# Gyroscope Angular Momentum Analysis

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?

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

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.