2. The model system
The model is a four-particle system, shown in figures 1 and 2. It consists of an axle of length \(L\). About the end points two perpendicular rods can rotate. These have particles attached at the end points, one heavy of mass \(M\) and one light of mass \(m\). Both rods have their centres of mass on the axle, so if the distance to \(m\) from the axle is \(\rho\), the distance to \(M\) is \(\mu\rho\) with \(\mu=m/M\).
We will assume that the relative angle of the two rods and the length of the rods can be changed by means of internal torques and forces, while there is no external torque acting on the system. The centres of mass of the rods will then remain on the axle. The only external forces are the weights of the particles and these can be represented by a resultant acting downwards at the common centre of mass \(G\). In the freely falling reference frame there are no external forces acting.
We describe the system using the angles \(\varphi_1\) and \(\varphi_2\) shown in figure 2. The distances of the lighter masses to the axis through the centres of mass are denoted \(\rho_1\) and \(\rho_2\) respectively. There are thus four degrees-of-freedom and the corresponding coordinates are \(\varphi_1,\varphi_2,\rho_1,\rho_2\).
2.1. Model angular momentum
We now introduce cylindrical (polar) coordinates \(\rho,\varphi,\) and \(z\) and corresponding unit vectors \(\mathbf e_{\rho_k}=\cos\varphi_k\mathbf e_x+\sin\varphi_k\mathbf e_y,\mathbf e_{\varphi_k}=-\sin\varphi_k\mathbf e_x+\cos\varphi_k\mathbf e_y\). In terms of these we have for the position vectors \(\boldsymbol r_k=\rho_k\mathbf e_{\rho_k}+z_k\mathbf e_z\) and for the velocities \(\dot{\boldsymbol r}_k=\dot\rho_k\mathbf e_{\rho_k}+\rho_k\dot\varphi_k\mathbf e_{\varphi_k}+\dot z_k\mathbf e_z\). The position vectors of the four particles of the model system, relative to the centre of mass \(G\), are
\[\begin{align} \boldsymbol r_1&=\rho_1\mathbf e_{\rho_1}-\left(\frac L2\right)\mathbf e_z, \\ \boldsymbol r_2&=-\mu\rho_1\mathbf e_{\rho_1}-\left(\frac L2\right)\mathbf e_z, \\ \boldsymbol r_3&=\rho_2\mathbf e_{\rho_2}+\left(\frac L2\right)\mathbf e_z, \\ \boldsymbol r_4&=-\mu\rho_2\mathbf e_{\rho_2}+\left(\frac L2\right)\mathbf e_z. \end{align} \]Here \(L\) is the distance between the centres of mass of the rods and \(\mu=m/M\) ensures that the centres of mass are on the \(z\)-axis. The velocities are then
\[\begin{align} \boldsymbol v_1&=\dot\rho_1\mathbf e_{\rho_1}+\rho_1\dot\varphi_1\mathbf e_{\varphi_1}, \\ \boldsymbol v_2&=-\mu\dot\rho_1\mathbf e_{\rho_1}-\mu\rho_1\dot\varphi_1\mathbf e_{\varphi_1}, \\ \boldsymbol v_3&=\dot\rho_2\mathbf e_{\rho_2}+\rho_2\dot\varphi_2\mathbf e_{\varphi_2}, \\ \boldsymbol v_4&=-\mu\dot\rho_2\mathbf e_{\rho_2}-\mu\rho_2\dot\varphi_2\mathbf e_{\rho_2}. \end{align} \]The angular momentum of the system with respect to the centre of mass has four contributions and these are
\[\begin{equation}\boldsymbol L_G=m\boldsymbol r_1\times\boldsymbol v_1+M\boldsymbol r_2\times\boldsymbol v_2+m\boldsymbol r_3\times\boldsymbol v_3+M\boldsymbol r_4\times\boldsymbol v_4.\end{equation} \]Straightforward calculation simply gives
\[\begin{equation}\boldsymbol L_G=m\left(1+\mu\right)\left(\rho_1^2\dot\varphi_1+\rho_2^2\dot\varphi_2\right)\mathbf e_z,\end{equation} \]since all other terms cancel.
2.2. Two new variables
Assume now that the there is no external torque and that angular momentum is zero initially. We then have the simple relationship
\[\begin{equation}0=\rho_1^2\dot\varphi_1+\rho_2^2\dot\varphi_2.\end{equation} \]Let us define the mean angular velocity \(\dot\varphi\) and relative angular velocity \(\dot\psi\) through
\[\begin{align} &\begin{array}{ll} \displaystyle\dot\varphi=\frac12(\dot\varphi_1+\dot\varphi_2),&\displaystyle\dot\psi=\dot\varphi_2-\dot\varphi_1, \end{array} \\ &\begin{array}{ll} \displaystyle\dot\varphi_1=\dot\varphi-\frac12\dot\psi,&\displaystyle\dot\varphi_2=\dot\varphi+\frac12\dot\psi. \end{array} \end{align} \]The zero angular momentum equation \((11)\) then gives
\[\begin{equation}0=\left(\rho_1^2+\rho_2^2\right)\dot\varphi-\frac12\left(\rho_1^2-\rho_2^2\right)\dot\psi.\end{equation} \]From this equation we see that if \(\dot\psi=0\), so is \(\dot\varphi\). It is also clear that if \(\rho_1=\rho_2\) then \(\dot\varphi=0\). There will thus only be a mean angular velocity if both \(\rho_1\ne\rho_2\) and \(\dot\psi\ne0\).
2.3. Differential equation of rotation
Equation \((14)\) gives the mean angular velocity of our model system
\[\begin{equation}\dot\varphi=\frac12\frac{\rho_1^2-\rho_2^2}{\rho_1^2+\rho_2^2}\dot\psi.\end{equation} \]We put
\[\begin{equation}f(t)\equiv\frac12\frac{\rho_1^2(t)-\rho_2^2(t)}{\rho_1^2(t)+\rho_2^2(t)},\end{equation} \]since only this combination of length changes will affect the mean angular velocity. This gives us the differential equation
\[\begin{equation}\frac{\mathrm d\varphi}{\mathrm dt}=f(t)\dot\psi(t).\end{equation} \]We now wish to find functions \(f(t),\psi(t)\) that are such that after some time \(T\), the duration of the fall, the mean angle \(\varphi\) has changed by \(\pi~(180^\circ)\),
\[\begin{equation}\varphi(T)-\varphi(0)=\int_0^Tf(t)\dot\psi(t)\mathrm dt=\pi.\end{equation} \]To model the cat the functions \(f(t),\psi(t)\) and \(\dot\psi(t)\) should all start and end with zero values
\[\begin{equation} \begin{array}{lll} \displaystyle f(0)=f(T)=0,&\psi(0)=\psi(T)=0,&\dot\psi(0)=\dot\psi(T)=0. \end{array} \end{equation} \]This corresponds to the cat starting and ending its fall with equally long front and hind legs, which are parallel and non-rotating. To model the cat the relative angle \(\psi(t)\) should change from zero to some maximum value and back to zero since this is the what photos and videos indicate. This means that the function \(\dot\psi(t)\) changes sign once. For the system to keep rotating in the same direction the function \(f(t)\) must therefore change sign at the same time.
2.4. Two solutions
The simplest, and probably optimal, solution of this kind is given by choosing piecewise constant functions \(f(t)\) and \(\dot\psi(t)\), with \(f(t)\) discontinuous. It is given by,
\[\begin{equation} \begin{matrix} t=0&f=0&\rho_1=\rho_2&\psi=0&\dot\psi=0&\dot\varphi=0 \\[1ex] 0<t<T/2&f=1/2&\rho_2=0&\psi=\omega t&\dot\psi=\omega&\dot\varphi=\omega/2 \\[1ex] t=T/2&f=0&\rho_1=\rho_2&\psi=\omega T/2&\dot\psi=0&\dot\varphi=0 \\[1ex] T/2<t<T&f=-1/2&\rho_1=0&\psi=\omega(T-t)&\dot\psi=-\omega&\dot\varphi=\omega/2 \\[1ex] t=T&f=0&\rho_1=\rho_2&\psi=0&\dot\psi=0&\dot\varphi=0. \end{matrix} \end{equation} \]The total rotation angle \(\varphi(T)\) will be \(\pi\) if \(\omega=2\pi/T\). The maximum relative (twist) angle is then \(\psi(T/2)=\pi\). The solution is illustrated to the left in figure 3 and in figure 4. There are, obviously, some problems with this solution, also apart from requiring infinite accelerations and neglecting the time taken to extend and retract the rods (legs). It is a bit dubious to put one of the lengths \(\rho_k\) equal to zero since then the relative angle becomes undefined. This, however, can be fixed rather easily by putting the shorter length to \(1/10\) times the longer, instead of putting it to zero. The values of \(f(t)\) will then only change by one per cent, from those given in \((20)\).
A smooth and differentiable solution can be achieved, e.g. by choosing
\[\begin{equation} \begin{array}{ll} \displaystyle\psi(t)=\Psi\frac{16}{T^4}[t(T-t)]^2,&\displaystyle f(t)=\frac{3\sqrt3}{T^3}t(T-t)(T-2t). \end{array} \end{equation} \]Physically one must have \(-1/2\le f(t)\le1/2\) and this choice for \(f(t)\) obeys this. Here \(\Psi=\psi(T/2)\) is the maximum relative angle, a so far unknown parameter. The angular velocity is then
\[\begin{equation}\dot\psi(t)=\frac{32\Psi}{T^4}t(T-t)(T-2t)\end{equation} \]and the mean angular velocity \((17)\) is
\[\begin{equation}\frac{\mathrm d\varphi}{\mathrm dt}=\frac{96\sqrt3\Psi}{T^7}[t(T-t)(T-2t)]^2.\end{equation} \]The rotation angle is then
\[\begin{equation}\varphi(T)-\varphi(0)=\int_0^T\dot\varphi(t)\mathrm dt=\frac{16\sqrt3}{35}\Psi.\end{equation} \]This will be the desired angle \(\pi\) if the maximum relative angle is
\[\begin{equation}\Psi=\frac{35}{16\sqrt3}\pi\approx1.263\pi.\end{equation} \]We note that this value, approximately \(227^\circ\), is substantially larger than for the solution in \((20)\). The explicit solution is thus
\[\begin{equation}\varphi(t)=\frac\pi{T^7}t^3\left(120t^4-420Tt^3+546T^2t^2-315T^3t+70T^4\right).\end{equation} \]The two solutions are compared in figure 3. There are of course an infinite set of solutions to the differential equation \((17)\) depending on the choice of control functions \(\psi(t)\) and \(f(t)\). Those presented here are chosen to resemble the observed behaviour of falling cats.
标签：psi,mathbf,equation,varphi,问题,落猫,rho,dot From： https://www.cnblogs.com/laoshan-plus/p/18365006