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Torque to rotate spinning gyroscope

  • Thread starter soffish335
  • Start date
  • #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 1.92×104 . 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:

Answers and Replies

  • #2
557
1
Check this again:
Δθ=1.3*10^-6deg
=7.4484*10^-5rad
 
  • #3
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
 
  • #4
I think 1.27*10^-11 is not the correct answer , how is the question being solved in the correct way ?
 
  • #5
1
0
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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