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Advanced Engineering Dynamics

R. Valéry Roy

Section 15.3 Dynamic Characterization of Gyroscopes

We first attempt to derive general properties pertaining to the motion of a gyroscope \(\cB\) relative to a Newtonian referential \(\cE\text{.}\) We can express the angular velocity \(\bOm = \bom _{\cB / \cE}\) of body \(\cB\) in terms of unit vector \(\bz\) and its time rate of change relative \(\cE\text{.}\) Since unit vector \(\bz\) is attached to \(\cB\) we have  1 
\begin{equation} \frac{d \bz}{dt} = \bOm \times \bz\tag{15.3.1} \end{equation}
This equation can be solved to give \(\bOm\) by taking the cross-product of both sides of (15.3.1) with \(\bz\text{:}\) using \(\bz\times (\bOm \times \bz)= \bOm - (\bz \cdot \bOm)\bz\) we find
\begin{equation} \bOm = \omega_z \bz + \bz \times \frac{d\bz}{dt}\tag{15.3.2} \end{equation}
where \(\omega_z = \bOm\cdot\bz\) is the axial angular velocity of \(\cB\text{.}\) Hence vector \(\bz \times \frac{d \bz}{dt}\) represents its transverse angular velocity \(\bOm_\perp\text{.}\) We are generally interested in the effect of large values of \(\omega_z\) on the general motion of \(\cB\text{,}\) that is, \(|\om_z| \gg |\bOm_\perp|\text{.}\)
From (15.2.1) and (15.3.2) we can determine the angular momentum \(\bH_{G} = \bH_{G, \cB / \cE}\) of body \(\cB\) about mass center \(G\text{:}\)
\begin{equation} \bH_{G} = \cI_{G} (\bOm) = A \bz \times \frac{d\bz}{dt} + C \omega_z \bz \tag{15.3.3} \end{equation}
The dynamic moment equation about the mass center \(G\) (Euler’s second principle) gives
\begin{equation} \frac{d}{dt} \bH_G = \bM_G\tag{15.3.4} \end{equation}
where we have denoted by \(\bM_G \equiv \bM_{G , \coB \to \cB}\) the moment of the external actions about \(G\text{.}\) Recall that the axial moment \(\bM_G\cdot \bz\) is assumed to be zero. This implies that
\begin{equation*} \bz \cdot \frac{d}{dt} \bH_{G} = \frac{d}{dt} (\bz \cdot \bH_{G} ) - \bH_{ G } \cdot \frac{d\bz}{dt} = C \frac{d \omega_z}{dt} = 0 \end{equation*}
where we have used (15.3.3) and the fact that \(\bz \cdot \tfrac{d\bz}{dt} = 0\text{.}\) We can conclude with the following result.
We can then simplify the dynamic moment about \(G\text{:}\)
\begin{equation*} \bD_{ G } = \frac{d}{dt} \bH_{G} = C \omega_z \frac{d \bz}{dt} + A \bz \times \frac{d^2 \bz}{dt^2} \end{equation*}
There are two ways of interpreting this equation:
  1. The equation (15.3.5) can be interpreted as the equation that would govern a slender rod directed along \((G, \bz)\) having the same mass center \(G\text{,}\) the same transverse moment of inertia \(A\) (about \(G\)), and subjected to the same external moment \(\bM_G\) as well as the fictitious gyroscopic couple \(\bGa_g = - C \omega_z \tfrac{d \bz}{dt}\text{.}\)

    Proof.

    The angular momentum of a slender rod directed along line \((G, \bz)\) and of transverse moment of inertia \(A\) about its mass center \(G\) is \(A \bz \times (d \bz / dt)\text{,}\) thus leading to (15.3.5) if the rod is subject to external moment \(\bM_G\) and gyroscopic couple \(\bGa_g\) by application of Euler’s second principle.
  2. The equation (15.3.5) can be interpreted as the equation of motion of a fictitious particle \(P\) defined by the position vector \(\br_{OP}= \bz\) (with respect to some arbitrary origin \(O\) of \(\cE\)), of fictitious mass \(A\text{,}\) and constrained to move on the unit sphere of center \(O\) according to the equation
    \begin{equation} A ( \ba_P - (\bz \cdot \ba_P ) \bz ) = \bM_G \times \bz + C \omega_z \, \bz \times \frac{d \bz}{dt} \tag{15.3.6} \end{equation}
    where \(\ba_P - (\bz \cdot \ba_P) \bz\) is the tangential acceleration of \(P\text{.}\) Hence particle \(P\) is subject to the two tangential “forces” \(\bM_G \times \bz\) and \(C \omega_z \bz \times \tfrac{d \bz}{dt}\text{.}\)

    Proof.

    Particle \(P\) is constrained to move on the unit sphere of center \(O\text{:}\) its velocity \(\vel_P = d\bz/dt\) is tangential to the sphere, its acceleration \(\ba_P = d^2 \bz/dt^2\) however has both a component tangential and normal to the sphere. Equation (15.3.6) is then obtained by taking the cross-product of both sides of (15.3.5) with unit vector \(\bz\text{.}\)
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