Precession of a Gyroscope — Question 5

Question 5

Equal spin energy leads to a square-root inertia dependence

Two gyroscopes have the same $M$, $\ell$ and spin kinetic energy $K_s$. Rotor B has four times the axial inertia of rotor A. Both satisfy the slow-precession approximation.

1. Compare their spin momenta and precession rates.

2. How would the rate comparison change if spin speeds rather than spin energies were held equal?

Question 5 - Solution

Equal spin energy leads to a square-root inertia dependence

(a) Derivation and calculation

At the same spin kinetic energy, $$K_s=\frac{L_s^2}{2I_s}\Rightarrow L_s=\sqrt{2I_sK_s}.$$ Therefore $$\boxed{\frac{L_B}{L_A}=\sqrt{\frac{I_B}{I_A}}=2},\qquad \boxed{\frac{\Omega_B}{\Omega_A}=\frac{L_A}{L_B}=\frac12}.$$

(b) Result and interpretation

At equal spin speeds instead, $$\frac{L_B}{L_A}=\frac{I_B\omega_s}{I_A\omega_s}=4.$$ Since the gravitational moments are equal, $$\boxed{\frac{\Omega_B}{\Omega_A}=\frac14}.$$ The slow precession depends on spin angular momentum; comparing inertia alone is insufficient without specifying the held spin quantity.

Physical check

At equal energy, rotor B spins at half the speed but carries twice the angular momentum.