Solution

Let

R

g

the uniform gravitational acceleration. The gravitational torque about

R

\sum_{i} (r_{i} - R) \times m_{i} g = (\sum_{i} m_{i} (r_{i} - R)) \times g = 0.

Choose body-fixed principal axes of inertia. The inertial derivative of the angular momentum is

N = (\frac{d L}{d t})_{body} + ω \times L .

For torque-free motion

N = 0

and

L_{i} = I_{i} ω_{i}

I_{1} \overset{ω}{˙}_{1} + (I_{3} - I_{2}) ω_{2} ω_{3} I_{2} \overset{ω}{˙}_{2} + (I_{1} - I_{3}) ω_{3} ω_{1} I_{3} \overset{ω}{˙}_{3} + (I_{2} - I_{1}) ω_{1} ω_{2} = 0, = 0, = 0.

I_{1} = I_{2} = I

and let

I_{3}

be the axial moment. Then

ω_{3}

is constant and

\overset{ω}{˙}_{1} = - Ω ω_{2}, \overset{ω}{˙}_{2} = Ω ω_{1}, Ω = \frac{I _{3} - I}{I} ω_{3} .

Thus

ω_{1}^{2} + ω_{2}^{2}

is constant and the transverse angular-velocity vector rotates uniformly about the body symmetry axis. Its rate is

∣Ω∣

; it precesses in the positive axial sense when

(I_{3} - I) ω_{3} > 0

and in the opposite sense when this product is negative.

Solved by gpt-5.6-sol high.

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