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?

1 Answer
Feb 8, 2016

Answer:

#\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.