Motion of a wheel with an added mass to its edge Imagine a wheel with homogeneous mass distribution, but there is a mass concentrated to its edge at a specific point. The wheel is rotating over a horizontal surface with a given initial velocity. How can I describe the motion of the wheel now mathematically?
 A: I think this is what you describe:

You need to find the new combined inertial properties. The center of mass distance from the center is $$c = \frac{m R}{M+m}$$ and the mass moment of inertia about the center of mass is found using the parallel axis theorem $$I_C = \left( M R^2+M c^2 \right) + \left(m (R-c)^2 \right) = \left(M+\frac{m M}{M+m}\right) R^2 $$
The contact point may or may not have slipping (not defined in problem) so the slipping speed is $v_{slip} = v - \omega R$. If this is a rough surface then $v_{slip}=0$ and thus $\dot{v}=-R \dot{\omega}$. Otherwise the problem has two degrees of freedom (translation and rotation) instead of only one with pure rolling.
At this point the acceleration of the center of mass needs to be defined in terms of our motion parameters $v$ and $\omega$ and their derivatives.
First the velocity of the center of mass is described as a 2D vector
$$v_C = \begin{pmatrix} v + c \omega \cos\theta \\ -c \omega \sin \theta \end{pmatrix}$$
By taking the total derivative of the above the acceleration vector of the center of mass is
$$ a_C = \begin{pmatrix} 
\dot{v} + c \left( \dot{\omega} \cos\theta - \omega^2 \sin\theta \right) \\ 
- c \left( \dot{\omega} \sin \theta + \omega^2 \cos\theta \right) 
\end{pmatrix}$$
here is where the discussion about slipping is important because I am assuming a rough surface and thus $\omega = \frac{1}{R} v$ and $\dot{\omega} = \frac{1}{R} \dot{v}$. Together with the definition for the center of mass $c$ the above is
$$ a_C = \begin{pmatrix}
  \dot{v} + \frac{m}{M+m} \left( \dot{v} \cos\theta - \frac{v^2}{R} \sin\theta \right)\\
 - \frac{m}{M+m} \left( \dot{v} \sin\theta + \frac{v^2}{R} \cos\theta \right)
\end{pmatrix} $$
If there is no driving force on the x axis, and no driving moment then the only forces acting is the ring gravity at the center, the mass gravity at the hoop edge and the contact force $N$ (shown below) and any friction $F$ if present.

The equations of motion are:


*

*Sum of forces equals mass times acceleration of the center of mass
$$ \begin{pmatrix} F \\ N - (M+m) g \end{pmatrix} = (M+m) \;a_C $$

*Sum of moments at the center of mass equals the mass moment of inertia times the rotational acceleration (with positive along the motion, or clockwise)
$$ c \sin\theta\, (N-M g)+(R-c) \sin\theta\, (m g)-(R+c \cos\theta) F = I_C \frac{\dot{v}}{R} $$


The above are three equations (two linear and one angular) to be solved for linear acceleration $\dot{v}$, normal force $N$ and friction $F$.
The solution I get is
$$ \begin{aligned}
  \dot{v} & = \frac{m \left(g+\frac{v^2}{R}\right) \sin\theta}{2 \left(M+m (1+\cos\theta)\right)} \\
  F &= \frac{m \left(g-\frac{v^2}{R}\right) \sin\theta}{2} \\
  N &= \mbox{too complex for here}
\end{aligned}$$
A motion is described if the above is integrated (numerically) with 
$$ \begin{aligned}
v &= \int \dot{v}\,{\rm d}t \\
x &= \int v\,{\rm d}t \\
\omega & = \frac{v}{R}\\
\theta = & \int \omega\,{\rm d}t = \int \frac{v}{R}\,{\rm d}t = \frac{x}{R}\\
\end{aligned}$$
