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I'm puzzled by the following questions on gyroscope in HRK physics 5ed (p. 220)

Basically the gravity torque is $~τ=Mg\,L\,\sinθ~$. The angular momentum $L_s=I_s\,ω_s~$ has a horizontal radial component ($I_s\,ω_s\,\sinθ~$) keeps rotating. So $\frac{dL_s}{dt} = I_s\,ω_s\,\sinθ\,ω_p~$, Substitute $τ=\frac{dL_s}{dt}$ we get $ω_p = \frac{M\,g\,L}{L_s}$

The question is the mass center is doing circular motion. So relative to the the axis at the bottom, it has an angular momentum and this angular momentum also has a horizontal radial component ($~M\,L\,\cosθ\,L\,\sinθ\,ω_p~$) keeps rotating with ωp. Do we need to consider this in the $τ=\frac{dL}{dt}$ equation?

Or it is because $ω_s \gg ω_p$, we ignore this term?

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  • $\begingroup$ Please render all formulas in Mathjax. $\endgroup$
    – Gert
    Nov 7, 2021 at 17:01

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You are right, the $ \overrightarrow{(\rm pos)} \times \overrightarrow{(\rm momentum)}$ terms should be included.

$$ \vec{L} = \mathrm{I}\, \vec{\omega} + \vec{r} \times \vec{p} $$ $$ \vec{L} = \mathrm{I}\, \vec{\omega} + \vec{r} \times M ( \vec{\omega} \times \vec{r}) $$

or in body-frame vectors

$$ \vec{L}_{\rm body} = \mathrm{I}_{\rm body}\, \vec{\omega}_{\rm body} + L \hat{\jmath} \times M ( \vec{\omega}_{\rm body} \times L \hat{\jmath}) $$

which as you point out is a $L^2 M \omega_p \sin \theta$ term.

It is not included though, if it is already incorporated into the mass moment of inertia, but evaluating it about the pivot

$$ \vec{L} = \mathrm{I}_{\rm pivot} \vec{\omega} $$ which should produce the same result.

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