Derivations
12.
(a) There is a relation between the inertial set and the primed set, $$
\dot{\vec{r}}=\dot{\vec{r}}\prime+\vec{\omega}\times \vec{r}\prime.
$$ Therefore the kinetic energy is, $$
T=\frac{1}{2}m_a{\dot{\vec{r}}_a}^2=\frac{1}{2}m_a\left( \dot{r}_{ai}\prime+\varepsilon _{ijk}\omega _jr_{ak}\prime \right) \left( \dot{r}_{ai}\prime+\varepsilon _{ilm}\omega _lr_{am}\prime \right)
.$$ This implies that, \begin{align*}
\Longrightarrow T=&\frac{1}{2}m_a\left( \dot{r}_{ai}\prime^2+2\dot{r}_{ai}\prime\varepsilon _{ilm}\omega _lr_{am}\prime+\varepsilon _{ijk}\omega _jr_{ak}\prime\varepsilon _{ilm}\omega _lr_{am}\prime \right) \\=&T\prime+m_a\left( \vec{r}_a\prime\times \dot{\vec{r}}_a\prime \right) \cdot \vec{\omega}+\frac{1}{2}\omega _i\left( r_a\prime^2\delta _{ij}-r_{ai}\prime r_{aj}\prime \right) \omega _j. \end{align*}
That is, $$
\Longrightarrow T=T\prime+\vec{\omega}\cdot \vec{L}\prime+\frac{1}{2}\vec{\omega}\cdot \vec{I}\cdot \vec{\omega}.
$$ Because the potential is only determined by the distance between the point and the origin, it has the same form in two sets of axes. Thus the Lagrangian is, $$
L=T\prime+\vec{\omega}\cdot \vec{L}\prime+\frac{1}{2}\vec{\omega}\cdot \vec{I}\cdot \vec{\omega}-V.
$$
(b) Choose the coordinate, so that we have, $$
I’=\left( \begin{matrix}
I_1& & \\
& I_2& \\
& & I_3\\
\end{matrix} \right)
$$ For $\omega_1=0$, $\phi=0$ and $I_1=I_2$. The Lagrangian is $$
L=\frac{1}{2}I_1\dot{\theta}^2+\frac{1}{2}I_3\dot{\psi}^2-I_1\omega _2\dot{\theta}\sin \psi +I_3\omega _3\dot{\psi}+\frac{1}{2}I_2{\omega _2}^2+\frac{1}{2}I_3{\omega _3}^2-V.
$$ We can derive the equations of motion, $$
I_1\ddot{\theta}-I_1\omega _2\dot{\psi}\cos \psi =0
,$$ $$
I_3\ddot{\psi}+I_2\omega _2\ddot{\theta}\cos \psi =0
.$$ Consider $\theta\to 0$ and $\psi\to 0$, we have, $$
\ddot{\theta}+\frac{I_1}{I_3}{\omega _2}^2\theta =0.
$$ Therefore the frequency of the small oscillation is $$\Omega=\sqrt{\frac{I_1}{I_3}}\omega_2.$$
Exercises
19. We can denote the length of the spring is $x$, the angle between the rod and the vertical is $\theta$. Then we can describe any point on the rod, $$
r\prime=x\,\,\hat{x}+l\prime\left( \cos \theta \,\,\hat{x}+\sin \theta \,\,\hat{y} \right).
$$ Then the velocity is, $$
v\prime=\left( \dot{x}-l\prime\dot{\theta}\sin \theta \right) \hat{x}+l\prime\dot{\theta}\cos \theta \,\,\hat{y}.
$$ So the kinetic energy is, $$
T=\frac{1}{2}\int\limits_0^{2l}{dl\prime\frac{M}{2l}\left( \dot{x}^2+l\prime^2\dot{\theta}^2-2\dot{x}l\prime\dot{\theta}\sin \theta \right) =\frac{1}{2}M\dot{x}^2+\frac{2}{3}Ml^2\dot{\theta}^2-Ml\dot{\theta}\dot{x}\sin \theta}.
$$ The Lagrangian is, $$
L=\frac{1}{2}M\dot{x}^2+\frac{2}{3}Ml^2\dot{\theta}^2-Ml\dot{\theta}\dot{x}\sin \theta -\frac{1}{2}kx^2-mg\left( x+l\cos \theta \right).
$$ Then we can get the equation of motion easily, $$
\ddot{x}-l\ddot{\theta}\sin \theta -l\dot{\theta}^2\cos \theta =-\frac{k}{M}x+g,
$$ $$
\ddot{\theta}-\frac{3}{4l}\ddot{x}\sin \theta =-\frac{3g}{4l}\sin \theta.
$$
23. By the relation $d\vec{L}/dt=\vec{a}\times\vec{F}$, we have $$\ddot{\theta}=-\frac{af}{r_0^2}\sin\theta.$$ We can solve this differential equation, so that, $$
t=\frac{r_0}{\sqrt{2af}}\int\limits_0^{\pi /2}{\frac{d\theta}{\sqrt{\cos \theta}}\approx 2.622}\frac{r_0}{\sqrt{2af}}.
$$ Then we take $I=\frac{1}{3}m(2a)^2$, $a=1.2m$ and $f=0.3m/s^2$, we can show, $$
t\approx 2.622\sqrt{\frac{2a}{3f}}=3.04s.
$$
25.
(a) It is same as in HW4. Set an O-xyz coordinate on the plane, denote three Euler angles are $\theta$, $\phi$ and $\psi$ respectively. As a result, we have $$\begin{cases}\omega_x=\dot{\phi}\sin{\theta}\sin{\psi}+\dot{\theta}\cos{\psi}\\\omega_y=\dot{\phi}\sin{\theta}\cos{\psi}-\dot{\theta}\sin{\psi}\\\omega_z=\dot{\phi}\cos{\theta}+\dot{\psi}\end{cases}.$$The rolling constraints are, $$
R\left[ \begin{array}{c}
\omega _y\\
-\omega _x\\
\end{array} \right] =\left[ \begin{array}{c}
\dot{x}\\
\dot{y}\\
\end{array} \right].
$$ $$\implies
\begin{cases}
R\left( \dot{\phi}\sin \theta \cos \psi -\dot{\theta}\sin \psi \right) =\dot{x}\\
-R\left( \dot{\phi}\sin \theta \sin \psi +\dot{\theta}\cos \psi \right) =\dot{y}\\
\end{cases}.
$$ That is $$
\begin{cases}
f_1=R\left( d\phi \sin \theta \cos \psi -d\theta \sin \psi \right) -dx\\
f_2=R\left( d\phi \sin \theta \sin \psi +d\theta \cos \psi \right) +dy\\
\end{cases}.
$$ The conditions are nonintegrable, for an integrable constraint $f(q_i)$ must satisfy that, $$\frac{\partial^2 f}{\partial q_i \partial q_j}=\frac{\partial^2 f}{\partial q_j \partial q_i}.$$ Therefore the constaint is nonholonomic.
(b) ???
27. From the previous section, we have $$
f(u)=\dot{u}^2=\left(1-u^2\right)(\alpha-\beta u)-(b-a u)^2
,$$ where $$
\begin{aligned}
\alpha &=\frac{2 E-I_3 \omega_3^2}{I_1} \\
\beta &=\frac{2 M g l}{I_1} \\
a &=\frac{I_3 \omega_3}{I_1}=\frac{I_3(\dot{\psi}+\dot{\phi} \cos \theta)}{I_1} \\
b &=a u_0+\dot{\phi} \sin ^2 \theta_0
\end{aligned}.
$$ Consider the initial conditions $$
t=0, \theta=\theta_0, \phi=\psi=0, \dot{\phi}=\dot{\phi_0}=c, \dot{\psi}=\omega_0
.$$ We can derive that $$
x_1\approx\left(\frac{\beta}{a^2}-\frac{2 \dot{\phi_0}}{a}\right) \sin ^2 \theta_0,
$$ And $$
\ddot{x}+a^2 x=\left(\frac{\beta}{2}-a c\right) \sin ^2 \theta_0
.$$ Therefore $$
\begin{aligned}
\omega_1 &=\dot{\phi} \sin \theta \sin \psi+\dot{\theta} \cos \psi \\
& \approx \dot{\phi} \sin \theta_0 \sin \left(\omega_3 t\right)+\dot{\theta} \cos \left(\omega_3 t\right) \\
&=\frac{a x_1}{2 \sin \theta_0}\left[(1-\cos a t) \sin \left(\omega_3 t\right)+\sin a t \cos \left(\omega_3 t\right)\right] \\
&=\frac{a x_1}{2 \sin \theta_0}\left[\sin \left(\omega_3 t\right)+\sin \left(a-\omega_3\right) t\right] \\
&=\frac{a x_1}{2 \sin \theta_0}\left[\sin \left(\omega_3 t\right)+\sin \Omega t\right]
\end{aligned}
.$$ So $$
\Omega=a-\omega_3=\frac{I_3-I_1}{I_1} \omega_3
,$$ and $$
\omega_2=\frac{a x_1}{2 \sin \theta_0}\left[\cos \left(\omega_3 t\right)-\cos \Omega t\right].
$$