Work-Energy Theorem but Work is $F.S_{COM}$ (Extended)

Question

From the definition of Work Done, we know it is $\vec F . \vec s$ , where $\vec s$ is displacement of point of application of force. Everything was alright until I read these texts from Kleppner and Kolenkow:

highlighted texts show clearly, they were considering the displacement of centre of mass. Also, below the highlighted text they clear say that "dR is the displacement of the centre of mass".
Also,

Moreover, In this example they have considered the work done by static friction, the point of contact is at rest because of pure rolling.

Now, I did read this answer: https://physics.stackexchange.com/a/121906 which is indirectly related to this and does somehow give intuition that it is the point of application of force (I held the same view until I read these), but the answer doesn't seem to clarify what they did.

Why did KK decide to take the centre of mass displacement? Is it correct? If so, when can we take like that?

I want to know whether the formula they used is valid or not. If so, doesn't it contradict with the answer i linked? If it is valid, in what cases? Only for rigid bodies?Also, how would you justify the definition of work here?... Pls explain all these points keeping in mind the answer that I linked.

Answer 1

The work-KE theorem is a mathematical consequence of finding the work due to the net force on a particle. Using the definition of work, as a line integral, we can calculate: $$ \begin{align} W_{F_\text{net}} &= \int_{t_i}^{t_f} \vec{F}_\text{net} \cdot \frac{d \vec{r}}{d t} \, dt\\ &= \int_{t_i}^{t_f} m \frac{d \vec{v}}{d t} \cdot \vec{v} \, dt\\ &= \int_{t_i}^{t_f} m \frac{1}{2} \frac{d }{d t}\left( \vec{v} \cdot \vec{v}\right) \, dt\\ & = \frac{m}{2} \left[ \vec{v} \cdot \vec{v} \right]^{t_f}_{t_i}\\ &= \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \end{align} $$ As Kleppner and Kolenkow point out, we can apply this exact same mathematical analysis to Newton's Second Law for a composite object of many particles: $$ \vec{F}_\text{net, ext} = M \frac{d \vec{V}}{dt}\\ $$ where $\vec{F}_\text{net, ext}$ is the sum of the external forces on a collection of $N$ particles; $M$ is the total mass, $\sum_n m_n$; and the center-of-mass velocity of the system of particles is defined as: $$ \vec{V} = \frac{d\vec{R}}{dt} \quad \text{with} \quad \vec{R} = \frac{m_1 \vec{r}_1 + \cdots + m_N \vec{r}_N}{M} $$ See my recent answer here for how to derive this equation. The same calculation for "net work" shown above can be accomplished using this equation, but in terms of the center-of-mass velocity. As a shorthand, let's call that the "work on the center of mass": $$ W_\text{com} := W_{F_\text{net, ext}} = \frac{1}{2} M V_f^2 - \frac{1}{2} M V_i^2 $$

So, in the first part of their example, they are simply finding $W_\text{com}$ for the composite object (a wheel) rolling down the incline. The net external force on the object is the sum of friction, normal, and gravity: $$ \vec{F}_\text{net, ext} = \vec{F}_{fs} + \vec{F}_N + \vec{F}_g $$ and the work due to any one of these forces can be found by the simpler expression for work, $\vec{F} \cdot \Delta \vec{R}$, because the force is constant over the path. For example, for friction: $$ W_{F_{fs}} = \int_{t_i}^{t_f} \vec{F}_{fs} \cdot \frac{d\vec{R}}{dt} \, dt = \vec{F}_{fs} \cdot \int_{t_i}^{t_f} \frac{d\vec{R}}{dt} \, dt = \vec{F}_{fs} \cdot \Delta \vec{R} = - F_{fs} \, \ell $$ With work done by normal zero, we find the total net work (on the center-of-mass of the composite object) to be: $$ W_\text{com} = W_{F_\text{net, ext}} = \left( F_g \sin \beta - F_{fs} \right) \ell $$

But, because this work was defined to be that on the center-of-mass, we have neglected kinetic energy of the rotation of this composite object, and the work done to achieve that "internal" energy of the object.

To find this "work associated to the rotational kinetic energy", we can write down "Newton's Second Law for rotation" (of a rigid body): $$ \tau_\text{net, $z$} = I \alpha_z = I \frac{d \omega_z}{d t} $$ And then we can manipulate this equation to get something in the form of a Work-KE theorem: $$ \begin{align} \tau_\text{net, $z$} &= I \frac{d \omega_z}{d t}\\ \tau_\text{net, $z$} \, \omega_z &= I \frac{d \omega_z}{d t} \, \omega_z \\ \tau_\text{net, $z$} \, \omega_z &= I \frac{1}{2} \frac{d}{d t}\left(\omega_z^2 \right) \\ \int_{t_i}^{t_f} \tau_\text{net, $z$} \, \omega_z \, dt &= \int_{t_i}^{t_f} I \frac{1}{2} \frac{d}{d t}\left(\omega_z^2 \right) \, dt \\ \int_{t_i}^{t_f} \tau_\text{net, $z$} \, \frac{d\theta_z}{dt} \, dt &= I \frac{1}{2} \left[\omega_z^2 \right]_{t_i}^{t_f} \\ \int_{\theta_i}^{\theta_f} \tau_\text{net, $z$} \, d\theta_z &= \frac{1}{2} I \omega_f^2 - \frac{1}{2} I \omega_i^2 \end{align} $$ We can then define the left-hand side to be the "work associated to the rotational kinetic energy" and get its relationship to rotational kinetic energy: $$ W_\text{rot} = \int_{\theta_i}^{\theta_f} \tau_\text{net, $z$} \, d\theta_z = \frac{1}{2} I \omega_f^2 - \frac{1}{2} I \omega_i^2 $$

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So, what is the connection between these things? It seems, and Kleppner and Kolenkow claim without proof, that the net work on an object should be equal to the "work on the center of mass" added to the "work associated to rotational kinetic energy": $$ W_\text{net, tot} = W_\text{com} + W_\text{rot} $$
Is that actually true?

Well, the laws of nature are Newton's Second Law (for a particle) and Newton's Third Law (for interaction between particles). So, if it is true, we should be able to show it from the total net work on particles, i.e., the sum of the net works for all particles in the object. Let's try to calculate that directly: $$ W_\text{net, tot} = \sum_{i=1}^N W_{F_\text{net, i}} = \sum_i^N \int_{t_i}^{t_f} \vec{F}_\text{net, i} \cdot \vec{v}_i \, dt $$ For an object that rotates rigidly about its center of mass, we can write the velocity of any particle as $\vec{v}_i = \vec{V} + \vec{v}_\text{rot, i}$, where $\vec{V}$ is again the center-of-mass velocity, and $\vec{v}_\text{rot, i}$ is the velocity of the particle relative to the center of mass (tangent to the circular path of that particle about the rotational axis). Then: $$ \begin{align} \sum_{i=1}^N W_{F_\text{net, i}} &= \sum_{i=1}^N \int_{t_i}^{t_f} \vec{F}_\text{net, i} \cdot \vec{v}_i \, dt \\ & = \sum_{i=1}^N \int_{t_i}^{t_f} \vec{F}_\text{net, $i$} \cdot \left(\vec{V} + \vec{v}_\text{rot, $i$}\right)\, dt \\ & = \sum_{i=1}^N \int_{t_i}^{t_f} \vec{F}_\text{net, $i$} \cdot \vec{V}\, dt + \sum_{i=1}^N \int_{t_i}^{t_f} \vec{F}_\text{net, $i$} \cdot \vec{v}_\text{rot, $i$}\, dt \\ & = \int_{t_i}^{t_f} \left( \sum_{i=1}^N \vec{F}_\text{net, $i$} \right) \cdot \vec{V}\, dt + \sum_{i=1}^N \int_{t_i}^{t_f} m_i \, \vec{a}_i \cdot \vec{v}_\text{rot, $i$} \, dt \end{align} $$ Now, in the first term, Newton's Third Law will lead to a cancellation of all of the (internal) forces of interaction between pairs of particles; the sum of the $\vec{F}_\text{net, i}$'s is thus the sum of the external forces on the object, $\vec{F}_\text{net, ext}$. Therefore: $$ \int_{t_i}^{t_f} \left( \sum_{i=1}^N \vec{F}_\text{net, $i$} \right) \cdot \vec{V}\, dt = \int_{t_i}^{t_f} \vec{F}_\text{net, ext} \cdot \vec{V} \, dt = W_\text{com} $$ So the first term gives us exactly what we needed.

To show that the second term is $W_\text{rot}$, recognize that: $$ \left|\vec{v}_\text{rot, $i$} \right| = \omega R_i \quad \text{and} \quad \vec{a}_i \cdot \vec{v}_\text{rot, $i$} = a_\text{tan, $i$} \,v_\text{rot, $i$} $$ where $R_i$ is the distance of the $i$th particle from the rotational axis and $a_\text{tan, $i$} = R_i \alpha$ is the tangential component of the acceleration of the $i$th particle. Therefore: $$ \begin{align} \sum_{i=1}^N \int_{t_i}^{t_f} m_i \,\vec{a}_i \cdot \vec{v}_\text{rot, $i$} \, dt &= \sum_{i=1}^N \int_{t_i}^{t_f} m_i \left(R_i \alpha\right) \left(\omega R_i \right) \, dt\\ & = \sum_{i=1}^N \int_{t_i}^{t_f} m_i R_i^2 \frac{d \omega}{dt} \omega \, dt\\ & = \int_{t_i}^{t_f} \left( \sum_{i=1}^N m_i \, R_i^2 \right) \frac{1}{2} \frac{d}{dt} \left(\omega^2\right) \, dt\\ & = \frac{1}{2} I \omega_f^2 - \frac{1}{2} I \omega_i^2 \\ & = W_\text{rot} \end{align} $$ Indeed, for an object that rigidly rotates about its center of mass: $$ W_\text{net, tot} = W_\text{com} + W_\text{rot} $$ as they were defined by Kleppner and Kolenkow.

Answer 2

The work-energy theorem is a difficult concept, and it is frequently misunderstood. Note that before equation 5.15 we are given equation 5.14, which is: $$\mathbf{ F} = M \ddot{\mathbf{R}}$$ This is Newton's 2nd law. The $\mathbf{F}$ in Newton's 2nd law is the net force $$\mathbf{F_{net}}=\sum_{i=1}^n \mathbf{F_i}$$ To me, this is a little confusing already. The authors should always (in my opinion) write the net force as $\mathbf{F_{net}}$ or with some other similar indication that it is a net force and not an individual force.

Similarly, the $\mathbf{R}$ in Newton's 2nd law is the position center of mass (COM). While it is not as critical as the net-force distinction, it would also help in clarity if it were written $\mathbf{R_{com}}$. So if we write 5.15 with full clarity then we get $$\int_{\mathbf{R_{com,a}}}^{\mathbf{R_{com,b}}}\mathbf{F_{net}}\cdot d\mathbf{R_{com}}=\frac{1}{2}M \mathbf{V_{com,b}}^2-\frac{1}{2}M \mathbf{V_{com,a}}^2$$

This is clearer, but less concise and more effort to write.

Why did KK decide to take the centre of mass displacement? Is it correct? If so, when can we take like that?

They took the center of mass displacement because that is the $\mathbf{R}$ in Newton's 2nd law. As such, it is correct whenever you are using Newton's 2nd law. Indeed, that is the meaning of the 2nd law.

However, although it is always correct in Newtonian mechanics, the point that causes confusion is the following. The quantity on the left of 5.15 is called the "net work" where the word "net" refers to the "net force". In turn, the word "net" in "net force" is used because the "net force" is the sum of all of the individual forces acting on the system. This inevitably leads to the confusion that students believe that the "net work" is also the sum of all of works done by each individual force acting on the system. This is false.

Let's use the term "total work" to refer to the sum of all works done by each individual force acting on the system. The "net work" is not necessarily equal to the "total work". They are two separate concepts, and they can even differ when there is only a single force acting on the system!

The total work gives the total change in energy. The net work gives only the portion of the change of total energy corresponding to a change in the kinetic energy of the center of mass, often called the translational kinetic energy.

For example, consider a spring being compressed at a constant rate, $v$, by a force $F$ from my hand while the other side is attached to a fixed wall. By Newton's 2nd law, the force from the wall is $-F$. From these we can calculate both the "net work" and the "total work". The "net work" is the net force times the displacement of the COM, and since the net force is $F-F=0$ the "net work" is also zero regardless of the displacement. The work energy theorem says that the spring is not gaining KE. The "total work" is the sum of the work from each force. The wall has a displacement of zero, so the wall's work is $0$. The hand has a work of $\int F \ ds$, which from Hooke's law is $1/2 \ k x^2$, assuming that the spring started at equilibrium. So the total work is $0+1/2 \ kx^2$. Both the "net work" and the "total work" are correct. The "net work" says that the spring is not gaining KE. The "total work" says that the spring is gaining total energy. So, in this case, the difference is the increase in internal energy, the elastic potential energy.

If it is valid, in what cases? Only for rigid bodies? Also, how would you justify the definition of work here?

In the question about the drum, they are also using the rotational version of the work energy theorem. This would be the "net rotational work" about the COM. It is also valid when the rotational version of Newton's 2nd law is valid. So, in this case $fb\theta$ is the "net torque" not the torque due to the single force $f$. Again, this "net rotational work" is a separate concept from the "total work" ("total work" is not divided into rotational and translational parts). Those two quantities differ, even in this case where there is a single force providing torque. For the "net rotational work" it doesn't matter that the point of application of the force providing the torque is not moving, that is relevant for the separate concept of "total work".