Dzhanibekov effect – handling the T-handle

Posted on by arkajad

Errors that you remember most are the errors you have made yourself. Toys that you remember most are the toys that you have made yourself. As a kid I was making all kind of toys. Planes that would fly. Slings that would shoot. But also this funny toy that I was not able to find on Youtube in English. I found it only in Polish. Here it is – the “worm”:

And here are more detailed instructions with narration in Polish, but it should not matter, because you see all the details:

The bearing ball inside, invisible, makes it to behave strangely, the “worm” flips and jumps, almost like being alive.
So much about kid’s toys. Now we are playing with another toy, that also looks like being alive. Except that it whirls and flips in space, on ISS space station:

Dzhanibekov effect is the name. It does not have a bearing ball inside. What is hidden inside are Jacobi elliptic functions. They make it to flip, for no apparent reason at all, and do it, as it seems, periodically. Why? Dzhanibekov, the Russian cosmonaut who has discovered the effect, didn’t know why. He wrote to me that it is still a puzzle for him, though he prefers now to concentrate on “quantum puzzles”.
The movie above shows a T-handle. So, in today’s post I will construct my own T-handle that will whirl and flip the same way, but without going to space. We will do it with graphics, with math, with numbers.

I want to build a T-handle with principal moments of inertia 1,2,3 (I like this sequence). It will look like this:

But I want to make it as simple as possible, so I will make it very-very thin. Mathematically: I will make it of lines rather than tubes:

One line is along the x-axes. I fix it to have length L_1=2, from x=-1, to +1. For the metal I choose the linear density to be \mu=3. So the mass of the horizontal tube is 6. The length of the vertical part is to be found. I want it to be such that the principal moments of inertia with respect to the center of mass are 1,2,3.

First we need to find the formula for the center of mass position y. It is clear, from symmetry, that the center of mass will have x-coordinate zero. We need a formula for y. The center of mass of the horizontal part, that has mass m_1=6, is at (0,0). The center of mass of the vertical part, which has mass 3L_2, is at (0,L_2/2). Thus the y coordinate of the center of mass of our T-handle is

    \[y=\frac{m_1\times 0 + m_2\times L_2/2}{m_1+m_2}=\frac{3L_2^2/2}{6+3L_2}= \frac{L_2^2}{4+2L_2}.\]

Now we need to calculate the moments of inertia. First the moment of inertia with respect to the y-axis.
Wikipedia has a List of moments of inertia. There we find that for a cylinder of length L and mass m the moment of inertia with respect to the axis through its center is mL^2/12. I checked, calculating it myself, and it is correct. Here is how you can calculate it:

    \[I=2\int_0^{L/2}x^2\mu dx=2\mu\frac{x^3}{3}|_{L/2}=2\mu\frac{L^3}{3\times 8}=m\frac{L^2}{12}.\]

In our case only L_1 contributes to the moment of inertia with respect to the y-axis, because we assume our cylinders to be of zero diameter. And the contribution from L_1 is 6\frac{2^2}{12}=2. This is our I_2:

    \[I_2=2.\]

Our task is now to choose the Length L_2 so that the moment of inertia I_1 of the whole body with respect to the horizontal axis through y is 1.
It will be convenient to use what is called “Parallel axis theorem“. From Wikipedia:

First consider the contribution from L_1. The mass is m_1=6, the distance is d=y, so the contribution is

    \[I_{1,1}=0+6y^2.\]

The first term here is 0 since the moment of inertia of L_1 with respect to the x-axis is zero.
Then the contribution from L_2:

    \[I_{1,2}=\mu L_2 \frac{L_2^2}{12}+\mu L_2(L_2/2-y)^2.\]

Substituting the expression for y, after some little calculation, we get

    \[I_1=I_{1,1}+\[I_{1,2}=\frac{L_2^3(8+L_2)}{4(2+L_2)}.\]

We now solve the equation L_2=1. This is fourth degree equation, we find that two roots are real, and only one is positive:

    \[L_2=1.10938.\]

This is our T-handle.
And here is the animation – with Jacobi elliptic functions inside. The T-handle is flipping, as if being alive, with elliptic functions inside:

Click on the image to open the gif animation.

24 thoughts on “Dzhanibekov effect – handling the T-handle”

  3. Arku,

    you have made a very nice animation.
    One thing confuses a little.
    Direction of axes do not correspond to animations from previous posts (for example from “More is different”), I suppose.

      1. Well, of course spaces of the animations are different. Former animation shows the point of view from spinning top. T-handle animation shows the point of view of inertial observer.
        But perhaps it would be better if T-handle in its almost steady state rotated around y axis. This way it would better correspond to the L2 axis.

      1. I see your point. On the other hand:

        We have complete freedom to orient the laboratory axes the way we like. It is important to realize this fact. I like my T-handle rotating about my x-axis, so that on my animation it looks similar to the T-handle on the movie.

        1. OK. So let’s leave it as it is.
          Now, I repeat my question.
          What is the value of d=2E/L in your T-handle animation? Is it greater or less than half?

          2. So please make animation with d = 0.50000001
            and place it near the former – for comparison.

          4. You’ve done the work as I wanted. Thank you very much.
            But I expected some kind of quality difference which I can’t see.
            I have to think it over.

  5. Ark,
    I suspect why I don’t see the quality difference between animations for d less and grater than 0.5.

    Your T-handle doesn’t have colors.
    Can you put on a table flatly your T-handle and paint the visible (half) side of it in red?

    1. OK. But please, wait until the paint will dry. It will take a couple of hours.
      Edit: Something is wrong with my code. It crashes now. Need more time. Perhaps tomorrow.

  7. Ark
    You are great.

    Now I can see the quality difference?
    Do you?

    On the right animation the leg of the T attacks “air” always by the grey side.
    On the left animation the leg of the T attacks “air” by the grey side and by the red side alternately.

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