Let's imagine we are on the top of some axisymmetric surface. Let it be $r = r(z)$ in cylindrical coordinates $\left(r,\varphi, z\right)$. It can be a sphere, which we have discussed in my previous question.

The mass particle can just slide down due to force of gravity without friction. In that case I probably could write the eq-s of motion. The normal force $N$ would depend, generally speaking, on the curvature of the surface (to equate gravity and take into account centripetal force). Since it's all axisymmetric, our velocity doesn't depend on $\varphi$, for finding curvature we can take derivatives $\frac{dr}{dz}$ as it is a 2D problem (like here).

Now, let's think about such initial conditions or additional forces actioned on a particle that it has $\hat\varphi$ component of velocity. Then, as I think, it's not so trivial to find a radius of curvature. How can I do it? Probably using directional derivative?

  • $\begingroup$ The linked answer assumes a fixed path. The problem described above is dynamic whereas the path depends in the initial conditions (like a marble rolling in a bowl). $\endgroup$ – ja72 Dec 29 '14 at 19:08
  • 1
    $\begingroup$ See Frenet Formulas on how to describe the path in 3D $\endgroup$ – ja72 Dec 29 '14 at 19:11

In the cylindrical coordinates described the position if parameterized as a surface (with two coordinates)

$$\vec{r}(\varphi,z) = ( r(z) \cos \varphi, r(z) \sin \varphi, z) $$

Using the chain rule you differentiate to get the velocity vector

$$ \vec{v}(\dot{\varphi},\dot{z}) = \begin{pmatrix} r'(z) \dot{z} \cos\varphi - r(z) \dot{\varphi} \sin\varphi \\ r'(z) \dot{z} \sin\varphi + r(z) \dot{\varphi} \cos\varphi \\ \dot{z} \end{pmatrix} $$

where $r'(z) = \frac{{\rm d}r(z)}{{\rm d}z}$ and $r''(z) = \frac{{\rm d}^2r(z)}{{\rm d}z^2}$

$$ \vec{a}(\ddot{\varphi},\ddot{z}) = \begin{pmatrix} \left(r''(z) \dot{z}^2+r'(z) \ddot{z}-r(z) \dot{\varphi}^2 \right)\cos\varphi-\left(r(z)\ddot{\varphi}+2\dot{z}\dot{\varphi} \right)\sin\varphi \\ \left(r''(z) \dot{z}^2+r'(z) \ddot{z}-r(z) \dot{\varphi}^2 \right)\sin\varphi+\left(r(z)\ddot{\varphi}+2\dot{z}\dot{\varphi} \right)\cos\varphi \\ \ddot{z} \end{pmatrix} $$

Looking at http://en.wikipedia.org/wiki/Normal_(geometry) you see that the surface normal is $$ \vec{n} = {\rm unit}\left( \frac{{\rm d}\vec{r}}{{\rm d}\varphi} \times \frac{{\rm d}\vec{r}}{{\rm d}z} \right) $$ $$\vec{n} ={\rm unit}\begin{pmatrix} \cos\varphi \\ \sin\varphi \\ -r'(z) \end{pmatrix} $$

So the equations of motion are

$$ \sum \vec{F} =N \vec{n} = m \vec{a} $$

which are three equations to be solved for the normal force magnitude $N$, and the parameter accelerations $\ddot{\varphi}$ and $\ddot{z}$.

The solution I get has

$$ \begin{aligned} N &= m \frac{r''(z) \dot{z}^2 - r(z) \dot{\varphi}^2}{\sqrt{1+r'(z)^2}} \\ \ddot{\varphi} & = - \frac{2 \dot{z} \dot{\varphi} r'(z)}{r(z)} \\ \ddot{z} & = \frac{r'(z) \left( r(z) \dot{\varphi}^2-r''(z) \dot{z}^2\right)}{1+r'(z)^2} \end{aligned} $$


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