Bicycle Questions: Static Friction & Gear Ratios

[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...

 

[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.

 

[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 doesn't contract.

 

[Jano L.]

...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.

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.

 

[Delta2]

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.

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?

 

[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).

 

[Jano L.]

... but since the wheel is rolling, the point of application of the static force is moving with respect to the ground (or pavement).

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.

 

[Delta2]

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.

The point of application of force has always zero velocity relative to the ground indeed however it is moving! Because let's 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.

 

[mattt]

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.

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.

 

[jbriggs444]

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.

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.

 

[Delta2]

Comeon guys you really can't 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.

 

[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

 

[Delta2]

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

The mistake on your line of thinking is that you don't 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 doesn't 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.

 

[256bits]

The mistake on your line of thinking is that you don't 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 doesn't 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.

There is no mistake in the thinking of Jano L.

 

[Jano L.]

The mistake on your line of thinking is that you don't 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 doesn't 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.

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.

 

[mattt]

The mistake on your line of thinking is that you don't 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 doesn't 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.

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).

 

[Delta2]

Apparently you don't 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.

 

[jbriggs444]

Apparently you don't 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.

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.

 

[A.T.]

Thus the work done is Tds.

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.

 

[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 doesn't increase the translational kinetic energy if no friction is present?

 

[Delta2]

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.

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

 

[jbriggs444]

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

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?

 

[rcgldr]

To eliminate the single point of contact, replace the bicycle with a continuous 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).

 

[A.T.]

Yes ok maybe the work of friction is zero.

The work done on the wheel is zero.

Can you tell me which work of which force increases the translational kinetic energy of the bike, and why this work doesn't increase the translational kinetic energy if no friction is present?

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 don't 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.

 

[Delta2]

- 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.

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.

- If you abstract away the wheels, the propulsion, and model the entire bike as one tranlsating body, then you don't 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.

This is not the case i wonder why you even mention this.

 

[Nugatory]

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 eliminate friction between wheel and ground while changing nothing else, the chain will still do work on the wheel - it will show up as an increase in the rotational speed of the now free-spinning wheel.

 

[Orodruin]

No friction, no translational kinetic energy increase no matter what, simple as that.

This does not mean that the friction is doing work. If an object is in a perfectly circular orbit with constant velocity, the centripetal force is changing the motion of the object without doing any work. In an analogous fashion, the rotational energy of the wheel is here transformed into translational kinetic energy of the entire bike due to friction without the friction doing any work.

With your argumentation, I should not have to pedal the bike since I could just use the work provided by the ground. I will be happy to try to apply this the next time I am biking uphill.

 

[Jano L.]

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 doesn't increase the translational kinetic energy if no friction is present?

It is the work of internal forces of the system. Force due to biker's feet does work on them, the pedals do work on the axis, the axis on the chain and the chain on the wheel.

Kinetic energy of the bike increases due to work of internal forces, and the energy necessary for that comes from energy in biker's muscles.

Presence of friction just allows redirection of this energy into translational kinetic energy of the bike, but is not supplying any additional energy.

 

[Delta2]

if you eliminate friction between wheel and ground while changing nothing else, the chain will still do work on the wheel - it will show up as an increase in the rotational speed of the now free-spinning wheel.

Basically that's the whole point. Why it is up to friction whether the work of chain will go solely as rotational KE (case of no friction) or partially to rotational and partially to translation KE (case with friction).

 

[Delta2]

This does not mean that the friction is doing work. If an object is in a perfectly circular orbit with constant velocity, the centripetal force is changing the motion of the object without doing any work. In an analogous fashion, the rotational energy of the wheel is here transformed into translational kinetic energy of the entire bike due to friction without the friction doing any work.

With your argumentation, I should not have to pedal the bike since I could just use the work provided by the ground. I will be happy to try to apply this the next time I am biking uphill.

Not a perfect analogy since changing the direction only of velocity doesn't need external work. However here we talk about changing the magnitude of the velocity, without doing work. Basically what i had in mind was that the ground "steals" rotational kinetic energy from the wheel simultaneously giving it back as translational kinetic energy.

 

[Orodruin]

Basically what i had in mind was that the ground "steals" rotational kinetic energy from the wheel simultaneously giving it back as translational kinetic energy.

Thus doing a grand total of zero work, just as the centripetal force does zero work to rearrange kinetic energy from being due to movement in the x-direction to being due to movement in the y-direction.

The x and y directions are separate degrees of freedom in the orbital example just as much as the wheel angle and bike travel distance are in the bike example. There then happens to be reaction forces constraining the x and y coordinates to a one-dimensional subspace, but this is also in complete analogy to the friction constraining the bike's travel distance to the wheel's rotation angle. In both cases, the reaction forces do not do any work but affect the movement of the system.

 

[Nugatory]

Basically that's the whole point. Why it is up to friction whether the work of chain will go solely as rotational KE (case of no friction) or partially to rotational and partially to translation KE (case with friction).

They're different physical systems, with different forces acting on them (static friction in one case but not the other). - it would be more surprising if they didn't behave differently.

This bit about a static friction and work has been discussed extensively before. Try this thread and the thread it in turn references: https://www.physicsforums.com/showthread.php?t=745038

 

[nasu]

Basically that's the whole point. Why it is up to friction whether the work of chain will go solely as rotational KE (case of no friction) or partially to rotational and partially to translation KE (case with friction).

The internal forces can change KE but they cannot change the total momentum or angular momentum.

In absence of any external forces not only the wheels will spin but the whole bicycle and bicyclist will rotate in opposite direction so that total angular momentum is zero.

You need an external force not to change KE but to change the momentum. Or the momentum distribution, if you wish. The external force will "produce" a linear momentum and its torque around the CM will prevent the bicycle to rotate as a whole. The normal force contribute too, not only the friction.

 

[A.T.]

What work is that of the wheel to the axis...

Think of the wheel as a lever.
- pivot : ground contact
- input (work done on the wheel) : the pull of the chain at the sprocket
- output (work done by the wheel) : the push of the axle at the bike frame

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.

The work done by the chain on the wheel is greater than the work done by the wheel on the bike frame. The reminder goes into accelerating the wheel.

This is not the case i wonder why you even mention this.

It is a valid way to model the system where the ground contact force is doing work. But it should not be confused with a model that includes the wheel as a separate body.

 

[mattt]

It is the work of internal forces of the system. Force due to biker's feet does work on them, the pedals do work on the axis, the axis on the chain and the chain on the wheel.

Kinetic energy of the bike increases due to work of internal forces, and the energy necessary for that comes from energy in biker's muscles.

Presence of friction just allows redirection of this energy into translational kinetic energy of the bike, but is not supplying any additional energy.

This.

It is not that difficult to understand.

 

[Delta2]

The work done by the chain on the wheel is greater than the work done by the wheel on the bike frame. The reminder goes into accelerating the wheel.

Thats unclear to me, it seems to me that the so called reminder goes into rotational acceleration.

 

[Delta2]

The internal forces can change KE...

I don't seem to get this sorry. Can u explain it abit more. Ok at first glance they seem they can change total rotational KE but i am not sure about the TOTAL translational KE.

You need an external force not to change KE but to change the momentum. Or the momentum distribution, if you wish. The external force will "produce" a linear momentum and its torque around the CM will prevent the bicycle to rotate as a whole. The normal force contribute too, not only the friction.

Well as you say we need external force to change linear momentum, thus the magnitude of velocity thus the translational KE. I ve to say this is the point where it gets hard for me to grasp, how is it possible for the external force to change the linear momentum but not the translational KE.

 

[Delta2]

This.

It is not that difficult to understand.

Well as i said in other posts for me personally the hard point to understand is how this external friction force does the redirection of energy without doing any work.

 

[mattt]

Well as you say we need external force to change linear momentum, thus the magnitude of velocity thus the translational KE. I ve to say this is the point where it gets hard for me to grasp, how is it possible for the external force to change the linear momentum but not the translational KE.

It is exactly the same when you start to walk. The exterior force (static friction) changes the total linear momentum (of you), but the work of this static friction force is zero.

The total work (which in this case is equal to the work done by internal forces) is not zero, and it is just equal to the change in total kinetic energy (from 0 when you are at rest, to some KE when you start to walk).

They are just two different theorems:

\vec{F}_{ext}(t) = \frac{d \vec{P}(t)}{dt}

(that is, the total exterior force is equal to the rate of change of the total linear momentum vector of the system of particles, assuming internal forces are "Newtonian").


\int_{t_0}^{t_1}\vec{F}(t)\cdot\vec{v}(t)dt = \Delta KE

(that is, the total work, sum of all works of all forces on all particles, is equal to increment in total kinetic energy of the system of particles).


They are two different mathematical theorems and they state different things.

 

[A.T.]

Thats unclear to me, it seems to me that the so called reminder goes into rotational acceleration.

Linear and rotational. Or just assume a massless wheel, which doesn't require energy to accelerate.

 

[A.T.]

Well as i said in other posts for me personally the hard point to understand is how this external friction force does the redirection of energy without doing any work.

Consider a stick falling over under gravity, with the lower end fixed on the ground.

 

[Delta2]

It is exactly the same when you start to walk. The exterior force (static friction) changes the total linear momentum (of you), but the work of this static friction force is zero.

The total work (which in this case is equal to the work done by internal forces) is not zero, and it is just equal to the change in total kinetic energy (from 0 when you are at rest, to some KE when you start to walk).

They are just two different theorems:

\vec{F}_{ext}(t) = \frac{d \vec{P}(t)}{dt}

(that is, the total exterior force is equal to the rate of change of the total linear momentum vector of the system of particles, assuming internal forces are "Newtonian").

Νοw that you mentioned this you made me think in terms of impulse. So from that relation follows \vec{p(t)}-\vec{p(0)}=\int_0^t{\vec{T(t)}}dt. So its the impulse of friction T(t) that changes the linear momentum of the bike (since there are no other external forces on the x-axis). So if friction is zero the impulse is zero hence p(t)=p(0)=constant, hence translational KE also constant.

However if the friction is not zero then it follows that the change in the linear momentum (hence the change in translational KE=p^2(t)/2m) is due to the impulse of the friction. We need not provide work from internal forces to increase translational KE.

 

[A.T.]

... changes the linear momentum of the bike...

If you have a body called "bike" in your model, which combines the entire bike and rider into one translating block, then the horizontal ground force will do work on this body, because the interface between "bike" and ground is not static anymore in this coarse model. It doesn't matter that it could be treated as static friction in a more detailed model with wheels, because in physics you do calculations based on your actual model, and not on some extra info that you didnt include in the model. And this coarse model discards the information about how wheels work, and the details of their interaction with the ground, so this interaction is not modeled as static friction anymore.

 

[Delta2]

If you have a body called "bike" in your model, which combines the entire bike and rider into one translating block, then the horizontal ground force will do work on this body, because the interface between "bike" and ground is not static anymore in this coarse model. It doesn't matter that it could be treated as static friction in a more detailed model with wheels, because in physics you do calculations based on your actual model, and not on some extra info that you didnt include in the model. And this coarse model discards the information about how wheels work, and the details of their interaction with the ground, so this interaction is not modeled as static friction anymore.

Τhat equation holds regardless of the modeling of the bike or whether you apply it only to wheel or the whole bike, since there arent any other external forces. To make you see it better, notice that the impulses due to internal forces will cancel each otther and only the impulse of external force will remain. To make u see it even better i ll apply it to the wheel and the rest of bike separately and i assume only the internal force between the wheel and the axis (you can also include the internal force of the chain but the treatment will be similar because one end of chain is on the wheel while the other end of chain is on the pedal which is on the rest of bike). So for the linear momentum of the wheel it will be

P_w(t)-P_w(0)=impulse(T,t)+impulse(F_internal,t)

and for the linear momentum of the rest of bike will be

P_b(t)-P_b(0)=impulse(-F_internal,t). Adding those two together the two impulses of F_internal will cancel out and what will remain will be the equation we want. ( I am really busy now and i avoid typing in tex, sorry anyway).

 

[mattt]

Νοw that you mentioned this you made me think in terms of impulse. So from that relation follows \vec{p(t)}-\vec{p(0)}=\int_0^t{\vec{T(t)}}dt. So its the impulse of friction T(t) that changes the linear momentum of the bike (since there are no other external forces on the x-axis). So if friction is zero the impulse is zero hence p(t)=p(0)=constant, hence translational KE also constant.

However if the friction is not zero then it follows that the change in the linear momentum (hence the change in translational KE=p^2(t)/2m) is due to the impulse of the friction. We need not provide work from internal forces to increase translational KE.

That statement is misleading (to say the least). If there were no internal forces (in the man + bicycle system) or just if the work of internal forces were zero in this system, then the bicycle and man would remain at rest and total KE would remain as zero.Exactly the same in the "man or robot starting to walk" case. If there were no internal forces or just the work of these internal forces (in the man or in the walking robot) were zero, then the man (also the robot) would not move. Total kinetic energy would remain as zero.The problem, more than anything, is "the way you state it". I think you already know that in this system (man + bicycle, or equivalently "man or robot starting to walk") the work of the static friction force is zero, that the static friction force is equal to the rate of change of total linear momentum vector, and that the work of all internal forces is equal to the increment of total kinectic energy. I think you understand all these facts, but you "decide" to state it in a very misleading way.

 

[Delta2]

To tell you the truth i just don't know. Maybe i am wrong on how i am thinking and how i am stating things. This thread has tired me. Deep down what i understand is that regardless if it is exactly one or more contact points, friction "steals" rotational kinetic energy from the back wheel and gives it as translational (and probably the opposite is happening in the front wheel where we don't have chain to do work, there friction counteracts the force of the axis and gives it back as rotational kinetic energy). Oh well i am just tired, maybe when i come back in this thread after some weeks and i read again i understand better.

 

[A.T.]

Τhat equation holds regardless of the modeling of the bike or whether you apply it only to wheel or the whole bike, since there arent any other external forces.

You equation is about momentum, not work.

 

[nearlynothing]

For Delta²:

Internal forces can change the total kinetic energy of your system. Whereas they can never change the total momentum of it, only external forces can do that. There's no contradiction here they are just two different theorems, as someone stated above. I think what's confusing you is the differentiation you make between "translational" and "rotational" kinetic energy, ultimately they are the same thing, namely energy due to motion of material particles.

Consider a system composed of a spring attatched to a block, with the spring against a wall, when you compress the spring and release it, the "translational" kinetic energy of the block + spring system increases, the only external force here is the normal from the wall, would you say that force does work?

 

[Delta2]

For Delta²:

Internal forces can change the total kinetic energy of your system. Whereas they can never change the total momentum of it, only external forces can do that. There's no contradiction here they are just two different theorems, as someone stated above. I think what's confusing you is the differentiation you make between "translational" and "rotational" kinetic energy, ultimately they are the same thing, namely energy due to motion of material particles.

Consider a system composed of a spring attatched to a block, with the spring against a wall, when you compress the spring and release it, the "translational" kinetic energy of the block + spring system increases, the only external force here is the normal from the wall, would you say that force does work?

Oh well i am tempted again to write in this thread...
Nice example. I could say again that the impulse of the normal changes the momentum of the system and since at least in this case the total translation kinetic energy is p^2/2M the impulse is what changes the kinetic energy of the system.

There is some sort of other thing that puzzles me also: What is happening to the work of the force from the block to the spring? It appears to me this work is negative yet it increases the kinetic energy of the spring (work from the normal is zero anyway). We have a violation of the work-energy theorem (applied to the spring only) or the theorem doesn't hold in the case of spring?

Also back to the bike . I wonder what is happening to the front wheel where there is no chain attached. How this weel is gaining rotational kinetic energy? Does the friction via its torque do work in the front wheel? I really don't understand the work of the torque of friction is different from the work of friction in this case?

 

[mattt]

Oh well i am tempted again to write in this thread...
Nice example. I could say again that the impulse of the normal changes the momentum of the system

Yes, that was the first of the two mathematical theorems I wrote in an earlier post.

and since at least in this case the total translation kinetic energy is p^2/2M the impulse is what changes the kinetic energy of the system.

After the interaction (when the spring + block leave the wall) the kinetic energy of the system will be \sum_{i}\left(\frac{1}{2}m_i v^2_i\right)

if all the particles move with the same velocity vector (which will be equal to the velocity vector of the center of mass) then that is the same as p^2/2M, but in this example it does not hold.

(the spring, if ideal, will for ever be compressing and stretching after it leaves the wall).

But you still are confused about the following:

You seem to think that the change in total kinetic energy of a system of particles is a function of the total linear momentum vector change. That is not correct in general. That is true only when (before the interaction, and also after the interaction) all the particles of the system have exactly the same velocity vector, which will be equal to the velocity vector of the center of mass.

In a given system of Newtonian particles, the same change in total linear momentum vector can result in totally different changes in total kinetic energy of the system.

Imagine this very simple example:

A system of two point particles at rest with different masses:

Particle 1 is at rest at P1(0,1) (for example), and particle 2 is also at rest at P2(0,0). Imagine that mass of particle 2 is = 2* mass of particle 1.

case 1: A constant force \vec{F} (horizontal, to the right) is exerted on particle 1 during a time interval [t_1, t_2].

case 2: That "same" force (only difference now will be the point of application) is exerted on particle 2 during the same time interval.

(for simplicity, imagine there are no internal forces in this example).


The change in total linear momentum vector of the system will be exactly the same in case 1 and case 2.

But the change in total kinetic energy of the system will be different in case 1 and case 2 ( in case 1 the total kinetic energy change will be = 2* total kinetic energy change in case 2 ).

So the "same" force (only difference being the point of application) acting during the same time interval, produces exactly the same change in total linear momentum vector of the system, but it produces (in this simple example) double of change in total kinetic energy of the system in case 1 wrt case 2, depending on the point of application of this "same" force.

I hope this helps you to understand the subtleties of these concepts.


In the "spring + block system", the exterior force (the normal force the wall exerts on one end of the spring) is equal to the rate of change of total linear momentum vector of the system (ignoring gravity for simplicity). But the work of this normal force is zero.

The total change in kinetic energy of the system is equal to the total work of all internal forces in this case.

There is some sort of other thing that puzzles me also: What is happening to the work of the force from the block to the spring? It appears to me this work is negative yet it increases the kinetic energy of the spring (work from the normal is zero anyway).

Try to think about the whole system made up of thousands of point particles (this always helps). Put your attention to one concrete given particle (it does not matter if it is a particle of the spring or a particle of the block). During the time interval when the spring is stretching and about to leave the wall (i.e., during the interaction wall-system), this particle you are putting your attention on, is accelerating to the right (imagine the wall is vertical and the system is at the right side). That is because the total force on this concrete particle during this time interval, is non-zero and pointing to the right (and this total force on this concrete particle is equal to the total INTERNAL force on this concrete particle, ignoring the weight for simplicity).

Obviously the work of this total force ( = total internal force ) on this concrete particle (of the spring or of the block, it is the same) is not zero, it is positive.

If you think exactly the same with ANY other particle of the spring or of the block, you realize the total work of internal forces is obviously non-zero, it is positive, and it is equal to the total change in kinetic energy of the system.

We have a violation of the work-energy theorem (applied to the spring only) or the theorem doesn't hold in the case of spring?

It is a mathematical theorem. It can not be violated :-)

It could not apply depending on the mathematical-physical hypothesis of our model, but in this example-model, it does apply.

Also back to the bike . I wonder what is happening to the front wheel where there is no chain attached. How this weel is gaining rotational kinetic energy? Does the friction via its torque do work in the front wheel? I really don't understand the work of the torque of friction is different from the work of friction in this case?

I will explain this later, I got to go now.