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非对称陀螺:详细的理论分析||刚体力学

2021-08-12 21:21 作者:湮灭的末影狐  | 我要投稿

//上次我们使用Mathematica数值模拟了自由刚体的转动过程。

//这里我们介绍相对完整的理论分析。

之前已经提过,对于固连在刚体上的Oxyz坐标系,角速度满足的微分方程组:

%5Cleft%5C%7B%5Cbegin%7Baligned%7D%0A%20%20%20%20I_%7B1%7D%5Cdot%5Comega_x%2B%20(I_%7B3%7D-I_%7B2%7D)%5Comega_y%5Comega_z%3D0%5C%5C%0A%20%20%20%20I_%7B2%7D%5Cdot%5Comega_y%2B%20(I_%7B1%7D-I_%7B3%7D)%5Comega_z%5Comega_x%3D0%5C%5C%0A%20%20%20%20I_%7B3%7D%5Cdot%5Comega_z%2B%20(I_%7B2%7D-I_%7B1%7D)%5Comega_x%5Comega_y%3D0%0A%5Cend%7Baligned%7D%5Cright.

结合欧拉运动学方程

%5Cbegin%7Baligned%7D%0A%5Comega_%7Bx%7D%20%26%3D%5Cdot%7B%5Cphi%7D%20%5Csin%20%5Ctheta%20%5Csin%20%5Cpsi%2B%5Cdot%7B%5Ctheta%7D%20%5Ccos%20%5Cpsi%20%5C%5C%0A%5Comega_%7By%7D%20%26%3D%5Cdot%7B%5Cphi%7D%20%5Csin%20%5Ctheta%20%5Ccos%20%5Cpsi-%5Cdot%7B%5Ctheta%7D%20%5Csin%20%5Cpsi%20%5C%5C%0A%5Comega_%7Bz%7D%20%26%3D%5Cdot%7B%5Cphi%7D%20%5Ccos%20%5Ctheta%2B%5Cdot%7B%5Cpsi%7D%0A%5Cend%7Baligned%7D

原则上可解非对称陀螺的运动。但是直接暴力求解相当困难,这里我们可以首先写出角动量和能量积分。总能量守恒:

I_1%5Comega_x%5E2%2BI_2%5Comega_y%5E2%2BI_3%5Comega_z%5E2%3D2E

虽然在固连在刚体的Oxyz系,角动量方向会变化,但模长仍然不变:

L_x%5E2%2BL_y%5E2%2BL_z%5E2%20%3D%20L

可以把动能也用角动量分量表示:

%5Cfrac%7BL_x%5E2%7D%7BI_1%7D%2B%5Cfrac%7BL_y%5E2%7D%7BI_2%7D%2B%5Cfrac%7BL_z%5E2%7D%7BI_3%7D%3D2E

为了进一步推进,我们引入相空间:坐标轴为角动量的三个分量,与真实空间没有关系。相空间的每个点表示刚体的角动量矢量的一个值。

我们注意到,角动量积分在相空间是一个半径L的球面,而能量积分则得到一个椭球面。球面和椭球面的交线就是刚体的整个运动过程中,角动量必须处在的相空间中的曲线。不同的曲线代表不同初值下刚体的运动。

我们不妨改变球面半径(这相当于改变初值),得到椭球面和不同球面的交线族:

相空间中的解曲线族

不难根据目前的结果做出一些推断:

结论1 非对称陀螺的运动总是周期性的,这是因为前面得到的交线总是封闭曲线。

结论2 (Dzhanibekov定理)非对称陀螺绕转动惯量最大、最小的轴转动在微扰下稳定;否则不稳定。从上图就可以看出,曲线族在其中两个主轴附近均能形成闭合曲线,这意味着当初始角动量在这两个轴附近时,整个运动过程角动量都只会指向轴附近,体现为稳定转动;而在剩下一个轴附近所有曲线都远离这个轴,所以任意微扰都会导致Dzhanibekov效应。

继续,我们尝试对方程进行求解。直接从能量和角动量积分中暴力消去其中两个角速度:

%5Cbegin%7Baligned%7D%0A%5Comega_%7B1%7D%5E%7B2%7D%20%26%3D%5Cfrac%7B%5Cleft%5B%5Cleft(2%20E%20I_%7B3%7D-L%5E%7B2%7D%5Cright)-I_%7B2%7D%5Cleft(I_%7B3%7D-I_%7B2%7D%5Cright)%20%5Comega_%7B2%7D%5E%7B2%7D%5Cright%5D%7D%7BI_%7B1%7D%5Cleft(I_%7B3%7D-I_%7B1%7D%5Cright)%7D%20%5C%5C%0A%5Comega_%7B3%7D%5E%7B2%7D%20%26%3D%5Cfrac%7B%5Cleft%5B%5Cleft(L%5E%7B2%7D-2%20E%20I_%7B1%7D%5Cright)-I_%7B2%7D%5Cleft(I_%7B2%7D-I_%7B1%7D%5Cright)%20%5Comega_%7B2%7D%5E%7B2%7D%5Cright%5D%7D%7BI_%7B3%7D%5Cleft(I_%7B3%7D-I_%7B1%7D%5Cright)%7D%0A%5Cend%7Baligned%7D

然后代入I_%7B2%7D%20%5Cdot%7B%5Comega%7D_%7B2%7D%2B%5Cleft(I_%7B1%7D-I_%7B3%7D%5Cright)%20%5Comega_%7B3%7D%20%5Comega_%7B1%7D%3D0

得到...

%5Csqrt%7BI_%7B1%7D%20I_%7B3%7D%7D%20I_%7B2%7D%20%5Cfrac%7Bd%20%5Comega_%7B2%7D%7D%7Bd%20t%7D%3D%5Csqrt%7B%5Cleft%5B%5Cleft(2%20E%20I_%7B3%7D-L%5E%7B2%7D%5Cright)-I_%7B2%7D%5Cleft(I_%7B3%7D-I_%7B2%7D%5Cright)%20%5Comega_%7B2%7D%5E%7B2%7D%5Cright%5D%5Cleft%5B%5Cleft(L%5E%7B2%7D-2%20E%20I_%7B1%7D%5Cright)-I_%7B2%7D%5Cleft(I_%7B2%7D-I_%7B1%7D%5Cright)%20%5Comega_%7B2%7D%5E%7B2%7D%5Cright%5D%7D

然后,分离变量,积罢!

当然,这是一个椭圆积分...还是稍微处理一下吧,不妨设L%5E2%3E2EI_2. 定义几个量如下:

%5Cbegin%7Baligned%7D%0A%5Ctau%20%26%3Dt%20%5Csqrt%7B%5Cfrac%7B%5Cleft(I_%7B3%7D-I_%7B2%7D%5Cright)%5Cleft(L%5E%7B2%7D-2%20E%20I_%7B1%7D%5Cright)%7D%7BI_%7B1%7D%20I_%7B2%7D%20I_%7B3%7D%7D%7D%20%5C%5C%0As%20%26%3D%5Comega_%7B2%7D%20%5Csqrt%7B%5Cfrac%7BI_%7B2%7D%5Cleft(I_%7B3%7D-I_%7B2%7D%5Cright)%7D%7B2%20E%20I_%7B3%7D-L%5E%7B2%7D%7D%7D%20%5C%5C%0Ak%5E%7B2%7D%20%26%3D%5Cfrac%7B%5Cleft(I_%7B2%7D-I_%7B1%7D%5Cright)%5Cleft(2%20E%20I_%7B3%7D-L%5E%7B2%7D%5Cright)%7D%7B%5Cleft(I_%7B3%7D-I_%7B2%7D%5Cright)%5Cleft(L%5E%7B2%7D-2%20E%20I_%7B1%7D%5Cright)%7D%0A%5Cend%7Baligned%7D

然后,就能得到...

%5Ctau%3D%5Cint_%7B0%7D%5E%7Bs%7D%20%5Cfrac%7Bd%20s%7D%7B%5Csqrt%7B%5Cleft(1-s%5E%7B2%7D%5Cright)%5Cleft(1-k%5E%7B2%7D%20s%5E%7B2%7D%5Cright)%7D%7D

这是椭圆积分的标准形式,得到雅可比椭圆函数:

s%20%3D%20%7B%5Crm%20sn%7D%20%5Ctau

再然后,就可以写出

%5Cbegin%7Baligned%7D%0A%26%5Comega_%7B1%7D(%5Ctau)%3D%5Csqrt%7B%5Cfrac%7B%5Cleft(2%20E%20I_%7B3%7D-L%5E%7B2%7D%5Cright)%7D%7BI_%7B1%7D%5Cleft(I_%7B3%7D-I_%7B1%7D%5Cright)%7D%7D%20%5Cmathrm%7Bcn%7D%20%5Ctau%20%5C%5C%0A%26%5Comega_%7B2%7D(%5Ctau)%3D%5Csqrt%7B%5Cfrac%7B%5Cleft(2%20E%20I_%7B3%7D-L%5E%7B2%7D%5Cright)%7D%7BI_%7B2%7D%5Cleft(I_%7B3%7D-I_%7B2%7D%5Cright)%7D%7D%20%5Coperatorname%7Bsn%7D%20%5Ctau%20%5C%5C%0A%26%5Comega_%7B3%7D(%5Ctau)%3D%5Csqrt%7B%5Cfrac%7B%5Cleft(L%5E%7B2%7D-2%20E%20I_%7B1%7D%5Cright)%7D%7BI_%7B3%7D%5Cleft(I_%7B3%7D-I_%7B1%7D%5Cright)%7D%7D%20%5Cmathrm%7Bdn%7D%20%5Ctau%0A%5Cend%7Baligned%7D

其中,%7B%5Crm%20cn%7D%20%5Ctau%20%3D%20%5Csqrt%7B1-%20%7B%5Crm%20sn%7D%5E2%20%5Ctau%7D%2C%5C%3B%20%7B%5Crm%20dn%20%7D%5Ctau%20%3D%20%5Csqrt%7B1-%20k%5E2%7B%5Crm%20sn%7D%5E2%20%5Ctau%7D.

前面已经定义了tau,所以就解完了。

以及,非对称陀螺运动的周期是

T%3D4%20K%20%5Csqrt%7B%5Cfrac%7BI_%7B1%7D%20I_%7B2%7D%20I_%7B3%7D%7D%7B%5Cleft(I_%7B3%7D-I_%7B2%7D%5Cright)%5Cleft(L%5E%7B2%7D-2%20E%20I_%7B1%7D%5Cright)%7D%7D

其中,

K%3D%5Cint_%7B0%7D%5E%7B1%7D%20%5Cfrac%7Bd%20s%7D%7B%5Csqrt%7B%5Cleft(1-s%5E%7B2%7D%5Cright)%5Cleft(1-k%5E%7B2%7D%20s%5E%7B2%7D%5Cright)%7D%7D

所以,解这玩意还真麻烦...

后面该进入分析力学了...点个关注即可持续获取最新白给级物理笔记!

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