Decompose a motion into its screw motion (mozzi-chasles' theorem)

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The theorem says that any rigid body movement can be described (for each point) as a set translation along a set axis and a rotation (by a set angle) around it.

It's easy to imagine what the movement looks like if only a rotation or only a translation takes place around a set axis.

However, I fail to imagine where would the axis be and which way pointing, and the rotation angle and the amount of translation, for one particular movement (I chose a slightly more complex one than those allowed in the trivial cases above).

The movement is as follows:

Looking from the top, I have a cube on the table, lying flat on one of its faces.

  1. I simply stand it on one of its corners reaching unstable equilibrium (by a rotation along one axis), leaving one corner on the table exactly where it was.
  2. Along the vertical axis, I rotate it clockwise by for example 30°.

Where is the screw axis of this movement? Is there any translation. Any rotation?

Being this a more complex movement, I suspect that there is both a non-zero translation (that is applied to every point) and none-zero rotation (along the screw axis). How come it cancels out at the corner point, that does not move?

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In short, Euler's rotation theorem says "any displacement of a rigid body such that a point on the rigid body remains fixed, is equivalent to a single rotation about some axis that runs through the fixed point."

That means, that if I choose the center point of the body as fixed for example and I rotate it many times around various axes (going through center), I could achieve the same displacement with a single rotation along an axis.

In the question cube example, the fixed point was the corner (both in rotation 1. and 2.). So, according to Euler, there exists an axis going through the corner and if I rotated the cube around it, I would achieve the displacement in the question. All of that without any translation along the rotation axis. In fact, if it was a true screw motion (with the translational component), indeed no point would be fixed.

For imagining the direction of the screw axis (we now know that just rotation axis): You can try grabbing a cube not by opposing corners but slide your fingers a little closer, on the opposing edges, forming a rotation axis, that way you can achieve similar (however not sure if the same) displacement to that in question with exception the center is fixed (if you managed to form an axis going through the center) -- just to illustrate compound rotations along two different axes might be substituted by a single rotation.


More formally, the results in the following link imply that chained rotations actually have a single equivalent rotation:

A proof that rotations in 3d form a group (and have at least one axis). (I would say that proves the Euler's theorem I cited above), https://www.math.purdue.edu/~arapura/algebra/algebra13.pdf

It is important that the rotation in proof above, in linear algebra sense, is meant the vectors rotate around an axis always going through the origin -- zero-vector. Fortunately, the twice rotated cube in question is always rotated around the same point -- corner --, that would be the zero-vector if we applied the theory on the example.