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For a free rotating symmetric top, where there is no torque, in the body fixed frame, am I correct in saying that the angular momentum $L$ is not constant because the angular velocity in the body fixed frame $$\omega = Acos(\Omega t+\phi )E1 + Asin(\Omega t+\phi )E2 + \omega _{3}E3$$ (where E1,E2 and E3 are the basis vectors in the principle axis direction) changes with respect to time and the angular momentum is also not parallel to the angular velocity because $$M=I\dot{\omega }+(\omega \times I\omega )$$ and so because the torque is 0 and the rate of change of angular velocity is also not 0 then the cross product is not 0? However the component $\omega _{3}$ is parallel to $L _{3}$ because Euler's equations say that for a free symmetric top with no torque, when $I_{1}=I_{2}$, $\dot{\omega _{3}}$ equals 0 and so $\omega _{3}$ is parallel to $L _{3}$. Is this correct? Thanks

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  • $\begingroup$ What you write is a bit unclear. Freely rotating – in which kind of frame? By "freely rotating" do you mean that the torque is zero (which seems to contradict what you write afterwards)? Also, in the frame of the (rigid) top the angular momentum should be zero. $\endgroup$
    – pglpm
    May 16 at 14:05
  • $\begingroup$ Thanks for you comment. Yeah I mean that the torque is 0, and I think I mean the lab frame because Euler's equations use the torque in the inertial frame. I don't see how I have contradicted that the torque is non 0. I was just stating Euler's equation and then setting the torque to 0. I guess my question is how does the angular momentum change in the rigid body frame. I don't see how it can be 0 because it is precessing? $\endgroup$
    – aP123
    May 16 at 14:32
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The angular momentum, seen from the inertial frame, is constant. The angular velocity is not constant.

The curve traced out by the endpoint of the angular velocity vector $\omega$ of a freely rotating rigid body is called the herpolhode The endpoint of the angular velocity moves in a plane in absolute space, called the invariable plane which is orthogonal to the angular momentum vector ${\bf L}$. The fact that the herpolhode is a curve in the invariable plane appears as part of Poinsot's construction.

As described in Goldstein's Classical Mechanics: "The polhode rolls without slipping on the herpolhode lying in the invariable plane."

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This is only an addendum to Mike's answer, which is correct.

This may clear up some confusion:

You may be following Taylor's Classical Mechanics, which seems to be the case from your wording. If so, he writes the Euler equations in terms of the body frame. In this frame, L (angular momentum) is moving in a loop about $ \hat e_3 $ much like $ \omega $ (the vector) and $ \hat e_3 $ move around in a circle about L in the lab/inertial frame.

Consequently, you are correct that $ L_3 $ is constant when you work out the equations, but that the $ L_2 $ and $ L_1 $ components are not, so $ L $ the vector consequently rotates around $ \hat e_3 $ in the body frame. $ \omega $ does too. (this is assuming body-fixed means "the body is fixed in this frame".)

Finally, saying $ \omega_3 \hat e_3 $ is parallel to $ L_3 \hat e_3 $ is correct, but I'm not sure why you bring it up.

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  • $\begingroup$ Thankyou very much, that makes more sense. Just one thing, by using the equation in the OP if $M=I\dot{\omega }+(\omega \times (I\omega ))$ then if torque is 0 does that mean that if we take the first and second components (as opposed to the 3rd) then because $\dot{\omega }_{1}$ and $\dot{\omega }_{2}$ don't equal 0 and $M_{1} $ and $M_{2} $ equal 0 then $\omega _{1}\times L_{1}$ and $\omega _{2}\times L_{2}$ are non 0 and therefore $\omega _{1}$ is not parallel to $L_{1}$ and likewise for the second component? Is that true-I'm not sure if that is correct mathematically. $\endgroup$
    – aP123
    May 16 at 18:34
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    $\begingroup$ I actually don't remember the M equation; I guess its the equation for torque but frankly I've forgotten. I'm unsure what you mean by $ \omega_1 \times L_1$. L is a vector, $ L = ( L_1, L_2, L_3) $ and likewise for $ \omega$. Consequently, saying $ \omega_1 \times L_1$ would be interpreted as $ \omega_1 e_1 \times L_1 e_1$, if you use $ \times $ to indicate cross product. However, I think you have the right picture in your head, because as a whole, $ L $ is not parallel to $ \omega$ since the moments of inertia differ for the $ e_3 $ versus $ e_2 $ and $ e_1 $ axes. $\endgroup$
    – anon.jpg
    May 16 at 18:47
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    $\begingroup$ to be clear, $ \omega \times L \ne \omega_1 \times L_1 + \omega_2 \times L_2 ... $ etc. you may be confusing dot product with cross product? $\endgroup$
    – anon.jpg
    May 16 at 18:48

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