# Nutation frequency for a weighted gyroscope

So, I've been led to believe that the frequency of nutation of a gyroscope can be calculated using the formula In which the I's are the moments of inertia around the principal axes and omega-3 is the angular velocity of the disk of the gyroscope. But the moments of inertia around the principal axes of a cilinder are

and working this out for a gyroscope with a disc of 1.5 kg, a radius of 11.5 cm and a height of 2,5 cm I get: (There was a math error here, it's gone now) Which would mean the frequency of nutation is 2 times higher than the angular velocity, which does not match observation. What am I doing wrong? Thank you in advance. The gyroscope used.

• You have misunderstood the formula. The correct version relates the nutation and precession frequencies, not the angular velocity. Your interpretation of what the three moments of inertia are also seems to be wrong. Jun 10, 2019 at 21:18
• Could you perhaps direct me to some place where this is explained further? I got that formula from Morin's "Introduction to Classical Mechanics" and this is the best I can make of it. Jun 10, 2019 at 21:23
• The moments of inertia that you are currently using are the ones for the case of a body that is in free fall. The torque-free precession in free fall is with respect to the center of mass of the body. (The spinning disk in free fall case is also known as Feynman's wobbling plate, which indeed displays that ratio of 2:1) However, the picture of your setup shows a wheel mounted on an axis, and the pivot point is far away from the wheel. You need to use the moment of inertia of that entire assembly: wheel, axis and counterweight. Jun 12, 2019 at 18:54