# As a gyroscope precesses, its center off mass moves in a circle

1. Dec 10, 2011

### aaaa202

my textbook says:

"As a gyroscope precesses, its center off mass moves in a circle with radius r in a horizontal plane. Its vertical component of acceleration is zero so the upward normal force exerted by the pivot must equal mg."

Now wouldn't this always be true. I mean if u have a flywheel attached to a pivot and at first dont let it spin. Then the weight of it produces a torque that makes the arm rotate till it hits something, e.g. itself or the table it stands on. But its not translating linearly, so isnt there also a normal force exerted upwards by the pivot in this case? It just doesn't produce a torque since it's distance to the rotation axis is zero.

2. Dec 10, 2011

### Staff: Mentor

Re: gyroscope

No. In your example of the non-spinning flywheel, the vertical component of acceleration would not be zero. It's falling. There would be an upward normal force at the pivot, but it wouldn't equal mg.

3. Dec 10, 2011

### aaaa202

Re: gyroscope

what would it equal then? And what is it that makes the pivot exert a normal force in the case, where the flywheel IS spinning?

4. Dec 10, 2011

### Staff: Mentor

Re: gyroscope

Something less than mg. You could figure it out by computing the acceleration of the falling flywheel.
The flywheel has weight.

5. Dec 11, 2011

### aaaa202

Re: gyroscope

I dont get it. The flywheel isnt exactly falling it's just being rotated by the torque of its weight.

6. Dec 11, 2011

### Staff: Mentor

Re: gyroscope

Its center of mass is accelerating downward.

7. Dec 11, 2011

### rcgldr

Re: gyroscope

If the gyroscope is not precessing at the exact rate required to keep it near horizontal, it's center of mass is accelerating vertically. It could be oscillating up and down at it's rate of precession also oscillates.

If the gyroscope's center of mass isn't accelerating downwards at 1 g, then there's some force applied at the post opposing gravity. If the center of mass isn't accelerating vertically, then the upwards force from the post is m g.