A game of boxes with rotating gears

Question

Setup:

Suppose you have 3 (identical looking) boxes named A,B,C [furthermore they have identical mass]. You are told the following about the boxes:

One of the boxes actually is a uniform mass. We can call that box $B_1$

One of the boxes contains a large wheel inside spinning extremely fast. We can call that box $B_2$ [That is this box encloses a system with a large non-zero angular momentum]

The third box contains two medium sized wheels, spinning in opposite directions, [That is this box encloses a system with net-$0$ angular momentum], this box is called $B_3$.

The problem:

Is it possible to devise an experiment to differentiate between the 3 boxes simply on the basis of their internal states?

What I've found so far:

Identifying $B_2$ is quite easy to do. We can take all three boxes and have 3 separate people/machines $P_1, P_2, P_3$ (each of which is sitting on a spinning chair or some structure with one degree of rotational freedom) hold them and then rotate them.

When this occurs one of $P_1, P_2, P_3$ will start to undergo rotation of their chair/structure because of conservation of angular momentum. That particular person/machine was holding box $B_2$.

How to differentiate boxes $B_1, B_3$ however?

A potential strategy but it does seem like overkill:

You could try to put the two remaining boxes into orbits around identical large masses. The hope is that the box with 2-rotating wheels might behave differently in its orbit due to lens-dragging but this seems like an excessively complicated strategy.

In a similar spirit, if we assumed the wheels were charged we could look for magnetic effects outside the box, but in this case we are adding an assumption to the problem that I don't want to take for granted.

Some techniques I considered that didn't work:

Collisions: Attempting to collide the boxes doesn't appear to give rise to anything interesting mainly because the internal contents don't have any unique quantity [ex: angular momentum] to transfer to an object that it collides with.

Rotations: Rotating the box by itself doesn't appear to have any interesting properties [Although I do plan on experimentally verifying this]. Since the angular momentum of the box is 0 before and after rotation, you can essentially argue that there won't be any torque acting back on the system rotating the box. It's exactly this type of back action that was exploited in the single-rotating wheel case.

Answer 1

Based on the comments, the mass in boxes $B_2$ and $B_3$ is not uniformly distributed throughout the box, whereas it is in box $B_1$. This means that the inertia tensor for boxes $B_2$ and $B_3$ will be different than the inertia tensor for box $B_1$; the inertia tensor can be discovered experimentally by applying a known torque for a set time on a few different axes (I think two would be sufficient) and measuring the angular velocity vector of the box after the torque is applied.

So now we have a procedure:

• If there is a strong torque when you try to rotate the box, it's $B_2$.
• If it has an inertia tensor corresponding to a uniform cube, it's $B_1$.
• Otherwise, it's $B_3$.

If the boxes did all have a uniform mass distribution, we can still distinguish $B_1$ and $B_3$ based on the fact that the two disks have to have centers of mass at different locations (even if they're only slightly different). This means that the lever arm on each of the two disks will be slightly different, so the two disks will feel a different torque for the same force on the outside of the box. So, if you exert a force on one of the sides of the box, and that side happens to be parallel to the plane of rotation of the disks, you will exert slightly different torques on each of the two disks; this means that the reaction torques will not completely cancel, and you will feel a small, but nonzero, net torque as you exert that force. If you test all of the sides of the box and you do not feel this torque, the box is $B_1$. Otherwise, it's $B_3$.