# Gyroscopes and the torque required for precession

## 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

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BruceW
Homework Helper
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.

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$\omega$

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