Gyroscopes and the torque required for precession

[physicsguy128]

Homework Statement

The Hubble Space Telescope is stabilized to within an angle of about 2 millionths of a degree by means of a series of gyroscopes that spin at 1.92×10^4 rpm. Although the structure of these gyroscopes is actually quite complex, we can model each of the gyroscopes as a thin-walled cylinder of mass 2.00kg and diameter 5.00cm , spinning about its central axis.

How large a torque would it take to cause these gyroscopes to precess through an angle of 1.30×10−6 degrees during a 5.00 hour exposure of a galaxy?

Homework Equations

Torque = $\frac{dL}{dt}$ (1) , Ω = $\frac{Δ∅}{Δt}$ (2)

The Attempt at a Solution

For a gyroscope torque is a a rate of change of momentum (1). So,

Torque = $\frac{Angular momentum before - Angular momentum after}{5X60X60}$

Torque = $\frac{I(ωbefore) - I(ωafter)}{18000}$

This reduces down to Torque = $\frac{∏}{90000}$ - $\frac{ωafter}{800X18000}$

I am at a loss of how to find ω after, could anyone shed some light? Cheers

 

[BruceW]

I think you need to look up precession. I think precession is when the object keeps rotating at the same speed, but the axis of rotation moves round in a cone-like shape as well. Much like a spinning top, when its axis is not perfectly vertical, its axis will wobble around the vertical direction, due to the torque applied by gravity.

 

[ngilmore]

If you still need help on this problem, I figured it out. I used the formula Ω = torque/angular momentum.
As we know, angular momentum is simply the moment of Inertia * the angular velocity. In the problem you are given the angular velocity as 1.92*10^4 rpm. convert that to radians/second.

Now you need to find the moment of Inertia. Since it's a thin-walled cylinder, you know I=mr^2

As for omega you are given the degrees and the time. you have to convert degrees into radians and time into seconds. Divide your radians/secs and then multiply your answer by the moment of Inertia and angular velociy. That will give you your answer as:
Ω = $\tau$/L
Ω = $\tau$/I$\omega$
$\tau$ = ΩI$\omega$

 

[physicsguy128]

Thanks, what i was doing wrong was to convert from revs / s to radians / s was dividing by 2∏ when actually i should have multiplied by ∏! oops! Cheers