# What are the equations of motion for a pendulum in $n$ dimensions?

Consider a pendulum formed of a light rod of length $$l$$ and a point mass $$m$$. At the pivot the pendulum experiences friction with coefficient $$b$$, and an agent can exert torque $$\tau$$.

In 2 dimensions the motion of this pendulum is given by (I hope): $$\ddot{\theta} = \frac{g}{l} \sin \theta + \frac{1}{m l^2} \tau - \frac{1}{m l^2} b \dot{\theta}$$ where $$\theta$$ is the angle with the vertical (the positive bit of the y axis).

I would like to extend this to $$n$$ dimensions.

My idea was to come up with general equations for $$n$$ dimensions, and then check they reduce to the correct solutions in 2D and 3D.

It seems easy to convert $$\theta$$, $$\tau$$ and $$b$$ into vectors (to get the second two terms of the eq above), but I am having trouble working out how to project gravity into each of the dimensions of $$\bf{\theta}$$ (the first term).

For a given dimension of $$\theta$$, I think I need to find the vector which is in the hyperplane that the dimension "moves" in and is orthogonal to the rod of the pendulum. We can then take the dot project between gravity and this vector. However I'm not sure how to find this vector. I am also struggling because gravity is naturally given in cartesian coordinates, whereas the pendulum is naturally given in spherical coordinates (with fixed r).

• The paper I'm following is: arxiv.org/abs/1803.08287. On page 7 they define it like this, but their friction parameter is in Nms/rad. I'll update the question to just $b\dot{\theta}$ for simplicity. Jul 11, 2019 at 15:28
• Ah no sorry, you are right. I got things mixed up for a bit Jul 11, 2019 at 15:34
• @AaronStevens you use hyperspherical coordinates, with one radius and $n-1$ angles. Jul 11, 2019 at 16:52
• @AaronStevens well, that is the main problem with the question. Only in a circle do you have a nonambiguous notion of a constant vector field. In even dimensional spheres, you can't even define a nonzero vector field. OP should think about how to define the torque on a sphere. Jul 11, 2019 at 16:57
• Honestly, using a Lagrangian perspective would probably be a lot easier here, particularly if you're willing to ignore external torques & dissipative forces. Jul 11, 2019 at 19:51