Bicycle Questions: Static Friction & Gear Ratios

[jbriggs444]

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?

If you look at center-of-mass work (force times displacement of the center of mass), the work-energy theorem applies. The net force on the spring from the wall and from the block combined is away from the wall. The center of mass moves away from the wall. Positive center-of-mass work is done and linear kinetic energy increases.

If you look at total work (force times displacement of the material at the point of contact), conservation of energy applies. The wall does zero work on the spring and the block does negative work on the spring. It must be so to balance the energy books. The block carried away kinetic energy and that energy had to have come from somewhere. The potential energy of the spring makes up for the resulting energy deficit.

You can easily see that the work done by the block is negative -- were it not for that interaction, the spring would *ZING* across the room with essentially all of its potential energy having been converted to kinetic and vibrational energy in the spring and none having been converted to kinetic energy in the block.

Always keep straight which kind of work you are talking about.

 

[A.T.]

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

The kinetic energy of the spring comes from internal work of the spring (stored potential energy), not from the work of the block on the spring.

I wonder what is happening to the front wheel where there is no chain attached. How this weel is gaining rotational kinetic energy?

How do you gain rotational kinetic energy, when somone pushes you over? In the rest frame of the ground, the bike frame is doing work on the front wheel. It applies a force at the axle, which creates a torque around the static ground contact point.

 

[Jano L.]

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.

You're right in what you mean, but you're saying it incorrectly. In general, kinetic energy changes as a result of work of some force. The interesting thing about this is that the force can be any force, even internal force. In the case of block with a spring, the kinetic energy of the block increases as a result of the force due to spring. The kinetic energy of the spring increases due to work of internal stress forces in the spring on the spring itself.

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).

This work is negative, so it cannot be the sole work done on the spring! There is additional work done on the spring due to internal elastic forces. The work-energy theorem is valid, we just have to include all mass points and all forces that act on them.

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?

In the frame of the road, the fork pushes the axis forward and since the axis moves in the direction of this force, the fork does work on the axis. Friction force enables transfer of part of this work to rotational energy, but it does not contribute any work of its own.

 

[mattt]

To Delta^2:

As you can see in the last answers, you can "translate" from a "microscopic view" (system of point particles, the way I tried to explain it to you) to "macroscopic view" (talking about "the spring", "the block", the work this subsystem "does" on this other subsystem,...the way the last few posters tried to explain it to you) and vice versa.

They are equivalent ways of stating the same things (the same results), if done correctly, obviously.

Depending on the person, some find it easier to "look at it" from the "microscopic=system of point particles" point of view, some others find it easier to "think about it" from the "macroscopic=subsystems doing work to each other" point of view.

As I said, if done correctly, it is totally equivalent.

Personally, when I start to have some doubts about a given problem (in Newtonian mechanics), I usually see it more clearly when I go down to the "microscopic"="system of point particles" point of view. But probably other people feel the other way around :-)

 

[Delta2]

Good answers regarding the kinetic energy of the spring, however i am not satisfied about the other two questions.

As even matt stated, changes in the linear momentum vector result in changes in translational kinetic energy. The change in translational KE is always a function of the momentum, which exact function depends in the case.
So still can say impulse of external forces changes momentum p which changes translational KE(p).

Jano.L sorry you have to explain better how the friction which is applied at the contact point in the ground, transforms the work of the force from the fork to the axle into rotational kinetic energy. What i see is that the torque of friction gives rotational acceleration and that the force from the axle gives no torque. Its kinda the same situation, we have the friction which changes the angular momentum of the wheel which changes the rotational kinetic energy of the wheel which is function of its angular momentum.

 

[mattt]

Good answers regarding the kinetic energy of the spring, however i am not satisfied about the other two questions.

As even matt stated, changes in the linear momentum vector result in changes in translational kinetic energy. The change in translational KE is always a function of the momentum, which exact function depends in the case.
So still can say impulse of external forces changes momentum p which changes translational KE(p).

Did you really read my post? :-)

I said change in total kinetic energy of a system of particles IS NOT a function of change of total linear momentum vector of this system of particles.

I even wrote an example of a system in which THE SAME change in total linear momentum vector give rise to two totally different changes in total kinetic energy of the system.

I even stated when it is and when it is not. :-(

Jano.L sorry you have to explain better how the friction which is applied at the contact point in the ground, transforms the work of the force from the fork to the axle into rotational kinetic energy. What i see is that the torque of friction gives rotational acceleration and that the force from the axle gives no torque. Its kinda the same situation, we have the friction which changes the angular momentum of the wheel which changes the rotational kinetic energy of the wheel which is function of its angular momentum.

More on this later on.

 

[Delta2]

Did you really read my post? :-)

I said change in total kinetic energy of a system of particles IS NOT a function of change of total linear momentum vector of this system of particles.

I even wrote an example of a system in which THE SAME change in total linear momentum vector give rise to two totally different changes in total kinetic energy of the system.

I even stated when it is and when it is not. :-(



More on this later on.

yes i read your post and jano.l's which both i found most usefull to me. I read also the other two posts but were rather confusing or i didnt find em satisfying.
Matt, can you find me a case where we have a change in the magnitude of the linear momentum, yet we have no change in the translational KE?

What i am saying is that

external impulse cause change in linear momentum which cause change in translational KE.

It doesn't matter if the same change of momentum in two different cases cause two different changes in KE , all it matters is that it causes the change.

 

[nearlynothing]

external impulse cause change in linear momentum which cause change in translational KE.

It doesn't matter if the same change of momentum in two different cases cause two different changes in KE , all it matters is that it causes the change.

If I understand correctly, you're saying that without this external force which changes the momentum, and so changes the velocity, there wouldn't be a change in traslational KE, so that the external force must be the factor that indeed changes the traslational KE, and so we can say that it performs work. Tell me if I'm wrong.

The problem here is that then you have to define work done by a force as "the amount of change in the kinetic energy that wouldn't happen without this particular force". That's not the definition we use in physics. It would be a pretty hard to apply definition also.

 

[Delta2]

If I understand correctly, you're saying that without this external force which changes the momentum, and so changes the velocity, there wouldn't be a change in traslational KE, so that the external force must be the factor that indeed changes the traslational KE, and so we can say that it performs work. Tell me if I'm wrong.

The problem here is that then you have to define work done by a force as "the amount of change in the kinetic energy that wouldn't happen without this particular force". That's not the definition we use in physics. It would be a pretty hard to apply definition also.

Hm no, i am saying an external force changes momentum which in the general case changes the KE. So i require some explanation how this happens whithout the external force doing work. Some answers are confusing some are satisfying, overall i can't say i have a clear picture. Maybe i can ask it more clear , is it possible for an external force to change the magnitude of the total linear momentum without doing work?

I guess it is all we need is impulse and not work such as in the case of the normal of wall in your example.

 

[mattt]

yes i read your post and jano.l's which both i found most usefull to me. I read also the other two posts but were rather confusing or i didnt find em satisfying.
Matt, can you find me a case where we have a change in the magnitude of the linear momentum, yet we have no change in the translational KE?

Yes.

Imagine two point masses:

Particle 1 of mass = 1 Kg is at rest at point (0,1) and particle 2 of mass = 2 Kg is moving horizontally to the left at constant velocity \vec{v}_2 = -\frac{1}{\sqrt{2}} \vec{i} m/s.

Suppose there are no internal forces in this example.

An exterior force \vec{F}_a = 1 \vec{i} N acts on particle 1 during 1 second of time.

Another exterior force \vec{F}_b = \sqrt{2}\vec{i} N acts on particle 2 also during 1 second of time.

Before the exterior forces are applied, we have:

\vec{p} = \vec{p}_1 + \vec{p}_2 = \vec{0} + m_2\vec{v}_2 = 2 \frac{-1}{\sqrt{2}}\vec{i} = -\frac{2}{\sqrt{2}}\vec{i} = -\sqrt{2}\vec{i} Kg m/s

After the exterior forces have been applied, we have (do yourself the computations) :

\vec{p} = \vec{p}_1 + \vec{p}_2 = 1 \vec{i} + \vec{0} = 1 \vec{i} Kg m/s

So the total linear momentum vector has changed in magnitude (and also in sense).

The total kinetic energy before the exterior forces have been applied was:

\frac{1}{2}m_1 v^2_1 + \frac{1}{2}m_2 v^2_2 = 0 + \frac{1}{2}2\frac{1}{2} = \frac{1}{2} J.

The total kinetic energy after the exterior forces have been applied is:

\frac{1}{2}m_1 v^2_1 + \frac{1}{2}m_2 v^2_2 = \frac{1}{2}.1. 1^2 + 0 = \frac{1}{2} J.


So the total linear momentum vector has changed in magnitude, but the total kinetic energy of the system remains the same.

What i am saying is that

external impulse cause change in linear momentum which cause change in translational KE.

It doesn't matter if the same change of momentum in two different cases cause two different changes in KE , all it matters is that it causes the change.

As you can see in this last example, that is not true. :-)

 

[mattt]

Anyway, the example of one single wheel rolling on an horizontal ground is even more interesting.

I guess you are perplexed that if the wheel is rolling (without sliding) and the only exterior forces acting on the wheel are the weight, the normal force and the static friction force, how is it that the wheel end up stopping if the only force that causes a non-zero torque (wrt COM) is the static friction force and it "should" increase the angular velocity (the angular momentum vector) ? :-)

Or imagine you are pushing the center of mass of the wheel so that the center of mass is moving with uniform rectilinear motion. Again, the only force (apparently) that causes a non-zero torque (wrt COM ) is the static friction force and, again, it "should" increase the angular velocity (angular momentum vector), how is it that the angular velocity remains constant? :-)

As I said, this example is good enough to test many things about Newtonian mechanics. I'll leave it to you to think about it and later on I'll comment on this example. :-)

 

[A.T.]

you have to explain better how the friction which is applied at the contact point in the ground, transforms the work of the force from the fork to the axle into rotational kinetic energy.

When you push a standing person at his center of mass, the friction at his shoes transforms the work you do into rotational kinetic energy. Do you understand that?

What i see is that the torque of friction...

You confuse reference frames. In the ground frame the instantanious center of rotation is the point where the friction acts, so it creates no torque around that point.

 

[Delta2]

When you push a standing person at his center of mass, the friction at his shoes transforms the work you do into rotational kinetic energy. Do you understand that?

No. What is the definition of "Force A transforms the work of Force B" there is no such definition in physics. There is Force A does work, or Force A creates Torque, or Force B gives impulse.

You confuse reference frames. In the ground frame the instantanious center of rotation is the point where the friction acts, so it creates no torque around that point.

i can choose whatever frame i want and a constant point in the frame to take the torques of the forces.

 

[Delta2]

Yes.


As you can see in this last example, that is not true. :-)

yes ok it might not be true in ur example but it still hold in the case of our interest with the bike.

 

[rcgldr]

The issue here is that rolling involves surfaces that move with respect to each other. Any point on the wheel is moving in a cycloid path, and over the period of time for each revolution of a wheel, the average velocity of any point on the wheel relative to the ground is the same as the average velocity of the bike relative to the ground.

There's no claim the ground is performing work (assuming a ground based frame of reference), it's the internal energy source that is doing the work, exerting a backwards force onto the ground, and with the Newton third law reaction of the ground exerting an equal but opposing forwards force onto the wheel. Here's a link to a web article about this:

http://electron6.phys.utk.edu/101/CH2/wheels.htm

Another example of this would be a solid uniform sphere rolling up or down an incline without slipping. The total change in energy of the ball equals m g Δh, composed of linear and angular energy. The change in the angular component of energy is due to the static friction force exerted by the incline onto the ball opposing the gravitational component of force, and equals the static friction force times the distance rolled. Similar to above, there's no claim that the incline is performing work, in this case gravity is responsible for the change in energy.

Yet another example would be a solid uniform sphere rolling on the accelerating surface of a very long treadmill. In this case, the treadmill is the source of energy. The center of mass of the sphere will accelerate at 2/7th's the rate of acceleration of the tread mill, relative to the non-moving ground. The static friction force times the distance moved relative to the non-moving ground will equal the change in linear kinetic energy of the sphere (relative to the ground). The static friction force times the sphere surface distance rolled (or torque times angular displacement) will equal the change in angular kinetic energy of the sphere.

 

[Delta2]

The issue here is that rolling involves surfaces that move with respect to each other. Any point on the wheel is moving in a cycloid path, and over the period of time for each revolution of a wheel, the average velocity of any point on the wheel relative to the ground is the same as the average velocity of the bike relative to the ground.

Ok, but sorry what is the point of this point?

There's no claim the ground is performing work (assuming a ground based frame of reference), it's the internal energy source that is doing the work, exerting a backwards force onto the ground, and with the Newton third law reaction of the ground exerting an equal but opposing forwards force onto the wheel. Here's a link to a web article about this:

http://electron6.phys.utk.edu/101/CH2/wheels.htm

Ok fine ground does not perform work. The internal energy is doing work and is keeping the friction force alive (thats what you saying if i understand correctly). Does the friction force gives impulse or the impulse is zero also? And if the impulse is zero how the momentum of the bike is changing?
Also in the link you gave when it says "Sliding friction converts (or transforms) most of the work you do on the file cabinet into thermal energy" it admits that the sliding friction does work. Anyway that's for sliding friction. What the intersting portion for me in this link is :

"##How can the force of static friction be responsible for accelerating the car and do no work? Internal forces between different parts of the car in the engine, transmission, etc, come in action-reaction pairs. These forces can do work. The frictional force can cancel one of the forces of an action-reaction pair with a direction opposite to the direction of the acceleration, leaving a net force in the direction of the acceleration."

Ok fine friction cancels one of the forces in an action-reaction pair, but how it cancels the work of that internal force, since the friction doesn't do work?

Another example of this would be a solid uniform sphere rolling up or down an incline without slipping. The total change in energy of the ball equals m g Δh, composed of linear and angular energy. The change in the angular component of energy is due to the static friction force exerted by the incline onto the ball opposing the gravitational component of force, and equals the static friction force times the distance rolled. Similar to above, there's no claim that the incline is performing work, in this case gravity is responsible for the change in energy.

Yet another example would be a solid uniform sphere rolling on the accelerating surface of a very long treadmill. In this case, the treadmill is the source of energy. The center of mass of the sphere will accelerate at 2/7th's the rate of acceleration of the tread mill, relative to the non-moving ground. The static friction force times the distance moved relative to the non-moving ground will equal the change in linear kinetic energy of the sphere (relative to the ground). The static friction force times the sphere surface distance rolled (or torque times angular displacement) will equal the change in angular kinetic energy of the sphere.

So though again the friction does no work you saying that the changes in energy equal to some sort of "pseudo work" of friction right?

 

[A.T.]

No. What is the definition of "Force A transforms the work of Force B" there is no such definition in physics. There is Force A does work, or Force A creates Torque, or Force B gives impulse..

When you push someone over, the friction at his feet acts as a constraint which enforces a certain movement (rotation), but does no work on him in the rest frame of the ground. Do you understand that?

i can choose whatever frame...

The rule that static friction does no work applies in the rest frame of the contact surfaces. In other frames it can do work.

 

[rcgldr]

Does the friction force gives impulse

Yes. It doesn't matter that the ground is not the source of energy when the ground exerts a forwards force in reaction to the wheel exerting a backwards force. The net impulse on the bike will be the forwards friction force x time and the net change in energy of the bike will be the forwards friction force x distance traveled relative to the ground (well almost, a tiny amount of the energy change goes into the earth).

Ok fine friction cancels one of the forces in an action-reaction pair, but how it cancels the work of that internal force, since the friction doesn't do work?

The friction doesn't cancel one of the forces. The backwards force exerted by the wheel onto the ground affects the earth. The forwards force exerted by the ground onto the wheel affects the bike. The change in momentum of Earth and bike cancel each other, since the Earth and bike can be considered a closed system with no external forces. Almost all of the change in energy goes into the bike since the Earth is so massive that the change in momentum (equal and opposite of that of the bike) results in a very tiny change in speed.

The rule that static friction does no work applies in the rest frame of the contact surfaces. In other frames it can do work.

Rolling motion complicates this. In the rest frame of the ground, the average velocity of any point on the surface of the wheel during the time it takes to make some integer number of revolutions is the same as the average velocity of the bike during that time period.

 

[Delta2]

Ok guys let me get this straight.

Total translational KE changes (in some cases) as the result of the work of only internal forces, however total linear momentum changes only by the impulse of external forces (impulse doesn't have work as consequence necessarily).

I have to say the more i think of this the more weird and strange it looks to me. I don't know why.

 

[Delta2]

The friction doesn't cancel one of the forces. The backwards force exerted by the wheel onto the ground affects the earth. The forwards force exerted by the ground onto the wheel affects the bike. The change in momentum of Earth and bike cancel each other, since the Earth and bike can be considered a closed system with no external forces. Almost all of the change in energy goes into the bike since the Earth is so massive that the change in momentum (equal and opposite of that of the bike) results in a very tiny change in speed.

But that's what it says at the link. The way you describe it now makes it look to me like the friction(force from ground to the wheel) changes the energy of the bike. Ok hm i guess the friction redirects most of the energy back to the bike via its impulse not work. (Problem here is that there is not official definition of a force that redirects or transforms work or energy but oh well wth)

 

[rcgldr]

But that's what it says at the link.

Now that I've read the article linked to before, it's worded awkwardly. Newton third law pair of forces do not act on the same object, so they don't "cancel". In this case, the third law pair of forces is the backwards force that acts upon the surface of the Earth (exerted by the wheel) and the forward force that acts upon the wheel (exerted by the surface of the earth). The article is either treating the Earth as if it had infinite mass, resulting in a force that is "canceled" because it's acting upon an object of infinite mass and has zero effect on the earth, or it's just ignoring the effect on the earth, so it's not a closed system where momentum of the system would be conserved.

The way you describe it now makes it look to me like the friction(force from ground to the wheel) changes the energy of the bike.

The origin of the friction force is the torque that the rider exerts onto the pedals. The momentum of Earth and bike are changed with equal magnitude but opposite direction. The energy of both Earth and bike are changed, but as mentioned before, since the Earth is massive compared to the bike, the velocity change of the Earth is tiny compared to the velocity change of the bike, so the bike gets most of the change in energy.

 

[A.T.]

Problem here is that there is not official definition of a force that redirects or transforms work or energy but oh well wth

There is an official definition of work. And forces that do no work are usually called "constraint forces".