Not sure whether this is an appropriate question for this site, but could anyone explain the physics behind how this skier is able to change his direction of rotation mid-air? https://www.youtube.com/watch?v=iCKNid-ZkIk
Answer
There are two and possibly three factors at work here:
One which I don't think has been mentioned is that unless a body is rotating about a so-called principal axis, in general the angular momentum and the angular velocity vectors do not point in the same direction: they are related by a nontrivial matrix (the inertia tensor) in Euler's second law of (rotational) motion. The best you can do is diagonalise the inertia tensor: this diagonalisation discovers three orthogonal principal axes and one rewrites Euler's second law in a frame that rotates with the principal axes in the so called Euler equations. The inertia tensor is still not proportional to the identity hence the difference in direction. The skier probably begins to "fly" (i.e. freefall) with an angular momentum not aligned with the angular velocity: it is the latter that catches your eye. So the direction of "apparent spin" can be quite different from that of the angular momentum.
To add to David Hammen's answer: The skier with his skis is not a rigid body but a deformable one, and by, smoothly cycling his shape through a shape configuration space, can change his orientation even though his angular momentum cannot change. As David says this is exactly how a cat flips over when falling or how an astronaut can shift his or her orientation in space.
I say a great deal more about astronauts in my answer here (I give a simple model system illustrating cyclic shape shifting) and about falling cats in my article "Of Cats and Their Most Wonderful Righting Reflex" my my website here. The Kane reference in David Hammen's answer is heavy going, but uses only elementary dynamical concepts. So it should be accessible a bright freshman willing to do a bit of work. If, however, you have a knowledge of fibre bundles, the Montgomery reference is a much clearer overview: we begin with the shape configuration space: this can be any of several reasonable assumptions, but in Kane's and Montgomery's simplified analysis, the cat's shape is defined by two angles and leads to a pinched cylinder, pinched torus or 2-sphere, depending on how you exactly model things. The exact topology is not important: what is important is that now we (or rather, Montgomery) kit the shape space with a fibration: at each point the fibre we add is a space of orientations in 3-space (i.e. $SO(3)$) for the cat and show that the conservation of angular momentum defines a nontrivial fibre connexion. Depending on the cat's symmetry assumed, the fibre bundle that results either has a structure group ("gauge group" if you will) of $SO(2)$ (the tailess, symmetrical cat can flip about one axis only) or the full $SO(3)$ acting on the fibre $SO(3)$. The computation of the structure group is the messy bit and where the conservation of angular momentum comes in to compute the connexion ("gauge covariant derivative") for the bundle - this is what is effectively done in the Kane paper to find the Kane dynamical equations. Of course, one can also think of these ideas in the language of anholonomy (curvature of the fibre bundle).
A wonderful quick summary is given by this video (NOT mine: I found it on Youtube)
https://www.youtube.com/embed/yGusK69XVlk
The analysis in this video is correct for a cat which is symmetrical about its transverse plane (in the anatomists Coronal/Sagital/Transverse convention), in particular for a tailess cat. Most domestic cats don't use their tail a great deal in the righing reflex and consequently their flip tends to be confined to one axis, but small tree-dwelling cats of South East Asia like the Marbled Cat and Clouded Leopard have a huge tail (more like a club) in relation to their body and use it for re-orientation in all axes as they divebomb their prey.
At the speed the skier is flying, air resistance means that the flight will not be torque free. However, this hopefully is not a major factor, and indeed one of the skills needed by the skier is to hold his or her body so that it "tumbles" (i.e. moves essentially torque free, but buffetted by perturbational torques) in a stable way and keeps the perturbative torques from becoming a problem. Tiny buffetting torques can otherwise make the apparent spin of the body flail about wildly indeed.
As discussed in this exposition here J. Peraire, S. Widnall, "Lecture L28 - 3D Rigid Body Dynamics: Equations of Motion; Euler’s Equations", we suppose that a body is at first rotating about a principal axis $z$ and is then buffetted by a tiny angular impulse. The linearised Euler equations for the torque free motion that follows the impulse described by the angular velocities $\omega_x,\,\omega_y,\,\omega_z$ relative to the rotating principal axis frame are:
$$\dot{\omega}_x - \frac{(I_{yy}-I_{zz})(I_{zz}-I_{xx})}{I_{yy}\,I_{xx}} \omega\,\omega_x = 0$$
with a like equation for $\omega_y$ (here $\omega$ is the initial angular speed before the impulse) The solutions to this DE are $e^{\pm \sqrt{a}\,t}$ where
$$a=\frac{(I_{yy}-I_{zz})(I_{zz}-I_{xx})}{I_{yy}\,I_{xx}} \omega$$
If $I_{zz}$ is either less than both or greater than both $I_{xx}$ and $I_{yy}$, $a$ is negative and both the terms $e^{\pm \sqrt{a}\,t}$ are oscillatory. The body wobbles a little bit, but is otherwise unaffected by the perturbation. If, however, our hapless skieer gets into a situation where $I_{zz}$ lies between $I_{xx}$ and $I_{yy}$, then $a$ is positive and one of the solutions grows exponentially. A large, torque free change in angular velocity follows. Such a body's attitude is essentially out of control, even in the presence of tiny, buffeting winds.
No comments:
Post a Comment