偏心轮问题

吹逼 物理

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假定偏心轮连同上方重物的质量为$m$，假定偏心轮半径$r$，偏心率$\beta$，两个磁铁的磁矩为$\Pi_1$，$\Pi_2$在$z$轴相聚$z_0$，$x$轴相距$x_0$，现在求该系统能转过$\pi$的条件

解 假定该系统转过$\theta$

根据条件，偏心率为$\beta$意味着转轴到轮心距离为$\beta r$，在转过$θ$后，轮心和重物均上升了$\beta r(1-{\rm Cos}[\theta])$，重力势能增加$W_{\rm g} = mg\beta r(1-{\rm Cos}[\theta])$

磁场能量：$W_{\rm m} = -\vec \Pi\cdot \vec B = -\vec \Pi_1 \cdot \vec B_2$，其中$\vec B_2$为由$\vec \Pi_2$产生的磁场，即
$\vec B_2[\vec r_1] = \frac{\mu_0}{4\pi}\left(\frac{(\vec \Pi_2\cdot \vec r_{21})\vec r_{21}}{r_{21}^5}-\frac{\vec \Pi_2}{r_{21}^3}\right)$，其中$\vec r_{21}=\vec r_1-\vec r_2$为两个磁矩之间的距离。

现计算：
$\vec r_2 = x_0 \vec e_x-z_0 \vec e_z$
$\vec r_1 = -\beta r {\rm Sin}[θ] \vec e_x + \beta r (1-{\rm Cos}[θ])\vec e_z$
$\vec r_{21} = -(\beta r {\rm Sin}[θ]+x_0)\vec e_x+(\beta r (1-{\rm Cos}[θ])+z_0)\vec e_z$
$r_{21} = \left(\left(\beta r {\rm Sin}[θ]+x_0\right)^2+\left(\beta r (1-{\rm Cos}[θ])+z_0\right)^2\right)^{1/2}$
$\vec \Pi_2 = -\Pi_2 \vec e_z$

$$\vec B_2[\vec r_1] = \frac{\mu_0}{4\pi}\left(\frac{-\Pi_2 \left(\beta r (1-{\rm Cos}[θ])+z_0\right)\left(-(\beta r {\rm Sin}[θ]+x_0)\vec e_x+(\beta r (1-{\rm Cos}[θ])+z_0)\vec e_z\right)}{\left(\left(\beta r {\rm Sin}[θ]+x_0\right)^2+\left(\beta r (1-{\rm Cos}[θ])+z_0\right)^2\right)^{5/2}}\\-\frac{-\Pi_2 \vec e_z}{\left(\left(\beta r {\rm Sin}[θ]+x_0\right)^2+\left(\beta r (1-{\rm Cos}[θ])+z_0\right)^2\right)^{3/2}}\right)\\=\frac{\mu_0}{4\pi \left(\left(\beta r {\rm Sin}[θ]+x_0\right)^2+\left(\beta r (1-{\rm Cos}[θ])+z_0\right)^2\right)^{3/2}}\\\left(\frac{-\Pi_2 \left(\beta r (1-{\rm Cos}[θ])+z_0\right)\left(-(\beta r {\rm Sin}[θ]+x_0)\vec e_x+(\beta r (1-{\rm Cos}[θ])+z_0)\vec e_z\right)}{\left(\beta r {\rm Sin}[θ]+x_0\right)^2+\left(\beta r (1-{\rm Cos}[θ])+z_0\right)^2}+\Pi_2 \vec e_z\right)\\=\frac{\mu_0 \Pi_2}{4\pi \left(\left(\beta r {\rm Sin}[θ]+x_0\right)^2+\left(\beta r (1-{\rm Cos}[θ])+z_0\right)^2\right)^{5/2}}\\\left(\left(\beta r (1-{\rm Cos}[θ])+z_0\right)\left(-(\beta r {\rm Sin}[θ]+x_0)\vec e_x+(\beta r (1-{\rm Cos}[θ])+z_0)\vec e_z\right)\\+\left(\left(\beta r {\rm Sin}[θ]+x_0\right)^2+\left(\beta r (1-{\rm Cos}[θ])+z_0\right)^2\right)\vec e_z\right)\\=\frac{\mu_0 \Pi_2}{4\pi \left(\left(\beta r {\rm Sin}[θ]+x_0\right)^2+\left(\beta r (1-{\rm Cos}[θ])+z_0\right)^2\right)^{5/2}}\\\left(-\left(\beta r (1-{\rm Cos}[θ])+z_0\right)(\beta r {\rm Sin}[θ]+x_0)\vec e_x+\left((\beta r {\rm Sin}[θ]+x_0)^2+2(\beta r (1-{\rm Cos}[θ])+z_0)^2\right)\vec e_z\right)$$

$\vec \Pi_1 = -\Pi_1 \left({\rm Sin}[\theta]\vec e_x-{\rm Cos}[\theta]\vec e_z\right)$

$$W_{\rm m}=-\vec \Pi_1 \cdot \vec B_2[\vec r_{21}] = \frac{\mu_0 \Pi_1 \Pi_2}{4\pi \left(\left(\beta r {\rm Sin}[θ]+x_0\right)^2+\left(\beta r (1-{\rm Cos}[θ])+z_0\right)^2\right)^{5/2}}\\\left({\rm Sin}[\theta]\left(\beta r (1-{\rm Cos}[θ])+z_0\right)(\beta r {\rm Sin}[θ]+x_0)+{\rm Cos}[\theta]\left((\beta r {\rm Sin}[θ]+x_0)^2+2(\beta r (1-{\rm Cos}[θ])+z_0)^2\right)\right)$$

为使其可以转至$\pi$ 角度，应该有

$$W_{\rm g}+W_{\rm m}\geq0,\forall \theta\in \mathbb R_{[0,\pi]}$$

即

$$\frac{\mu_0 \Pi_1 \Pi_2}{4\pi \left(\left(\beta r {\rm Sin}[θ]+x_0\right)^2+\left(\beta r (1-{\rm Cos}[θ])+z_0\right)^2\right)^{5/2}}\left({\rm Sin}[\theta]\left(\beta r (1-{\rm Cos}[θ])+z_0\right)(\beta r {\rm Sin}[θ]+x_0)\\+{\rm Cos}[\theta]\left((\beta r {\rm Sin}[θ]+x_0)^2+2(\beta r (1-{\rm Cos}[θ])+z_0)^2\right)\right)+mg\beta r(1-{\rm Cos}[\theta])\geq0$$ $$(\forall \theta\in \mathbb R_{[0,\pi]})$$

通过上式可以解得关于$\Pi_1\Pi_2$的不等式。