# Gyroscopic precession on a friction-less surface

I am having trouble understanding the total energy for a heavy spinning symmetric top (Gyroscope) on a friction-less surface. I am trying to understand it via the Lagrangian of the gyroscope. My understanding from my university lecture notes is that on a normal surface with friction, the Lagrangian takes the form of:

$$L=\frac{1}{2}I(\dot{\theta}^2 + \dot{\phi}^2sin^2(\theta)) + \frac{1}{2}I_3(\dot{\psi}+\dot{\phi}cos(\theta)) - Mgz$$

And that on a frictionless surface it takes the form:

$$L=\frac{1}{2}M(\dot{x}^2+\dot{y}^2+\dot{z}^2)+\frac{1}{2}I(\dot{\theta}^2 + \dot{\phi}^2sin^2(\theta)) + \frac{1}{2}I_3(\dot{\psi}+\dot{\phi}cos(\theta)) - Mgz$$

where the difference now is that the translational kinetic energy of the COM has also been taken into account.

This is causing me great confusion. The main questions with which I would like help understanding are:

• On a friction-less surface, if the COM is staying still, shouldn't the translational kinetic energy of the Gyroscope be zero given that it is not moving? So shouldn't its Lagrangian be the first equation?
• On a surface with friction, the COM is moving, so I can understand if the latter equation was used in that instance, but I am told it is not.

I have been told that in the first instance we take the moments of inertia to be about the Apex of the gyroscope (Point of contact with surface) and in the second instance on a friction-less surface the moments of inertia are taken about the centre of mass. But i'm still unclear how this makes the Lagrangian what they are in both instances.

Any explanation to aid my understanding will be greatly appreciated.

In the case of a surface with friction the constraint is simply that $$x,y=const.$$ and so their time-derivatives vanish. Similarly, in the case of a frictionless surface where we assume the COM of the top stays still the time-derivatives of $$x$$ and $$y$$ vanish in which case the Lagrangian reduces to the Lagrangian for a top on a surface with friction (our assumption can be taken as a constraint if you like, but generally we should find the equations of motion for $$x$$ and $$y$$ and use this assumption as the initial condition to solve for the trajectory). However, the motion along $$z$$ is not constrained and will generally be non-zero. This leads us to the next point.
$$T = (\text{KE of motion of COM}) + (\text{KE of rotation about the COM})$$