Derivations
7. It is obvious that we can write A as, $$
A=\left[ \begin{matrix}
-1& 0& 0\\
0& -1& 0\\
0& 0& 1\\
\end{matrix} \right]
.$$ Therefore, it is straightforward that, $$
P_+=\left[ \begin{matrix}
0& 0& 0\\
0& 0& 0\\
0& 0& 1\\
\end{matrix} \right] \Longrightarrow {P_+}^2=P_+
,$$ and $$
P_-=\left[ \begin{matrix}
1& 0& 0\\
0& 1& 0\\
0& 0& 0\\
\end{matrix} \right] \Longrightarrow {P_-}^2=P_-
.$$ The geometric interpretation of operator $P_+$ and $P_-$ is the projection on $x-y$ plane and z axis respectively.
20. 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.
Exercises
22. To a first approximation, we can regard the projectile as moving horizontally, then the Coriolis force is $$F=|-2m\omega\times v|=2m\omega v \cos{\theta}.$$ It equals to the centripetal force, $$2m\omega v \cos{\theta}=m\Omega v.$$ Therefore the angular deviation with time at rate, $$\Omega=2\omega\cos{\theta}.$$
23. By Newton’s second law, we have, $$
\ddot{\vec{r}}=\frac{\vec{T}}{m}+\vec{g}-2\vec{\omega}\times \vec{v}.
$$ Using the approximation of small oscillation, we have $$
\vec{T}=-\frac{mg}{l}\left( x\hat{i}+y\hat{j} \right) -m\vec{g}.
$$ This implies, $$
\Longrightarrow \begin{cases}
\ddot{x}=-\frac{g}{l}x+2\omega \dot{y}\cos \theta\\
\ddot{y}=-\frac{g}{l}y-2\omega \dot{x}\cos \theta\\
\end{cases}.
$$ Denoting $\xi=x+iy$, we have $$
\ddot{\xi}=-\frac{g}{l}\xi -2i\omega \cos \theta \dot{\xi}
.$$ Denote $\frac{g}{l}=\omega_0^2$, we can solve this equation, $$
\xi \left( t \right) =e^{-it\omega \cos \theta}\left( Ae^{i\sqrt{{w_0}^2+\omega ^2\cos ^2\theta}t}+Be^{-i\sqrt{{w_0}^2+\omega ^2\cos ^2\theta}t} \right)
.$$ For $\omega_0\gg \omega$, we have approximately, $$
\xi \left( t \right) =e^{-it\omega \cos \theta}\left( Ae^{iw_0t}+Be^{-iw_0t} \right)
.$$ Therefore the plane of oscillation rotates by the rate of $\omega \cos\theta$, that is rotating uniformly $2\pi\cos\theta$ radians per day.