What is the power while accelerating a bicycle?

I decided to post an answer because of an idea that I had today, and it is too long for a comment.

In the system bike + biker, the only external horizontal force is the friction with the ground. I am neglecting air drag and other friction forces.

During the acceleration, the wheels are increasing its angular velocity. The required torque only for that is $T_i = Ialpha$. where $I$ is the moment of inertia of the wheel, and $alpha$ the angular acceleration. It would be there even if there was no ground friction, and the bike was stationary, slipping in the same location.

But, because of the ground friction, the torque is greater. $T_{tot} = Ialpha + Fr$. The biker may provide less than $Fr$, depending on the total gear ratio, but that is the main torque in the wheels (normally well above $T_i = Ialpha$).

The power during the acceleration is: $W = T_{tot}omega$. Most of it to accelerate the translational movement of the system, and a minor part to increase the angular velocity of the wheels.

My mistake was to take $Fromega = Fv$ as an additional power to increase the rotation of the wheels, only because it has the form: $Tomega$. In reality that is already the power to accelerate the bike. That is why it can be expressed in the form $W = Fv$.

Consider the bicycle and cyclist as a system. If the bike is on a flat road, there are two relevant external forces:

• Friction with the ground pushes the bicycle forward. Because the velocity of the wheel's contact point with the ground is zero, the power supplied by friction is exactly zero.
• Sources of dissipation, such as air resistance, do negative work on the bike.

Therefore, the net power of external forces acting on the bicycle-cyclist system is always negative. (The force that the cyclist exerts on the pedals is an internal force, so it doesn't count in this analysis.) This makes sense, because over a long period of time, the cyclist's lunch is used up in pedalling. The chemical energy of the lunch leaves the system and ends up dissipated as heat.

The reason you're confused is probably because you think that the friction force with the ground ought to contribute positive work, since it's responsible for moving the bike forward. But it doesn't do any work at all, because it doesn't change the energy of the bicycle-cyclist system. It only converts rotational energy of the wheels to translational motion of the bike.

The energy produced by the cyclist's legs go four places:

1. Overcoming friction in gearing and between wheels and pavement.
2. Overcoming change in elevation (when cycling up a hill).
3. Overcoming wind resistance.
4. Adding momentum to bike and rider (acceleration).

Generally, the friction in gearing (including chain) is relatively small and can be ignored, but, depending on the style of tires, their inflation, the weight of bike and rider, and the pavement surface, the friction between wheels and pavement can be substantial (especially with wide, heavily-treaded tires that are not highly inflated).

Wind resistance, on reasonably level surfaces and at typical biking speeds (10-25 mph), is by far the biggest "drain" (and hence cyclists try to reduce their "profile" in various ways, from crouching to using fairings).

Energy that goes into elevation gain is (potentially) recovered on the downhill run (though the difference in speed means that more energy is lost to wind resistance going downhill).

Energy used to increase momentum is "preserved", and, after accounting for wind resistance and braking loses, is reclaimed when the cyclist stops pedaling.

I'll note that, in addition to the power that goes into increasing the momentum of bike and rider, there is a tiny bit that goes into increasing the angular momentum of the wheels. This was addressed fairly well in this question. The net-net is that, if each bike rim and tire weighs 1kg, then the two wheels add the equivalent of 4kg to the weight of bike and rider, when considering the overall momentum of the system.

There's only one external force on the bicycle. That one force can produce translational momentum changes/energy and rotational momentum changes/energy. But the sum of all translational and rotational energy will be equal to $Fd$, or the power equal to $Fv$.

If the bike has some fixed total mass $M$, then we can distribute the mass between the frame of the bike and the wheels. The more mass is in the wheel, the greater the rotational moment of inertia and the greater the rotational energy at a given $v$.

This means that if you could keep $F$ and $v$ fixed (same power) and total $M$ fixed, the bike with less mass in the wheels will go faster since more of the power is going into accelerating the bike and less is going into the energy of spinning the wheels.

It's a matter of identifying your systems. As you mentioned, power is derived from force; hence, just like force, power is delivered from one system to another system.

You can talk about the power from the person or the ground or both together, and to the wheels or the bicycle as a whole.

Since the wheels are rotating, you apply the torque formula to them, as you did above. But note that this is actually a vector formula:

$W_r = overrightarrowtau cdot overrightarrowomega$

If you picture the bicycle traveling to the right, the wheel rotation ($overrightarrowomega$) is clockwise. The ground is applying friction to the right, which makes for a counterclockwise torque ($overrightarrowtau$), so that's actually a negative power. Meanwhile the person, via the chain, applies a clockwise torque and hence a positive power. So actually these two powers are subtracted, not added. In fact, the difference of the torques determines the angular acceleration of the wheels, which must match the linear acceleration of the bike, so you can solve this system of equations:

$tau_{person} - tau_{ground} = I_{wheels}alpha_{wheels}\ F_{ground} = m_{bike}a_{bike}$

If you want to talk about the bike as a whole, its motion is translational, not rotational. So you use the force formula, but again keep in mind it's actually a dot product:

$W = overrightarrow{F} cdot overrightarrow{v}$

Here, as you already figured out, the force from the ground is all acting on the rotating wheels, so you will get the same value as with the rotational formula - but it's just an alternative way of analyzing the same interaction force.

But when you talk about the power "to accelerate the bicycle", it sounds like you're thinking of the power of the person on the bicycle as a whole. But this is a little illusory. Here, you would need the total force between the person and the bicycle - but they are moving together. If they are coasting at constant speed with no wind resistance, that force is zero. And if there is wind resistance, or they are still getting up to speed, the person is effectively being pulled by the bike, and hence they are putting a negative force back on the bike! Even though they are putting power into the rotation of the wheels, the person cannot put translational power into the whole bike because they are sitting on it! Instead, the ground must convert that rotational power into translational via friction.

Basically, since the person and bike move together, it's easiest to treat them as one system together. Then the only external horizontal force on the system, and therefore the only power, is from the friction with the ground.

EDIT: Or better yet, the ground supplies the only translational power, assuming that's what you want to calculate. But, as knzhou points out, just remember that net power from the ground is zero, because it's negative rotational power cancels its positive translational power (it converts the former to the latter). So the net power on the bike as a whole is the positive rotational power from the person, just as we would intuitively expect.