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Consider a symmetric, non-nutating precessing top with one point fixed (the tip if you will). It's symmetry axis is at an angle $\theta$ to the vertical and it steadily precesses at some angular velocity $\Omega$. The top's angular momentum component along its symmetry axis is:

$$L_z=I_z\dot{\psi}=const.$$

Now, given these conditions I wanted to see if the mass could be obtained. Note that Gravity is acting on the top. I found an expression by taking the time derivative of the angular momentum and equating that to the torque induced by gravity. My answer is:

$$m=\frac{L_z\Omega}{g\ell}.$$

Where $\ell$ is the distance from the fixed point to the center of mass. However this seems a bit strange because all the explicit dependence on the angle $\theta$ disappeared. My question is, does this make sense, and if so why? Or is my answer garbage? I'm thinking it has something to do with the fact that the initial assumptions make the top unable to nutate, and the $\theta$ dependence is somehow implicit.

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The expression you derived seems quite correct to me. I'd say the reason why you don't have an explicite dependence on the angle $\theta$ is that if there is no observable nutation (that is, if the top’s angular momentum due to precessional motion is small compared to its spin angular momentum), then the torque due to the earth’s gravitational field is always at right angles to the top’s spin angular momentum vector.

This diagram might be helpful.

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  • $\begingroup$ What I find strange is that it seems to work for arbitrary values of angular momentum and precession frequency because we impose the no nutation convention $\endgroup$
    – ClassicStyle
    May 18, 2016 at 4:19
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    $\begingroup$ @TylerHG Why exactly do you find it strange? What the expression is telling you is how $\Omega$ and $L_z$ must be related. Further, if you replace $L_z$ by $I \omega$, where $\omega$ is the spin angular velocity, you'll get a fixed relation between the precession angular velocity and the spin speed. See p. 6 of this document, where your expression is derived: physnet.org/modules/pdf_modules/m77.pdf $\endgroup$
    – dahemar
    May 18, 2016 at 8:37
  • $\begingroup$ Cool. That document cleared a lot up! $\endgroup$
    – ClassicStyle
    May 18, 2016 at 13:54

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