Question 13.8: A simple gyrocompass consists of a gyroscope that rotates ab...
A simple gyrocompass consists of a gyroscope that rotates about its axis with the angular velocity ω. Let the moment of inertia about this axis be C and the moment of inertia about a perpendicular axis be A. The suspension of the gyroscope floats on mercury, hence the only acting torque forces the gyroscope axis to stay in the horizontal plane. The gyroscope is brought to the equator. Let the angular velocity of the earth be Ω. What is the response of the gyroscope?
Since the earth rotates with angular velocity Ω, the angular momentum in the earth system satisfies
\frac{dL}{dt}= D− Ω×L,where D is the total torque. At the equator Ω points along the y-axis, and the z-axis is perpendicular to it.
The components of the angular momentum are
L_{x} = C ω sin \varphi,L_{y} = C ω cos \varphi, \frac{ω}{Ω} \ll 1,
L_{z} = A \dot{\varphi}.
We suppose that \varphi is small. Then
L_{x} ≅ C ω \varphi,L_{y} ≅ C ω , \frac{ω}{Ω} \ll 1,
L_{z} ≅- A \dot{\varphi}.
Since there are no forces acting in the x,y-plane, D_{z} = 0. Hence, the equation for L_{z} is
A \ddot{\varphi} =−CωΩ\varphior
\ddot{\varphi} + \frac{C}{A} ωΩ\varphi = 0; i.e., \ddot{\varphi} + ω^{2}_{r} \varphi = 0, ω^{2}_{r}= \frac{C}{A} ωΩ.
\varphi oscillates with the frequency
ω_{r} =\left(\frac{C}{A} ω Ω\right)^{1/2}in the north–south direction!
