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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.

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Since Lagrangian mechanics only deals with conservative forces, friction here is only meant to convey a constraint on the system. Namely, whether the COM has translational motion along the plane on which the top sits.

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.

The point about which we calculate the moment of inertia matters here in the following way. Generally a top on a surface with friction will precess and nutate. In this case, despite the apex remaining a rest, the COM will move and therefore will have kinetic energy. In the case of a frictionless surface, we calculate the moment of inertia about the COM so that the kinetic energy splits into two parts:

$$ T = (\text{KE of motion of COM}) + (\text{KE of rotation about the COM}) $$

whereas, in the case of a surface with friction, calculating the moment of inertia about the apex will allow us to write the kinetic energy in terms of rotation about the apex (which will capture the kinetic energy due to the rotations of the COM).

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  • $\begingroup$ Thanks to you, I think I understand why the moment of inertia is calculated about the apex first and then the centre of mass, because we are considering the motion about those particular points. I am still finding it difficult as to why the KE under frictionless circumstances splits. I would of thought the KE of motion of COM would have gone to zero as the COM is stationary. $\endgroup$ – C.Pic Dec 6 '18 at 17:05
  • $\begingroup$ In the frictionless case the entire top can slide across the surface as it is spinning, so generally the KE will have contributions from both the COM translational motion and the rotational motion about the COM. If the surface has friction there may still be movement of the COM due to precession (rotation of the spin direction about the normal) and nutation (bobbing of the spin direction up and down from the spin direction) of the top. So generally, even if the apex position is fixed there will be COM motion. $\endgroup$ – astromm Dec 7 '18 at 3:17

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