# How to calculate the moment of inertia of a Möbius strip?

Well, I was thinking about how to calculate the MOI of a Möbius Strip. But it's very difficult (seems to me) to find a symmetry in this scenario.

Edit: (Some people have commented to add some details about the strip.) The strip has mass M, width b , length a and uniform arial mass density. I want to calculate the MOI from any axis in which the mathematics is easy.

Any hint?

• The moment of inertia about what axis? Mar 24, 2017 at 3:53
• wouldn't depend on the spatial details of the strip, i.e. if it is "stretched" rather "plump", the thickness of the strip etc? Mar 24, 2017 at 4:10
• You would have to specify the geometry of the particular Mobius strip that you're considering and probably have to use numerical integration to get the moment of inertia. BTW, seems like a pretty wacky question: Why would anyone be interested in the moment of inertia of a Mobius strip?
– user93237
Mar 24, 2017 at 4:15
• This problem seems too open ended. A Mobius strip can take many different forms. Specifying the mass, width, length, and mass distribution still doesn't specify the geometry. And even if you do find the moment of inertia the result will most likely not be as remarkable as you're expecting it to be. It will be a good math exercise though. Mar 24, 2017 at 7:45

WARNING : I am not convinced that my answer below is correct.

I have followed the tilted sticks definition of the Mobius Strip given by Herman Jaramillo in answer to the cited Mathematics SE question $^1$. However, I think that this definition does not conform with the definition of a paper strip joined with a half twist. The sticks are closer together where the ends tilt inwards and further apart where the ends tilt outwards; as a result the strip varies in density. This cannot happen on a paper strip, which maintains its density. It cannot expand or contract anywhere : parallel lines ruled across the width of the flat, pre-twisted strip must remain parallel (ie the same distance apart at all points) after the strip is twisted, when measured along the surface perpendicular to the lines.

One consequence is that the mid-line of the twisted strip is not a plane circle. Stresses force the mid-line to bend out of the plane into what looks from the side like a figure-of-eight. Furthermore, I suspect that the exact shape differs depending on the aspect ratio $a/b$.

CONCLUSION : My definition of the Mobius Strip is only a poor approximation to the real thing.

I shall define the mid-line of the strip to be a circle of radius $a$, located in the xy plane with origin as centre. The strip is made up of infinitesimally thin sticks of length $b$ and uniform mass centred on the mid-line. The stick parallel to the z axis has its centre at the point $(+a,0,0)$. The sticks tilt around the circle, making an angle $\phi$ with the z axis which is $0^{\circ}$ at $(+a,0,0)$, $90^{\circ}$ at $(-a,0,0)$ and $180^{\circ}$ again at $(+a,0,0)$. The MI of each stick about an axis through its centre parallel to the z axis is $I_0\sin^2\phi$ where $I_0$ is the MI about an axis through its centre perpendicular to the stick $^2$. The MI about the z axis is found by applying the Parallel Axes Theorem. The MI of the whole figure is found by integrating around the circle, ie through all tilt angles from $\phi=0$ to $\pi$.
The distance of the CM of each stick from the x axis is $a\sin\theta$ where $\theta=2\phi$ is angular distance around the circle. The MI of each stick about an axis through its CM parallel to the x axis is $I_0\sin^2\phi \sin^2\theta$ (?). Again the Parallel Axis Theorem is required to transfer this axis through the origin.