2
$\begingroup$

I am trying to figure out how to supplement the equations of motion of a free particle on a sphere

$$\displaystyle \ddot{\theta} = \dot{\phi}^2 \sin \theta \cos \theta \\ \displaystyle \ddot{\phi} = - 2 \dot{\phi} \dot{\theta} \frac{1}{\tan \theta}$$

with a drag term

$$ \dot{\mathbf{v}} = -\gamma \mathbf{v}$$

It seems pretty clear to me that the equation for $\ddot{\theta}$ will be supplemented like so

$$ \ddot{\theta} = \dot{\phi}^2 \sin \theta \cos \theta -\gamma \dot{\theta} $$

But I am getting really confused about what happens with $\ddot{\phi}$.
As I've worked it out, I get

$$ \displaystyle \ddot{\phi} = - 2 \dot{\phi} \dot{\theta} \frac{1}{\tan \theta} - \gamma \dot{\phi} $$

But intuitively, it seems to me that $\sin\theta$ ought to be accounted for in that term somehow.
Does anyone know the correct way to do this?

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ The position vector of a sphere is $\overrightarrow{R}=\overrightarrow{R}\left( \theta ,\phi \right) $ and the velocity $\overrightarrow{v}=\dfrac{\partial \overrightarrow{R}}{\partial \theta }\dfrac{d\theta }{dt}+\dfrac{\partial \overrightarrow{R}}{\partial \phi }\dfrac{d\phi }{dt}$ so the Drag Force is $F= \left( \dfrac{\partial \overrightarrow{R}}{\partial \overrightarrow{q}}\right) ^{T}\left( -\gamma \overrightarrow{v}\right) $ $\endgroup$ – Eli Jan 29 at 13:16
  • $\begingroup$ Hi bob.sacamento. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Feb 9 at 2:01
2
$\begingroup$

from http://mathworld.wolfram.com

http://mathworld.wolfram.com

the particle position vector $\vec{R}$ is :

$$\vec{R}=r\,\left[ \begin {array}{c} \cos \left( \theta \right) \sin \left( \phi \right) \\ \sin \left( \theta \right) \sin \left( \phi \right) \\ \cos \left( \phi \right) \end {array} \right] \tag 1$$

from equation (1) you can obtain the kinetic $T$ and potential energy $U$

$$T=\frac{m}{2}\vec{\dot{R}}\cdot\vec{\dot{R}}=$$ and $$U=m\,g\,\vec{R}_z$$

the drag force is :

$$\vec{F}_D=-d\,\vec{\dot{R}}$$

if you calculate the equations of motion with Euler Lagrange method, the drag force is a generalized external force $\vec{F}_Q$ on the RHS of the E.L equations

$$\vec{F}_Q=\left(\frac{\partial \vec{R}}{\partial \vec{q}}\right)^T\,\vec{F}_D $$

where $\vec{q}=[\theta,\phi]^T$

$$\vec{F}_Q= d\,r^2\begin{bmatrix} \sin^2(\phi)\,\dot{\theta} \\ \dot{\phi} \\ \end{bmatrix} $$ thus the equations of motion are:

$$\ddot{\theta}+2\,{\frac {\cos \left( \phi \right) {\it \dot{\theta}}\,{\it \dot{\phi}}}{\sin \left( \phi \right) }}+{\frac {{\it \dot{\theta}}\,d}{m}}=0 $$

$${\it \ddot{\phi}}+{\frac {{\it \dot{\phi}}\,d}{m}}-{{\it \dot{\theta}}}^{2}\cos \left( \phi \right) \sin \left( \phi \right) -{\frac {g\sin \left( \phi \right) }{r}} =0$$

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ Thanks! I take it that in the $\ddot{\phi}$ equation, you mean $\sin(\theta)$? $\endgroup$ – bob.sacamento Jan 31 at 14:49
  • $\begingroup$ @bob.sacamento this is because i use different position vector $\vec{R}$ then you did? $\endgroup$ – Eli Jan 31 at 15:46
  • $\begingroup$ Ah! I see! I think in terms of $\phi$ being azimuthal, and you think in terms of $\phi$ being zonal. A common misunderstanding, I have found! Thanks again! $\endgroup$ – bob.sacamento Jan 31 at 16:45
2
$\begingroup$

This can be solved most elegantly (but may be less intuitively) by the Lagrangian method with dissipation. See Lagrangian mechanics - Extensions to include non-conservative forces.

The Lagrangian function of your free particle (mass $m$) on a sphere (constant radius $R$) is $$L = \frac{m}{2}\mathbf{v}^2 = \frac{m}{2}R^2(\dot{\theta}^2+\sin^2\theta\ \dot{\phi}^2) $$

The Rayleigh dissipation function $D$ is chosen so that it generates the drag force $\mathbf{F}_d=-\frac{\partial D}{\partial\mathbf{v}} =-\gamma\mathbf{v}$. That is $$D = \frac{\gamma}{2}\mathbf{v}^2 = \frac{\gamma}{2}R^2(\dot{\theta}^2+\sin^2\theta\ \dot{\phi}^2) $$

The Lagrangian equations (including dissipation) for $\theta$ and $\phi$ are $$ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right) -\frac{\partial L}{\partial \theta} + \frac{\partial D}{\partial\dot{\theta}} = 0 $$ $$ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\phi}}\right) - \frac{\partial L}{\partial\phi} + \frac{\partial D}{\partial\dot{\phi}} = 0 $$

Doing the calculus is straight-forward, and you get the equations of motion $$\ddot{\theta}=\sin\theta\cos\theta\ \dot{\phi}^2 - \frac{\gamma}{m}\dot{\theta}$$ $$\ddot{\phi}=-\frac{2}{\tan\theta}\dot{\theta}\dot{\phi} - \frac{\gamma}{m}\dot{\phi}$$

| cite | improve this answer | | | | |
$\endgroup$

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.