# Equations of motion on sphere with drag term

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?

• 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 $F= \left( \dfrac{\partial \overrightarrow{R}}{\partial \overrightarrow{q}}\right) ^{T}\left( -\gamma \overrightarrow{v}\right)$ – Eli Jan 29 at 13:16
• 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. – Qmechanic Feb 9 at 2:01 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:

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

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

• Thanks! I take it that in the $\ddot{\phi}$ equation, you mean $\sin(\theta)$? – bob.sacamento Jan 31 at 14:49
• @bob.sacamento this is because i use different position vector $\vec{R}$ then you did? – Eli Jan 31 at 15:46
• 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! – bob.sacamento Jan 31 at 16:45

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