# Ring moving on paraboloid?

Using Newton's law and rotational mechanics it is easy to write motion equations for a ring that moves on the flat surface, but for me hard thing is to write equations for a paraboloid. for example, the normal constantly changes due to the geometry of the paraboloid. I had 3 angles for a flat surface, but I think I need another angle here and it is hard to connect this angle to the previous 3. what pieces of advice can you give me?

• you mean disk rolling without friction on a paboloid? – Eli Mar 29 at 12:01
• yeah kind of, but it isn't a disk it is a ring, kind of empty disk – math boy Mar 29 at 13:40
• I know you insist on Newton's Laws, but in this case, Lagrangian mechanics would be far easier. Also, your equation of motion would be a non-linear ODE which would most likely require a computer to solve it – user256872 Mar 29 at 17:55
• I just need equations, then i will solve it graphically using computer. – math boy Mar 29 at 19:05
• I'm thinking that a ring rolling inside of a paraboloid is going to be unstable. It will fall over. Two possible exceptions: given an initial angular velocity it might roll down one side, through the bottom, and up the other side (all in a vertical plane through the center); or around a horizontal circle with just the right speed and tilt. – R.W. Bird Apr 4 at 16:50

this animation is result of simulation the equation of motion, I used Euler- Lagrange with non holonomic constraint equation (rolling condition).

Ring rolling on paraboloid surface

Paraboloid parameter equation

$$\boldsymbol R= \begin{bmatrix} x(\lambda~,\vartheta) \\ y(\lambda~,\vartheta) \\ z(\lambda~,\vartheta) \\ \end{bmatrix}= \left[ \begin {array}{c} a\sqrt {{\frac {\lambda}{h}}}\cos \left( \vartheta \right) \\ a\sqrt {{\frac {\lambda}{h}}} \sin \left( \vartheta \right) \\ \lambda \end {array} \right]$$

• $$a~$$ Paraboloid radius
• $$h~$$ Paraboloid height
• $$\lambda~> 0$$ Paraboloid parameter
• $$\vartheta~\in [0~,2\,\pi]~$$ Paraboloid parameter

the contact point is located at the plane that created by the tangential vector to the line $$~\lambda~$$, vector $$~\boldsymbol t_\lambda~$$ and to the line $$~\vartheta~$$ ,vector $$~\boldsymbol t_\vartheta$$

with:

\begin{align*} &\boldsymbol t_\lambda=\frac{\partial \boldsymbol R}{\partial \lambda}=\left[ \begin {array}{c} {\frac {a\cos \left( \vartheta \right) }{ \sqrt {4\,\lambda\,h+{a}^{2}}}}\\ {\frac {a\sin \left( \vartheta \right) }{\sqrt {4\,\lambda\,h+{a}^{2}}}} \\ 2\,{\frac {\sqrt {\lambda}\sqrt {h}}{\sqrt {4\, \lambda\,h+{a}^{2}}}}\end {array} \right]~, \boldsymbol t_\vartheta=\frac{\partial \boldsymbol R}{\partial \vartheta}=\left[ \begin {array}{c} -\sin \left( \vartheta \right) \\ \cos \left( \vartheta \right) \\ 0\end {array} \right]~, \boldsymbol t_n=\boldsymbol t_\lambda\times\boldsymbol t_\vartheta \end{align*}

we put at the ring center the coordinate system $$~\boldsymbol\xi_n~,\boldsymbol\xi_\lambda~,\boldsymbol\xi_\varphi$$ the disk can rotate about $$~\boldsymbol\xi_\varphi~$$ axes with the angle $$~\varphi(t)~$$ , and additional rotate with constant angle $$~\psi~$$ about the $$~\boldsymbol\xi_n~$$ axes.

obtaining the $$~\xi~$$ coordinate system

\begin{align*} &\boldsymbol\xi_\lambda=\boldsymbol S_\Psi\,\boldsymbol t_\lambda\\ &\boldsymbol\xi_\varphi=\boldsymbol S_\Psi\,\boldsymbol t_\vartheta\\ &\boldsymbol\xi_n=\boldsymbol S_\Psi\,\boldsymbol t_n\\ &\text{with}\\ &\boldsymbol S_\Psi= \left[ \begin {array}{ccc} \cos \left( \psi \right) &-\sin \left( \psi \right) &0\\ \sin \left( \psi \right) &\cos \left( \psi \right) &0\\ 0&0&1\end {array} \right]\\ \end{align*}

thus the position vector to the center of the ring is

$$\boldsymbol R_c=\boldsymbol R+\rho\,\boldsymbol \xi_n~$$ where $$~\rho~$$ is the disc radius.

The rolling condition (non holonomic )

\begin{align*} &g_n=\left[(\boldsymbol\omega\times\boldsymbol r)-\boldsymbol v\right]\cdot \boldsymbol \xi_\lambda=\left[\left(\dot\varphi\,\boldsymbol\xi_\varphi\times\ \rho\,\boldsymbol\xi_n\right)-\boldsymbol v\right]\cdot \boldsymbol \xi_\lambda =0\tag A\\ &\text{where}\\ &\boldsymbol v=\frac{\partial \boldsymbol R_c}{\partial \boldsymbol q}\,\boldsymbol{\dot{q}}\\ &\text{and}\\ &\boldsymbol q=\begin{bmatrix} \lambda \\ \vartheta \\ \varphi \end{bmatrix} \end{align*}

Kinetic and potential energy

\begin{align*} &\text{kinetic energy}\\ &T=\frac{m}{2}\,\boldsymbol v\cdot\,\boldsymbol v+\frac{I_\varphi}{2}\dot{\varphi}^2\\ &\text{potential energy}\\ &U=-m\,g\,\left(\boldsymbol{R}_c\right)_z\\ \end{align*}

Euler Lagrange with non-holonomic constraint equations

\begin{align*} &\mathcal{L} =T-U\\ &\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{\boldsymbol{q}}}\right)- \frac{\partial \mathcal{L}}{\partial \boldsymbol{q}} =\left[\frac{\partial \boldsymbol{R}_c}{\partial \boldsymbol{q}}\right]^T\boldsymbol{f}_a+ \left(\frac{\partial \boldsymbol{g}_n}{\partial \dot{\boldsymbol{q}}}\right)^T\boldsymbol{\chi}\tag 1 \end{align*} \begin{align*} &\ddot{\boldsymbol{g}}_n=\frac{\partial \boldsymbol{g}_n}{\partial \dot{\boldsymbol{q}}}\ddot{\boldsymbol{q}}+\frac{\partial \boldsymbol{g}_n}{\partial \boldsymbol{q}}\dot{\boldsymbol{q}}\tag 2 \end{align*}

Where $$~\boldsymbol g_n~$$ ist the non holonomic equation (Eq. A).

equations (1) and (2) are 4 equations for 4 unknowns, the solution give you the differential equations $$\ddot\lambda~,\ddot\vartheta~,\ddot\varphi~$$ and the constraint force $$~\chi$$.

• can you send me the code for this? it's great solution – math boy Apr 7 at 20:20
• I am using Maple symbolic program ,du you know thus program, if so i can send you Maple program code? – Eli Apr 7 at 20:30
• okay it would be nice, thanks a lot – math boy Apr 7 at 20:35
• can you show your Euler-Lagrange equations, it's kind of vague for me. – math boy Apr 8 at 8:55
• @math boy i put the equations of motion. I am sure if you still need the MAPLE program ? – Eli Apr 8 at 13:02