I see. So the normal forces do no translational work. However - do they do rotational work on the ladder?

@AvivCohn No, the normal forces do no work at all on the ladder. Gravity does all the work; the normal forces only constrain the ladder so that rotation is possible much like the case of friction induces rolling without slipping for a wheel.

So where is the error with the following reasoning? 1. The ladder (if of uniform mass) is rotating around its CoM. 2. The ladder starts from rest. So to gain angular momentum, it must have torque (as the derivative of the angular momentum is the torque). 3. If each normal force does no work, that means it exerts no torque on the ladder. 4. So the torque must come from the force of gravity. 5. The torque by a force $\vec{F}$ around the CoM is computed by $\vec{r} x \vec{F}$, where $\vec{r}$ points from CoM to the point of application. 6. The torque by gravity is $\vec{0}x\vec{W} = \vec{0}$

So according to this reasoning, gravity does no torque, and so do the normal forces. So what causes the rotation?

@AvivCohn The ladder is not rotating about it's center of mass, rather the center of mass is rotating about the contact point on the ground, which also happens to be changing with time. It is important to pay attention to the center of rotation.

Oh I see! makes sense. Another point that confuses me about what you're saying is: the CoM, as you say, travels not only horizontally but also vertically. In order to gain both vertical and horizontal velocity, we need forces acting in these directions. Gravity acts only vertically so of course the horizontal normal from the wall is required to make the CoM move horizontally. So far, correct?

So here's what I find confusing: we agree (?) that the normal force from the wall is what makes the CoM travel horizontally. How is this possible, along with the fact that this force does no work?

@AvivCohn No. Gravity makes the ladder travel horizontally. The normal force is just a force of constraint that does no work. Forces of constraint limit the kinds of motion possible, but do not have to do active work, i.e. gravity can't just pull the ladder straight down, it is limited to making it slide down the wall.

Oh, I see. It's kind of similar to how with a ball swinging as a pendulum, the tension string is a constraint force which makes the thing go in a circle. However - it does no work, as it operates in perpendicular to the velocity of the point of contact of the string and the ball (even though it does sometimes make the thing go "up" relative to its current position). Correct?

A few follow up points if you don't mind. 1. In my current course, I never ran into the term "constraining force". Is there some simple guideline/definition I can use to spot these in my problems? 2. Is it true that a constraint force doesn't do work?

3. You wrote above: "In each case of the normal force, the motion of the ladder at the contact points is perpendicular to the ladder's motion, thus the normal forces do no work on the ladder...". You meant to write the following, correct? "In each case of the normal force, the motion of the ladder at the contact points is perpendicular to the direction of the normal force, thus the normal forces do no work on the ladder...".

@AvivCohn Yes, that is a typo, thanks for catching that and deducing the proper correction.

So if a force is acting on a point which is moving perpendicular to the direction of the force, I get that the force will do no translational work. However is this also a guarantee that the force does no rotational work?

@AvivCohn Yes, that is correct, it does no work translational or rotational.

I'd appreciate it if you can also clarify the following thought. I wonder if the following two things can be true at the same time. 1. Force F acts on a body B, at a point of contact which is moving in a direction perpendicular to the direction of F. And thus F does no work on B. 2. However, the center of mass of B is moving in a direction which is not perpendicular to the direction of F.

It seems to me that there should be a contradiction in this case - meaning that these two are not possible at the same time. Because point 1 implies that F is doing no work. However, we know that total translational work is computed like so: $W = \int \vec{F_{net}} \cdot d\vec{r_{com}}$. And since F is included in $\vec{F_{net}}$, and is not perpendicular to the path traveled by the CoM, it will do work. Is my reasoning correct?

@AvivCohn No. The formula you are using for work is incorrect. It is the formula for pseudowork which is not the same as translational work, excepting the case of point particles.

Oh, interesting. My textbook (An Introduction To Mechanics by Kleppner and Kolenkow) seems to say that the general work done for a given rigid body is calculated like so: $W = \int \vec{F_{net}} \cdot d\vec{r_{com}} + \int_{\phi1}^{\phi2} \tau d\phi$. Is this wrong?

And if it is wrong, how would you say we should compute total work done? Assuming the rotational part above is correct, do we compute translational work by summing the line integrals of each applied force along the path of its respective point of application? E.g. $W_{rotational} = \int_{\phi1}^{\phi2} \tau d\phi$ and $W_{translational} = \Sigma \int_{c_i} \vec{F_i} \cdot d\vec{r_{app_i}}$

Kleppner and Kolenkow are simplifying things for the sake of pedagogy. You need to see the paper "Pseudowork and Real Work" by Bruce Sherwood to understand what I am saying. You wan to use $$W=\sum\int_{c_i}\vec F_i\cdot d\vec r_{\text{appi}}.$$ From that it should be obvious that the normal forces do no work.

@AvivCohn The work done on the ladder is :$$W=\int_C -mg\mathbf{\hat{j}}\cdot d\vec r_{\text{com}}={1\over 2}mV_{\text{com}}^2+{1\over 2}I\omega^2.$$ Where $C$ is the contour followed by the center of mass. And $${1\over 2}MV_{\text{com}}^2=\int\vec F_{\text{net}}\cdot d\vec r_{\text{com}}.$$

I see. In the first equation in your last comment, did you mean to write the following? $W = \int_c -mg\hat{j} \cdot d\vec{r_{com}} + \int_{\phi{1}}^{\phi{2}} \tau d\phi = ...$ ? I mean it seems like you forgot the expression for the rotational work.

@AvivCohn No, I meant what I wrote. The work-energy theorem says that the work done equals the change in kinetic energy $$W=\int_C-mg\mathbf{\hat{j}}\cdot d\vec r_{\text{com}}.$$ And change in kinetic energy assuming starting from rest: $$\Delta T={1\over 2}mV_{\text{com}}^2+{1\over 2}I\omega^2.$$ Note that:$$W=\sum_i\int_{c_i}\vec F_i\cdot d\vec r_{\text{app}_i}=\int_C-mg\mathbf{\hat{j}}\cdot d\vec r_{\text{com}}$$

@AvivCohn I really like that Kleppner and Kolenkow book, is that the text your university uses?

@AvivCohn Could you tell me where in Kleppner and Kolenkow that they say that work for rigid bodies is defined as you state in your earlier comment?