Why does the work-energy theorem need to include internal forces?

Question

Can anyone kindly explain me why work energy theorem must also include internal forces?

The proof of work energy theorem is derived from Newton's laws of motion, but Newton's laws of motion don't take internal forces into account, so why should internal forces be taken into account in the work energy theorem?

Answer 1

This is a bit of a strange question, because Newton's laws do include internal forces.

However, Newton's third law happens to cancel out their overall effect on a center of mass. But, if you want to understand the motions of the constituent parts of the system, then you do have to understand their internal forces.

So let's assume that we have a collection of particles $\{i\}$ with masses $m_i$ and (vector) positions $x_i$ each feeling external forces $F_i$ and internal forces $V_{ij} = -V_{ji}.$

We usually describe the system using the overall mass $M = \sum_i m_i$ at the center-of-mass position $X = \sum_i \frac{m_i}M x_i.$ Newton's laws say that the EOM for the center-of-mass are (with dots as time-derivatives) $$M \ddot X = \sum_i m_i \ddot x_i = \sum_{i}\left(F_i + \sum_j V_{ij}\right) = \sum_i F_i = F.$$Here $F$ is the "effective force" on the center of mass. In the above we found out that the $V_{ij}$ term disappeared, why? In a little more detail the argument looks like this:

1. We know $V_{ij} = -V_{ji}$, this is called “antisymmetry”.
2. This means that $V_{ij}+V_{ji}=0$, and since we can add zero to anything without changing it, $$V_{ij} = V_{ij} + k (V_{ij}+V_{ji}) $$ for any k. We choose the average between the equally terms, k = -½ so that $$V_{ij} = {V_{ij} - V_{ji}\over2}.$$
3. Then when we calculate $\sum_{ij} V_{ij}$ we expand it out into these two terms, $\left(\sum_{ij} V_{ij} - \sum_{ji} V_{ji}\right)/2.$ In the second term we relabel $i \leftrightarrow j$ simultaneously and we find $\sum_{ij} (V_{ij} - V_{ij})/2 = \sum_{ij} 0 = 0$ directly.

In a paragraph or two I will call this the "antisymmetric cancellation trick."

Similarly we can use the usual work-energy trick and multiply both sides by $\dot X,$ yielding$$ \begin{align} M\ddot X\cdot \dot X &= F\cdot\dot X\\ \frac d{dt} \left( \frac 12 M \dot X ^2 \right) &= F \cdot \dot X = P, \end{align} $$ and then we can define $K=\frac 12 M \dot X ^2$ as "the kinetic energy of the center of mass" and $P$ as "the power of the effective force on the center of mass." Same work-energy trick as for a particle, but now applied to an aggregate of particles.

However there is a bunch of kinetic energy in the system which is not seen in this definition of $K$! The easiest way to think about this is to think of a gyroscope which is spinning but standing still: $K$ as we have defined it is zero, and all of that rotational kinetic energy is being ignored by this picture, because the center of mass isn't moving.

If we instead want the total kinetic energy, then we find that the power exerted on it is $$T = \sum_i\frac 12 m_i \dot x_i^2 = \sum_i \left( F_i \cdot \dot x_i+ \sum_j V_{ij}\cdot \dot x_i \right)$$The $V_{ij}$ terms here do not vanish via the antisymmetric cancellation trick! That is because when you do it, you get $\sum_{ij} V_{ij} \cdot(\dot x_i - \dot x_j) / 2$ after the relabeling, but there is no guarantee that $\dot x_i = \dot x_j.$

Answer 2

I'm just going to give a couple of examples: cases where it is obvious the internal forces change the kinetic energy state of the whole system.

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• Consider a system of two masses resting on a frictionless, horizontal surface with a light spring held between them but not connected to either mass. If the initial state the spring is held tightly coiled by a bar-and-latch mechanism and is permanently affix to one of the masses. The initial kinetic energy and momentum is zero. When the spring is released the two masses are thrown apart (by internal forces that do positive net work) and move apart. The momentum of the final state is zero, but the kinetic energy (all the energy) is positive.

• Consider a toroidal, spinning space station. Give it two elevators for symmetry and have each simultaneously lift a mass $m$ from the rim to the hub. Compute the change in angular kinetic energy as this happens, and compare to the work done in lifting the masses. Again, internal forces do positive work resulting in an increase in overall kinetic energy.

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Newton's 3rd tells you that the system conserved momentum, not energy. This is in part because the kinetic energy is a positive definite quantity.

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I'm catching some flack for users who interpret the work-energy theorem as excluding the internal kinetic-energy of the system. That is not the rule that Goldstein or Marion & Thornton use.

In particular Goldstein writes (in section 1.2 (about systems of particles) of the 2nd editions, page 9 in my copy)

Hence the work done can still be written as the difference of the final and initial kinetic energies $$W = T_2 - T_1$$ where $T$, the total kinetic energy of the system, is $$T = \frac{1}{2}\sum_i m_i v_i^2 \,.$$

The emphasis here is mine. This definition clear include the internal kinetic energy of the system in the work-energy theorem and that requires including internal force as outlined above.

I suppose it is possible that there are two camps on this (I don't have an references that give the other form, so I can't say for sure), but if so the OP's professors is clearly in the same camp as Goldstein and Marion & Thornton. The other interpretation runs into severe problems as soon as you allow systems that roll. In that case external work gets divided into translational and rotational kinetic energies, so we must include the internal motions in the accounting to have the work-energy theorem work as advertised.