I’ve started with lagrange equations:

[itex]

x(t)=l(t) \sin (\theta) \cos (\phi)\\

y(t)=l(t) \sin (\theta) \sin (\phi)\\

z(t)=h(t)+l(t) \cos(\theta)\\ \\

T = \frac{1}{2}m(\dot x^2 +\dot y^2+\dot z^2)\\

U=m g l(t)(1-\cos(\theta)) + mg(S-(h(t)+l(t)))

[/itex]

where [itex]l(t)[/itex] is the length of the string. Here only I am assuming the radius of the pole is really small compared to [itex]l[/itex]. The polar angle is [itex]\theta(t)[/itex]. [itex]h(t)[/itex] is the change in height due to the string wrapping on the pole. [itex]S[/itex] is the length of string when unwrapped.

The change in length along is given by:

[itex]

\dot l(t) = -\frac{r\dot\phi}{ \sin(\theta)}

[/itex]

where [itex]r[/itex] is the radius of the pole. And the sliding pivot point is given by:

[itex]

\dot h(t) = \frac{r\dot\phi}{ \tan(\theta)}

[/itex]

After plugging those in [itex]T[/itex] I apply the Lagrange derivative to [itex]L = T-U[/itex] and solve for [itex]\ddot \theta[/itex] and [itex]\ddot \phi[/itex]

Now when I simulate the results I get a linear velocity that is some how increasing which is not supposed to happen for a tetherball because no new energy has been introduced to the system and the angular momentum is not conserved.

I would appreciate some feedback

http://ift.tt/1mdruPb