Is Mechanical Energy Conservation Free of Ambiguity?

[etotheipi]

Yes, I would prefer to stress the generality of the work-energy theorem in Newtonian mechanics, rather than inferring its failure from macroscopic examples

Yes I think the main take-away from this discussion is that, whilst the work-energy theorem holds conceptually for any arbitrary body (i.e. assuming that we take all works [including internal] into account), it is only useful as a problem-solving tool so long as we have some means of computing these internal works.

For a simple system, like two bodies interacting gravitationally, it's very easy to apply the work energy theorem to the entire system because the total internal work is simply $W = \int \vec{F}_{12} \cdot d\vec{x}_1 + \int \vec{F}_{21} \cdot d\vec{x}_2 = -\Delta U_{\text{system}}$. The theorem yields $-\Delta U_{\text{system}} =\Delta T_{\text{system}} \implies \Delta U_{\text{system}} + \Delta T_{\text{system}} = 0$ immediately.

For a more complicated system, like a ball of putty, or a spring, the work energy theorem is not so easily applied. With the spring, for instance, if we compress it slowly from its unstretched length through a distance $\delta$, we do external work $\frac{1}{2}k \delta^2$ on the spring. However it's change in kinetic energy is zero! If we step back and think about general energy considerations, it's clear that this is because the work we did is now locked up in elastic potential energy, and not kinetic.

What we infer is that exactly $-\frac{1}{2}k\delta^2$ of internal work was done by the spring on itself (by the many internal forces between the many particles in the spring, as they were moved closer together), so that the spring has no change in kinetic energy. We could go one step further to note that this internal work is done by conservative forces, which means the change in potential energy of the spring is $\Delta U_{\text{spring}} = -\left( - \frac{1}{2} k \delta^2 \right) = \frac{1}{2}k\delta^2$, as we expect.

So everything works out if we are careful

 

[atyy]

The general work-energy theorem holds for any point charges and arbitrary forces acting on them:
$$\dot{T}=\sum_{i=1}^n m_i \dot{\vec{x}}_i \cdot \ddot{\vec{x}}_i=\sum_{i=1}^{n} \dot{\vec{x}}_i \cdot \vec{F}_i.$$

That equation is correct but note that it is not what is referred to as the work energy theorem in this article. What he is talking about is (in your notation):

$$\dot{T}=\dot{\vec{x}}_{COM} \cdot \sum_{i=1}^{n} \vec{F}_i$$

Ah ok, now that I read carefully, I realize there are 2 work-energy theorems being discussed (I only saw the claim that the work-energy theorem is violated, which I thought can't be true).

So the true mystery is, if the general work-energy theorem is true, why is the CM work-energy "theorem" false in general, but true in special cases?

 

[etotheipi]

So the true mystery is, if the general work-energy theorem is true, why is the CM work-energy "theorem" false in general, but true in special cases?

The CM work-energy theorem is always true, you just need to be careful what it's telling you. König theorem guarantees that the total kinetic energy of a body is the sum of the kinetic energy of its centre of mass (as if all mass were a point particle, at the centre of mass) plus the kinetic energy in the centre of mass frame.

The general work-energy theorem gives you the change in total kinetic energy, the CM work-energy theorem gives you only the kinetic energy of the centre of mass.

 

[atyy]

The CM work-energy theorem is always true, you just need to be careful what it's telling you.

Hmmm, I thought the claim of this insight is that the CM-work-energy "theorem" fails for the example of the gymnast jumping off the floor?

 

[etotheipi]

Hmmm, I thought the claim of this insight is that the CM-work-energy "theorem" fails for the example of the gymnast jumping off the floor?

No as I understand it, the concept of centre-of-mass work isn't included in the article, and was first introduced by @Dale in post #2. The example of the gymnast is pointing out that you need to be careful when applying the general work energy theorem to the gymnast, in that it won't work unless you take into account internal works.

There is derivation of the centre-of-mass work concept in my post #52, which shows that it only yields the term $\frac{1}{2}mv_{\text{cm}}^2$.

 

[atyy]

No as I understand it, the concept of centre-of-mass work isn't included in the article, and was first introduced by @Dale in post #2. The example of the gymnast is pointing out that you need to be careful when applying the general work energy theorem to the gymnast, in that it won't work unless you take into account internal works.

There is derivation of the centre-of-mass work concept in my post #52, which shows that it only yields the term $\frac{1}{2}mv_{\text{cm}}^2$.

But you think the CM-work-energy theorem is ok, because if I understand your post #52 you allow the internal force to be the cause of the CM displacement, whereas from @Dale's remarks in #65 he thinks the theorem requires the force to be the normal force if the boundaries are drawn to include the legs, in which case the theorem fails since the normal force does not move.

 

[vanhees71]

That equation is correct but note that it is not what is referred to as the work energy theorem in this article. What he is talking about is (in your notation):

$$\dot{T}=\dot{\vec{x}}_{COM} \cdot \sum_{i=1}^{n} \vec{F}_i$$

Hm, what's the minimal requirement for that being true? It's for sure true in a homogeneous constant gravitational field, which is why sometimes the center of mass is sometimes also called center of gravity, but is this the most general case?

 

[etotheipi]

But you think the CM-work-energy theorem is ok, because if I understand your post #52 you allow the internal force to be the cause of the CM displacement, whereas from @Dale's remarks in #65 he thinks the theorem requires the force to be the normal force if the boundaries are drawn to include the legs, in which case the theorem fails since the normal force does not move.

The gymnast example has caused a lot of confusion here, because I think we've all been talking about slightly different things.

I interpreted your example as that there is a normal force $\vec{N}$ acting from the floor on his feet, and a weight $m\vec{g}$ acting through his centre of mass. In that case, if his centre-of-mass rises by $\Delta y$ the CM work-energy theorem yields that $\Delta T_{\text{CM}} = (N - mg)\Delta y$. This does not include the kinetic energy in the CM frame.

@Dale intepreted it slightly differently, in that now the system is just the upper body, and the legs are applying a force of $\vec{N}$ on the upper body. Then, since the point of application of $\vec{N}$ as well as thee point of application of $m\vec{g}$ rise by $\Delta y$, he uses the general work-energy theorem to write that $(N- mg)\Delta y = \Delta T_{\text{upper}}$, where $\Delta T_{\text{upper}}$ is the total change in kinetic energy of his upper body only.

Everything works out, but we just need to be on the same page in what system we're using and what version of the theorem we're using.

Hm, what's the minimal requirement for that being true? It's for sure true in a homogeneous constant gravitational field, which is why sometimes the center of mass is sometimes also called center of gravity, but is this the most general case?

This version is only true if the $T$ in $\dot{T}$ is the kinetic energy of the centre of mass only, without the kinetic energy in the CM frame being added on. You can see derivation in lower part of #52

 

[etotheipi]

By the way, Bruce Sherwood has written an article about the difference between the general work-energy theorem and the centre-of-mass work-energy theorem (which he calls 'pseudowork'). He also discusses a third option, which is to use the first law of thermodynamics (i.e. just standard conservation of energy). I think it nicely summarises the key points here:

https://brucesherwood.net/wp-content/uploads/2017/06/Pseudowork1983.pdf

 

[atyy]

I interpreted your example as that there is a normal force $\vec{N}$ acting from the floor on his feet, and a weight $m\vec{g}$ acting through his centre of mass. In that case, if his centre-of-mass rises by $\Delta y$ the CM work-energy theorem yields that $\Delta T_{\text{CM}} = (N - mg)\Delta y$. This does not include the kinetic energy in the CM frame.

However, personally I would prefer approaches that can work regardless of how you choose the system boundary. It has trouble if you make the choice to consider the legs as part of the system.

Frankly, I like to keep my legs attached to my torso, even when just assigning system boundaries.

So it is fine to use the normal force in the CM-work-energy theorem, even though the normal force doesn't move, since we are already moving the force vectors for each particle all over the place to make a "net force". In this case, one can draw the system boundary to include the legs, and the CM-work energy theorem doesn't fail for the case of the gymnast jumping off the floor. Interesting concept, I think it works, the main danger is in the naming (pseudowork seems a good suggestion), since generally for work, we want to consider the force applied to the particle at each point in the particle's path.

 

[etotheipi]

Yeah. To be honest, I don't really like the example with the gymnast, because humans are complicated physical systems and it's not really clear what is meant by 'normal force applied by the legs, on the upper body'. It might be clearer to think about a massless spring positioned upright on the floor, and a massive particle attached to the top. If we compress the particle and spring, and then release the system, we have several choices for how to analyse it using energy. A few might be:

• let the system be just the particle, and use the work energy theorem on the particle, considering the tension force from the spring, and weight force on the particle
• let the system be the spring and the particle, and use the CM work energy theorem, considering the weight force on the particle and the normal force from the floor, and using the displacement of the centre of mass of the system. [The force pair between the spring and particle are internal to this system, so they cancel (we can ignore it).]

Since the spring is massless by construction in this problem, the normal force from the floor is equal in magnitude to the tension in the spring [net force on spring must be zero]. Also, since the spring is massless, the CM of the spring-mass system is identical to the position of the mass. It's easy then to see that these two approaches are completely equivalent, but conceptually they're a little different.

 

[Dale]

I would prefer to stress the generality of the work-energy theorem in Newtonian mechanics, rather than inferring its failure from macroscopic examples.

Hmm, that doesn't make sense to me. If something is general then it should not fail in general. It should be able to be applied to general situations, not limited to specific (e.g. microscopic) situations. The conservation of energy is thus the much more general principle since it works macroscopically and microscopically and even beyond Newtonian mechanics.

I think you must mean something different that what I think of when using the word "generality". Perhaps you are thinking something along the lines that microscopic processes are the underlying phenomena and that macroscopic phenomena arise from lack of knowledge about the details of the microscopic processes. I can see that, but I would probably use the term "fundamental" rather than "general" to describe that relationship.

I think the approach in this insight fails the work-energy theorem too early, and introduces thermodynamics as too fundamental.

I wouldn't introduce thermodynamics as fundamental, since it is derived from statistical mechanics. I would introduce the conservation of energy as fundamental. But I would introduce the first law of thermodynamics as a more general tool than the work-energy theorem for solving a wide variety of problems using the conservation of energy.

Ah ok, now that I read carefully, I realize there are 2 work-energy theorems being discussed (I only saw the claim that the work-energy theorem is violated, which I thought can't be true).

And, to be fair, even the COM-based work-energy theorem is still a theorem and so it is always true provided the assumptions are met (e.g. Newton's laws hold). The problem isn't that it is wrong, just that it is confusing. People use it thinking that the center of mass work (which is confusingly called the net work) represents the total amount of work done by the external forces that make up the net work. That is not correct, and it is also not what the COM-based theorem claims, but it is an understandable confusion.

 

[Dale]

The CM work-energy theorem is always true, you just need to be careful what it's telling you.

Yes, I agree completely. It is a theorem, so it is always true provided the assumptions are met (e.g. Newton's laws hold). The issue is not that the theorem is wrong, but that it is easy to misuse. The problem is when people incorrectly consider $F_{net} \cdot \Delta x_{com}$ to be the same as $\Sigma (F_{i}\cdot \Delta x_{i})$. The mistake is exacerbated by calling the former quantity "net work" (I have been converted to the "center-of-mass work" terminology) when people are much more likely to mentally associate the second quantity with that term.

 

[cianfa72]

• let the system be the spring and the particle, and use the CM work energy theorem, considering the weight force on the particle and the normal force from the floor, and using the displacement of the center of mass of the system. [The force pair between the spring and particle are internal to this system, so they cancel (we can ignore it).]

Since the spring is massless by construction in this problem, the normal force from the floor is equal in magnitude to the tension in the spring [net force on spring must be zero].

Just to be clear: here you are considering the 'spring' as subsystem of the 'spring + particle' system, I believe.

From the point of view of this subsystem the normal force from the the floor and the force from the particle on the spring (the tension in the spring in your words) are really the two external equal magnitude forces yielding zero net force of course.

Do you agree ?

 

[kuruman]

Just to be clear: here you are considering the 'spring' as subsystem of the 'spring + particle' system, I believe.

From the point of view of this subsystem the normal force from the the floor and the force from the particle on the spring (the tension in the spring in your words) are really the two external equal magnitude forces yielding zero net force of course.

Do you agree ?

It seems to me that if the system is the mass + spring, then the external agents that exert a force on this system are the Earth (weight of mass) and the floor (normal force). These two don't need to be equal.

 

[cianfa72]

It seems to me that if the system is the mass + spring, then the external agents that exert a force on this system are the Earth (weight of mass) and the floor (normal force). These two don't need to be equal.

Sure, my point was around the reason why the normal force from the floor is equal in magnitude to the tension in the spring (= force upon the spring provided by the particle according Newton's third law).

 

[kuruman]

Sure, the point I was to highlight is the reason why the normal force from the floor is equal in magnitude to the tension in the spring (= force upon the spring provided by the particle according Newton's third law).

Yes, the normal force is equal to the spring force even if the mass is accelerating. Say the spring is compressed and then released. After release, the spring exerts force $F_{SM}=kx$ on the spring mass so that $kx-mg=ma~\Rightarrow~kx=m(g+a)$.

At the two ends of the spring are the mass exerting force $-F_{SM}$ by Newton's third and the floor exerting force $N$. The net force on the spring is $F_{net,S}=N+(-F_{SM})=N-kx=N-m(g+a).$ For a massless spring, the net force is zero so that the normal force is $N=m(g+a)=kx.$ This may seem a bit odd, but we have to remember that this is a vertical spring and $x$ here is the displacement from the equilibrium position, not from the relaxed length. If the spring were part of a spring scale, $N$ would be its displayed reading.

 

[jbriggs444]

force upon the spring provided by the particle according Newton's third law

If we consider the spring as an entity then the force of particle on spring and the force of floor on spring are a "second law pair".

 

[cianfa72]

Say the spring is compressed and then released. After release, the spring exerts force $F_{SM}=kx$ on the spring so that $kx-mg=ma~\Rightarrow~kx=m(g+a)$.

I believe it should read "After release, the spring exerts force $F_{SM}=kx$ on the particle so that $kx-mg=ma~\Rightarrow~kx=m(g+a)$"

If we consider the spring as an entity then the force of particle on spring and the force of floor on spring are a "second law pair".

What do you mean with "second law pair" ?

My point was that the two actions --- namely the force of the particle on the spring and the force of the spring on the particle -- are equal and opposite as established by Newton's third law (action & reaction)

 

[jbriggs444]

What do you mean with "second law pair" ?

My point was that the two actions --- namely the force of the particle on the spring and the force of the spring on the particle -- are equal and opposite as established by Newton's third law (action & reaction)

Right. Those are a third law pair. Particle on spring and spring on particle.

The point was that the force of particle on spring and floor on spring are not a third law pair. The fact that those are equal and opposite is a consequence of the second law: $\sum F = ma$. If $ma$ is negligible, it follows that the two forces that contribute to $\sum F$ must be equal and opposite. I like to use the term "second law pair" to refer to this argument.

 

[kuruman]

I believe it should read "After release, the spring exerts force $F_{SM}=kx$ on the particle so that $kx-mg=ma~\Rightarrow~kx=m(g+a)$"

Yes, good catch. It is fixed, thanks.