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?
 A: 

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