Derivations
5. Because there’s no movement along the axle, we have $v_x \cos{\theta}+v_y \sin{\theta}=0$, i.e.,$$\bbox[#FFF, 5px, border: 2px solid orange]{\cos{\theta}dx+\sin{\theta}dy=0.}$$ And for a single wheel, we have $v_{\tau 1}=\omega_1 a$. For the other wheel, we have $v_{\tau 2}=\omega_2 a$. The velocity of mass center is therefore, $$v_c=v_x \sin{\theta}-v_y \cos{\theta}=\frac{1}{2}(v_{\tau 1}+v_{\tau 2})=\frac{1}{2}a(\omega_1+\omega_2).$$ Thus, $$\bbox[#FFF, 5px, border: 2px solid orange]{\sin{\theta}dx-\cos{\theta}dy=\frac{1}{2}a(d\phi+d\phi’).}$$ Notice that the vertical component (perpendicular to the plane) of angular velocity of the rotating axle is $\dfrac{d\theta}{dt}=-\dfrac{v_{\tau 1}-v_{\tau 2}}{b}=-\dfrac{a}{b}\left(\omega_1-\omega_2\right)$. Integrating it, $$\bbox[#FFF, 5px, border: 2px solid orange]{\theta=C-\frac{a}{b}\left(\phi-\phi’\right).}$$
Exercises
16. We want to find $U$ satisfying, $$\frac{d}{dt}\frac{\partial U}{\partial \dot{r}}-\frac{\partial U}{\partial r}=F.$$ We can appropriately assume that $U(r,\dot{r})=f(r)g(\dot{r})$, thus $$\frac{\partial U}{\partial r}=f'(r)g(\dot{r})$$ $$\frac{d}{dt}\frac{\partial U}{\partial \dot{r}}=\frac{d\left[f(r)g'(\dot{r})\right]}{dt}=f'(r)g'(\dot{r})\dot{r}+f(r)g”(\dot{r})\ddot{r}.$$ Therefore,$$\frac{d}{dt}\frac{\partial U}{\partial \dot{r}}-\frac{\partial U}{\partial r}=f'(r)g'(\dot{r})\dot{r}+f(r)g”(\dot{r})\ddot{r}-f'(r)g(\dot{r})=\frac{1}{r^2}-\frac{\dot{r}^2-2\ddot{r}r}{r^2 c^2}$$ Comparing the terms with $\ddot{r}$, we find that $f(r)g”(\dot{r})=\dfrac{2}{r c^2}$. So let $g”(\dot{r})=C$, $g'(\dot{r})=C\dot{r}+D$, $g(\dot{r})=\dfrac{1}{2}C \dot{r}^2+D\dot{r}+E$, $f(r)=\dfrac{2}{rc^2 C}$, $f'(r)=-\dfrac{2}{r^2 c^2 C}$. Comparing the term with $\dot{r}$, we find $$-\frac{2}{r^2 c^2 C}\left(\frac{1}{2}C\dot{r}^2-E\right)=\frac{1}{r^2}\left(1-\frac{\dot{r}^2}{c^2}\right).$$ Then we have $C=1$, $E=\dfrac{1}{2}c^2$. Finally we get $$\bbox[#FFF, 5px, border: 2px solid orange]{U(r,\dot{r})=f(r)g(\dot{r})=\frac{1}{r}\left(\frac{\dot{r}^2}{c^2}+k\frac{\dot{r}}{c}+1\right).}$$ in which $k$ is an arbitary constant. The Lagrangian is $$\bbox[#FFF, 5px, border: 2px solid orange]{\mathscr{L}=T-U(r,\dot{r})=\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)-\frac{1}{r}\left(\frac{\dot{r}^2}{c^2}+k\frac{\dot{r}}{c}+1\right).}$$
20. We have $$\frac{\partial \mathscr{L}}{\partial x}=m\dot{x}^2 V'(x)-2V(x)V'(x),$$ $$\frac{d}{dt}\frac{\partial \mathscr{L}}{\partial \dot{x}}=\frac{d}{dt}\left(\frac{m^2 \dot{x}^3}{3}+2m\dot{x}V(x)\right)=m^2 \dot{x}^2 \ddot{x}+2m\ddot{x}V(x)+2m\dot{x}^2V'(x).$$ By the Lagrange’s equation $$m\cancel{\left(m\dot{x}^2+2V(x)\right)}\ddot{x}=-\cancel{\left(m\dot{x}^2+2V(x)\right)}V'(x).$$ That is $$\bbox[#FFF, 5px, border: 2px solid orange]{m\ddot{x}=-V'(x)}$$ which is just Newton’s second law.
21. We can take the length of string remaining on the table and the angle between string and a fixed direction as generalized coordinates $s$ and $\theta$. Then the Lagrangian is $$\bbox[#FFF, 5px, border: 2px solid orange]{\mathscr{L}=\frac{1}{2}m_1(\dot{s}^2+s^2\dot{\theta}^2)+\frac{1}{2}m_2 \dot{s}^2-m_2gs.}$$ We can calculate that $\dfrac{\partial\mathscr{L}}{\partial s}=m_1 s\dot{\theta}^2-m_2g$, $\dfrac{d}{dt}\dfrac{\partial\mathscr{L}}{\partial \dot{s}}=m_1 \ddot{s}+m_2\ddot{s}$, $\dfrac{\partial\mathscr{L}}{\partial \theta}=0$, $\dfrac{d}{dt}\dfrac{\partial\mathscr{L}}{\partial \dot{\theta}}=\dfrac{d}{dt}\left(m_1 s^2 \dot{\theta}\right)=2m_1 s \dot{s}\dot{\theta}+m_1 s^2 \ddot{\theta}$. Therefore by Lagrange’s equation, we have $$m_1(2 \dot{s}\dot{\theta}+s \ddot{\theta})=0\qquad\implies \text{conservation of angular momentum}$$ $$m_1(\ddot{s}-s\dot{\theta}^2)+m_2\ddot{s}=-m_2g\quad\implies\text{Newton’s second law in polar coordinate}.$$ We can integrate the first equation as $s^2\dot{\theta}=\dfrac{L}{m_1}$, put in the second equation, we get $$\bbox[#FFF, 5px, border: 2px solid orange]{(m_1+m_2)\ddot{s}-\frac{L^2}{m_1s^3}=-m_2g.}$$ We can also obtain the first integral $$\bbox[#FFF, 5px, border: 2px solid orange]{\frac{1}{2}(m_1+m_2)\dot{s}^2+\frac{L^2}{2m_1 s^2}+m_2gs=E.}$$ This is the conservation of energy.
24.
(a) The Lagrangian is $$\mathscr{L}=\frac{1}{2}m\left(\dot{l}^2+(L_a+l)^2\dot{\theta}^2\right)-\frac{1}{2}kl^2+mg(L_a+l)\cos{\theta}.$$ We have $\dfrac{\partial\mathscr{L}}{\partial l}=m (L_a+l)\dot{\theta}^2-kl+mg\cos{\theta}$, $\dfrac{d}{dt}\dfrac{\partial\mathscr{L}}{\partial \dot{l}}=m \ddot{l}$, $\dfrac{\partial\mathscr{L}}{\partial \theta}=-mg(L_a+l)\sin{\theta}$, $\dfrac{d}{dt}\dfrac{\partial\mathscr{L}}{\partial \dot{\theta}}=\dfrac{d}{dt}\left(m (L_a+l)^2 \dot{\theta}\right)=2m (L_a+l) \dot{l}\dot{\theta}+m (L_a+l)^2 \ddot{\theta}$. Therefore we have $$\bbox[#FFF, 5px, border: 2px solid orange]{\ddot{l}-(L_a+l)\dot{\theta}^2=g\cos{\theta}-\frac{kl}{m}}\tag{*}$$ $$\bbox[#FFF, 5px, border: 2px solid orange]{2\dot{l}\dot{\theta}+(L_a+l)\ddot{\theta}=-g\sin{\theta}.}\tag{#}$$ That’s the Newton’s second law in polar coordinate.
(b) When stretching and angular displacements are small (to first order), the equations are $$\ddot{l}=g-\frac{kl}{m}\,\implies\frac{d^2}{dt^2}(l-\frac{m}{k}g)+\frac{k}{m}(l-\frac{m}{k}g)=0$$ $$\implies \bbox[#FFF, 5px, border: 2px solid orange]{l=\frac{mg}{k}+A\sin{\omega_1 t}+B\cos{\omega_1 t}}\tag{1}$$ in which $\omega_1=\sqrt{\dfrac{k}{m}}.$ $$L_a\ddot{\theta}=-g\theta$$ $$\implies \bbox[#FFF, 5px, border: 2px solid orange]{\theta=C\sin{\omega_2 t}+D\cos{\omega_2 t}}\tag{2}$$ in which $\omega_2=\sqrt{\dfrac{g}{L_a}}.$
(c) When stretching and angular displacements are to second order, the equations are $$\ddot{l}-L_a\dot{\theta}^2=-\frac{1}{2}g\theta^2+g-\frac{kl}{m}\,\implies \ddot{l’}+\omega_1^2 l’-L_a\theta^2+\frac{1}{2}g\theta^2=0$$ in which $l’=l-\dfrac{mg}{k}.$ $$2\dot{l}\dot{\theta}+(L_a+l)\ddot{\theta}=-g\theta\,\implies \ddot{\theta}+\frac{2}{L_a+l}\dot{l}\dot{\theta}+\frac{g}{L_a+l}\theta=0$$ $$\implies \ddot{\theta}+\frac{2}{L_a}\dot{l}\dot{\theta}+\omega_2^2\theta-\frac{g}{L_a^2}l\theta=0$$ We use the perturbation method, so that $$l'(t)=\epsilon l’_0(t)+\epsilon^2 l’_1(t),\, \theta(t)=\epsilon\theta_0(t)+\epsilon^2 \theta_1(t)$$ satisfying $$\ddot{l’_0}+\omega_1^2 l’_0=0,\, \ddot{\theta_0}+\omega_2^2\theta_0=0.$$ So that $$\epsilon(\ddot{l’_0}+\omega_1^2l’_0)+\epsilon^2(\ddot{l’_1}+\omega_1^2 l’_1-L_a\dot{\theta_0}^2+\frac{1}{2}g\theta_0^2)=0$$ $$\implies \bbox[#FFF, 5px, border: 2px solid orange]{\ddot{l’_1}+\omega_1^2 l’_1-L_a\dot{\theta_0}^2+\frac{1}{2}g\theta_0^2=0}\tag{3}$$ And $$\epsilon(\ddot{\theta_0}+\omega_2^2\theta_0)+\epsilon^2(\ddot{\theta_1}+\omega_2^2\theta_1+\frac{2}{L_a}\dot{l_0}\dot{\theta_0}-\frac{g}{L_a^2}l_0\theta_0)=0$$ $$\implies \bbox[#FFF, 5px, border: 2px solid orange]{\ddot{\theta_1}+\omega_2^2\theta_1+\frac{2}{L_a}\dot{l_0}\dot{\theta_0}-\frac{g}{L_a^2}l_0\theta_0=0}\tag{4}$$ Using equation (1) for (3), we have $$\ddot{l’_1}+\omega_1^2 l’_1=\alpha\cos{2\omega_2 t}+\beta\sin{2\omega_2 t}+\gamma.$$ Solve this equation, we have
\begin{align}l(t)=\epsilon&(A\sin{\omega_1 t}+B\cos{\omega_1 t})+\frac{mg}{k}\\+&\epsilon^2\left[\frac{\omega_2^2}{\omega_1^2-4\omega_2^2}\left(\frac{3}{4}(C^2-D^2)\cos{2\omega_2 t}+\frac{1}{2}gL_a C^2D^2\sin{2\omega_2 t}\right)+\frac{3(C^2+D^2)L_a\omega_2^2}{4\omega_1^2}\right].\end{align}
For the same reason, we have
\begin{align}\theta(t)=\epsilon&(C\sin{\omega_2 t}+D\cos{\omega_2 t})\\+&\epsilon^2\frac{A\omega_2}{2L_a\omega_1}\left[\frac{\omega_2C+2\omega_1D}{2\omega_2-\omega_1}\cos{(\omega_1-\omega_2)t}+\frac{\omega_2C+2\omega_1D}{\omega_1+2\omega_2}\cos{(\omega_1+\omega_2)t}\right.\\-&\left.\frac{\omega_2D+2\omega_1C}{\omega_1+2\omega_2}\sin{(\omega_1+\omega_2)t}+\frac{\omega_2D-2\omega_1C}{2\omega_2-\omega_1}\sin{(\omega_2-\omega_1)t}\right].\end{align}
It is easy to tell that the resonance happens when $\omega_1=\pm2\omega_2$, $l(t)\to\infty$ and $\theta(t)\to\infty$. For $\dfrac{k}{m}=\dfrac{g}{L_a}$ is approachable, the resonance is likely.
(d) We adjust (*) and (#) to the first order and introduce the mass of spring, we have $$\bbox[#FFF, 5px, border: 2px solid orange]{\ddot{l}+\frac{k}{M}l-g=0}$$ $$\bbox[#FFF, 5px, border: 2px solid orange]{\ddot{\theta}+\frac{g}{L_a}\left(1-\frac{mg}{2kL_a}\right)\theta=0.}$$
(e) Take $L_a=0.2$, $m=0.1$, $g=9.8$ and $k=50$. And the initial condition is $l(0)=0.05$, $\theta(0)=\dfrac{\pi}{20}$, $l'(0)=0$ and $\theta'(0)=0$. We can plot the graph by $Mathematica$,
in which the blue line represent $l(t)$ and the orange line represents $\theta(t)$.
Code:
really good！
tql！