<s>[INST] Number degrees of freedom for Sphere on a inclined plane How many generalized coordinates are required to describe the dynamics of a solid sphere is rolling without slipping on an inclined plane?

What I think is that there two translational degrees of freedom and three rotational degrees of freedom. But I'm not able to write the rolling constraint to see if they are integrable or not? So please help me through this.
 [/INST] 
you have one holonomic  constraint equation which is $~z=a~$ and two nonholonomic
constraint equations
the relative velocities componets , at the contact point between the sphere and the incline plane are zero thus:
$$\begin{bmatrix}
  \omega_x \\
  \omega_y \\
  \omega_z \\
\end{bmatrix}\times \begin{bmatrix}
 0 \\
  0\\
  a \\
\end{bmatrix}-\begin{bmatrix}
 \dot{x} \\
  \dot{y}\\
  0 \\
\end{bmatrix}=\begin{bmatrix}
0\\
  0\\
  0 \\
\end{bmatrix}
 $$
you obtain two equations :
$$\omega_x\,a+\dot x=0\tag 1$$
$$\omega_y\,a-\dot y=0\tag 2$$
where :
$$\omega_x=\dot\varphi -\sin \left( \vartheta  \right) \dot\psi $$
$$\omega_y=\cos \left( \varphi  \right) \dot\vartheta +\cos \left( \vartheta 
 \right) \sin \left( \varphi  \right) \dot\psi 
 $$
from the six possible velocities  of the sphere $~\dot x\,,\dot y\,,\dot z\,,\dot\varphi\,,\dot\vartheta\,\,,\dot\psi~$
just three of them are independent.
solve Eq. (1) (2) for $~\dot x,~\dot y$ you obtain with $~\dot z=0$
$$ \left[ \begin {array}{c} {\dot x}\\{\dot y}
\\\dot \varphi \\\dot \vartheta
\\\dot \psi \end {array} \right]=\left[ \begin {array}{ccc} \sin \left( \psi \right) \cos \left(
\vartheta  \right) a&\cos \left( \psi \right) a&0\\
-\cos \left( \psi \right) \cos \left( \vartheta  \right) a&\sin
 \left( \psi \right) a&0\\ 1&0&0
\\ 0&1&0\\ 0&0&1\end {array}
 \right]\,\underbrace{\left[ \begin {array}{c} \dot\varphi \\ \dot\vartheta 
\\ \dot \psi \end {array} \right] }_{\vec{\dot{q}}}
$$
thus you have 3 generalized velocities coordinates $~,\dot\varphi~,\dot\vartheta~,\dot\psi$
you can also solve Eq. (1) (2) for $~\dot \varphi,~\dot \vartheta$
$$\left[ \begin {array}{c} {\dot x}\\{\dot y}
\\\dot \varphi \\\dot \vartheta
\\\dot \psi \end {array} \right]=\left[ \begin {array}{ccc} 1&0&0\\ 0&1&0
\\ {\frac {\sin \left( \psi \right) }{a\cos \left(
\vartheta  \right) }}&-{\frac {\cos \left( \psi \right) }{a\cos
 \left( \vartheta  \right) }}&0\\ {\frac {\cos
 \left( \psi \right) }{a}}&{\frac {\sin \left( \psi \right) }{a}}&0
\\ 0&0&1\end {array} \right]\,\begin{bmatrix}
  \dot x \\
  \dot y \\
  \dot\psi \\
\end{bmatrix}$$
again you have 3 generalized velocities  coordinates $~\dot x~,\dot y,~\dot \psi$
Edit
the equations of motion for a ball rolled on a plane
I use the NEWTON- EULER approach
\begin{align*}
 &\boldsymbol J^T\,\boldsymbol M\,\boldsymbol J\,\boldsymbol{\ddot{q }}=\boldsymbol J^T\left(\boldsymbol f_A-\boldsymbol M\,\underbrace{\boldsymbol{\dot{J}}\,\boldsymbol{\dot{q}}}_{\boldsymbol f_Z}\right)\tag 1\\
 &\text{where}\\
 &\boldsymbol{\dot{J}}=\frac{\partial \left(\boldsymbol J\,\boldsymbol{\dot{q}}\right)}{\partial \boldsymbol q}
\end{align*}
\begin{align*}
&\boldsymbol q=\left[ \begin {array}{c} x\\  y\\
\psi\end {array} \right]
\\
&\boldsymbol M=\left[ \begin {array}{cccccc} m&0&0&0&0&0\\ 0&m&0&0
&0&0\\  0&0&m&0&0&0\\  0&0&0&\Theta
k&0&0\\  0&0&0&0&\Theta k&0\\  0&0&0
&0&0&\Theta k\end {array} \right]
~,
  \boldsymbol J= \left[ \begin {array}{ccc} 1&0&0\\  0&1&0
\\  0&0&0\\  {\frac {\sin \left(
\psi \right) }{a\cos \left( \vartheta  \right) }}&-{\frac {\cos
 \left( \psi \right) }{a\cos \left( \vartheta  \right) }}&0
\\  {\frac {\cos \left( \psi \right) }{a}}&{\frac {
\sin \left( \psi \right) }{a}}&0\\  0&0&1\end {array}
 \right]\\
 &\boldsymbol f_z=\left[ \begin {array}{c} 0\\  0\\ 0
\\  {\frac { \left( \cos \left( \psi \right) {\it \dot{x}}
+{\it \dot{y}}\,\sin \left( \psi \right)  \right) \dot{\psi} }{a\cos \left(
\vartheta  \right) }}\\  -{\frac { \left( -\cos
 \left( \psi \right) {\it \dot{y}}+{\it \dot{x}}\,\sin \left( \psi \right)
 \right) \dot{\psi} }{a}}\\  0\end {array} \right]~,
 \boldsymbol f_A=\left[ \begin {array}{c} F_x\\  0\\
-mg\\  0\\  0\\
\tau_\psi \end {array} \right]
\end{align*}
and from the nonholonomic constraint equations  you get additional two differential equations
\begin{align*}
&F_{c1}=\left( \sin \left( \psi \right) \cos \left( \vartheta  \right) \dot
\varphi +\cos \left( \psi \right) \dot \vartheta  \right) a-{ \dot x}
\\
&F_{c2}=- \left( \cos \left( \psi \right) \cos \left( \vartheta  \right) \dot
\varphi -\sin \left( \psi \right) \dot\vartheta  \right) a-{\dot y}
\\
&\Rightarrow\\
  &\dot{\varphi}={\frac {-\cos \left( \psi \right) {\it \dot{y}}+{\it \dot{x}}\,\sin \left( \psi
 \right) }{a\cos \left( \vartheta  \right) }}
\tag 2\\
&\dot{\vartheta}={\frac {\cos \left( \psi \right) {\it \dot{x}}+{\it \dot{y}}\,\sin \left( \psi
 \right) }{a}}\tag 3
\end{align*}
altogether you have eight first order differential equations (Eq. (1),(2),(3)) to solve this problem
</s>