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Gyroscopes and the torque required for precession

  • #1

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 = [itex]\frac{dL}{dt}[/itex] (1) , Ω = [itex]\frac{Δ∅}{Δt}[/itex] (2)

The Attempt at a Solution



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

Torque = [itex]\frac{Angular momentum before - Angular momentum after}{5X60X60}[/itex]

Torque = [itex]\frac{I(ωbefore) - I(ωafter)}{18000}[/itex]

This reduces down to Torque = [itex]\frac{∏}{90000}[/itex] - [itex]\frac{ωafter}{800X18000}[/itex]

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

Answers and Replies

  • #2
BruceW
Homework Helper
3,611
119
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.
 
  • #3
1
0
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 = [itex]\tau[/itex]/L
Ω = [itex]\tau[/itex]/I[itex]\omega[/itex]
[itex]\tau[/itex] = ΩI[itex]\omega[/itex]
 
  • #4
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
 

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