# Non-constant angular velocity in orbit

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)

• 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. Apr 25, 2018 at 20:41
• 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. Apr 25, 2018 at 20:43