Right-angle lever paradox: where does the angular momentum go? I was looking at the explanation to the 'right-angle lever paradox' as explained in Franklin, 2008 (link to arXiv paper). He argues that the reason behind the lack of rotation of the lever in a moving frame is due to a distinction between $\newcommand{\p}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\f}[2]{\frac{ #1}{ #2}} \newcommand{\l}[0]{\left(} \newcommand{\r}[0]{\right)} \newcommand{\mean}[1]{\langle #1 \rangle}\newcommand{\e}[0]{\varepsilon} \newcommand{\ket}[1]{\left|#1\right>} \vec{r} \times \p{\vec p}{t}$ and $\vec{r} \times \vec{a}$ with the former determining the rate of change of angular momentum and the latter determining the tendency of rotation.
Assuming that this interpretation is correct (objections welcome), then my question is given that in the lever paradox the total $\vec{r} \times \p{\vec p}{t}$ is non-zero but the total $\vec{r} \times \vec{a}$ is, we seem to have a rate of change of angular momentum but with no rotation. Therefore my question is:
Where does the change in angular momentum go?
 A: When the two ends of the lever are pulled by rods, then there are two levers that can absorb angular momentum from each other.
When there are rockets attached to the ends of the lever, then the system consisting of the levers and rockets and exhaust gases, but not including the angular momentum of the lever, is gaining momentum all the time.
I have not checked that the above works.
A: I have an objection to the paper, the objection is that there is a better explanation:
Let's imagine a right-angled triangle sliding on a frictionless horizontal surface. One of the triangle's legs touches the surface, the other leg is vertical. 
As the triangle is moving very fast, the leg touching the surface is contracted. 
Now if the surface changes so that it has some friction, the triangle either falls forwards or doesn't. 
The triangle is losing kinetic energy when it is sliding on the surface. Let's say the kinetic energy becomes thermal energy of the sliding leg. 
Let's say the hypotenuse transmits kinetic energy from the vertical leg to the horizontal leg. The energy must travel through the hypotenuse with some velocity relative to the hypotenuse. Energy has momentum, it exerts a force on something when it changes its velocity. In our case energy exerts a force on the upper end of the triangle when starting to move towards the other end of the hypotenuse, that force prevents the triangle to topple over too easily. (The energy also exerts a force at the other end of the hypotenuse, that force does nothing very interesting.)
So although the contracted leg exerts a decreased torque on the surface, the stability of the triangle is not decreased.
