# Gyroscopic Precession

1. Jan 26, 2009

### csnsc14320

1. The problem statement, all variables and given/known data
A gyroscope consists of a rotating disk with a 48.7cm radius suitably mounted at the midpoint of a 12.2cm long axle so that it can spin and precess freely. Its spin rate is 975 rev/min. The mass of the disk is 1.14kg and the mass of the axle is 130g. Find the time required for one precession if the axle is supported at one end and is horizontal.

2. Relevant equations
$$\omega_p = \frac{m g r}{L}$$
$$\L = I \omega$$

3. The attempt at a solution
Using the equation, I plugged in (1.14kg+.130kg)(9.8m/s^2)(.112/2m)/(.5(1.14kg)(.487m)^2(32.5pi)) and got that $$\omega_p$$ = .05 rad/s, which will take 2.074 min to complete one full precession. However, the back of the book says the answer is 1.90 min. Am I forgetting something?

2. Jan 26, 2009

### chrisk

You must include the moment of inertia for the rod about the pivot point.

3. Jan 26, 2009

### csnsc14320

I can't seem to figure out where to include it. Including it on the bottom (L = (Idisk + Irod)w) doesn't work, and trying to add the angular momentum of the precession doesn't seem to work either?

4. Jan 26, 2009

### chrisk

Think of the problem in two parts. Ignore the mass of the rod in the first part. This will give you the torque created in the azimuthal direction. This torque is applied to the rod. Find I for the rod and this will give you the net angular velocity or precession of the system.

5. Jan 26, 2009

### csnsc14320

I of rod = ML^2/3
Torque = (M of disk + M of rod)gR, since the rod's weight would also be contributing to the torque?)

If I use torque = (1.14 + .13)(9.8).112/2 = .697

then I have an equation torque = angular velocity of precession x angular momentum

if this is the right eq. to use, which angular momentum would I use?

6. Jan 26, 2009

### chrisk

I believe this problem takes into consideration the rod rotating. Then the moments of inertia of the rod and the wheel would add since they're rotating about the same axis and this would give the total moment of inertia

$$L=I_{wheel}\omega\mbox{ + }I_{rod}\omega$$

So calculate the moment of inertia of the rod about an axis the same as the wheel.

7. Jan 26, 2009

### csnsc14320

I don't think it does take into account the rod's inertia - I just plugged into $$torque = \omega_p I_d \omega_d$$ and got the right answer. I don't think the rod's moment of inertia is counted since it would produce an angular momentum up, and therefore a torque up, but I think the precession is stable in a flat plane, so the bobbing motion is not taken into account.

Plus, if we were taking into account all the moment of inertias, wouldn't we add the moment of inertia of the disk about the axle, the moment of inertia of the rod about the pivot, AND the moment of inertia of the disk about the pivot?

8. Jan 27, 2009

### chrisk

Yes, the problem does not take into account the other moments of inertia because the rod is much lighter than the wheel so appoximations are made; the angular momentum vector due to the wheel is parallel with the axis of rotation. A rigorous analysis would show nutation, and fast and slow precession solutions. I'm glad you figured it out!