Torque to rotate spinning gyroscope

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Homework Help Overview

The problem involves calculating the torque required to cause precession in gyroscopes used in the Hubble Space Telescope. The gyroscopes are modeled as thin-walled cylinders with specific mass and diameter, and the discussion revolves around the necessary torque to achieve a certain angular displacement over a given time period.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculations for angular displacement and torque, with some questioning the accuracy of the initial values used, particularly in the conversion of degrees to radians and the moment of inertia formula.

Discussion Status

There is ongoing clarification regarding the calculations, with participants pointing out potential errors in the original poster's approach. Some guidance has been offered about the correct formula for moment of inertia, but no consensus has been reached on the final answer.

Contextual Notes

Participants are addressing the need for precise conversions and the correct application of formulas, indicating that assumptions about the geometry of the gyroscope may need to be revisited. The original poster expresses uncertainty about the correctness of their calculations and seeks further understanding.

soffish335
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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 1.92×104 . Although the structure of these gyroscopes is actually quite complex, we can model each of the gyroscopes as a thin-walled cylinder of mass 2.00 and diameter 5.00 , 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 degree during a 5.00 hour exposure of a galaxy?

Homework Equations


L=I*ω
torque = Ω*L
I=mr^2
Ω=Δθ/Δt

The Attempt at a Solution


ω=1.92*10^4rpm
=2010.62rad/s

I=mr^2
=2(0.005)^2
=0.005

Δθ=1.3*10^6deg
=7.4484*10^-5rad

Δt=5 hours
=18000s

torque = Ω*I*ω
=(7.4484*10^-5/18000)*0.005*2010.62
=4.1*10^-8 Nm

This is the wrong answer and I feel as if the torque shold be rather larger by intuition. Any help would be appreciated.

Edit: 3.17*10^-12 Nm turned out to be the correct answer, but i'd still like to know how to do this question.
 
Last edited:
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Check this again:
Δθ=1.3*10^-6deg
=7.4484*10^-5rad
 
sorry, i made a mistake in my original post, the given angle is 1.3^-6deg.

ok, so now i get
Δθ=2.2*10^-8deg

torque=(2.2*10^-8/18000)*0.005*2010.62
=1.27*10^-11
 
I think 1.27*10^-11 is not the correct answer , how is the question being solved in the correct way ?
 
The problem with your solution, in case you are still wondering,
is I= mr^2
not I=md^2
therefore you take the diameter of 5 divide it by 2.
So I=2(0.025^2)
and that would be the last error after the conversion of degrees to radians which you fixed.
I just had to do the same problem so I decided to answer your question.
 

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