Why is the velocity on the top of a wheel twice the velocity of its axle? When a wheel is rolling, not skidding, and its axle moves at velocity $\vec{v}$, then a point on the top of its circumference will move at velocity $2\vec{v}$, shown below.

I really don't understand this. I'm quite familiar with the geometry of a circle, but I don't understand how it's being applied on this case. Also:


*

*Why doesn't the relation between the velocities depend on the radius of the wheel? 

*When is this relation valid? Does it take friction into account or it doesn't matter? If it doesn't matter, why?

*A consequence of all this is that the point in the bottom has no velocity. Why? That makes no sense to me.


This is a very confusing topic to me. Here is a very nice page full of pictures and animations about the physics of a wheel but I still don't get it.
 A: Put on your "instantaneous" glasses.
i.e. think of what happens in a really, really, short moment of time.
Either that, or suppose the wheel is turning really, really, slowly.
Either that, or suppose it is not turning at all, except you move the top of it forward 1 millimeter.
Now suppose it is not a wheel at all.
Suppose it is just a stick or rod from the bottom where its in no-slip contact with the road to top where you have moved it a little bit.
Or suppose it is the wheel, but you have drawn a line on it from bottom to top, so that line is just like the stick.
OK, you moved the top 1 millimeter.
How far did the bottom move? (zero - no slip)
How far did the axle move, considering that it's half-way up?
You take it from there...
A: I'll tackle your questions in reverse:
3. The contact point is stationary because the wheel is not slipping. This happens when the force of static friction is able to counter the force of the wheel on the ground. This is what you want for controllable transport. If the wheel starts slipping (because of low friction) that's a skid and you are no longer able to steer or brake. If you like, imagine getting your car stuck in mud. You spin the wheel and fling mud all over the place without getting anywhere - not enough friction. If that doesn't help try taking a wheel, marking a spot on it, and slowly rolling it while carefully watching the point of contact.
2. You need sufficient static friction to enforce the no-slip condition. The relation between the velocity at the top, centre and bottom of the wheel is geometrical and is not affected by friction per-se. If the car wheel spins with angular frequency $\omega$, has a radius $R$ and velocity at the axle of $v$ then the velocity of the wheel at the top and bottom is $$ v_{top} = v + \omega R, v_{bottom} = v - \omega R $$
1. If the no slip condition holds then $v_{bottom}=0$, so $ v = \omega R $. Using this in the top equation gives $ v_{top} = v + v = 2v $, independent of $R$. This is because both $v$ and $v_{top}$ increase in the same proportion as $R$ increases.
(Aside: Is there a way to make markdown do reverse lists?)
A: A wheel rolling without slipping means, any point on the circumference of the wheel will have both rotatory motion and translatory motion. Due to translation, if the velocity of the vehicle is v m/s, then all the points of the wheel will have v m/s linear velocity in the forward direction. Then due to the rotatory motion, each point will have a linear velocity of v = rω, where ω - is the angular velocity. For the top most point at a particular instant, the linear velocity will be horizontal forward. So at any instant, a top most point on the wheel will have two linear velocities of same magnitude $v_\mathrm{total} = v+v = 2v$.
Slipping means all the points of the wheel will be moving with same velocity. So, there is no rotatory motion and hence no additional linear component of velocity. So, $v_\mathrm{total} = v+0 = v$. Hence the above equation is valid only when the wheel is rolling without slipping. (Slipping or skidding of wheels of a vehicle occurs when the brakes are fully applied and still the vehicle is moving but the wheels are not rotating).
Also, this condition is valid for a particular point at a particular instant only. In the next instant the point we considered will have velocity and another point which is very next to the previous one will have zero velocity because, otherwise if the point touching the ground is having velocity, that means the wheel is slipping not rolling.
A: Let's start with a freely moving body. Imagine a solid rod of radius $r$ attached perpendicularly to an axle (like the rods of a bike to which the pedals are attached). Let's assume the axle is oriented horizontally to a horizontal plane so the rod can in a plane perpendicular to this horizontal plane.
Now let the rod rotate and let's imagine that the outer part of the rod is moving with velocity $\vec v$ (which obviously has a different direction on each point of the circle traced out by the rod). At this moment the whole structure isn't moving with respect to the horizontal plane.
Then we let the rod and axle (by whatever means) move with a velocity $\vec v$ in the direction of the velocity the rod has when it's at its highest point.
It's clear that the upper part of the rod has now velocity $2\vec v$ and that the lower part has velocity (speed) $0$ (for the wheel this means that there is a friction force that opposes the imposed velocity $\vec v$ on the whole structure).
So the speed of the outer part of the rod (the wheel) varies between $0$ (when it reaches its lowest point) and $2v$ (when it reaches the highest point). So no point on the outer rod ever moves backward.
I think the answer to your first question has to do with the infinitesimal $2ds$ being depicted as a finite distance (and $2ds$ is on the circle), which can make one think that the velocities of the wheel are indeed radius dependent. But in practice, it's impossible to depict an infinitesimal quantity.
