To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.
[tex]I = MR^2[/tex]
Here,
M = Mass
R = Radius of the hoop
The precession frequency is given as
[tex]\Omega = \frac{Mgd}{I\omega}[/tex]
Here,
M = Mass
g= Acceleration due to gravity
d = Distance of center of mass from pivot
I = Moment of inertia
[tex]\omega[/tex]= Angular velocity
Replacing the value for moment of inertia
[tex]\Omega= \frac{MgR}{MR^2 \omega}[/tex]
[tex]\Omega = \frac{g}{R\omega}[/tex]
The value for our angular velocity is not in SI, then
[tex]\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})[/tex]
[tex]\omega = 104.7rad/s[/tex]
Replacing our values we have that
[tex]\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}[/tex]
[tex]\Omega = 1.17rad/s[/tex]
The precession frequency is
[tex]\Omega = \frac{2\pi rad}{T}[/tex]
[tex]T = \frac{2\pi rad}{\Omega}[/tex]
[tex]T = \frac{2\pi}{1.17}[/tex]
[tex]T = 5.4 s[/tex]
Therefore the precession period is 5.4s
