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This is an excerpt from an elementary Physics textbook ("Fundamentals of Physics" by Halliday), which describes what happens when a wheel that is "rolling without slippging" doesn't have a constant speed. This part doesn't make sense:

Figure 11-7 shows an example in which a wheel is being made to rotate faster while rolling to the right along a flat surface, as on a bicycle at the start of a race. The faster rotation tends to make the bottom of the wheel slide to the left at point P. A frictional force at P, directed to the right, opposes this tendency to slide.

Why does this frictional force go the same direction as the force that's accelerating the wheel? I've seen some numerical proofs of this on StackExchange, but I couldn't find a conceptual proof for it.

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    $\begingroup$ Can you link to the "numerical proofs" on StackExchange you've seen? $\endgroup$
    – Kyle Kanos
    Dec 28, 2018 at 13:07

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Why does this frictional force go the same direction as the force that's accelerating the wheel?

Because it is the force that's accelerating the wheel. There are no other forces pushing forward (to the right). Without static friction the bike would not move forward at all.

Try to lift the wheel off of the ground - you can spin the pedals all you want, but you won't move forward. Or try cycling on slippery ice with no friction. You won't be able to move anywhere.

So, obviously static friction does push forward - but why?

  • Consider the act of running. You push your foot backwards on the ground. So, your foot exerts a backwards force on the ground through static friction. And this pushes your body forward. There is namely a responding static friction acting on you the opposite way.

This is Newton's 3rd law. You apply a force and a responding force (or a reaction force) acts on yourself the opposite way. This is what happens for a rolling wheel at the contact point.

  • With the pedals you create a torque about the wheel center. This makes the wheel rotate. Was the wheel kept stationary, it would be sliding over the ground at the contact point. It is trying to push the ground backwards through friction. The ground responds by applying that same friction forward.

And this is how a bicycle is able to push itself forward by exerting a backwards push in the ground.

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  • $\begingroup$ If the static friction force for the rear wheel is going forward and the rotational accelerating force is going backward, wouldn't that mean that the net force on the rear wheel is zero, and is static? $\endgroup$
    – APerson
    Dec 31, 2018 at 20:15
  • $\begingroup$ @APerson The "rotational acceleration force" (which is also static friction) is acting on the ground, and the forward static friction force is acting in the wheel. They do not act on the same object. Otherwise nothing in this world would ever move, since every force has an opposite reaction force. The point is that this reaction force always acts on another object. So it doesn't cancel anything out on the same object. The wheel only feels the forward force, not the other force, and therefor it has a net acceleration. $\endgroup$
    – Steeven
    Dec 31, 2018 at 22:34
  • $\begingroup$ Thanks for the quick reply; from this, I can tell that I've been thinking about this the wrong way. Just to reiterate, there are two static friction forces, both acting in opposite directions and on different objects, right? If so, which object causes the "backward" static friction force and which one causes the "forward" static friction force. And which one "comes first" (i.e. which one causes the other). Sorry if I'm repeating earlier questions, just want to make sure I get this. $\endgroup$
    – APerson
    Dec 31, 2018 at 22:55
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Think about a bicycle accelerating from left to right in the diagram. The page in your image doesn't say that the diagram shows the rear wheel of the bike not the front wheel, but that is important. (See the last paragraph of the answer for what happens to the front wheel.)

If there is no friction, the rotation of the wheel is accelerated by a moment applied to the axle of the wheel by the bike chain. That will make the wheel spin clockwise as shown by the arrow, so the bottom of the wheel is slipping from right to left relative to the ground.

The friction force that opposes the slipping acts on the bike from left to right, as shown in the picture. This is "the force that is accelerating the linear motion of the bike".

The force the rider applies to the pedals does not accelerate the bike, it only spins the rear wheel faster. If there was no friction force, the bike wouldn't move at all. Think about the similar situation of trying to start a car moving on an icy road - if there is no friction between the wheels and the ice, the engine will spin the wheels but the car won't move.

If you think about the front wheel, the situation is different. Now, there is a horizontal force applied to the axle of the wheel by the bike frame. If there was no friction, the wheel would slide to the right without rotating at all. The friction force on the sliding wheel acts to the left, and increases the rotation speed of the wheel so that it doesn't slip.

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  • $\begingroup$ Why is there a torque being applied to the axle? Isn't the axle on the axis of rotation? Also, why does the force that the rider applies not accelerate the bike? If the rear wheel is spinning faster, doesn't the bike also have to be moving faster to "keep up"? $\endgroup$
    – APerson
    Dec 26, 2018 at 15:02
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Figure 11-7 shows an example in which a wheel is being made to rotate faster while rolling to the right along a flat surface, as on a bicycle at the start of a race.

To simplify matter assume that it is the wheel of a unicycle.
The passage implies that a clockwise torque about the axle has been applied on the bicycle wheel by the pedals attached to the bicycle wheel.
(All a chain and gear arrangement does is to transfer and modify the applied torque).

That clockwise torque will produce a tendency for the bicycle wheel to slip relative to the ground.
To prevent this a friction force acts to the right which applies an anti-clockwise torque on the bicycle wheel ie in the opposite direction to that produced by the pedals to reduce the tendency for the wheel to rotate faster.

That frictional force to the right also produces a translational acceleration of the centre of mass of the wheel to the right which will increase the translational speed of the wheel to the right thus reducing the tendency of the wheel to slip as the wheel is forced to rotate faster by the applied torque.

In terms of the no slip condition $v=r\omega$, the frictional torque is trying to prevent $\omega$ increase “too rapidly” and the frictional force is making $v$ increase to keep up with the increase in $\omega$.

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The idea of the torque due to friction and the force due to friction being “opposed” to one another occurs very often in rotational dynamics.
If you have a cylinder rolling down a slope the frictional force is up the slope and is trying to reduce the translational acceleration of the centre of mass of the cylinder down the slope whilst that same frictional force is providing a torque on the cylinder which is trying to increase the rotational speed of the cylinder.

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To understand this we must understand that friction is the force that is actually driving the wheel.

When you apply force on the wheel by peddling the wheel moves in a clockwise direction from 'right to left'.

As the motion is just starting we see that the wheel is slipping relative to ground which means friction is acting from 'left to right'.

And this is the force that starts the rotation of the wheel.

Friction is the very force that accelerates the wheel by opposing its rolling motion in backwards direction.

The concept here is same, ie- Friction is still opposing the motion of wheel. This is opposition is leading to a net acceleration in front direction.

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