Gear ratio in bicycles using rotational motion When we change the gears of the bicycle we are riding, we change the the disc we are currently at (which are located at the place where we pedal) to some other disc. This means the radius of the circular disc we were pedaling/rotating changes. So that means if we were rotating the disc with angular velocity $ω$, if $r$ changes (radius of the disc) $rω$ changes. And that means the speed with with the chain which rolls over the disc, i.e. $v=rω$ changes. But how does that make the bike move faster with the same angular velocity we were providing it as before? If we want it to move faster, then the velocity of COM of the rear/front tire should increase, but how does changing gears do that?
 A: The same angular velocity of the pedal do not means same angular velocity of the wheel.
Assume a chairing with radius $r_1$ and angular speed $\omega_1$, and the cassette with angular speed $\omega_2$ and radius $ r_2$ (considering the cassette or the wheel do not make any difference). The speed the chain rolls reads:
$r_1 \omega_1=v=r_2 \omega_2$.
From this equation is clear that $\omega_1\neq\omega_2$ (otherwise $r_1=r_2$). Let's now change the gear i.e. going from $r_1$ to $r_1'$ (assume $r_1'>r_1$) by keeping the same angular speed $\omega_1$.
$r_1'\omega_1=v'=r_2\omega_2'$. Since $r_1'>r_1$ we have $v'>v$ therefore $r_2\omega_2'>r_2\omega_2\rightarrow\omega_2'>\omega_2 $ 
A: When you are changing gears you are trading speed for torque (or vice versa). The overall power transmitted maintains the same so $P=\omega_I T_I = \omega_O T_O$.
The way this works is by the chain forcing the same tangential velocity on the two sprockets (input and output sprocket) from which their angular velocity is found $\omega_I = \frac{v}{r_I}$ and $\omega_O = \frac{v}{r_O}$. Use these relationships in the power above to get
$$ \left. P = \frac{v}{r_I} T_I = \frac{v}{r_O} T_O \right\} \left. \frac{T_O}{T_I} = \frac{r_O}{r_I} \right\} \frac{\omega_I}{\omega_O} = \frac{r_O}{r_I} $$
Now consider that your legs make pretty much constant torque $T_I$ for a variety if pedal speeds $\omega_I$ in order to move faster you are going to need a smaller output sprocket (small $r_O$). This is because $\omega_O = \frac{v}{r_O}$. The effect of that to the output torque is $T_O =  T_I \frac{r_O}{r_I}$ and because the input torque $T_I$ is constant and the input sprocket hasn't changed the output torque (to the wheel) is reduced.
With less torque available for acceleration it becomes harder to resist hills and air drag and hence it feels you are doing more work in higher gears (small output sprocket).
A: I will first discuss how the system is like and then elaborate how the rotations are connected.
The paddles of a bicycle is attached to the front gear, which you directly supply a force to. The back gears are attached to the back wheel which you are able to swap the gear size.
Central to the question is that:
$$v=r\omega$$
When you paddle (the front gear), you supply a certain $\omega$. The speed, of the chain which goes around the gear, will then travel with velocity $v$.
We note that since the chain (should be!) taut, the same chain that wraps around the gear at the back wheel is also be moving at the same $v$. Looking back at the expression, we see that since for the same $v$, a smaller gear (smaller $r$) would mean a higher $\omega$, meaning to say that the gear (and wheel) spins faster.
This would mean then that now, in the time you make 1 rotation on the front gear, the back gear and thus back wheel can potentially spin more than once, depending on how small the gear is. This means that the total circumference the wheel makes now is larger in the same time, covering a larger distance. Another way to see this is to consider $v_{wheel}=r_{wheel}\omega_{backgear}$. In other words, if you paddle at the same rate, the back wheel can potentially rotate faster if the back gear is smaller.
However, this only translates to theoretically faster speeds. In actuality the whole system is also constraint by how much torque/force that is supplied. If you notice it is also harder to start from rest if the back gear that is used is smaller.
While my post has gone out of the way to describe another configuration, you should notice something: that the velocity that one can (potentially) achieve is increased if, all else being equal:
1) You change the front gear to a bigger one
2) You change the back gear to a smaller one
This is where the concept of gear ratio come in slightly more handy.
