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:

1. Why doesn't the relation between the velocities depend on the radius of the wheel?
2. When is this relation valid? Does it take friction into account or it doesn't matter? If it doesn't matter, why?
3. 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.

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The really simple way to get #3 is to start from a frame of reference tied to the axle. In this frame, the bottom of the wheel has velocity -v, while the top of the wheel has velocity v. To convert back into the ground's frame, add v to each of these, so they become 0 and 2v. – Ben Crowell Sep 10 '13 at 4:36

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?)

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@MichaelBrown "The relation between the velocity at the top, centre and bottom of the wheel is geometrical". Could you please elaborate a little bit more on the geometry? I know that an inscribed angle in a circumference determines an arc measuring twice its length, for example --and I fully comprehend the triangular similarity involved in that. Does that have anything to do with the wheel? I don't understand how. – BeetleTheNeato Jan 3 '13 at 20:35
@MichaelBrown I'm also struggling to understand that "The contact point is stationary because the wheel is not slipping.". Isn't the contact point ever changing? If I mark a point on the bottom of the wheel with a chalk, it'll move upwards as the wheel rolls, right? So how is the contact point stationary? Or does that mean that at every moment the one contact point at that time is stationary? – BeetleTheNeato Jan 3 '13 at 20:41
@BeetleTheNeato The part of the wheel in contact with the ground is stationary for the instant it is in contact. It moves before and after that, of course, as the wheel turns. You can resolve the velocity of any point on the wheel surface into horizontal and vertical components. If $\theta$ is the angle from the top of the wheel, then for a free spinning wheel with no overall forward velocity you have $v_h =v \cos\left(\theta\right), v_v=- v \sin\left(\theta\right)$. When the wheel moves without slipping add $v$ to the horizontal velocity: $v_h= v \left[\cos\left(\theta\right)+1\right]$ – Michael Brown Jan 4 '13 at 1:49
Notice that at the bottom of the wheel, $\theta=\pi$, $v_h = v_v = 0$. The contact point is stationary. I arranged that by adding just the right amount to the horizontal velocity. As a consequence, at the top of the wheel $v_h = 2v$. For any given piece of the wheel these equations are instantaneous: the velocity changes as it moves around the wheel. This means the material in the wheel is accelerating, which means there must be a force towards the centre (centripetal force), which means there must be spokes or something connecting it to the axle or the wheel must be sufficiently strong. :) – Michael Brown Jan 4 '13 at 1:57
If the geometry of the similar triangles isn't helping you try thinking of it this way: in the rest frame of the axle (moving along with the car) the top of the wheel and bottom of the wheel are moving in opposite directions at the same speed $\omega R$. When you go to the frame of the ground you add $v$ to the horizontal component of the velocity. Unless $v=\omega R$ the bottom will slip, so take $v=\omega R$. This kills the velocity of the contact point because it's going the opposite direction, gives the axle a velocity of $v$ and doubles the velocity of the top. – Michael Brown Jan 4 '13 at 2:04