Finding angular velocity from tangential speed and radius I can't wrap my head around this problem I'm trying to solve. I'm given the tangential speeds of two objects undergoing uniform circular motion and the difference in their radiuses and I'm supposed to somehow find angular velocity. The problem is similar to something like this: Two cars are moving on two separate race tracks. The radius of the track for car 2 is 0.9 times the radius of the track for car 1. Both cars start side by side, along a radial line, and drive at the same speed in a counterclockwise direction. How many revolutions has car 2 made when car 1 and car 2's velocity point in opposite directions for the first time.
Any help would be appreciated...
 A: Well first recall that velocity is given by 
$$v(t)=\frac{dl}{dt}$$
and in the case of circular motion $l=r\ \theta$ so $v=r\frac{d\theta}{dt}$. Now, recall that $\frac{d\theta}{dt}=\omega$. Then we find $\omega=v/r$.
Now, in regards to the car question, we know that $v$ is equal for both cars. Let car 1 be at a radius $R$ . Then its angular velocity will be $\omega_1=v/R$. On the other hand, car 2 will have an angular velocity of $\omega_2=v/0.9R$. We now know the angular velocities for both cars.
The only thing left to do is to figure out when their velocities will point in opposite directions. It should be obvious that this occurs when the two cars are on opposite ends of the circle. In particular we will want $$\theta_2(t)-\theta_1(t)=\pi$$
Then we write:
$$\omega_2t-\omega_1t=\frac{v}{R}(1-0.9)t=0.1\frac{v}{R}t=\pi$$
so the time at which this happens is $t_0=10\pi R/v$.
Finally to find the angular distance that car 2 will have covered by then we take $R\omega_2t_0$ to find that it is $9\pi R$. This is equal to 4.5 circumferences of the circle so the second car will have completed $\mathbf{4.5\ revolutions}$ by the time that the two have opposite velocities.
