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.
 A: 
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
A: The constraints for a sphere rolling without slipping on a flat plane are well-known to be non-integrable.    The presence of a potential energy function on the plane does not affect whether or not the constraints are integrable, so a sphere rolling on an inclined plane is non-integrable as well.
A proof that the constraints cannot be holonomic can be found in these lecture notes by Michael Fowler:

To see this, imagine a sphere placed at the origin in the $(x,y)$ plane. Call the point at the top of the sphere the North Pole. Now roll the sphere along the $x$ axis until it has turned through ninety degrees. Its NS axis is now parallel to the $x$ axis, the N pole pointing in the positive $x$ direction. Now roll it through ninety degrees in a direction parallel to the $y$ axis. The N pole is still pointing in the positive $x$ direction, the sphere, taken to have unit radius, is at $(\pi/2,\pi/2)$.
Now start again at the origin, the N pole on top. This time, first roll the sphere through ninety degrees in the $y$ direction. The N pole now points along the positive $y$ axis. Next, roll the sphere through ninety degrees in the $x$ direction: we’re back to the point $(\pi/2,\pi/2)$ but this time the N pole is pointing in the $y$ direction.
The bottom line is that, in contrast to the cylindrical case, for a rolling sphere the no-slip constraint does not allow us to eliminate any dynamical variables—given that initially the sphere is at the origin with the N pole at the top, there is no unique relationship between orientation $(\theta,\phi)$ and position $(x,y)$ at a later point, we would have to know the rolling history, and in fact we can roll back to the origin by a different route and in general the N pole will not be at the top when we return.

A: You start with the 6 DOF of the sphere. Then the contact condition removes one DOF to get to 5. The no-slip conditions mean translation and rotation are coupled and this happens about an arbitrary axis or about 2 fixed axes. So we can expect 3 DOF to describe the dynamics.
Consider an inertial coordinate system aligned with the inclined plane, with the z axis pointing out of the plane.
The velocities of the sphere is described as follows in term of the DOF
$$ \begin{aligned} \boldsymbol{v} & = \pmatrix{ \dot{x} \\ \dot{y} \\ 0 } \\
\boldsymbol{\omega} & = \pmatrix{ -\frac{\dot{x}}{R} \\ \frac{\dot{y}}{R} \\ \dot{\theta} } \end{aligned} $$
with DOF $x$  the center of the sphere along the x axis, $y$ the center of the sphere along the y axis, and $\theta$ the spin angle about the x axis.
