前人的工作多假设椭球整体质量均匀(不含气室), 而实际情况, 诸如鸡蛋等, 由于存在气室, 故而质心同形心不重合^[23].

在此, 本文研究偏心距为 \(\rho \) 的椭球体在不光滑水平面上的旋转运动. 本文采用的模型如图1所示.

对称轴\({Gz} (Oz)\)和轴\({GZ}\)确定了\({\Pi}\)平面, 椭球关于\({GZ}\)轴的进动角速度为\(\varOmega \). \(GXYZ\)为地面坐标系, \({Gxyz}\)为体轴系, \(GX\)在\(\Pi\)平面内水平, \({GY}\)垂直于纸面向里, 与\({Gy}\)重合.质心\(G\)相对于水平面的高度为\(h(\theta )\), 物体的质量为\(M\), 作用在物体上的力有重力\({\pmb F}_{g}=(0, 0, -Mg)\), 作用在\(P\)点的支反力\({\pmb N}\)及\(Y\)向的摩擦力\({\pmb F}\), 其中\(F_{x}\)为\({\pmb F}\)在\(X\)轴方向的分量.

图1

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	图1 偏心椭球竖直截面图：其中\(O\)为形心, \(G\)为质心, 椭球体与水平面的接触点为\(P\), \(z\)是极轴, \(Z\)是垂直轴, \(\theta \)为垂直轴\(Z\) 与椭球极轴\(z\)的夹角

在图1所示参考系中, 椭球体的角速度\({\pmb\omega}\)的表达式为^[2]

\[{\pmb \omega } = \left( {\left( {n - \varOmega \cos\theta } \right)\sin \theta , \dot {\theta }, \varOmega \sin ^2\theta + n\cos\theta } \right) \ \ (1)\]\(n(t)\)为物体关于极轴\({Oz}\)的转速.

椭球体关于\(O\)点的对\(X, Y, Z\)轴的三个主惯性矩为\(A\), \(A\), \(C\), 故角动量\({\pmb L}\)为\[{\pmb L} \!=\! \left( {\left( {Cn - A\varOmega \cos \theta } \right)\sin \theta , A\dot {\theta }, A\varOmega\sin ^2\theta + Cn\cos \theta } \right) \ \ (2)\]在坐标系\(GXYZ\)中, \(P\)点的位置坐标表示为^[2]

\[{\pmb R}_{P} = \left( {X_P , Y_P , Z_P } \right) \ \ (3a)\]

\[X_P=\dfrac{d h }{d \theta} \ \ (3b)\]

\[Y_P=0 \ \ (3c)\]

\[Z_P=-h(\theta) \ \ (3d) \]

\(P\)点的速度为\({\pmb V}_{ P} = \left( { - h\dot {\theta }, V_P , 0}\right)\), 当摩擦力很微弱时, \(\dot {\theta }\)也相应很小^[2], 可得\[{\pmb V}_{p}=(0, V_{P}, 0) , \ \ {\pmb F}=(0, F, 0)\]同时, 作用于\(P\)点的支反力\({\pmb N}=(0, 0, N)\), \(N\)的大小取决于\({Mg}\)的值.

则欧拉角动量方程^[24]可写为\[\dfrac{\partial {\pmb L}}{\partial t} +{\pmb \varOmega }\times {\pmb L} = {\pmb R}_{P} \times\left( {{\pmb N} + {\pmb F}} \right) \ \ (4)\]其在\(Y\)方向上的分式为\[A\ddot {\theta } + \varOmega \sin \theta \left( {Cn -A\varOmega \sin \theta } \right) = - NX_P \ \ (5)\]因\(\theta \)的变化较缓慢, 则\(| \ddot \theta | \ll \varOmega^2\)^[2], 且任意稳定状态\(d h/d\theta =0\), 即\(X_{P}=0\), 故式(5)可化为\(\left( {Cn - A\varOmega \cos \theta } \right)\varOmega \sin \theta = 0\)另假定\(\sin\theta \ne 0\), 得\[Cn = A\varOmega \cos \theta \ \ (6)\] \( \theta \to 0\) 时, 式(4)在\(X\)方向和\(Z\)方向的分式分别为\(A\dot {\varOmega } = FX_P \), \(A\varOmega \dot {\theta } = FZ_P \), 据此可得转动惯量J \[{J} = - {\pmb L} \cdot {\pmb R}_{P} = A\varOmega h \ \ (7)\]将式(3d)和式(7)代入\(A\varOmega \dot {\theta } = FZ_P \), 可得\[\dot {\theta } = - Fh^2\left( \theta \right) / J \ \ (8)\]

由标准椭球体\(A = M\left( {a^2 + b^2} \right) / 5\), \(C = 2Mb^2 /5\), 借助平行移轴定理, 可得偏心椭球的主惯性矩为

\[A = \dfrac{M\left( {a^2 + b^2} \right)}{5} + \pi ab\rho ^2 , \ C = \dfrac{2Mb^2}{5}\]

又

\[ V_P = \left( {\varOmega \sin ^2\theta + n\cos \theta } \right)d h / d\theta + \\ \qquad \left( {n- \varOmega \cos \theta } \right)h\left( \theta \right)\sin \theta \]

借助式(6)和式(7)以及定义式

\[\varsigma \left( \theta \right) \equiv\left( {\sin ^2\theta + \dfrac{A}{C}\cos ^2\theta } \right)^{ -\tfrac{1}{2}}\]

可化简得

\[V_P = \left( {J / A} \right)\varsigma ^{ - 3}h^{ - 1}d\left( {\varsigma h}\right) / d\theta \ \ (9)\]

假定\(F\)满足黏性摩擦定律, 则\(F = - \mu MgV_P \), 与式(8)和式(9)联立, 可得

\[ \dot {\theta } = \dfrac{5\mu Mgh \sin 2\theta }{4M\left( {a^2 + b^2}\right) + 20\pi ab\rho ^2} \cdot \]

\[\qquad \left\{ {\left[ {2 + \cos ^2\theta \left( {\dfrac{a^2 - b^2}{b^2} + \dfrac{5\rho \pi a}{Mb}}\right)} \right]} \right.\cdot \]

\[\qquad \left( {\dfrac{b^2 - a^2}{h + \rho \cos \theta } + \dfrac{\rho }{\cos \theta }} \right) + \]

\[\qquad \left.{h\left( {\dfrac{a^2 - b^2}{b^2} + \dfrac{5\rho \pi a}{Mb}} \right)} \right\}\ \ (10)\]

考虑到此时^[25]

\[h\left( \theta \right) = \left( {a^2\cos ^2\theta + b^2\sin ^2\theta }\right)^{\tfrac{1}{2}} - \rho \cos \theta\]

则

\[h^2 = \left( {a^2 - b^2} \right)\cos ^2\theta + b^2\sin ^2\theta - 2\rho h\cos \theta \ \ (11)\]

当\(\theta \to 0\)时, \(\cos \theta \approx 1\), \(h(\theta ) \approx a-\rho \), 同式(11)一并代入式(10), 得

\[\dot {\theta } = r\sin 2\theta + s\sin 4\theta \ \ (12)\]

其中

\[ r = \dfrac{5\mu Mg}{4M\left( {a^2 + b^2} \right) + 20\pi ab\rho ^2}\left\{ {\left({\dfrac{b^2 - a^2}{a} + \rho } \right)}\right.\cdot \]

\[ \qquad\left( {a - \rho } \right)\left( {\dfrac{a^2 + 3b^2}{2b^2} + \dfrac{5\rho \pi a}{2Mb}} \right) + \]

\[ \qquad \left. { \left( {\dfrac{a^2 - b^2}{b^2} + \dfrac{5\rho \pi a}{Mb}}\right)\left[ {\dfrac{a^2 - \rho ^2 + b^2}{2} - 2\rho \left( {a - \rho }\right)} \right]} \right\} \]

\[ s = \dfrac{5\mu Mg}{4M\left( {a^2 + b^2} \right) + 20\pi ab\rho ^2}\left[ {\dfrac{a - \rho}{4}\left( {\dfrac{b^2 - a^2}{a} + \rho } \right)\cdot }\right.\]

\[\qquad \left( {\dfrac{a^2 - b^2}{b^2} + \dfrac{5\rho \pi a}{Mb}} \right) + \dfrac{a^2 -\rho ^2 - b^2}{4}\cdot \]

\[\qquad \left.{\left( {\dfrac{a^2 - b^2}{b^2} + \dfrac{5\rho \pi a}{Mb}} \right)} \right] \]

通过解式(12)得

\[ \dfrac{s}{\left( {r + 2s} \right)\left( {r - 2s} \right)}\ln \left| {\left({r - 2s} \right)\tan ^2\theta + }\right. \\ \qquad \left.{ \left( {r + 2s} \right)} \right| + \dfrac{1}{2\left( {r + 2s} \right)}\ln \left| {\tan \theta } \right| = t -t_0 \ \ (13)\]

\(t_{0}\)为一常量. 特别地, 偏心距\(\rho =0\)情况下, 由式(13)可得与Moffatt一致的研究结论^[2]

\[\tan \theta = {\rm e}^{ - \mu q\left( {t - t_0 } \right)} , \ \ q = \dfrac{5g\left( {a^2 - b^2} \right)}{2\left( {a^2 + b^2} \right)}\]