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Gyroscopes and the torque required for precession

  1. Nov 13, 2011 #1
    1. The problem statement, all variables and given/known data

    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?

    2. Relevant equations

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

    3. 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: Nov 13, 2011
  2. jcsd
  3. Nov 13, 2011 #2


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    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.
  4. Nov 14, 2011 #3
    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]
  5. Nov 14, 2011 #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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