I Is Mechanical Energy Conservation Free of Ambiguity - follow up

[cianfa72]

TL;DR

Clarification about the calculation of thermodynamic work done by kinetic friction in "Work done against kinetic friction" example.

Reading again this Insight, I have a question regarding "I. Work done against friction" section. In that problem (billiard ball of radius $R$ initially spinning in the air then falling on a pool table) the system under analysis is "billiard ball".

The work-energy theorem (which would be better to call Center-of-Mass work theorem) applies regardless the specific type of system's internal forces (the key requirement is that they are supposed to be Newton's 3rd law pairs as they are). It states that the change in system CoM's kinetic energy (i.e. system's bulk kinetic energy) equals the "CoM work", i.e. the work calculated as done by the net external force thought as applied to the system's CoM.

In that specific example, forces acting vertically cancel out, and the work tracked from "Bridgman's sentries" (i.e. the real external "thermodynamic" work done on the system) turns out to be $$f_k \Delta s_{cm}-f_k R\Delta \theta$$ where $f_k$ is the kinetic friction from the pool table on the "system", $R$ the billiard ball's radius and $\theta$ the rotated angle until the ball rolls without slipping on the pool table.

Where does that latter expression come from ? Thanks.

 

[anuttarasammyak]

$R\triangle\theta$ is the distance whose product with force is work.

 

[cianfa72]

$R\triangle\theta$ is the distance whose product with force is work.

Ah ok, basically $R\triangle\theta$ is the "total displacement" along which the kinetic friction $f_k$ acts (that negative work term accounts for the rolling part of the ball). Furthermore the term $+f_k \Delta s_{cm}$ has to be included to account for slipping (up to billiard ball stop slipping and rolls w/o slipping).

Quoting that Insight

What else there is to say is that Bridgman’s sentries report that additional mechanical work $\Delta W=\text{-}f_k R\Delta \theta$ is converted by friction into heat on the periphery of the system boundary.

I would say that some of the system's (i.e. billiard ball) kinetic energy in the quantity $f_k R\Delta \theta$ is converted by friction into heat on the periphery of the system boundary. Or, in other words, that the mechanical work $\Delta W=\text{-}f_k R\Delta \theta$ "leaving" the system is actually converted by friction into heat, this because the Newton's 3rd law pair from the system (billiard ball) on the surrounding (pool table) doesn't do any mechanical work on the pool table (zero displacement of the pool table).

 

[kuruman]

I would say that some of the system's (i.e. billiard ball) kinetic energy in the quantity $f_k R\Delta \theta$ is converted by friction into heat on the periphery of the system boundary.

What you are saying is implicit in the equation found in the insight that you quoted.
$$\Delta K=f_k \Delta s_{cm}-f_k R\Delta \theta.$$ In this expression,

• If there is rolling without slipping, $\Delta K=0$ because $f_k=0.$ There is no loss of kinetic energy.
• If the sphere spins in place about a vertical axis, $\Delta s_{cm}=0,~\Delta \theta\neq 0.$ Kinetic energy is lost to spinning friction only.
• If the sphere rolls and simultaneously slips, $\Delta s_{cm} \neq 0,~\Delta \theta\neq 0.$ Kinetic energy is lost to both sliding and spinning friction.
• If you cut the sphere in half, put the flat face on the surface and give it a push, $\Delta s_{cm} \neq 0,~\Delta \theta= 0.$ Kinetic energy is lost to sliding only.

The kinetic energy conversion into heat occurs at the interface between the sphere and the surface. Some it flows into the sphere and raises it internal energy and the rest goes into the surface. If you haven't done so already, you may wish to take a look at example "Sliding with rolling (revisited)" in this followup insight article.

 

[kuruman]

Sorry, in the first case (rolling without slipping) it should be $f_k \neq 0$, however $\Delta s_{cm} = R\Delta \theta$ hence $\Delta K=0$.

No. It is static friction $f_s$ that is non-zero. That is the case when the points of contact on the sphere and on the surface are instantaneously at rest relative to each other. That is the definition of rolling without slipping in which case $f_s \neq 0$ and $f_k = 0.$

If there is rolling with slipping, the points of contact on the sphere and on the surface move relative to each other. That is the definition of rolling with slipping in which case $f_s = 0$ and $f_k \neq 0.$

 

[cianfa72]

No. It is static friction $f_s$ that is non-zero. That is the case when the points of contact on the sphere and on the surface are instantaneously at rest relative to each other. That is the definition of rolling without slipping in which case $f_s \neq 0$ and $f_k = 0.$

Ah ok, so in case of pure rolling the billiard ball's CoM moves horizontally at constant velocity (w.r.t. the pool table inertial rest frame) however the kinetic friction $f_k$ is zero.

What about the non-zero static friction $f_s$, does it do any mechanical work on the "system" ?

 

[jbriggs444]

Ah ok, so in case of pure rolling the billiard ball's CoM moves horizontally at constant velocity (w.r.t. the pool table inertial rest frame) however the kinetic friction $f_k$ is zero.

What about the non-zero static friction $f_s$, does it do any mechanical work on the "system" ?

One assumes that we are using the inertial rest frame of the ground.

Since the "system" consists of the ball and since the material at the point of contact of ball on ground is motionless, the force of static friction does zero mechanical work in this case.

If we adopted a different frame, we could get a different result.

 

[cianfa72]

One assumes that we are using the inertial rest frame of the ground.

Since the "system" consists of the ball and since the material at the point of contact of ball on ground is motionless, the force of static friction does zero mechanical work in this case.

Ah ok so, in rolling with slipping phase, since the ball's material at the point of contact moves w.r.t. the pool table's inertial rest frame, kinetic friction $f_k \neq 0$ does work on the "system" (the billiard ball).

Then, in pure rolling phase (i.e. rolling without slipping), $f_k=0$, however $f_s \neq 0$ doesn’t do any work ($\Delta K =0$).

 

[erobz]

So, when we are walking on the ground without slipping we can have static friction doing work? I never really can wrap my head around this. Is that that somehow adopting a different frame? Or is this relative motion criteria specific to rolling? You'll run into other examples of a car accelerating down the road etc... where we model "static friction" as doing the work.

The models for walking on the floor (the devil is probably in the gate of the walk) and the car accelerating down the road (you probably have to take off the wheels or its something with the tire deformation) you'll undoubtedly see in intro physics textbooks are apparently too simplistic to capture the reality. I think the experts many have oversimplified things in these cases in order for us to be able to put pencil to paper in these seemingly trivial problems. So, if you get confused about it sometime, you are in good company.

Also static friction should be $f_s \geq 0$. you can have pure rolling without acceleration and $f_s = 0$

 

[cianfa72]

Just a couple more things: first we assume the mass of pool table much more greater than billiard ball's mass, otherwise it would start to accelerate hence its rest frame would be no longer inertial.

Second: in rolling with slipping phase, having chosen as inertial frame the pool table's rest frame, the ball's material at point of contact on ground has (horizontal) velocity equals to $$\dot \Delta s_{cm} - R\dot \Delta \theta$$

 

[kuruman]

Then, in pure rolling phase (i.e. rolling without slipping), $f_k=0$, however $f_s \neq 0$ doesn’t do any work ($\Delta K =0$).

You have to be careful here. One can turn this around and argue

1. The force of static friction $f_s$ acts on the sphere.
2. The sphere is displaced by $\Delta s_{cm}.$
3. Therefore, static friction does work $W=f_s \Delta s_{cm}$ on the sphere.

Just because the surface is rough does not mean that $f_s \neq 0.$ The key point about static friction is that it adjusts itself to provide the observed acceleration. When you have a sphere rolling at constant velocity on a horizontal plane, the horizontal acceleration is zero. Since the only horizontal force that can act on sphere is static friction and the acceleration is zero, Newton's second law says that the force of static friction must be zero. That is why static friction does zero work in this case.

 

[kuruman]

Just a couple more things: first we assume the mass of pool table much more greater than billiard ball's mass, otherwise it would start to accelerate hence its rest frame would be no longer inertial.

Why stop at the pool table? It is firmly attached to the pool hall which is firmly attached to the Earth.

Second: in rolling with slipping phase, having chosen as inertial frame the pool table's rest frame, the ball's material at point of contact on ground has (horizontal) velocity equals to $$\dot \Delta s_{cm} - R\dot \Delta \theta$$

That is correct. When $\dot \Delta s_{cm} = R\dot \Delta \theta$, the point of contact on the ball has zero velocity with respect to the inertial frame and we have rolling without slipping.

 

[jbriggs444]

So, when we are walking on the ground without slipping we can have static friction doing work?

Which flavor of work? Judged from which reference frame? Across which interface in which direction?

Let us adopt the reference frame of the ground and talk about the force from ground on shoes.

"Center of mass" work is judged by the dot product of force with the motion of the center of mass. Walking has non-zero work done by the ground on the person.

"Mechanical" work is judged by the dot product of force with the motion of the material at the contact point. Walking has zero work done by the ground on the unmoving soles of the person's shoes.


Let us adopt the reference frame of the person and continue to talk about the force from ground on shoes.

"Center of mass work" is zero this time. The center of mass is unmoving in the person's frame.

"Mechanical work" is negative this time. The ground is pushing forward on shoes that are retreating backward. This turns out to make sense since in this frame, we are transferring mechanical energy from shoes to ground, not from ground to shoes.


I am pleased to see that @cianfa72 is being careful to distinguish between center of mass work and mechanical work.

 

[cianfa72]

One assumes that we are using the inertial rest frame of the ground.

Since the "system" consists of the ball and since the material at the point of contact of ball on ground is motionless, the force of static friction does zero mechanical work in this case.

If we adopted a different frame, we could get a different result.

Sorry, thinking again about it: it is true that in pure rolling the static friction $f_s$ does zero external mechanical/thermodynamic work. However, by work-energy theorem the "system CoM or net work" isn't zero (static friction $f_s$ is the only external force since external forces acting vertically actually cancel out). Why the system CoM (i.e. the ball's center) actually doesn't accelerate but moves with constant (horizontal) velocity w.r.t. the pool table's inertial rest frame?

 

[jbriggs444]

Sorry, thinking again about it: it is true that in pure rolling the static friction $f_s$ does zero external mechanical/thermodynamic work.

In the frame of reference where the material at the current interface point is motionless, the force of static friction across that interface does zero mechanical work across that interface.

This would hold, in particular, for the rear wheels of a rear wheel drive car if we use the road frame and assume the absence of slipping.

If we adopted a different frame, the material at the contact point (on both sides) would be co-moving at a non-zero rate and non-zero mechanical work would flow across the same interface.

However, by work-energy theorem the "system CoM or net work" isn't zero (static friction is the only external force since external forces acting vertically actually cancel out).

Do you understand that there are two work-energy theorems? One for each flavor of work. But you have made it clear that you are using the CoM version.

I am going to need to review the entire thread to understand how we can have non-zero static friction as the only external force on an object rolling without slipping on level ground. Unless we have air resistance, rolling resistance or a slope, it sounds contradictory...

The OP (#1) talks about a spinning billiard ball that spins down to a rolling without slipping condition while picking up linear velocity. That's kinetic friction. Once rolling without slipping is achieved there is zero force of static friction and the work done by static friction is manifestly zero. So that cannot be what you are talking about...

But @kuruman in #5 talks about non-zero static friction:

No. It is static friction $f_s$ that is non-zero. That is the case when the points of contact on the sphere and on the surface are instantaneously at rest relative to each other. That is the definition of rolling without slipping in which case $f_s \neq 0$ and $f_k = 0.$

So somewhere between #1 and #5, the scenario must have changed...

My best guess is that @kuruman has generalized from the case of a freely rolling wheel to a driven wheel.

Yes, in this case, the CoM work done by the road on a car with a driven wheel is non-zero. The force is static friction, the direction is typically forward and the CoM is typically in forward motion.

In the absence of air resistance, the CoM version of the work energy theorem says that the car with a driven wheel gains energy and, hence, velocity. A momentum analysis would say the same thing. The car gains momentum due to an unbalanced net force. This means an increase in velocity and, hence, energy.

Why the system CoM (i.e. the ball's center) actually doesn't accelerate but moves with constant (horizontal) velocity w.r.t the pool table's inertial rest frame?

Because we have carelessly shifted from free rolling to a driven wheel. If you shift the car into neutral, it will coast at [nearly] constant speed.

Which is when the conversation turns to rolling resistance, axle friction, brake pads and air resistance.

 

[erobz]

I think we should focus on center of mass work. Lets say we have zoomed into the point of contact:

Does the instantaneous center of zero velocity exist and is it on the meshing surfaces?

The blue vectors are "micro-Normals". Do we actually have that the effect of "static friction" is actually just the summation of the "micro Normals" in the parallel direction?

Are there also "micro kinetic frictional forces" at the all the points of contact, because they aren't actually at rest? They are plastically deforming each other, and exchanging mass etc... Add them to the summation?

Is this what "the effect" of static friction really is that we just "give up" and call $f_s$?

 

[jbriggs444]

I think we should focus on center of mass work.

We should use whichever concept helps us deal with the problem at hand.

If one adopts center of mass work on the car then one does not notice that the drive wheels are rotating, the drive shaft is under torque and the engine is running. The source of the car's motion appears to be the asphalt. Which is true enough, as far as it goes.

For some problems, this might be the appropriate level of detail to use.

If one adopts mechanical work then one might choose to idealize the tires as having zero slip. This will not be exactly correct. We will miss the fact that the drive wheels are slipping slightly while accelerating. But for moderate load, such as on a dynamometer, this idealization will work fine.

Center of mass work would be utterly useless on a dynamometer, of course.

We might choose to acknowledge the existence of slip and quantify it in terms of percent slip versus applied force versus normal force. This would allow us to calculate how much engine power is wasted during acceleration and how much side slip should be expected on turns.

If a racing team is not down to at least this level of detail, they are not doing their jobs.

But I agree that no one wants to go down to the detail of micro-normal or micro-kinetic friction, even though that may well be a more precisely accurate model of what really goes on.

Is this what "the effect" of static friction really is that we just "give up" and call $f_s$?

Yes. Static friction is an engineering approximation.

 

[cianfa72]

My best guess is that @kuruman has generalized from the case of a freely rolling wheel to a driven wheel.

Ah ok, therefore in case of freely pure rolling wheel/ball both static $f_s$ and kinetic friction $f_k$ are null.

 

[cianfa72]

Ok, therefore the rotation of ball's material points about the center (CoM) is due to internal forces only.

 

[jbriggs444]

Ok, therefore the rotation of ball's material points about the center is due to internal forces only.

As long as we are ignoring gravity, yes. They are in maintained in uniform circular motion by the rigidity of the ball.

If we factor in gravity, we also have a balance between the upward normal force at the contact point, the downward force of gravity spread throughout the ball and the pattern of internal stresses that maintains rigidity in spite of this.

 

[erobz]

The source of the car's motion appears to be the asphalt. Which is true enough, as far as it goes.

I was under the impression that static friction doing work is a bit heretical.

Another thing:
In my text by Giancoli there is a section "Why Does a Rolling Sphere Slow Down" conclusion "A rolling wheel does not slow down under the force of static friction" paraphrasing
^*
...However, if we take a model for $f_s$ that assumes these micro forces are underlying, it's an inevitability.

I see it at the very least complimentary (if not competing in some cases - steel on steel) explanation for why a rolling sphere slows down that is what we actually call "static friction" as opposed to purely deformation (displacement of normal force toward leading edge - rubber ball explanation), as outlined in the Giancoli textbook.

About the instantaneous center, is it the case that it is definitely "one of the points" in contact in the micro normal/kinetic friction model? You didn't touch on that question.

* I can produce an image of the section of you want to read it for yourself.

 

[cianfa72]

As long as we are ignoring gravity, yes. They are in maintained in uniform circular motion by the rigidity of the ball.

If we factor in gravity, we also have a balance between the upward normal force at the contact point, the downward force of gravity spread throughout the ball and the pattern of internal stresses that maintains rigidity in spite of this.

In OP we said that forces acting vertically are supposed to cancel out (i.e. upward normal force at the contact point balances downward force from gravity). Their action at the ball's center of mass is null, however, as you pointed out, the downward force of gravity and the internal forces/stresses inside it are responsable to maintein uniform circular motion w.r.t the road rest frame.

 

[jbriggs444]

I was under the impression that static friction doing work is a bit heretical.

Not at all. If, for instance, we have a box of bricks sitting in the bed of our pickup truck and we accelerate from a stop, the force of static friction from the bed of the pickup on the box does work.

[In the frame of the ground anyway. In other frames, the work done will vary. Neither "energy" nor "work" are invariant quantities. If, for instance, we adopted the accelerating reference frame in which the pickup truck is at rest we would see that zero work is done by static friction]

In my text by Giancoli there is a section "Why Does a Rolling Sphere Slow Down" conclusion "A rolling wheel does not slow down under the force of static friction" paraphrasing

One can regard the off-center normal force from the increased up-force on the front and the reduced up-force on the back as resulting in a torque on the rolling sphere. That torque retards the rotation of the sphere. As the sphere tends to slow in response to this torque, the result is a force of static friction between the sphere and the ground. A force of static friction that tends to slow the linear motion of the sphere.

We need not invoke microforces to understand this macroscopic behavior. It is enough to invoke hysteresis in the compression and rebound of the sphere's material. Or the material of the ground.

 

[A.T.]

"A rolling wheel does not slow down under the force of static friction" paraphrasing*

Static friction doesn't dissipate energy into heat. But it might convert linear motion into rotation, and vice versa.

However, if we take a model for
that assumes these micro forces are underlying, it's an inevitability.

The above also applies to your micro normal forces.

 

[erobz]

If, for instance, we adopted the accelerating reference frame in which the pickup truck is at rest we would see that zero work is done by static friction]

The box in the case of the ground frame, work by static friction seems obvious. If you move to the frame that is accelerating with the box what do we see that is moving us forward?

 

[erobz]

Static friction doesn't dissipate energy into heat. But it might convert linear motion into rotation, and vice versa.

The above also applies to your micro normal forces.

I also think there are micro kinetic friction forces baked into static friction. When imagine that very zoomed in jagged wheel, it has a point of contact that is its instantaneous center of zero velocity for the "wheel". The other jagged surfaces that are in contact simultaneously are thus sliding past each other, because they are not actually on the instantaneous center.

So I think static friction, as we bake all of this microscopic stuff together to devine "$f_s$" does dissipate energy.

I understand that in perfect circle meets perfect line fantasy land $f_s$ wouldn't generate heat, but in that world a ball probably wouldn't even "roll" to begin with. Everything would just slide and rotate independently.

The idea of static friction surely demands this reality of imperfection to make sense.

 

[kuruman]

We need not invoke microforces to understand this macroscopic behavior. It is enough to invoke hysteresis in the compression and rebound of the sphere's material. Or the material of the ground.

I think you are talking about rolling resistance. Introducing it changes the simple model of rolling in which there is static friction $f_s$ (never mind how it arises) that adjusts itself to provide the observed acceleration. In this simple model, a ball rolling without slipping on a horizontal plane keeps on rolling along like ol' man river. It's a simplification used to introduce the main ideas behind rolling to physics students. Rolling friction is to rolling-without-slipping motion what air resistance is to projectile motion. So I think it is fair to say that, within this simplified model, when a ball rolls without slipping with constant velocity, $f_s=0.$

 

[jbriggs444]

The box in the case of the ground frame, work by static friction seems obvious. If you move to the frame that is accelerating with the box what do we see that is moving us forward?

We are not moving forward. If you are going to adopt a frame, adopt the frame!

If you want to ask what is holding the pickup truck in place against the gravity-like rearward pseudo-force, that would be the static friction of [rearward-moving] road on tires.

If you want to ask what is holding the box in place against the gravity-like rearward pseudo-force, that would be the static friction of stationary pickup truck bed on stationary box.

 

[jbriggs444]

So I think static friction, as we bake all of this microscopic stuff together to devine "$f_s$" does dissipate energy.

To the extent that there is "creep" across an interface where there seems to be no relative motion, there is energy loss due to the creep. We can calculate that energy loss by multiplying force times creep.

We usually ignore the creep. We can leave the car parked on a hill with the parking brake set and not expect to find it at the bottom the hill in the morning.

Or we can model it.

 

[erobz]

We are not moving forward. If you are going to adopt a frame, adopt the frame!

If you want to ask what is holding the pickup truck in place against the gravity-like rearward pseudo-force, that would be the static friction of [rearward-moving] road on tires.

I'm in the frame of the bricks, I can see the stuff in the ground frame passing by me and I cannot conclude that I am moving. So I can turn reality upside down and its perfectly valid, and also suddenly useless? So basically, the problem of the truck bed "dragging me and the bricks along for a ride" suddenly becomes the truck is on a world sized treadmill...the entire question becomes pointless there if I don't know about the wheels on the truck.