}$$
The answer is said to be $\dfrac{m_1}{m_2} = \dfrac{r_2}{r_1}$.
Why are the angular velocities equal? And what happened to the negative sign? I would greatly appreciate it if people would please take the time to clarify this.
 A: Your core question appears to be:

Why are the angular velocities equal? 

For some reason, Wikipedia's articles about the two body problem do not clarify this important point. Here are a couple of diagrams from that article:


Two bodies with similar mass orbiting a common barycenter external to both bodies, with elliptic orbits—typical of binary stars. 



Two bodies with a "slight" difference in mass orbiting a common barycenter. The sizes, and this type of orbit are similar to the Pluto–Charon system (in which the barycenter is external to both bodies), and to the Earth–Moon system—where the barycenter is internal to the larger body.

According to Newton's law of gravitation, the gravitational force between two bodies acts along the straight line connecting their centres of mass. This is the key to your question about angular velocities.
(It can also be shown that the gravity of a spherically symmetric body acts as if all of the body's mass were concentrated at its centre, so we can treat the body as a point particle).
So we have two stars, $S_1$ and $S_2$, exerting gravitational force on each other. The centre of mass of this system must be located on the line $S_1S_2$ that connects the two stars' centres. We can choose a reference frame with the centre of mass as its origin, like in the above diagrams. (As Rob Jeffries says, we can do this because of conservation of momentum).  So I'll call the centre of mass $O$.
Now, as the stars orbit around $O$ the only forces they exert on each other always act along the line
$S_1OS_2$, so the stars and the centre of mass must remain collinear, although the line $S_1OS_2$ rotates, and may change in length (as it does in the elliptical orbit example).
The only way for that to happen  is for the angular velocities of the two stars, $\omega_1$ and $\omega_2$, to always be equal to each other. Otherwise, $S_1OS_2$ turns into a triangle, not a straight line.


And what happened to the negative sign?

That negative sign merely tells us that the position vectors of the two stars, $OS_1$ and $OS_2$ point in opposite directions. That is, the two stars are on opposite sides of $O$.
A: In the absence of external forces, the total momentum of the system is constant. That means the stars must arrange themselves so that the centre of mass stays in the same place.
The only way they can do this is if they orbit with the same period and hence the same angular velocity, as shown below.
If that were not the case, then the centre of mass would "wobble about", which is clearly unphysical.
That also means you can immediately write
$$m_1 r_1 = m_2 r_2\ .$$
A: Expressing for force magnitudes :
$$
m_1a_1 = m_2a_2
$$
Gravitational force gives centripetal acceleration, substituting it gives :
$$ m_1 \omega_1 ^2r_1 = m_2 \omega_2 ^2r_2
$$
For double star $\omega_1 = \omega_2$, because it is a rotating dipole, which every point achieves $2\pi$ rotation in same period $T$.
I.e. imagine a couple of people holding each other hands and rotating around their COM. Can their angular speed differ ? No, they have same $\omega$, as well as each dipole point along the line connecting them.
Check this pic :

$$\omega_1 = \omega_2 = \omega_3 = \omega_4 = \omega_5 = \ldots = \omega_n$$
Where $\omega_n$ is n-th point along the line connecting pair of bodies and going through their barycenter angular speed.
So, given above equation reduces to :
$$ m_1r_1 = m_2r_2 $$
Which gives mass ratio :
$$ \frac{m_1}{m_2} = \frac{r_2}{r_1} $$
