Question

A gyroscope, a uniform disc of radius `R=5.0cm` at the end of a rod of length `l=10cm` (figure), is mounted on the floor of an elevator car going up with a constant acceleration `w=2.0m//s^2`. The other end of the rod is hinged at the point O. The gyroscope precesses with an angular velocity `n=0.5 rps`. Neglecting the friction and the mass of the rod, find the proper angular velocity of the disc.

Answer 1

The moment of inertia of the disc about its symmetry axis is `1/mR^2`. If the angular velocity of the disc is `omega` then the angular momentum is `1/2mR^2omega`. The precession frequency being `2pin`,
we have `|(dvecM)/(dt)|=1/2mR^2omegaxx2pin`
This must equal `m(g+w)l`, the effective gravitational torques (g being replaced by `g+w` in the elevator). Thus,
`omega=((g+w)l)/(piR^2n)=300rad//s`