Quote:
Originally Posted by Brett
Supplementary: anyone know off hand the limits for stability for a hollow cylinder rotating about its axis?
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To be stable, the cylinder bust be rotating about the principle axis with the largest moment of inertia (this gives it the lowest rotational kinetic energy for a given angular momentum, meaning it cannot shed energy into other rotational modes).
A thin cylindrical shell - a hollow cylinder without endcaps - with mass M and radius R has a moment of inertia of MR^2 for rotation around its center axis.
A uniform circular disk - one of the endcaps of the cylinder - with mass M' and radius R has a moment of inertia of M'R^2/2 for rotation around its center axis. The cylinder will have two of these endcaps.
Thus, the moment of inertia for the entire hollow cylinder is MR^2 + 2 * M'R^2/2 = R^2 (M + M').
A hollow cylinder of length L, radius R and mass M without endcaps rotating around an axis perpendicular to its primary axis has a moment of inertia of M(L^2/12 + R^2/2).
A disk M' of radius R oriented perpendicular to its axis of rotation at a distance of L/2 - the endcap - has a moment of inertia of M'(L/2)^2 + M'R^2/4. Again, there are two endcaps.
Thus, for a hollow cylinder tumbling end over end, we have a total moment of inertia of ML^2/12 + MR^2/2 + 2 * (M'(L/2)^2 + M'R^2/4) = L^2 (M/12 + M'/2) + R^2 (M + M')/2
If the cylinder and endcap both have a uniform areal density D (probably equal to 1 ton/m^2, as this is sufficient to cut the dose from cosmic radiation down to levels without known long term health risks), then M = 2 * pi * R * L * D, M' = pi * R^2 *D.
For rotation about the cylindrical axis, this gives
I_z = pi * R^3 (2 * L + R) * D.
For end-over-end tumbling, on the other hand
I_x,y = pi * L^2 R (L / 6 + R / 2) * D + pi * R^3 (L + R/2) D
And we need
I_z > I_x,y
for stability. This gives us the condition
R^3 + 2 * L R^2 - L^2 R - L^3 / 3 > 0
Since it is late, I'm not going to solve this cubic inequality now (or check my work, for that matter - others may wish to look it over for accuracy), but will note that cubic equations do have closed form solutions, so you can find the allowed values of R in terms of L that give you a cylinder that rotates stably about its axis rather than tumbling end over end.
http://en.wikipedia.org/wiki/Cubic_equation
Luke