Question

A gyroscope consists of a uniform disk with a 49 cm radius at the center of an axle that is 9 cm long and of negligible mass. It is horizontal and supported at one end. The disk is spinning around the axle at 990 rev/min, what is the precession rate?

Answer 1

`\omega_p = \frac{2gr}{\omegaR^2}=0.03148` rad/s `=0.3007` rev/min.

Explanation:

If `\vec{F_{}}` is the force generating the torque `\vec{\tau_{}}` on the gyroscope, the precession frequency `\omega_p` is related to the angular momentum `\vec{L_{}}` as follows. The angular momentum itself is related to the moment-of-inertia `I` and spin frequency `\omega`.

Torque : `\vec{\tau_{}} \equiv \vec{r_{}}\times\vec{F_{}}=\frac{d\vec{L_{}}}{dt}`,
Angular Momentum: `\vec{L_{}}\equiv I\omega`

Precission Frequency: `\omega_p = 1/L\frac{dL}{dt}=\tau/L`

The torque generating force is the weight ( `F=Mg`) of the spinning disk which acts vertically down. Since the gyroscope's axle is horizontal to the ground, `\vec{r_{}}` and `\vec{F_{}}` are perpendicular to each other.
`\tau = r.F.sin90^o=r.Mg.sin90^o=Mgr`

Moment of inertia of a disk of mass `M` and radius `R` is `I=\frac{MR^2}{2}`

Angular momentum `L=I.\omega=1/2 MR^2\omega`

Precessional frequency `\omega_p = \tau/L = \frac{Mgr}{1/2MR^2\omega}=\frac{2gr}{\omegaR^2}`

Given : `R=49cm=0.49m`;`\quad` `r=4cm=0.04m` `\quad` `g=9.8ms^{-2}`
`\omega=990` rev/min `=33\pi` rad/s,

`\omega_p = \frac{2gr}{\omegaR^2}=0.03148` rad/s `=0.3007` rev/min.