10.5: Example: Spinning Book

Consider a homogeneous rectangular block with \(I_{xx} < I_{yy} < I_{zz}\) and all off-diagonal moments of inertia are zero. The linearized angular momentum equations, with no external forces or moments, are

\begin{align} I_{xx} \dfrac{dp}{dt} + (I_{zz} - I_{yy})rq \, &= \, 0 \\[4pt] I_{yy} \dfrac{dq}{dt} + (I_{xx} - I_{zz})pr \, &= \, 0 \\[4pt] I_{zz} \dfrac{dr}{dt} + (I_{yy} - I_{xx})qp \, &= \, 0. \end{align}

We consider in turn the stability of rotations about each of the main axes, with constant angular rate \(\Omega\). The interesting result is that rotations about the \(x\) and \(z\) axes are stable, while rotation about the \(y\) axis is not. This is easily demonstrated experimentally with a book or a tennis racket.

10.5.1: \(x\)-axis

In the case of the \(x\)-axis, \(p = \Omega + \delta p\), \(q = \delta q\), and \(r = \delta r\), where the \(\delta\) prefix indicates a small value compared to \(\Omega\). The first equation above is uncoupled from the others, and indicates no change in \(\delta p\), since the small term \(\delta q \delta r\) can be ignored. Differentiate the second equation to obtain

\[ I_{yy} \dfrac{\partial^2 \delta q}{\partial t^2} + (I_{xx} - I_{zz}) \Omega \dfrac{\partial \delta r}{\partial t} \, = \, 0. \]

Substitution of this result into the third equation yields

\[ I_{yy} I_{zz} \dfrac{\partial^2 \delta q}{\partial t^2} + (I_{xx} - I_{zz})(I_{xx} - I_{yy}) \Omega^2 \delta q \, = \, 0. \]

A similar expression is \(\delta \ddot{q} + \alpha \delta q = 0\), which has the response \(\delta q (t) = \delta q (0) e^{\sqrt{- \alpha} t}\) when \(\delta \dot{q}(0) = 0\). For spin about the \(x\)-axis, both coefficients of the differential equation are positive, and hence \(\alpha > 0\). The imaginary exponent indicates that the solution is of the form \(\delta q(t) = \delta q(0) \cos \sqrt{\alpha} t\), that is, it oscillates but does not grow. Since the perturbation \(\delta r\) is coupled, it too oscillates.

10.5.2: \(y\)-axis

Now suppose \(q = \Omega + \delta q\): differentiate the first equation and substitute into the third equation to obtain \[ I_{zz} I_{xx} \dfrac{\partial^2 \delta p}{\partial t^2} + (I_{yy} - I_{xx})(I_{yy} - I_{zz}) \Omega^2 \delta p = 0. \]

Here the second coefficient has negative sign, and therefore \(\alpha < 0\). The exponent is real now, and the solution grows without bound, following \(\delta p(t) = \delta p(0) e^{-\sqrt{\alpha}t}\).

10.5.3: \(z\)-axis

Finally, let \(r = \Omega + \delta r\): differentiate the first equation and substitute into the second equation to obtain \[ I_{yy} I_{xx} \dfrac{\partial^2 \delta p}{\partial t^2} + (I_{xx} - I_{zz})(I_{yy} - I_{zz}) \Omega^2 \delta p = 0.\]

The coefficients are positive, so bounded oscillations occur.