Is the rotational KE of a rigid body considered as internal energy?

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  • #1
etotheipi
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I'm inclined to say no, but am by no means certain. The total kinetic energy of a system of particles is $$T = \sum_{i} \frac{1}{2} m_{i}\vec{v_i}\cdot\vec{v_i} = \frac{1}{2} m_{i}(\vec{v_{COM}} + \vec{v_{i}^{'}})^{2} = \frac{1}{2}M\vec{v_{COM}}^{2} + \frac{1}{2}\sum_{i} m_{i} \vec{v_{i}^{'}}^{2}$$ where ##\vec{v_{i}^{'}}## is the velocity of the particle relative to the centre of mass.

When we speak of non-rigid systems of particles, the second term has a well understood interpretation: the microscopic contribution to the kinetic energy. If we add this to the sum of the microscopic internal potential energies (between the particles), we obtain the internal energy.

However in the context of rigid bodies (systems of particles which retain fixed relative positions), the kinetic energy relative to the centre of mass, ##\frac{1}{2}\sum_{i} m_{i} \vec{v_{i}^{'}}^{2}##, is simply the rotational kinetic energy, ##\frac{1}{2} I\omega^{2}##. Since rotational energy appears to be macroscopic, it seems wrong to lump this into an internal energy term. However, the internal energy is defined as the total energy minus the potential energy in external fields and the kinetic energy of the centre of mass (that is, the kinetic energy in the rest frame of the body). This might lead one to believe rotational KE does not contribute to the internal energy.

I was wondering which interpretation is more correct?
 

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  • #2
PeroK
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I'm inclined to say no, but am by no means certain.
The total KE of a rigid body is, indeed, the KE of its CoM plus the rotational KE about its CoM.

Internal (thermal) energy is something different.
 
  • #3
etotheipi
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Thanks, that's slightly reassuring!

I know rest energy is a non-classical concept, however I remember in a thread a while ago someone said that the internal energy is identical to the rest energy.

The rest energy is defined as the total energy measured in a frame where the body is at rest, which I take to mean neither translating nor rotating. This indeed appears consistent with rotational kinetic energy not contributing to internal energy, since otherwise rotation would contribute to one and not the other!

So the relativistic relation goes something like ##mc^{2} = U_{micro/pot} + U_{micro/kin} + ...##, plus a bunch of other stuff (like the other terms in the liquid drop model...), but the rotational KE is not included in that sum. Everything seems OK...
 
  • #4
PeroK
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Thanks, that's slightly reassuring!

I know rest energy is a non-classical concept, however I remember in a thread a while ago someone said that the internal energy is identical to the rest energy.

The rest energy is defined as the total energy measured in a frame where the body is at rest, which I take to mean neither translating nor rotating. This indeed appears consistent with rotational kinetic energy not contributing to internal energy, since otherwise rotation would contribute to one and not the other!

So the relativistic relation goes something like ##mc^{2} = U_{micro/pot} + U_{micro/kin} + ...##, plus a bunch of other stuff (like the other terms in the liquid drop model...), but the rotational KE is not included in that sum. Everything seems OK...
Again, it's better not to mix concepts from relativity and classical mechanics. Rigid body motion is a classical concept. ##KE = \frac 1 2 mv^2## and ##RKE = \frac 1 2 I \omega^2## are classical formulae.

There is no concept of "rest energy" in classical mechanics (CM). In CM an elastic collision is one where KE is conserved. In SR it's one where rest mass is conserved. You have to stick with one or the other.

Even centre of mass is a classical concept. In SR it's centre of momentum. They are the same in CM.

I'm not really sure how you would study a structure rotating at relativistic speed. I suspect that's quite an advanced/specialist topic.
 
  • #5
etotheipi
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I'm not really sure how you would study a structure rotating at relativistic speed. I suspect that's quite an advanced/specialist topic.
I suppose that if you don't have rigid bodies any more, things get a little more fiddly...

There is no concept of "rest energy" in classical mechanics (CM). In CM an elastic collision is one where KE is conserved. In SR it's one where rest mass is conserved. You have to stick with one or the other.
But if we had a box of particles, set it in motion and did some calculations, if we could hypothetically work out the absolute internal energy classically then wouldn't we get the same value as what we calculate for the rest energy in SR? And if we add an amount of heat ##Q##, the rest energy/internal energy would both increase by ##Q##, etc.?
 
  • #6
PeroK
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But if we had a box of particles, set it in motion and did some calculations, if we could hypothetically work out the absolute internal energy classically then wouldn't we get the same value as what we calculate for the rest energy in SR? And if we add an amount of heat ##Q##, the rest energy/internal energy would both increase by ##Q##, etc.?
That's wishful thinking. In SR most of the energy of a body is in its rest mass. There's no justification in CM to calculate this as ##Mc^2##.

In general, trying to extend CM into SR, QM or GR is a bad idea. You have to go the other way round: CM is a special case of SR, QM and GR.

We had a long thread a while back from someone who wanted to justify ##E = \gamma mc^2## from CM. This is wrong. You can get ##\frac 1 2 m v^2## as a first-order approximation of ##\gamma mc^2##, but there's no viable justification for going the other way round and developing SR from CM.
 
  • #7
etotheipi
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Ah, I see. Then classically, ##T+U = T_{macro} + V_{ext} + (T_{micro} + V_{micro}) = T_{macro} + V_{ext} + U_{int}## is the total (mechanical) energy. Where ##T_{macro} = T_{translational} + T_{rotational}## and ##U_{int}=0## for a rigid body, or just ##T_{macro} = T_{translational}## for random systems of particles (since rotational energy is undefined for non-rigid bodies!).

Am I right in saying rigid bodies have no internal energy? Perhaps it would be more accurate to say that they have constant internal energy - that is, e.g. contributions from inter-particle potential energies might exist but since the distances are not changing these are, likewise, constant.
 
  • #8
PeroK
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Ah, I see. Then classically, ##T+U = T_{macro} + V_{ext} + (T_{micro} + V_{micro}) = T_{macro} + V_{ext} + U_{int}## is the total (mechanical) energy. Where ##T_{macro} = T_{translational} + T_{rotational}## and ##U_{int}=0## for a rigid body, or just ##T_{macro} = T_{translational}## for random systems of particles (since rotational energy is undefined for non-rigid bodies!).

Am I right in saying rigid bodies have no internal energy? Perhaps it would be more accurate to say that they have constant internal energy - that is, e.g. contributions from inter-particle potential energies might exist but since the distances are not changing these are, likewise, constant.
You just don't consider internal energy in rigid body mechanics. The next step is continuum mechanics and fluid mechanics etc. And thermodynamics, of course.
 
  • #9
etotheipi
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You just don't consider internal energy in rigid body mechanics. The next step is continuum mechanics and fluid mechanics etc. And thermodynamics, of course.
Awesome, thanks for the help. I think I've got it now.
 
  • #10
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...but am by no means certain.
And you will never be, about any is-considered-question.
 
  • #11
Mister T
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Am I right in saying rigid bodies have no internal energy?
A rigid body in newtonian physics might be a flywheel, for example. If it's spinning about an axis and there's friction involved, then mechanical energy is converted to internal energy. The flywheel gets warmer.
 
  • #12
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A rigid body in newtonian physics might be a flywheel, for example. If it's spinning about an axis and there's friction involved, then mechanical energy is converted to internal energy. The flywheel gets warmer.
But I suppose then the flywheel wouldn't be a perfect rigid body, since for it to get warm its molecules would have to start jiggling around and the relative positions of the particles would no longer be fixed.

Actually, now that I think about it, aren't all rigid bodies at absolute zero?
 
  • #13
etotheipi
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The flywheel problem got me thinking, since certainly for non-rigid bodies can still define a rotational kinetic energy (?). If we split the body up into loads of little ##dm##'s, then ##dE_{k} = \frac{1}{2}(dm)r^{2}\omega^{2} \implies E_{k} = \frac{1}{2}I\omega^{2}##, it's just now ##I## is also function of time!

That begs the question, what do we do with the term $$\frac{1}{2}\sum_{i} m_{i} \vec{v_{i}^{'}}^{2}$$ On the one hand, we might treat the non-rigid rotator as undergoing pure rotation with a variable moment of inertia. On the other hand, could we split this term up into a rotational term and another term encapsulating the random thermal motion of the molecules so that we end up with a rotational KE in addition to internal energy - or is this untenable?

I had settled on the notion that rigid bodies can have rotational KE and no internal energy, and non-rigid bodies can have internal energy but no rotational KE, but this appears to be perhaps a little restrictive.
 
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  • #14
A.T.
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I had settled on the notion that ...
You don't have to settle on anything. Decompose the total energy in the way, that is most practical for a given problem.
 
  • #15
etotheipi
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You don't have to settle on anything. Decompose the total energy in the way, that is most practical for a given problem.
I think that's right, from thinking about it a little more there's nothing stopping you from treating a non-rigid rotator as having rotational KE etc.

The only thing that seems wrong is to decompose the kinetic energy of a rotator into both internal kinetic energy (wrt. COM) and rotational kinetic energy. In that we either use one or the other.
 
  • #16
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But I suppose then the flywheel wouldn't be a perfect rigid body, since for it to get warm its molecules would have to start jiggling around and the relative positions of the particles would no longer be fixed.
In newtonian physics an object is rigid if its deformations are negligible. The random thermal motions of its molecules would not be relevant to ts consideration as a rigid body.

To make this more quantitative you can look at the sizes of the molecules compared to the sizes of the deformations. You can divide the object into pieces that are macroscopically small, meaning that they are small enough to be considered differential elements. They don't move relative to each other. But in terms of molecular size, these pieces are microscopically large.
 
  • #17
jbriggs444
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The only thing that seems wrong is to decompose the kinetic energy of a rotator into both internal kinetic energy (wrt. COM) and rotational kinetic energy. In that we either use one or the other.
There is nothing wrong with considering an object that...

1. Is sliding across the floor.
2. Is rotating as it does so.
3. Is warm.
4. Has chemical energy.
 
  • #18
etotheipi
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There is nothing wrong with considering an object that...

1. Is sliding across the floor.
2. Is rotating as it does so.
3. Is warm.
4. Has chemical energy.
So if we suppose we have a rigid disk that is sliding and rotating on a rough surface (such that it's not yet performing rolling), we need to find an expression for its total energy.

What you mention makes it appear that we could write ##E = KE_{trans} + KE_{rot} + U_{int}##

This is now assuming that rigid bodies can indeed have internal energy. Which terms would contribute to this? If ##U_{int} = U_{micro/kin} + U_{micro/pot}##.

But it doesn't make sense to me to decompose the total kinetic energy into a translational term, a rotational term and a microscopic kinetic term. Since##\frac{1}{2}\sum m_{i} \vec{v_{i}^{'}}^{2}## would then somehow have to equal ##KE_{rot} + U_{micro/kin}## and I have no idea how to derive that relation. I don't know how to think of it either, since each particle would be "approximately" rotating about the centre of mass along with strange random peculiar velocities to that motion.
 
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  • #19
jbriggs444
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But it doesn't make sense to me to decompose the total kinetic energy into a translational term, a rotational term and a microscopic kinetic term.
Nonetheless, I can go to the gym, pick a disc-shaped weight up off the rack, take it into the sauna, warm it up, take it back out onto a smooth section of floor and slide it with a bit of spin.

For practical purposes, that weight is rigid, has a measurable translational velocity, a measurable rotation rate and a measurable temperature. Refusing to determine its internal energy by using those measurements would be childish.

Do not get so attached to theory that it blinds you to practicality.
 
  • #20
etotheipi
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Nonetheless, I can go to the gym, pick a disc-shaped weight up off the rack, take it into the sauna, warm it up, take it back out onto a smooth section of floor and slide it with a bit of spin.

For practical purposes, that weight is rigid, has a measurable translational velocity, a measurable rotation rate and a measurable temperature. Refusing to determine its internal energy by using those measurements would be childish.

Do not get so attached to theory that it blinds you to practicality.
Would such a weight need to be modelled by something other than rigid body mechanics, like continuum mechanics?

Since @PeroK mentioned that rigid body mechanics doesn't really deal with internal energies?

And we'd need to be able to describe translational, rotational and internal energies.
 
  • #21
PeroK
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Would such a weight need to be modelled by something other than rigid body mechanics, like continuum mechanics?

Since @PeroK mentioned that rigid body mechanics doesn't really deal with internal energies?

And we'd need to be able to describe translational, rotational and internal energies.
The point is that you don't need a theory of everything to study some aspect of a system. If you want to study gyroscopes, say, and explan their motion, then it is not critical to consider air resistance and internal heat in the gyroscope. The critical mechanics is rigid body angular momentum. You don't gain a lot by insisting that the gyroscopes internal energy be added to the model.

And, in general, in some physical or engineering applications you can focus on the net external force and treat a component as a rigid body. I.e. you can ignore internal energy, forces, stresses and strains.

If, however, you want to study the strength of a component, then looking at external forces alone is not enough. You need a model for the internal forces and how the component behaves internally under various external forces.

Eventually, also, you may need to consider long-term changes to a component: oxidation, gradual weakening etc. But, if you want to study the motion of a rigid body, you certainly don't need to add a factor of gradual oxidation to your model.

In general, everything in physics and engineering involves selecting a relevant subset of the ultimate model of your system. In many cases, the internal thermal energy is simply not relevant to what is being studied.
 
  • #22
etotheipi
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Thank you, that's helpful.

I'm working from Tong's dynamics notes where he defines a rigid body to be
a collection of ##N## particles, constrained so that the relative distance between any two points, ##i## and ##j##, is fixed: ##|\vec{x_{i}}-\vec{x_{j}}| = \text{fixed}##
If you then use this property, as well as some properties of the centre of mass of a body, then you can get out König's theorem for the decomposition of kinetic energy into a translational CM and a rotational term.

But then, as @Mister T and Wikipedia mention, a rigid body might also be defined as something that has negligible deformations. I'm struggling to understand this, since it appears that a rigid body as per Tong has quite a specific definition (even though it's purely theoretical).

Intuitively, I can understand why if we tried to find the total energy of a weight, we might say it has this much linear KE and this much rotational KE (i.e. it's approximately this rigid body, so we calculate its moment of inertia) and it's this hot so it's also got some internal energy.

But nonetheless, it still doesn't appear to be a rigid body. Surely it doesn't make sense to split the velocity of any particle in the weight into ##\vec{v} = \vec{v_{CM}} + \vec{v_{rot|CM}} + \vec{v_{pec|rot}}##, where ##\vec{v_{pec|rot}}## is a sort of peculiar velocity relative to the ##\vec{\omega} \times \vec{r}## rotational velocity, which is itself relative to an axis moving with the centre of mass. In fact, if we take Tong's definition then I don't understand how the thing can have internal energy at all - the particles are fixed relative to one another.

So in order to model a weight with internal energy, some other framework must be employed surely?
 
  • #23
PeroK
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Intuitively, I can understand why if we tried to find the total energy of a weight, we might say it has this much linear KE and this much rotational KE (i.e. it's approximately this rigid body, so we calculate its moment of inertia) and it's this hot so it's also got some internal energy.
You might very well not be interested in how hot something is.

I'm working from Tong's dynamics notes where he defines a rigid body to be

But then, as @Mister T and Wikipedia mention, a rigid body might also be defined as something that has negligible deformations.
These are the same, surely?
 
  • #24
etotheipi
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These are the same, surely?
Perhaps the microscopic thermal motions of the molecules are such that ##|\vec{x_{i}}-\vec{x_{j}}|## is approximately constant? Is that the distinction between a perfectly rigid body and a general rigid body?

But even then, if I do the second layer of the decomposition, $$\frac{1}{2}\sum_{i} m_{i} \vec{v_{i}^{'}}^{2} = \frac{1}{2}\sum_{i} m_{i} (\vec{v_{rot, i}}+\vec{v_{pec, i}})^{2}$$I still don't have a way of dealing with the cross terms.
 
  • #25
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then I don't understand how the thing can have internal energy at all - the particles are fixed relative to one another.
Which is an assumption made so that the CHANGE in internal energy is negligible for a problem involving kinematics of rigid bodies in the general sense.
 

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