Question 13.7: What torque is needed to rotate an elliptic disk with the ha...
What torque is needed to rotate an elliptic disk with the half-axes a and b about the rotation axis 0A with constant angular velocity ω_{0}? The rotation axis is tilted from the large half-axis a by the angle α.
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)
I_{2} = σ \int\limits_{−a}^{+a}{ \int\limits_{−\varphi(z)}^{\varphi(z)}{z^{2} dz dy}} with \varphi(z) = b \sqrt{1 − \frac{z^{2}}{a^{2}}}
from the ellipse equation z^{2}/a^{2} +y^{2}/b^{2} = 1.
I_{2} = σ \int\limits_{−a}^{+a}{z^{2}y|^{b \sqrt{1−z^{2}/a^{2}}}_{−b\sqrt{1−z^{2}/a^{2}}}dz} = 2σb \int\limits_{−a}^{+a}{z^{2} \sqrt{1 − \frac{z^{2}}{a^{2}}} dz}
= 2σ \frac{b}{a} \left\{\frac{z}{8} (2z^{2} −a^{2})\sqrt{a^{2} −z^{2}} + \frac{a^{4}}{8}arcsin \frac{z}{|a|}\right\}|^{+a}_{-a}
= \frac{1}{4} σba^{3}π = \frac{1}{4}Ma^{2}; (13.21)
accordingly,
I_{3} = σ \int\limits_{−b}^{+b}{ \int\limits_{−\overline{\varphi}(y)}^{\overline{\varphi}(y)}{y^{2} dy dz}} with \overline{\varphi}(y) = a \sqrt{1 − \frac{y^{2}}{b^{2}}}
⇒ 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
ω = 0 · e_{1} −ω_{0} sin α · e_{2} + ω_{0} cos α · e_{3}.We insert into the Euler equations of the top:
D_{1} = I_{1} \dot{ω}_{1} + (I_{3} − I_{2})ω_{2}ω_{3}
=−\frac{1}{4} M(b^{2} −a^{2}) sin α cos α · ω^{2}_{0}, (13.24)
D_{2} = I_{2} \dot{ω}_{2} + (I_{1} − I_{3})ω_{1}ω_{3}=0,
D_{3} = I_{3} \dot{ω}_{3} + (I_{2} − I_{1})ω_{1}ω_{2}=0.
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_{1} = I_{1}\dot{ω}_{1} + (I_{3} −I_{2})ω_{2}ω_{3} = 0,D_{2} = I_{2}\dot{ω}_{2} + (I_{1} −I_{3})ω_{1}ω_{3} = 0,
D_{3} = I_{3}\dot{ω}_{3} + (I_{2} −I_{1})ω_{1}ω_{2} = 0. (13.26)