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I have been trying to solve the motion of a gyroscope with friction added, so that as in reality, its spin will slow down with time. My first naive attempt to add friction has run into trouble and I could use some help identifying what I am doing incorrectly. I present the issue, below, in zero g where I can use rotational symmetry to best demonstrate the problem. The Lagrangian in zero g is, given below, where the angles are the standard Euler angles used for gyroscopic motion:

$L=\frac{1}{2}I_{1}(sin\theta^{2} \dot{\phi}^{2}+\dot{\theta}^{2})+\frac{1}{2}I_{3}(\dot{\psi}+cos\theta \dot{\phi})^{2}$

The motion I want to consider is where there is ONLY spinning; that is $\dot{\psi}\neq0$, but $\dot{\theta}=\dot{\phi}=0$. For such pure spinning, I consider two cases $\theta=90^{0}$ and $\theta\neq 90^{0}$. By rotational symmetry of the laws of physics, these cases should give rise to equivalent equations of motion; but they don't, probably due to an oversight of mine, although I cant see what it might be.

Case 1:$\theta=90^{0}$ and $\dot{\theta}=\dot{\phi}=0$ with an initial spin of $\dot{\psi}\neq0$

Under these conditions, the Lagrangian becomes:

$L=\frac{1}{2}I_{3}\dot{\psi}^{2}$

whence we have

$\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial L}{\partial \dot{\psi}}=-Torque$

Note: Torque is a resistive torque applied by fiction to the axis of the gyroscope to slow it down.

and thus I obtain the correct equation of motion:

$I_{3}\ddot{\psi}=-Torque$

(1)

So far so good. Next, chose and angle when $\theta \neq 90^{0}$.

Case 2:$\theta\neq90^{0}$ and $\dot{\theta}=\dot{\phi}=0$ with an initial spin of $\dot{\psi}\neq0$

Like before I apply a resistive torque to the axis of the gyroscope to slow it down. By symmetry of the laws of physics we should get the same answer as Case 1, above. The trouble is I DON'T and the equation of motion I get is not equivalent to (1). I will now demonstrate this. Under the conditions of Case 2, the Lagrangian will the general form, listed above and using:

$\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial L}{\partial \dot{\psi}}=-Torque$

gives

$\frac{\mathrm{d} }{\mathrm{d} t}(I_{3}(\dot{\psi}+cos\theta \dot{\phi}) )=-Torque$

(2)

$\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial L}{\partial \dot{\phi}}=0$

gives

$I_{1}sin\theta^{2} \dot{\phi}+I_{3}(\dot{\psi}+cos\theta \dot{\phi})cos\theta=const$

(3)

and $\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial L}{\partial \dot{\theta}}-\frac{\partial L }{\partial \theta}=0$

gives

$I_{1}\ddot{\theta}+I_{3}(\dot{\psi}+cos\theta \dot{\phi})sin\theta=0$

(4)

To show the issue, I proceed commence with $\dot{\theta}=\dot{\phi}=0$ and assume that they will stay equal to zero.

Then equation (2) shows that $\dot{\psi}$ will start to slow down, which is what we would expect. But then, with $\dot{\phi}$, I encounter the first problem. Since $\dot{\psi}$ is decreasing, equation (3) tells us that $\dot{\phi}$ cannot remain zero; this is a problem as the torque should be along the spin axis only! The second problem is with $\dot{\theta}$. Since $\dot{\psi}$ is slowing down, and since $\dot{\phi}$ is initially zero, equation (4) tells us that that $\dot{\theta}$ cannot remain zero; again this is a problem as the torque should be along the spin axis only. The equations of motion are thus not equivalent to (1) as would have been expected.

I believe my error is in how I am applying the torque. I mean to apply the resistive torque along the axis of the gyroscopes spin, $\dot{\psi}$, but I suspect that I am somehow not doing that correctly. Possibly there is a transformation law I need to apply to my torque vector to point it correctly?

My question is: Can anybody tell me what I am doing wrong and how to fix it?

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Unsurprisingly, there have been no answers so far, as it is a rather niche area. In the meantime I think I may have found the answer myself. It occurs to me the following. Let us define angular momentum in the following way (using Einstein index notation):

$L_{i}=I_{ir}\omega^{r}$

Thus here I have written angular momentum is a contravariant tensor. We can then define the contravriant tensor that is the torque

$T_i=I_{iq} \dot{\omega}^{q}$

Above, $I_{ij}$ is the moment of inertia tensor, not to be confused with the identity matrix. Also, the $\omega^{j}$ are the orthogonal components of the rotation vector $\underline{\omega}$ in the selected orthogonal basis $\underline{i}$, $\underline{j}$ and $\underline{k}$ in the x, y and z directions.

Now lets say we transform our coordinate system to another one (which may not be orthogonal), then the transformation can me written:

$\omega'^{j}=T{^{j}}_{k}\omega^{k}$

and our moment of inertia tensor would transform as

$I'_{ij}=\left (T^{-1} \right ){^{p}}_{i}\left (T^{-1} \right ){^{q}}_{j}I_{pq}$

So that

$L'_{i}=I'_{ij}\omega'^{j}=\left (T^{-1} \right ){^{p}}_{i}\left (T^{-1} \right ){^{q}}_{j}I_{pq}T{^{j}}_{k}\omega^{k}=\left (T^{-1} \right ){^{p}}_{i}I_{pq}\left (T^{-1} \right ){^{q}}_{j}T{^{j}}_{k}\omega^{k}$

$L'_{i}=\left (T^{-1} \right ){^{p}}_{i}I_{pq}\delta{^{q}}_{k}\omega^{k}=\left (T^{-1} \right ){^{p}}_{i}I_{pk}\omega^{k}$

$L'_{i}=\left (T^{-1} \right ){^{p}}_{i}L_{p}$

Thus our contravariant angular moment transforms in the correct manner. With these definitions we can calculate the scalar equal to the kinetic energy

$\mathcal{L} =\frac{1}{2} L_{i}\omega^{i}=\frac{1}{2} L'_{i}\omega'^{i}$

whence we have that

$\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial }{\partial \omega^{i}} \mathcal{L} =\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial }{\partial \omega^{i}}\frac{1}{2}L_{p}\omega^{p}=\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial }{\partial \omega^{i}}\frac{1}{2}I_{pq}\omega^{p}\omega^{r}=\frac{\mathrm{d} }{\mathrm{d} t}I_{iq}\omega^{q}=I_{iq} \dot{\omega}^{q}$

Here I have used the fact that the moment of inertia tensor, I, is symmetric. We thus see that

$\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial }{\partial \omega^{i}} \mathcal{L}=T_{i}$

So therefor, if we need to transform from the cartesian coorinate system to a new coordinate system where we are using $\underline{\omega}'$ instead of $\underline{\omega}$ I think we would need to transform our torques according to:

$T'_{i}=\left (T^{-1} \right ){^{p}}_{i}T_{p}$

(*)

whence our equations of motion will transform to

$\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial }{\partial \omega'^{i}} \mathcal{L}=T'_{i}$

(**)

I don't know for sure, as I have not had to consider this type of problem before. But I believe, based on the mathematics of vector spaces, that the above is probably correct.

So I think, to answer my own question, I would write the axial torque out, contravariantly in x,y,z components. I can then easily find the transformation tensor T. It is just the linear transformation matrix that takes the cartesian rotation vectors to the vectors associated with the Euler angles. All I need to do then is use (*), above, to obtain the components of torque in the Euler angle coordinate system. And finally, the equations of motion that I want, in terms of the Euler angle coordinate system, are then just the equations(**)

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