24 Apr 08
Originally posted by mtthwWhich means...
But...if it's elastic it doesn't make sense to solve in terms of the mass/unit length any more, as this will vary as it stretches.
If we reformulate in terms of the overall mass: m = 2.pi.r.rho
=> T = mrw^2/2.pi
If the rest length is L, we also have T = k(2.pi.r - L) by Hooke's Law
Equating these, and rearranging, I get:
r = L/[2.pi(1 - m.w^2/4.pi^2.k)]
If this is correct (comments welcome!), I'd also note the following.
1. If L = 0 (the original problem), r = 0. Not very exciting!
2. If k < mw^2/4.pi^2, the solution breaks down. I'd assume this means that the spring constant is not large enough to stop the loop expanding regardless of the radius.
3. Of course, if the loop expands while freely rotating, it will slow its rotation (by the conservation of angular momentum). So there will probably be an equilibrium radius at a new angular momentum. Finding this is a different problem again...
25 Apr 08
Originally posted by mtthwI was thinking about this one a bit, and I have a different solution that doesn't use the linear density of the rope.
Which means...
If we reformulate in terms of the overall mass: m = 2.pi.r.rho
=> T = mrw^2/2.pi
If the rest length is L, we also have T = k(2.pi.r - L) by Hooke's Law
Equating these, and rearranging, I get:
r = L/[2.pi(1 - m.w^2/4.pi^2.k)]
If this is correct (comments welcome!), I'd also note the following.
1. If L = 0 (the original problem) ...[text shortened]... n equilibrium radius at a new angular momentum. Finding this is a different problem again...
From the previous analysis, Fc = 2.T.sin(a) = m.w^2.r. If we use the relation m = (2.a/2.pi)*M instead of a linear density, we come up with:
T = M.w^2/2.pi
Now if we equate the forces developed as per Hooke's law with the tension, we get:
F = kx = k(2.pi.r) = M.w^2/2.pi
r = M.w^2/k.(2.pi)^2 = M/k * (w/2.pi)^2
If we assume that the rope/spring has a non-zero rest length, then the expression becomes:
r = r0 + (M/k * (w/2.pi)^2)