Is Mechanical Energy Conservation Free of Ambiguity?

[etotheipi]

In the gymnast example, the work energy theorem does not fail.

I think that is not quite correct. Really the work energy theorem only applies to rigid bodies (and this is also assuming the strong form of Newton III holds!). Since the theorem is always applicable to particles, we can find the total work by all forces on all particles in a general extended body via. a summation,$$W = \sum_i \int \vec{F}_i \cdot d\vec{x}_i = \sum_i \int (\vec{F}_i^{\text{ext}} + \vec{F}_i^{\text{int}}) \cdot d\vec{x}_i = \sum_i (W_i^{\text{int}} + W_i^{\text{ext}})$$here $\vec{F}_i = \vec{F}_i^{\text{ext}} + \vec{F}_i^{\text{int}}$ is a sum of both the internal forces and external forces on that particle. If we consider a rigid body, then we have the constraint that $|\vec{x}_i - \vec{x}_j| = \text{constant}$, which means that$$|\vec{x}_i - \vec{x}_j|^2 = (\vec{x}_i - \vec{x}_j) \cdot (\vec{x}_i - \vec{x}_j) = \text{constant}$$ $$2(\vec{x}_i - \vec{x}_j) \cdot (\vec{v_i} - \vec{v}_j) = 0$$Also notice that, for any pair of particles, the total power of internal forces on each of them due to their mutual interaction ($\vec{F}_{ij} = -\vec{F}_{ji}$) is$$P = \vec{F}_{ij} \cdot \vec{v}_i + \vec{F}_{ji} \cdot \vec{v}_j = \vec{F}_{ij} \cdot (\vec{v}_i - \vec{v}_j)$$If the strong form of Newton III holds, then it is guaranteed that $\vec{F}_{ij} = k(\vec{x}_i - \vec{x}_j)$ where $k$ is some arbitrary scaling constant, in which case due to the aforementioned orthogonality, the power of these internal forces is zero. Hence the total work on the rigid body due to internal forces is zero, and the total work on the rigid body is solely due to external forces.

This is the important point; for deformable bodies, we often have no way of computing the work done by internal forces. Of course, hypothetically, if we could determine this number, then the work energy theorem would work fine. For rigid bodies, this internal work is exactly zero, so we can always use the work energy theorem.

This is what I believe @kuruman was explaining with the example of the gymnast. The normal contact force due to the floor does zero work, since the floor is fixed. The gravitational force in fact does negative work when the gymnast is jumping. However his/her change of kinetic energy is positive, which means there must have been positive work done by internal forces that exceeded in magnitude that done by gravity! It is of course easiest to use energy conservation, and attribute this to a transfer of chemical energy to kinetic energy.

Jumping can be treated with the work energy theorem as follows:
https://openoregon.pressbooks.pub/bodyphysics/chapter/comparing-work-energy-and-energy-conservation/

This is not the work energy theorem, this is another construct entirely. It deals with so-called centre-of-mass work (or what @Dale called 'net work' near the start of the thread), which is not real (thermodynamic) work in the sense of energy. It is derived from Newton's second law, and computes the integral of the scalar product of the force and the displacement of the centre-of-mass. That correctly gives the change in the kinetic energy of the centre-of-mass only:$$\begin{align*}

\left( \sum_i \vec{F}_i \right) = \vec{F} = m\vec{a}_{\text{cm}} \implies \int \vec{F} \cdot d\vec{x}_{\text{cm}} &= \int_{\vec{x}_1}^{\vec{x}_2} m \frac{d\vec{v}_{\text{cm}}}{dt} \cdot d\vec{x}_{\text{cm}} \\

& = \int_{t_1}^{t_2} m \frac{d\vec{x}_{\text{cm}}}{dt} \cdot \frac{d\vec{v}_{\text{cm}}}{dt} dt \\

& = \frac{1}{2} m \int_{t_1}^{t_2} \frac{d}{dt} \left( \vec{v}_{\text{cm}} \cdot \vec{v}_{\text{cm}} \right) dt \\

& = \Delta \left( \frac{1}{2}m v_{\text{cm}}^2 \right) = \Delta T_{\text{cm}}
\end{align*}$$We can imagine a cylinder rolling down a rough inclined plane. The work done by the force of friction is zero, since the friction force has zero power (the relative velocity at the point of contact is zero). Hence, the proper work energy theorem yields$$W_{\text{grav}} = \Delta T$$Here $\Delta T = \Delta T_{\text{cm}} + \Delta T^*$ is the total change in kinetic energy, which is the sum of the kinetic energy of the centre of mass and that in the frame of the centre of mass (König theorem). We can also use the centre-of-mass work concept that you linked, which would give$$\mathcal{W}_{\text{grav}} + \mathcal{W}_{\text{friction}} = \Delta T_{\text{cm}}$$here $\Delta T_{\text{cm}}$ is the change of linear kinetic energy only, and not the total change. N.B. I use $\mathcal{W}$ to distinguish it from thermodynamic work, but AFAIK there is no standard notation.

 

[cianfa72]

I personally think that the main culprit is the concept of "Net Work" introduced in the work energy theorem. There are many instances where if you have N mechanical forces F→i acting on a system, each force displacing the point of contact by Δr→i then Wnet=F→net⋅Δxcm≠Σi=1NF→i⋅Δr→i as you would expect. The quantity on the right is what matters with respect to conservation of energy and is the thermodynamic work (for a system acted on by N mechanical forces).

A good example is a spring being compressed by two equal and opposite mechanical forces. The "Net Work" is 0, but thermodynamic work is being done on the system and Σi=1NF→i⋅Δr→i is non-zero. People tend to get in trouble when they assume that "Net Work" is the thermodynamic work or is at all relevant for the conservation of energy.

Going back to the beginning...I believe the work-energy theorem as introduced in this insight actually establishes the following:

$\Delta K_{cm} = W_{net} = \vec F_{net} \cdot \vec \Delta x_{cm} = \Sigma_{i=1}^{N} \vec F_i \cdot \Delta \vec r_i$

$\vec F_{net}$ is just the sum of all external forces acting on the system. 'Net Work' is then defined by the dot product between net force and center of mass displacement

It basically evaluates the change of kinetic energy of the 'bulk' system alone (no internal kinetic energy as measured relative to the system center of mass rest frame) using the concept of 'Net Work'. In other words from W-E perspective the system has not an internal structure for storing energy

 

[cianfa72]

We can imagine a cylinder rolling down a rough inclined plane. The work done by the force of friction is zero, since the friction force has zero power (the relative velocity at the point of contact is zero). Hence, the proper work energy theorem yields $$W_{\text{grav}} = \Delta T$$

Basically the force of friction does zero work just because it 'moves' its point of application along the point of contact during the rolling down, right ?

Here $\Delta T = \Delta T_{\text{cm}} + \Delta T^*$ is the total change in kinetic energy, which is the sum of the kinetic energy of the centre of mass and that in the frame of the centre of mass (König theorem). We can also use the centre-of-mass work concept that you linked, which would give$$\mathcal{W}_{\text{grav}} + \mathcal{W}_{\text{friction}} = \Delta T_{\text{cm}}$$here $\Delta T_{\text{cm}}$ is the change of linear kinetic energy only, and not the total change.

Thus, what we're really doing, is compute the sum of all forces (gravity + friction) acting on the cylinder -- as if they were applied to the center of mass -- and multiply it for the center-of-mass displacement.

It is worth to highlight that the 'system' taken in account is the cylinder thus gravity and friction are actually external forces.

 

[etotheipi]

Basically the force of friction does zero work just because it 'moves' its point of application along the point of contact during the rolling down, right ?

The point of application moves as the cylinder rolls down, yes, but the important part is really that the point of the cylinder in contact with the ramp has zero velocity w.r.t. the ramp (due to the rolling condition), i.e. the power $P = \vec{F}_{\text{f}} \cdot \vec{v}_{\text{edge}} = \vec{F}_{\text{f}} \cdot \vec{0}$ of the friction force on the cylinder is zero in the rest frame of the ramp.

Thus, what we're really doing, is compute the sum of all forces (gravity + friction) acting on the cylinder -- as if they were applied to the center of mass -- and multiply it for the center-of-mass displacement.

It is worth to highlight that the 'system' taken in account is the cylinder thus gravity and friction are actually external forces.

Yes, exactly

 

[vanhees71]

I think that is not quite correct. Really the work energy theorem only applies to rigid bodies (and this is also assuming the strong form of Newton III holds!).

This in turn is also not correct. If you have an arbitrary system of point particles with arbitrary interaction and/or external forces derivable from a potential, i.e.,
$$\vec{F}_i=-\vec{\nabla}_i V(\vec{x}_1,\ldots,\vec{x}_n)$$
the total energy
$$E=\sum_{i=1}^n \frac{m_i}{2} \dot{\vec{x}}_i^2+V$$
is conserved.

Proof:
$$\dot{E}=\sum_{i=1}^{n} \dot{\vec{x}}_i \cdot (m_i \ddot{\vec{x}}+\vec{\nabla}_i V)=\sum_{i=1}^{n} \dot{\vec{x}}_i \cdot (m_i \ddot{\vec{x}}-\vec{F}_i)=0.$$

 

[etotheipi]

@vanhees71 I agree, but I don't know, how this is different to what I write. When I say that work-energy theorem only applies to rigid bodies, I mean it in the sense that for rigid bodies, the total work done is identical to the total work done by external forces (because the internal forces do zero work).

If we can figure out the work done by the internal forces of an arbitrary system of particles, then sure the work-energy theorem works again. (E.g. we could keep track of all of the potential energy changes like you did, if these interactions are purely conservative!),

If you have e.g. non-conservative internal interactions, then I think it's not going to be as straightforward to compute total internal work. A gymnast is quite a complicated system of particles, with all sorts of dissipative internal interactions for instance, so finding internal work will be difficult

 

[vanhees71]

The point is that the work-energy theorem applies not only to rigid bodies but to arbitrary systems of point masses. It's even more general than what I wrote, because what I wrote is energy conservation for arbitrary systems of point masses with all forces being derivable from a time-independent potential. The general work-energy theorem holds for any point charges and arbitrary forces acting on them:
$$\dot{T}=\sum_{i=1}^n m_i \dot{\vec{x}}_i \cdot \ddot{\vec{x}}_i=\sum_{i=1}^{n} \dot{\vec{x}}_i \cdot \vec{F}_i.$$

 

[etotheipi]

@vanhees71 it is definitely true if you look at it with that level of granularity, since for any particle $\dot{T}_i = \dot{\vec{x}}_i \cdot \vec{F}_i$. And clearly if you use summation, it holds for the entire body.

But to do this you would need to know what $\vec{F}_i$ is exactly, for every particle. It's possible conceptually, but if I try to apply work-energy theorem to a complicated macroscopic object like a car, or something, then it's going to be impossible to work out all of the internal forces and works done by those internal forces.

That's why, for non-rigid bodies, I don't think it's so useful a construction because it relies on you knowing all of these internal works.

But for rigid bodies, you only need to worry about the external forces (which are a lot easier to keep track of), and everything will work out okay!

 

[vanhees71]

Sure, but a rigid body is only a special case. The energy-work theorem holds much more generally. E.g. it hplds also in fluid dynamics.

 

[etotheipi]

Sure, but a rigid body is only a special case. The energy-work theorem holds much more generally. E.g. it hplds also in fluid dynamics.

That is a very good point! Though we must still be careful, because sometimes we must also take into account heat flux as well as work done by surface and body forces in order to find the time derivative of energy of a fluid element!

 

[cianfa72]

but the important part is really that the point of the cylinder in contact with the ramp has zero velocity w.r.t. the ramp (due to the rolling condition), i.e. the power $P = \vec{F}_{\text{f}} \cdot \vec{v}_{\text{edge}} = \vec{F}_{\text{f}} \cdot \vec{0}$ of the friction force on the cylinder is zero in the rest frame of the ramp.

ok, thus friction force does not actually any thermodynamic work. It basically accounts for the 'transformation' of the gravity thermodynamic work (due to the force of gravity as applied to the center-of-mass ) into the kinetic energy of the cylinder w.r.t. its center-of-mass.

 

[etotheipi]

ok, thus friction force does not any thermodynamic work. It basically accounts for the 'transformation' of gravity thermodynamic work (due to the force of gravity as applied to the center-of-mass ) into the kinetic energy of the cylinder w.r.t. its cnter-of-mass.

I'm not quite sure how to interpret that; I would just say it like - friction stops the cylinder from slipping, and gravitational work is responsible for the (rotational and linear) kinetic energy gained by the cylinder.

 

[cianfa72]

friction stops the cylinder from slipping, and gravitational work is responsible for the (rotational and linear) kinetic energy gained by the cylinder.

ok sorry for my english. Anyway the main point in the process as described here is that no heat transformation takes place, I believe.

 

[Dale]

Jumping can be treated with the work energy theorem as follows:
https://openoregon.pressbooks.pub/bodyphysics/chapter/comparing-work-energy-and-energy-conservation/

That is an odd approach. There they are considering the legs as external to the system. I agree that their approach successfully uses the work energy theorem in the context of jumping.

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.

 

[Dale]

It basically evaluates the change of kinetic energy of the 'bulk' system alone (no internal kinetic energy as measured relative to the system center of mass rest frame) using the concept of 'Net Work'. In other words from W-E perspective the system has not an internal structure for storing energy

Yes, which is why it can differ from the total thermodynamic work.

The more I consider it, the more I like the term “center of mass work”. It is more descriptive and makes it less likely to get confused between the work energy theorem and the conservation of energy.

 

[Dale]

The general work-energy theorem holds for any point charges and arbitrary forces acting on them:

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):

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

 

[atyy]

If we can figure out the work done by the internal forces of an arbitrary system of particles, then sure the work-energy theorem works again. (E.g. we could keep track of all of the potential energy changes like you did, if these interactions are purely conservative!),

The point is that the work-energy theorem applies not only to rigid bodies but to arbitrary systems of point masses. It's even more general than what I wrote, because what I wrote is energy conservation for arbitrary systems of point masses with all forces being derivable from a time-independent potential.

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.

Yes, I would prefer to stress the generality of the work-energy theorem in Newtonian mechanics, rather than inferring its failure from macroscopic examples. For the conversion of internal energy to mechanical energy, there are classical models. An example of a simple classical model is two hard spheres attached to each other by a compressed spring and held in check by a constraint, with the motion initiated by removal of the constraint. We should stress that Newtonian mechanics with conservative forces is in use even for regimes that "common sense" would consider microscopic, such as molecular dynamics simulations.

Of course, ultimately there is a failure of the work-energy theorem, but that is because there is the failure of the whole Newtonian concept of force (and definite position and momentum) in quantum mechanics. Here I would still stress the primacy of mechanics, and say that although force is no longer a concept, energy and momentum conservation remain valid. In fact, I would stress that at this fundamental level, all "forces" (in the generalized sense of electromagnetic force, weak force, strong force in quantum theory) are "conservative". There are no dissipative forces at the fundamental level.

Philosophically, most physicists would say that it is the conservative forces of mechanics (classical or quantum) that undergirds the energy conservation of the first law of thermodynamics. However, in thermodynamics there is the additional idea that because of our ignorance and lack of control of microscopic motion, there is a subjective division of energy into work and heat.

As for the second law of thermodynamics, I would say that philosophically, we still believe in the primacy of mechanics, and the second law of thermodynamics arises from the initial conditions of the universe.

So overall, I think the approach in this insight fails the work-energy theorem too early, and introduces thermodynamics as too fundamental. Of course, thermodynamics is fundamental in the sense that we don't expect it to fail, but it is not fundamental in the sense that it is more general than mechanics. Mechanics is fundamental, and thermodynamics emerges from mechanics together with coarse graining and the initial conditions of the universe.

 

[atyy]

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.

I found this article which has a simple model with many of the features in the scenario of the gymnast jumping off the floor.

Springbok: The physics of jumping
The Physics Teacher 39, 109 (2001); https://doi.org/10.1119/1.1355171
Robert J. Dufresne, William J. Gerace, and William J. Leonard
https://srri.umass.edu/sites/srri/files/dufresne-2001spj/index.pdf

The next article also looks relevant, although it's not peer reviewed.

Simple Model of a Standing Vertical Jump
Chris L. Lin
https://arxiv.org/abs/2007.08706

 

[cianfa72]

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):

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

Yes, exactly. I believe the main source of confusion in this thread is due to naming.

Work-Energy theorem and center-of-mass work theorem are really two different things...

 

[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.