# What is the power while accelerating a bicycle?

With respect to this question, the power on a accelerated body, that moves in the same direction of the acceleration, can be calculated by $$W = Fv$$, where $$F$$ is the external force and $$v$$ the velocity.

In the case of a bicycle (or any wheeled car), some power is also necessary to turn the wheels. The power of a rotating body subject to a torque $$T$$ is: $$W_r = T\omega$$, where $$\omega$$ is the angular velocity.

For a wheel with radius $$r$$ and torque generated by a friction force $$F_f$$ with the ground: $$T = F_fr$$.

For the 2 wheels of the bicycle, if the weight is equally distributed on both: $$W_r = 2F_fr\omega$$, and if there is no slippage: $$W_r = 2F_fv = Fv$$, where $$F$$ is the total friction force, the total external force on the bicycle, and $$v$$ its velocity.

My question is: the power on the bicycle is the sum of the power to rotate the wheels and to accelerates the bicycle, that happens be the equal ($$Fv$$), or I am counting 2 times the same thing?

I think that it is necessary to sum the 2 terms. If the bicycle was accelerating on a moving conveyor, that happened to keep it in the same place, certanly we had to add the power to accelerate the conveyor.

But the fact that the values are the same seems strange.

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.

• Neglecting all frictions (except wheels to ground), the biclycle is accelerating, and with an instantaneous velocity $v$. I understand that the friction force $F_f$ is static. But numerically, can the power be expressed by $Fv$ at that instant? Or $2Fv$, due to the torque on the wheels? Or there is no relation between power and the magnitude of the friction force? Commented Jul 25, 2020 at 22:46
• @ClaudioSaspinski The power is defined as the rate of change of energy. What system's energy do you want to consider? Commented Jul 25, 2020 at 22:48
• The system bicycle and driver. Power can be calculated by the change of kinetic energy. But also by $Fv$ when $F$ has the same direction of $v$. Commented Jul 25, 2020 at 22:52
• @ClaudioSaspinski Power is $F v$, but here the letter $v$ is the velocity of the point at which the force is applied. The friction force contributes no power at all, because its point of contact has $v = 0$. Commented Jul 25, 2020 at 22:55
• The friction force from the ground to the wheels causes an horizontal force of the same magnitude from the wheels to the axis (and the rest of the system), through the bearings. And that force does work during the acceleration. Commented Jul 26, 2020 at 2:15

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:

$$W_r = \overrightarrow\tau \cdot \overrightarrow\omega$$

If you picture the bicycle traveling to the right, the wheel rotation ($$\overrightarrow\omega$$) is clockwise. The ground is applying friction to the right, which makes for a counterclockwise torque ($$\overrightarrow\tau$$), 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.

• But if the bike is fixed on the ground, and the wheels are gears that move a conveyor? The force between gear teeth turns the wheels and move the conveyor. The power on the system bike + conveyor is $2Fv$, isn't it? Commented Jul 25, 2020 at 17:09
• @ClaudioSaspinski, F is the total frictional force on the bicycle. Where is the 2 coming from? Commented Jul 25, 2020 at 18:03
• @BowlOfRed $F$ the the tangential force that rotates the wheel. But $F$ also acts moving the bicycle as a whole, with a velocity $v$. Or in the comment, the reaction $F$ acts moving the conveyor with velocity $v$. Commented Jul 25, 2020 at 18:37
• "Then the only external horizontal force on the system, and therefore the only power, is from the friction with the ground." The ground doesn't provide any power to the bicycle at all. It's totally inert. If we really could extract unlimited power from a flat piece of concrete, then we wouldn't need power plants. Commented Jul 25, 2020 at 22:02
• @ClaudioSaspinski The friction $F$ puts power $Fv$ into the wheels, but it is NEGATIVE power, because the frictional torque is going AGAINST the rotation of the wheels, so the dot product of $\tau$ and $\omega$ is negative. The positive power on the wheels only comes from the person, and therefore the person's power must be greater than $Fv$ if the wheels are accelerating. Meanwhile $F$ puts positive power $Fv$ into the conveyor, which means the total power on the system by $F$ is zero! And that makes sense because $F$ is internal to the system, so it can't generate any energy. Commented Jul 26, 2020 at 3:17

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.

• Your second paragraph seems intuitive, but the math shows the opposite, as I comment to Hot Licks Commented Jul 26, 2020 at 1:58
• You are right about $Fv$ must be counted once. I posted an answer anyway to clear my understanding. Commented Jul 26, 2020 at 16:18

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.

• What the math shows is that independent of the proportion of mass or diameter of the wheels and the rest of the bicycle (including the biker), the power to turn the wheels ($T\omega$)is the same to that to linearly accelerate the bike ($Fv$) Commented Jul 26, 2020 at 1:51
• My comment above is wrong. I posted an answer. Commented Jul 26, 2020 at 16:22

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 $$T_i = I\alpha$$. 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. $$T_{tot} = I\alpha + Fr$$. 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 $$T_i = I\alpha$$).

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 $$Fr\omega = Fv$$ as an additional power to increase the rotation of the wheels, only because it has the form: $$T\omega$$. In reality that is already the power to accelerate the bike. That is why it can be expressed in the form $$W = Fv$$.