Question 13.7: What torque is needed to rotate an elliptic disk with the ha...

We choose the $e_{1}-axis$ orthogonal to the plane of the drawing, $e_{2}$ along the small half-axis b, and $e_{3}$ along the large half-axis. The principal moments of inertia are then $(M = σπab, dM = σdF)$

from the ellipse equation $z^{2}/a^{2} +y^{2}/b^{2} = 1$.

$= \frac{1}{4} σba^{3}π = \frac{1}{4}Ma^{2};$                                      (13.21)

accordingly,

$⇒ I_{3} = \frac{1}{4} Mb^{2}.$                                      (13.22)

We can immediately write down $I_{1}$ (because $I_{1} = I_{2} +I_{3}$ for thin plates):

$I_{1} =\frac{1}{4}M(a^{2} + b^{2}).$                                 (13.23)

For ω, we obtain

We insert into the Euler equations of the top:

$=−\frac{1}{4} M(b^{2} −a^{2}) sin  α   cos  α · ω^{2}_{0},$                                                    (13.24)

Thus, we obtain for the desired torque D

$D=−ω^{2}_{0}· \frac{M}{8} (b^{2} −a^{2}) sin  2α · e_{1}.$                                               (13.25)

It is obvious that

(1) for α = 0,π/2,π, . . . , the torque vanishes, since the rotation is performed about a principal axis of inertia; and

(2) for $b^{2} = a^{2}$, i.e., the case of a circular disk, the torque vanishes for all angles α.

We will consider once again the last conclusions: Given an elliptic disk with half-axes a and b, for α = 0°, 180°, or for α = 90°, 270°, the rotation axis coincides with one of the principal axes of inertia along the half-axes. In this case the orientation of the angular momentum is identical with the momentary rotation axis.  Because $ω_{0}$ = constant, we also have L = constant, and therefore the resulting torque vanishes. This also results by insertion into the Euler equations of the top: $\dot{ω}= (0, 0, 0), ω = (0, 0,ω_{0})   or   ω = (0,ω_{0}, 0)$

$D_{3} = I_{3}\dot{ω}_{3} + (I_{2} −I_{1})ω_{1}ω_{2} = 0.$                                                                         (13.26)