Question 12.4: Determine the ellipsoid of inertia for the rotation of a qua...
Determine the ellipsoid of inertia for the rotation of a quadratic disk about the origin, as described in Example 12.1. Find the moments of inertia of the disk for rotation about (a) the x-axis, (b) the y-axis, (c) the z-axis, (d) the three principal axes, and (e) the axis {cos 45°, cos 45° , cos 45°}.
The ellipsoid of inertia reads
\frac{Θ_{0}}{3} \varrho^{2}_{x}− \frac{Θ_{0}}{2} \varrho_{x} \varrho_{y} + \frac{Θ_{0}}{3} \varrho^{2}_{y} + \frac{2Θ_{0}}{3} \varrho^{2}_{z} = 1. (12.46)
(a) For rotation about the x-axis n = {1, 0, 0}, and thus \varrho= \left\{1/\sqrt{Θ_{x}} , 0, 0\right\}. Insertion into (12.46) yields
\frac{Θ_{0}}{3} · \frac{1}{Θ_{x}}= 1 ⇒ Θ_{x} = \frac{Θ_{0}}{3},as expected.
(b) Here, n = {0, 1, 0}, and following the procedure in (a), we find
Θ_{y}= \frac{Θ_{0}}{3}.(c) Here, n = {0, 0, 1}, and following the procedure in (a), we find
Θ_{z} = \frac{2}{3} Θ_{0}.
(d) The third principal axis is identical with the z-axis, which corresponds to (c). The first two principal axes are given by
n_{1} =\left\{\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}, 0\right\} and n_{2} =\left\{-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}, 0\right\},respectively. Therefore,
\varrho_{1} = \frac{n} {\sqrt{Θ_{1}}} = \left\{ \frac{1}{\sqrt{2Θ_{1}}},\frac{1}{\sqrt{2Θ_{1}}}, 0\right\}and
\varrho_{2} = \frac{n_{2}} {\sqrt{Θ_{2}}} = \left\{ -\frac{1}{\sqrt{2Θ_{2}}},\frac{1}{\sqrt{2Θ_{2}}}, 0\right\}Insertion into (12.46) yields
\frac{Θ_{0}}{3} \frac{1}{2Θ_{1}}− \frac{Θ_{0}}{2} \frac{1}{2Θ_{1}}+ \frac{Θ_{0}}{3} \frac{1}{2Θ_{1}}+0 = 1 ⇒ Θ_{1} = \frac{Θ_{0}}{12},
and
\frac{Θ_{0}}{3} \frac{1}{2Θ_{2}}+ \frac{Θ_{0}}{2} \frac{1}{2Θ_{2}}+ \frac{Θ_{0}}{3} \frac{1}{2Θ_{2}}+0 = 1 ⇒ Θ_{2} = \frac{7}{12}Θ_{0}.
These are the principal moments of inertia, as was expected.
(e) In this case, n is proportional to {cos 45°, cos 45°, cos 45°}. Thus,
n =\left\{\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right\},
and therefore,
\varrho_{1} = \frac{n} {\sqrt{Θ}}=\left\{\frac{1}{\sqrt{3Θ}},\frac{1}{\sqrt{3Θ}},\frac{1}{\sqrt{3Θ}}\right\}Insertion into (12.46) yields
\frac{Θ_{0}}{3} \frac{1}{3Θ}− \frac{Θ_{0}}{2} \frac{1}{3Θ}+ \frac{Θ_{0}}{3} \frac{1}{3Θ}+\frac{2}{3} Θ_{0} \frac{1}{3Θ}= 1,,
from which we find
Θ = \frac{10}{36} Θ_{0}.This problem demonstrates the simple handling and the usefulness of the ellipsoid of inertia.