Dynamics equations for unbalanced wheel rolling without slipping on flat surface

Question

I would like to develop a system of differential equations describing dynamics of an unbalanced wheel rolling without slipping on a flat surface. The difficulty i'm running into is that there is no steady state, like, for example, in the case of a balanced wheel rolling down an inclined plane. As the unbalanced wheel rolls, the friction force $F$, which determines the torque on the wheel, changes. And I don't understand how to compute $F$ from the no-slip condition and the state of the wheel, i.e., angle of the center of gravity relative to the center of the wheel and wheel angular velocity.

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Both @JohnAlexiou and @Eli solutions are correct. @Eli solution is a bit easier to generalize, in my opinion, since in it relies on Lagrangian formalism, which does not require understanding the detailed balance of forces.

Answer 1

Very crude free-body diagram

Note that the angle $\phi$ is drawn in a positive fashion, but for the wheel to roll to the right (along the blue velocity vector), then $\dot{\phi}<0$. To be consistent, the applied torque at the center of the wheel should also be $\tau_O < 0 $ and so it is drawn in the clockwise direction, which has a negative sense. Positive angles/torques are CCW in this FBD.

The center of mass kinematics are

• Position place the coordinate origin at the contact point at this instant

  $$\begin{pmatrix}x=c\sin\phi\\ y=R-c\cos\phi \end{pmatrix} \tag{1}$$

• Velocity roll the wheel to get

  $$\begin{pmatrix}\dot{x}=-R\dot{\phi}+c\dot{\phi}\cos\phi\\ \dot{y}=c\dot{\phi}\sin\phi \end{pmatrix} \tag{2}$$

  Note that the wheel center has velocity $v = - R \dot{\phi}$ for a no-slip condition. But since $\dot{\phi}<0$ then velocity is a positive value (as drawn above).

• Acceleration of the center of mass is used in dynamics

  $$\begin{pmatrix}\ddot{x}=-R\ddot{\phi}+c\ddot{\phi}\cos\phi-c\dot{\phi}^{2}\sin\phi\\ \ddot{y}=c\ddot{\phi}\sin\phi+c\dot{\phi}^{2}\cos\phi \end{pmatrix} \tag{3}$$

And the dynamics are

• Force/torque balance at the center of mass

  $$\begin{pmatrix}F=m\ddot{x}\\ N-mg=m\ddot{y}\\ -\tau_{O}+y\,F-x\,N=I_{C}\ddot{\phi} \end{pmatrix} \tag{4}$$

where $m$ is the mass of the wheel, $I_C$ is the mass moment of inertia of the wheel about the center of mass, $N$ is the contact normal force, and $F$ is the traction required to keep the wheel from slipping.

The solution of the above is

$$ \boxed{ \ddot{\phi} = - \frac{m\,c\left(g+R\dot{\phi}^{2}\right)\sin\phi+\tau_{O}}{I_{C}+m\left(R^{2}+c^{2}-2Rc\cos\phi\right)}} \tag{5}$$

and to confirm, if the eccentricity is zero, $c=0$, then $\ddot{\phi} = - \frac{\tau_O}{I_C + m R^2}$ which matches what is expected.

Analytical solution to (5) does not exist, because it is inhomogeneous and it depends on $\phi$ and $\dot{\phi}$ at the same time. But it is just a 1D ODE in terms of the angle of the wheel $\phi$, which means it is well suited for a numerical simulation.

There are some interesting situations that you need to check. For example, if at any point $N \leq 0$ it means the wheel is no longer in contact with the ground. And when $|F| > \mu |N|$ it means the wheel is slipping.

For both of those situations, the equations of motion change and a new set needs to be implemented in a simulation environment to get good results.

Appendix I

Calculation of moment arm of contact forces about the center of mass

Answer 2

$\def\b {\mathbf}$ Starting with the position vector to the center of mass:

$$\b R_p= \left[ \begin {array}{c} x-c\sin \left( \varphi \right) \\ R-c\cos \left( \varphi \right) \end {array} \right] \tag 1$$

from (1) and the rolling condition , $~x=R\varphi~$ the velocity

$$\b v=\frac{\partial\b R_p}{\partial \varphi}\,\dot\varphi$$

thus the kinetic energy T

$$T=\frac 12 M\,\b v\cdot\b v+ \frac 12 I_c\,\dot\varphi^2\\ T=\frac 12\,{\dot\varphi }^{2} \left( M{R}^{2}-2\,M\,R\,c\cos \left( \varphi \right) +M{c}^{2}+I_{{C}} \right) $$

the potential energy U $$U=M\,g\,\left(\b R_p\right)_y=M\,g \left( R-c\cos \left( \varphi \right) \right) +\tau _{{ \varphi }}\varphi $$

where $~\tau_\varphi~$ is external torque at the wheel center

with EL $~(\mathcal L=T-U)~$ you obtain the equation of motion

$$\ddot\varphi=-{\frac { \left( {\dot\varphi }^{2}R+g \right) \sin \left( \varphi \right) M\,c+\tau_\varphi }{ \left( {R}^{2}-2\,Rc\cos \left( \varphi \right) +{c}^{2} \right) M+{\it I_C}}} $$

Notice

If the position vector $~\b R_p~$ is a function of the generalized coordinate $~\varphi~$ you don't need to obtain the constraint force $~F~$ to get the equation of motion

Answer 3

I think ideally you'd solve this with Lagrangian mechanics but I am going to give a more elementary approach.

The no-slip condition relates horizonal position to rotation. If the wheel radius is $R$, and it has rolled $\phi$ radians counterclockwise, then the center has moved $R\phi$ radians to the left. So we can write $x=-R\phi$

You can differentiate this constraint to one on velocity: $\dot{x}=-R\dot{\phi}$ and again for acceleration: $\ddot{x}=-R\ddot{\phi}$

Separately, you can write the horizontal center of mass position $c_x=x+\sin(\phi)$ (careful with signs here). That, too, can be differentiated: $\dot c_x=\dot x+\dot \phi \cos(\phi)$ and $\ddot {c}_x=\ddot{x}+\ddot{\phi}\cos(\phi) - \dot{\phi}^2 \sin(\phi)$

This times mass will equal the combination of friction and any external horizontal forces. We can substitute in our equation for $\ddot{x}$ to get this all in terms of $\phi$.

If there are no forces besides gravity and the contact, then you can use this to calculate the force of friction. And that lets you calculate the torque that friction applies to the wheel.

So eventually, you'll have a messy equation that you can solve for $\ddot \phi$.