Consider a pair of objects in elliptical orbits around a common center of mass. For all considerations of angular motion and torque, the pivot point of interest is the center of mass in this discussion.

The only forces occurring point directly towards the center of mass, and cannot cause a torque. The system experiences no net torque, and so the angular momentum should be conserved.

When considering a particular object in this elliptical orbit, its moment of inertia, I, varies as the radius varies. This is looking at the objects in orbit in a $L = I \omega $ lens. This can also be translated into the lens of $L = r \times p$, but the challenge arises in the first lens (perhaps it is illegal for me to discuss the angular momentum in an $I\omega$ way). As the objects get nearer in their elliptical path, the moment of inertia decreases ($I=mr^2$), so the angular velocity must increase to keep angular momentum constant.

Therefore, the angular velocity is not constant, which means that about the center of mass, the system experiences angular acceleration. However, we know that $\alpha = \tau _{net} /I$. If $\alpha$ is non-zero, it seems to show that there must be a net torque, because it is definitely the case that about the center of mass, the angular speed of either object is non-constant.

Where is the break in this logic?

Context: I am a teacher of high school algebra based physics (AP Physics 1). Students made the reasonable link that changing angular speed seems to imply a nonzero net torque, given the rotational analogue of Newton's second law, which we teach to new students as $\alpha = \tau _{net} /I$. I know that the key is that the moment of inertia is non-constant, but it seems like no matter what, with that expression, net torque being zero will force $\alpha$ to be zero.

My gut: that the "rotational Newton's second law analogue" doesn't hold for non-constant I. (We are likely a little beyond scope for this course to tackle that)

  • 1
    $\begingroup$ No time to write an answer now, but you are on exactly the right track. Make the analogy to $F_\text{net} = \frac{\mathrm{d}p}{\mathrm{d}t}$ (rather than $F_\text{net} = ma$) to get $\tau_\text{net} = \frac{\mathrm{d}L}{\mathrm{d}t}$ and find the total derivative on the RHS. $\endgroup$ – dmckee --- ex-moderator kitten Apr 25 '18 at 20:41
  • 1
    $\begingroup$ BTW: We have MathJax running on the site so that $L = r \times p$ renders as $L = r \times p$ and $\omega$ renders as $\omega$ and the like. The syntax is essentially LaTeX math-mode. $\endgroup$ – dmckee --- ex-moderator kitten Apr 25 '18 at 20:43
  • $\begingroup$ Got it, I think I have arrived there. Is there any non-calculus way I can break this down, or will I mostly just have to argue to students "Turns out when I is non constant we have this second term in that product rule, so you can in fact have acceleration without torque"? I suppose the core of this funk is that there is no such problem in the linear case, where you typically do not have a second term on the derivative. (and thanks, I was trying for those symbols but didn't know that I needed the $.) $\endgroup$ – Steinkamp Apr 25 '18 at 20:47
  • $\begingroup$ Does your book do any linear momentum problems where a rail car passes under a chute that adds mass to it? I suppose that by looking at that case first you can show them that $F = ma$ is incomplete and needs a mass variation term as well. $\endgroup$ – dmckee --- ex-moderator kitten Apr 25 '18 at 20:49
  • 1
    $\begingroup$ You can do the product rule without calculus. Just draw a rectangle with height I and width $\omega$. Show how when we keep I constant, the change in area is just I times the change in $\omega$, but when both are allowed to vary, the change in area is (change in $\omega$)*I + (change in I)*$\omega$ $\endgroup$ – Acccumulation Apr 25 '18 at 21:01

Let's take this in two parts: first an exhibition of how this works for a trained physicist who has access to the tools of multivariate calculus, and second an examination of how you might explain this to students in an introductory class based on algebra and trigonometry (no calculus).

Sophisticated view

Just as the proper formulation of the Newtonian dynamical rule is $\vec{F}_\text{net} = \frac{\mathrm{d}\vec{p}}{\mathrm{d}t}$ rather than $\vec{F}_\text{net} = m \vec{a}$, the proper formulation of the dynamical rule for rotations is (treating the fized axis case so we can dispense with vector notation): \begin{align} \tau_\text{net} &= \frac{\mathrm{d}L}{\mathrm{d}t} \\ &= \frac{\partial L}{\partial \omega} \frac{\mathrm{d}\omega}{\mathrm{d}t} + \frac{\partial L}{\partial I} \frac{\mathrm{d}I}{\mathrm{d}t} \\ &= I \alpha + \omega \frac{\mathrm{d}I}{\mathrm{d}t}\;. \end{align} Of course, in the case of rigid objects in free rotation we have $\frac{\mathrm{d}I}{\mathrm{d}t} = 0$ so that this becomes $$ \tau_\text{net} = I \alpha \;,$$ but for mutable objects or cases where the axis of rotation is moving we need both terms.

Further when the net external torque is zero we can write $$ I \alpha = -\omega \frac{\mathrm{d}I}{\mathrm{d}t} \;. $$

Classroom view

The students don't have the mathematical tooling to parse the argument above in the form written, so we need to provide a scaffold of some kind.

Work it as a conservation problem

(Following a suggestion from Acccumulation in the comments.)

If you have the rule of conservation of angular momentum, you just go with \begin{align} L_f &= L_i \\ I_f \omega_f &= I_i \omega_i \\ \end{align}

Introduce the idea that you need a term for changes in the inertial tendency

I took a crack at this in the comments, and as you say it is less than satisfactory because those "dumps a load straight down" problems really involve multiple parts of a system in a way that is not exactly analogous to the question at hand.

You're clever student is likely to land right on the difference if they are presented together.

Tackle it at the level of forces and torques to motivate the need to a second term

(This is what you asked for in your follow-up comment.)

The key observation here is to track a single mass element through a change in radius from $r_1$ to $r_2 \ne r_1$.1 During the time that radial change occurs the mass continue to move "around" the center of rotation, but the path of the mass element is not a circle centered on the axis. That means that the net forces acting on the mass element are not centripetal, and therefore they exert non-zero work on the mass element: $\vec{F}_\text{net} \cdot \vec{s} \ne 0$.

Have the student(s) check this themselves.

But that makes the translational kinetic energy of the mass element increase when coming closer to the center or decrease when moving outward. Either way there is no way for $\omega$ to remain constant.

None the less for central forces we still have $\tau = 0$. But that leads to a contradiction if you insist that $\tau_\text{net} = 0$ is the complete rule for these system. As a result we must introduce a part that depends on varying $I$.

1This is completely natural in the orbital problem that you offer, but is worth saying explicitly so we'll remember when we work on mutable solid objects in rotation.

| cite | improve this answer | |
  • $\begingroup$ Great, thanks! Can you help with the definition of "centripetal" in your second to last paragraph? It is all good that there is a tangential component of the force along the path of motion, but if I have defined my "pivot" around which I am assessing torque as the center of mass of the system, I can only see that the force is parallel to the position vector. I can see that this argument leads to work being done on the system (due to the displacement along the direction of the force), but am not fully sold on the torque. Thanks for all your time so far! $\endgroup$ – Steinkamp Apr 26 '18 at 0:01
  • $\begingroup$ Hmmm ... now I am doubting my own argument. I'm going to have to think a bit to figure out how to clarify this, if—indeed—it holds together. I'm sure I can show that the mass element picks up increased linear velocity, but I may have jumped too far with the text above. $\endgroup$ – dmckee --- ex-moderator kitten Apr 26 '18 at 1:15
  • 1
    $\begingroup$ @DWade64 I think you are exactly right, and it remains for me to re-frame this in a way that can satisfy Steinkamp's students. I think I see the way I want to proceed but need some undisturbed time to formulate it correctly. It must be rather different from what I have written above. $\endgroup$ – dmckee --- ex-moderator kitten Apr 26 '18 at 2:26

I don't have the ability to simply comment, so I'll try to turn this to an answer.

It seems that the "break in the logic" comes when trying to apply $\tau_{ext}=I/\alpha$ to this problem. It got me to looking at textbooks I have and only one (Halliday and Resnick from back in the day) was actually explicit in saying that this equation only applies to rigid bodies even though that's how it was derived. This is overlooked often just as is the warning that $F=ma$ only applies to constant mass problems. This seems like it could be a good learning point for the students that they have to understand the limits of equations they are given.

The direct analogy between the mentioned coal/chute problem is the dropping of a concentric, non-rotating hoop onto a rotating disk. Here you have changed $I$ in a simple-to-calculate way and have no external torques (if your system is the two objects).

A very simple case of no external torques (no force period!) and a changing angular velocity (thus $\alpha\neq 0$) that is similar to your problem is a single particle moving in the x-y plane parallel to the x-axis with y=1. The angular momentum about the origin is constant but the angular velocity gets very large near the axis and goes to zero far from the origin. Since $I$ is changing one can't use the Newton's analogy equation.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.