# Many Bicycle Questions

1. Jun 11, 2014

### biq1

1. Static friction cannot do work (correct?). Yet that is the only external force acting on an accelerating bicycle. How can these two ideas jibe? My thought is that the human is like a battery with potential energy, which, while being depleted, is *redirected* in some sense by the static friction. But I do not know if that is correct, or how it could be rigorously expressed.

2. (And this is *totally* unrelated to 1, other than that it has to do with bicycles). Is there a way to consider gear ratios/mechanical advantage, only in terms of torque/forces and not with energy? (I think there must be.) The only way I know how to think about the gear ratios is to say that if X force operates over A distance, and that causes Y force over B distance, AX = BY (if no energy leaves the system), by conservation of energy.

Thanks for the help! These problems have really be frustrating me during my bike rides recently...

2. Jun 12, 2014

### dauto

Question 1: you have to be careful when applying the work-energy theorem to situations like this because the bike is not a particle. Different parts of the bike move at different speeds so while it's true that the static friction force doesn't do work on the bottom of the tire since that part of the tire is not moving, the center of mass of the bike is moving and yes the static friction force is doing work on the bike. Sometimes that second kind of work is called pseudowork to avoid confusion with the work as defined for a point particle.

3. Jun 12, 2014

### Delta²

1.Friction can do work though we are used to see the work of friction to be transformed to thermal energy. In the case of the bicycle some of the work of friction is going to thermal energy(because friction resists the rotation of the wheels, applying an external torque to the wheels) and some to increasing the translational kinetic energy of the bicycle.

2. If F is the force we apply to the pedal, T is the friction on the back wheel with the gears and A is the tension of the chain that connects the pedal gear with the back wheel gears, then

$FR-AR'=Ia (1)$ and
$Ar-Tr'=I'a' (2)$ and $aR'=a'r (3)$ ,

where R is the radius of pedal, R' the radius of the gear attached to the pedal, r the radius of the gear of the back wheel and r' the radius of the back wheel, a and a' the angular acceleration of the pedal gear and the back gear and I and I' the moments of inertia of the pedal gear and the back wheel respectively.

Equations (1) and (2) is from the direct application of the well known theorem that

Net_Torque=(moment of inertia) x (angular acceleration)

while (3) follows from the fact that the linear velocities (but not the angular velocities) of the pedal gear and and the back wheel gear have to be equal as long as the chain doesnt contract.

Last edited: Jun 12, 2014
4. Jun 12, 2014

### Jano L.

That is not correct. Friction force due to the road acts only at the bottom of the tire, which does not move, so the work done by this force is zero. It is also clear that the road does not lose any energy because of the friction force, which it would have to if the force did some work on the bicycle.

Dot product of the displacement of the center of mass of the bicycle with the friction force is positive, but it is not work of the friction force, because friction force does not act at the center of mass.

Bicycle is an example of the case where body can begin to move from rest without any external force doing work on it. In such cases the kinetic energy necessary is not supplied by the external work, but from internal energy of the system - from the person riding the bicycle.

5. Jun 12, 2014

### Delta²

However we have to admit that the friction is the force that gives translational acceleration to the bicycle (with no friction wheels just keep spinning around). Can you elaborate abit more on how it is possible a force to give translational acceleration yet producing no work?

6. Jun 12, 2014

### rcgldr

Static friction performs no work in the reference frame at the interface where neither surface is moving, but since the wheel is rolling, the point of application of the static force is moving with respect to the ground (or pavement). The work done equals the static friction force times the distance the application of force moves with respect to the ground. In this case, the source of the energy is the person applying a torque on the rotating pedals. The work done equals the torque times the angular displacement (amount of rotation) of the pedals.

For a simpler example, imagine a box resting on the floor of an accelerating car. From the reference frame of the car, no work is done, but from the reference frame of the ground (or pavement), work is performed on the box. To eliminate the issue of an accelerating frame, the car could be moving up an incline at constant speed, increasing the potential energy of the box with respect to some fixed point on the ground (or pavement).

7. Jun 12, 2014

### Jano L.

In the idealized picture where the wheel is solid and is rolling without sliding, the point where the force acts does not move with respect to the ground. In reality, there is always some wheel deformation and sliding, but this opposes the motion already established, rather than drives it. There is some small work done by the ground to stop the bicycle. But in the ideal case where nothing slides there is no work by the friction force.

8. Jun 12, 2014

### Delta²

The point of application of force has always zero velocity relative to the ground indeed however it is moving! Because lets say at start the bicycle c.o.m is at point A and the wheel contact point is at B after some pedal work and translation movement the c.o.m will be at point C and the wheel contact point will be at D, such that vector(AB)=vector(CD). This seems like a little paradox to me i ve to say.

9. Jun 12, 2014

### mattt

Agree with this.

If you think about the whole system as being made up of tiny particles (i.e. as a system of particles in Newtonian Mechanics), when the system starts to move, the work done by all internal forces (man + bicycle) is not zero, and it is exactly the increment of kinetic energy of the system.

The work of the exterior force (between system and floor, i.e. the static friction force) is zero.

It is exactly the same when a person or a robot starts to walk. The work of the exterior force (the force between the robot's feet and the ground) is zero. The increase in kinetic energy (when the robot or the person goes from rest to be walking) is equal to the work of all internal forces.

10. Jun 12, 2014

### jbriggs444

This is incorrect. The motion of the point of application of the force in this case is like the motion of the point of intersection of the blades of a pair of scissors. It is irrelevant to work done.

What is relevant is the instantaneous state of motion of the material at the point of application. As has been pointed out, both tire and road are (ideally) motionless at that point as long as we adopt a frame of reference in which the road is at rest.

11. Jun 12, 2014

### Delta²

Comeon guys you really cant see that the point of application of friction is actually moving together with the c.o.m? We have a typical force that moves its point of application so we have work.

12. Jun 12, 2014

### Jano L.

You confuse the material point that the force acts on with the geometrical point of contact that moves in a straight line with the velocity of the bicycle. Work is defined as force x displacement of the material point the force acts on in the direction of the force. Since this point moves down and up just before and after the contact, its displacement along the direction of friction force is zero:

https://en.wikipedia.org/wiki/File:Cycloid_f.gif

13. Jun 12, 2014

### Delta²

The mistake on your line of thinking is that you dont notice the fact that the material point of application of friction is continuously changing. If we consider the time t where the material point A of application is indeed the lower point of contact, then at the time interval (t,t+dt) friction doesnt touch point A (because in order to touch it its vertical displacement has to be zero) but we accept that during (t,t+dt) the vertical displacement is not zero.

14. Jun 12, 2014

### 256bits

There is no mistake in the thinking of Jano L.

15. Jun 12, 2014

### Jano L.

I do not understand what's your point.

Look at the animated picture I posted. The material point experiences friction force at one instant only, before and after that it moves perpendicularly to that force. When it experiences the force, it stands still. The material point performs zero displacement along the horizontal friction force during the time it experiences this force.

16. Jun 12, 2014

### mattt

It is better for you to think in a robot (or just a person) that goes from rest to walking

which physically is analogous to the man+bicycle system.

The kinetic energy increase is equal to the work of the internal forces. The work of the only exterior force, static friction, is zero. (The foot of the robot that is on the ground, does not move while it is on the ground).

Last edited by a moderator: Sep 25, 2014
17. Jun 12, 2014

### Delta²

Apparently you dont understand that no matter how big or small one chooses dt, there cannot be a time interval (t-dt,t+dt) such that the friction point continuously touches a specific material point A of the wheel.

It touches point A only at discrete times t0(A),t1(A)(=t0(A)+time needed for a complete rotation),t2(A),.... Therefore the reasoning of Jano.L does not apply.

What we can surely say is that the friction point in the time interval (t,t+dt) gradually touches an infinitesemal portion of the wheel ds. Thus the work done is Tds. In the time needed for a complete rotation, the total work is $\sum{Tds}=T2 \pi R$. As long as there is rolling without sliding 2piR is the distanced traveled by the c.o.m of the wheel.

18. Jun 12, 2014

### jbriggs444

By your reasoning, a car with its accelerator floored and its wheels sitting on the rollers of a dynamometer is doing no work on the dynamometer even if the brakes on the dynamometer are glowing red with the energy being dissipated.

19. Jun 12, 2014

### A.T.

Let`s say the static fricton acts forward (bike accelerates). By your logic the ground is doing positve work on the wheel, in the rest frame of the ground. But the ground has no energy in this frame, so it cannot do positive work (supply energy).

Does static friction do work on a non slipping shoe, in the rest frame of shoe and ground?

Just make the shoes infitisimal, and you have a wheel, but still no work done.

20. Jun 12, 2014

### Delta²

Yes ok maybe the work of friction is zero. Can you tell me which work of which force increases the translational kinetic energy of the bike, and why this work doesnt increase the translational kinetic energy if no friction is present?

21. Jun 12, 2014

### Delta²

It is impossible to perform this experiment with wheels that are touching the rollers on exactly one point.

22. Jun 12, 2014

### jbriggs444

Indeed. So for realistic tires touching realistic rollers, do you think that the power delivered to the dynamometer rollers by the tires is more closely approximated by multiplying the tangential velocity of the rollers by the tangential force exerted by the tires? Or by multiplying the [near-zero] velocity of the centroid of the contact patch by the tangential force exerted by the tires?

23. Jun 12, 2014

### rcgldr

To eliminate the single point of contact, replace the bicycle with a continous track vehicle, such as a tank or caterpillar tractor with tank treads.

Assume a lossless drive train and a closed system, consisting of an initially non-rotating earth and the tank at rest. The power source is the engine inside the tank, and this results in a newton third law pair of static friction forces, a backwards force exerted by the tread onto the surface of the earth, and a forwards force exerted by the earth onto the tread. Assuming a constant force and no other forces (such as aerodynamic drag or rolling resistance), then for some period of time Δt, the impulse exerted onto the earth ΔJearth = -F x Δt, and the impulse exerted onto the tank is ΔJtank = +F x Δt. The change in momentum of the earth and tank equals the impulse exerted onto the tank and the earth. The change in energy of the earth plus tank system equals the force times distance that the force is applied over, with almost all of the change in energy going into the tank since the earth is so massive and the change in velocity is so tiny.

There is no claim that the earth is the source of energy, but the only external force acting on the tank (other than gravity), is the forwards static friction force from the surface of the earth, and it's this force time the distance that this force is applied that corresponds to the change in energy of the tank (a very tiny amount of the change in energy goes into the earth).

24. Jun 13, 2014

### A.T.

The work done on the wheel is zero.

Here you talk about the bike as whole, as one translating body. You have to be decide how you model the system, and what your bodies are:

- If you model the wheel as a separate body, the static ground contact force does no work on it. The chain is doing work on the wheel, and the wheel is doing work on the axis, adding
KE to the bike.

- If you abstract away the wheels, the propulsion, and model the entire bike as one tranlsating body, then you dont have a static ground contact force in your model. Your one body bike is like a sliding block, and the external force from the ground is doing work.

25. Jun 13, 2014

### Delta²

What work is that of the wheel to the axis and anyway why this work becomes zero when the friction between the wheel and the ground becomes zero?? No friction, no translational kinetic energy increase no matter what, simple as that.

If you mean the work from some sort of internal force between the wheel and the axis, those are opposite and equal (if the wheel increases the translational KE of the axis then the axis reduces the translational KE of the wheel by the same amount) and cannot change the total translational KE of the system wheel+axis that we know its happening when a friction force is present.
This is not the case i wonder why you even mention this.

Last edited: Jun 13, 2014