GSO002 Problem 19
Problem Statement
Free Rotation of a Rigid Body and Flip Period of the Dzhanibekov Effect
Problem Statement
In zero-gravity, vacuum conditions such as those in the International Space Station, when an asymmetric rigid body (like a T-shaped wrench) is rotated, the rotation about a specific axis becomes unstable and the body periodically flips. This is known as the "Dzhanibekov effect" or the "Tennis Racket Theorem."
Let the center of mass of the rigid body be the origin, with a body-fixed orthogonal coordinate system O-123 aligned with the principal axes. The principal moments of inertia are I1,I2,I3, with I1>I2>I3>0. The center of mass is fixed and no external torques act.
At t=0, the body rotates about the intermediate axis (principal axis 2) at angular velocity ω0>0, with an additional small perturbation ϵ (0<ϵ≪ω0) about principal axis 3. The initial angular velocity about principal axis 1 is 0.
Let ω1(t),ω2(t),ω3(t) be the angular velocities about the three principal axes.
ω2(t) first decreases from ω0, eventually reverses sign to reach −ω0, then returns to ω0 — a periodic behavior (flip).
Find the time T from t=0 until ω2(t) first reaches −ω0.
Approximate higher-order terms in the small quantity ϵ appropriately, and use the asymptotic formula for the complete elliptic integral of the first kind to find T.
The complete elliptic integral of the first kind K(m)=∫0π/21−msin2ϕdϕ satisfies, as m→1−0:
K(m)≈21ln(1−m16)Constraints
- I1=5 kg⋅m2
- I2=4 kg⋅m2
- I3=2 kg⋅m2
- ω0=10 rad/s
- ϵ=1.0×10−4 rad/s
Input Format
The flip time T [s] takes the form:
T=BAln(DC×1010)where A is a square-free positive integer, B is a positive integer, and C/D is an irreducible fraction of coprime positive integers.
Give the value of A+B+C+D as a positive integer.
Solution
1. Euler's Equations and Conservation Laws
The torque-free rotation satisfies Euler's equations. The conserved energy and angular momentum magnitude (from initial conditions) are:
2E=I2ω02+I3ϵ2,L2=I22ω02+I32ϵ22. Expressing ω12 and ω32 in Terms of ω22
Eliminating ω12 and ω32 from the conservation equations:
ω32=I3(I1−I3)I2(I1−I2)(ω02−ω22)+ϵ2 ω12=I1(I1−I3)I2(I2−I3)(ω02−ω22)3. ODE for ω2
Substituting into Euler's second equation:
ω˙2=−α(ω02−ω22)(ω02−ω22+k2)where:
α=I1I3(I1−I2)(I2−I3),k2=I2(I1−I2)I3(I1−I3)ϵ24. Reduction to Elliptic Integral
By variable separation and the substitution ω2=ω0cosθ, then ϕ=π/2−θ:
T=αω02+k22K(m),m=ω02+k2ω025. Asymptotic Approximation
Since k2≪ω02: 1−m≈k2/ω02 and ω02+k2≈ω0:
T≈αω01ln(k216ω02)6. Numerical Substitution
- α=5×2(5−4)(4−2)=51
- k2=4(5−4)2(5−2)ϵ2=23ϵ2
- αω01=105
- k216ω02=23×10−816×100=332×1010
Matching the format T=BAln(DC×1010): A=5, B=10, C=32, D=3.
A+B+C+D=5+10+32+3=50Final Answer: 50