Power Generation in a Bicycle

Question

Consider a person pedaling a bicycle, if we consider the system consisting of the rider and cycle as a total and apply work energy conservation we can see that whatever force the rider applies on the pedal will also have an equal and opposite reaction and as both the pedal and the foot of the rider will have equal velocity the net power cancel outs? I don't know where the mechanical energy is generated in this process. I have attached a picture of a question from my engineering textbook (Engineering Mechanics by P.C Dumir) from which I got this question.

I have attached a portion of the solution as well, I am not sure why It has considered the reaction force to be acting at point B and not the foot of the rider, due to which it has calculated the power by taking the difference of velocities of the subsequent parts.

Answer 1

But that force travels through a distance which means work is performed. The work is provided by your muscles, which propels the bicycle against the retarding force of air and road friction.

Answer 2

we can see that whatever force the rider applies on the pedal will also have an equal and opposite reaction and as both the pedal and the foot of the rider will have equal velocity the net power cancel outs?

Basically correct. Any internal force pair will cancel out as far as work done on the system. But that doesn't mean the forces just disappear. The pedals are able to provide an opposing force on the foot, but only because that force is mostly transferred to other components.

I don't know where the mechanical energy is generated in this process.

The pedals are just one component of a force/torque/power train. The forces on the pedals are communicated through the crank/chainring/chain/gear/hub/wheel/tire until it reaches the ground which is outside the system.

So you can't look at just the pedals, you have to look at the entire system.

Chemical energy in the rider is transformed into mechanical energy in the leg muscles. This mechanical energy is transferred through forces/torques in the drivetrain to the ground. There the forces on the ground increase the kinetic energy of the bike/rider system as it accelerates from rest.

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Imagine your school-age child comes home and says that it's picture day and it would cost \$20 for a portfolio. You give them a \$20 bill and then think "The \$20 is still in my family. How does the school benefit from me giving my child the money?" In this case that particular transaction doesn't do anything, but it allows the child to deliver the money to school.

The drivetrain is doing the same job of delivering the generated power to the ground, with the pedals mostly just shuffling it along.

Answer 3

For a rotary system, the formula for work is the dot product of applied torque and angle traversed:

$$W = \vec \tau \cdot \vec \theta$$

which means that the power (Watts) produced is the dot product of torque and angular velocity (or "rpm" in popular usage):

$$W = \vec \tau \cdot \vec \omega$$

This follows from taking a differential turning of the axle over a distance $d\vec s = r ~d\vec \theta$ due to an applied force $F$, and integrating over the full circle (or full arc traversed), i.e.

$$\int \vec F \cdot d\vec s = \int \vec F \cdot r ~d\vec \theta$$ $$=\int \vec \tau \cdot ~d\vec \theta$$

Answer 4

as both the pedal and the foot of the rider will have equal velocity the net power cancel outs?

Yes, that is correct. There is no net power. But the interesting thing is not the net power but the flow of the power.

The foot is moving in the opposite direction as the force on the foot, which is negative so the foot is losing mechanical power. The pedal is moving in the same direction as the force on the pedal, which is positive so the pedal is gaining mechanical power. This is a flow of mechanical power from the foot to the pedal.

A similar analysis at the gear and chain shows that mechanical power flows from the pedal gear, to the chain, to the wheel gear, to the axle, and to the frame of the bicycle.

I don't know where the mechanical energy is generated in this process.

The mechanical power is generated in the muscles of the legs. The muscles are not rigid, they change their length. They get shorter while under tension, so the motion is in the opposite direction as the force on them. That gives a negative power, so that is mechanical power leaving the muscle.

Answer 5

The work energy theorem applies: The work of the net force on the system equals the increase of translational kinetic energy.

The velocity of the bicycle only increases because there is a friction force $F$ between tires and soil. That is the net force: $$F = ma \implies F.dl = ma.dl \implies dW = m\frac{dv}{dt}dl = mdv.\frac{dl}{dt} = mv.dv = d(\frac{1}{2}mv^2)$$