1
$\begingroup$

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?

$\endgroup$
6
  • 1
    $\begingroup$ you mean disk rolling without friction on a paboloid? $\endgroup$ – Eli Mar 29 at 12:01
  • $\begingroup$ yeah kind of, but it isn't a disk it is a ring, kind of empty disk $\endgroup$ – math boy Mar 29 at 13:40
  • $\begingroup$ 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 $\endgroup$ – user256872 Mar 29 at 17:55
  • $\begingroup$ I just need equations, then i will solve it graphically using computer. $\endgroup$ – math boy Mar 29 at 19:05
  • $\begingroup$ 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. $\endgroup$ – R.W. Bird Apr 4 at 16:50
2
$\begingroup$

enter image description here

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

enter image description here

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 $.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.