Precession of a Gyroscope — Question 1

Question 1

Gravity turns a fast gyroscope sideways through its torque

A fast symmetric gyroscope has $M=2\,\mathrm{kg}$, center offset $\ell=.15\,\mathrm m$, axial inertia $I_s=.020\,\mathrm{kg\,m^2}$ and spin $\omega_s=300\,\mathrm{rad/s}$. Its axis points along $+x$ from the pivot and spin angular momentum points along $+x$. Gravity is $-10\hat z\,\mathrm{m/s^2}$. Assume steady slow precession and spin-dominated angular momentum.

1. Find the gravity torque direction and precession rate.

2. Find the precession period and its direction viewed from above.

Question 1 - Solution

Gravity turns a fast gyroscope sideways through its torque

(a) Derivation and calculation

The pivot-to-center vector is $\ell\hat x$ and gravity is $-Mg\hat z$. Hence $$\vec\tau=\ell\hat x\times(-Mg\hat z) =\boxed{3.00\hat y\ \mathrm{N\,m}}.$$ The spin momentum is $\vec L_s=I_s\omega_s\hat x=6\hat x$ kg m$^2$/s. In slow steady precession, $$\vec\tau=\vec\Omega\times\vec L_s\Rightarrow \boxed{\vec\Omega=\frac36\hat z=.500\hat z\ \mathrm{rad/s}}.$$

(b) Result and interpretation

The precession period is $$\boxed{T_p=\frac{2\pi}{|\Omega|}=4\pi=12.57\ \mathrm s}.$$ Because $\vec\Omega$ points along $+z$, the precession is counterclockwise viewed from above. Initially, $$\frac{d\vec L_s}{dt}=+3\hat y,$$ so the spin-momentum tip turns toward $+y$, not toward the downward gravitational force.

Physical check

The vector identity $\hat z\times\hat x=\hat y$ verifies the precession sign.