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 $O$O-$123$123 aligned with the principal axes. The principal moments of inertia are $I_1, I_2, I_3$I1​,I2​,I3​, with $I_1 > I_2 > I_3 > 0$I1​>I2​>I3​>0. The center of mass is fixed and no external torques act.

At $t = 0$t=0, the body rotates about the intermediate axis (principal axis 2) at angular velocity $\omega_0 > 0$ω0​>0, with an additional small perturbation $\epsilon$ϵ ($0 < \epsilon \ll \omega_0$0<ϵ≪ω0​) about principal axis 3. The initial angular velocity about principal axis 1 is $0$0.

Let $\omega_1(t), \omega_2(t), \omega_3(t)$ω1​(t),ω2​(t),ω3​(t) be the angular velocities about the three principal axes.

$\omega_2(t)$ω2​(t) first decreases from $\omega_0$ω0​, eventually reverses sign to reach $-\omega_0$−ω0​, then returns to $\omega_0$ω0​ — a periodic behavior (flip).

Find the time $T$T from $t = 0$t=0 until $\omega_2(t)$ω2​(t) first reaches $-\omega_0$−ω0​.

Approximate higher-order terms in the small quantity $\epsilon$ϵ appropriately, and use the asymptotic formula for the complete elliptic integral of the first kind to find $T$T.

The complete elliptic integral of the first kind $K(m) = \int_0^{\pi/2}\frac{d\phi}{\sqrt{1-m\sin^2\phi}}$K(m)=∫0π/2​1−msin2ϕ​dϕ​ satisfies, as $m\to 1-0$m→1−0:

$$K(m) \approx \frac{1}{2}\ln\left(\frac{16}{1-m}\right)$$K(m)≈21​ln(1−m16​)

Constraints

• $I_1 = 5 \text{ kg}\cdot\text{m}^2$I1​=5 kg⋅m2
• $I_2 = 4 \text{ kg}\cdot\text{m}^2$I2​=4 kg⋅m2
• $I_3 = 2 \text{ kg}\cdot\text{m}^2$I3​=2 kg⋅m2
• $\omega_0 = 10 \text{ rad/s}$ω0​=10 rad/s
• $\epsilon = 1.0\times10^{-4} \text{ rad/s}$ϵ=1.0×10−4 rad/s

Input Format

The flip time $T \text{ [s]}$T [s] takes the form:

$$T = \frac{\sqrt{A}}{B}\ln\left(\frac{C}{D}\times10^{10}\right)$$T=BA​​ln(DC​×1010)

where $A$A is a square-free positive integer, $B$B is a positive integer, and $C/D$C/D is an irreducible fraction of coprime positive integers.

Give the value of $A + B + C + D$A+B+C+D as a positive integer.