Calculating Torque for Gyroscopes to Precess in Hubble Telescope

[Physics Dad]

Question
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 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 2.00 kg and diameter 5.00 cm , spinning about its central axis.

How large a torque would it take to cause these gyroscopes to precess through an angle of 1.20×10^-6 degree during a 5.00 hour exposure of a galaxy?

Equations
I=mr²
L=Iω
Ω=dΦ/dt
τ=ΩL

Variables and Conversions
m=2.00kg
r=0.025m
dΦ=(1.20×10^-6×2π)/360 = 2.09×10^-8 rad
dt=5×60×60=1800s
ω=(1.92×10⁴×2π)/60 = 640π rad/s

Attempt at Solution

I=(2.00kg)(0.025)²=1.25×10^-3kgm²
L=(1.25×10^-3kgm²)(640π)=4π/5
Ω=(2.09×10^-8)/(1800)=1.16×10^-12

τ=(1.16×10^-12)(4π/5)

=2.92×10^-12Nm

My answer is incorrect. Any help would be greatly appreciated. This is the ridiculous "mastering physics" website. It is the most pointless thing ever. I would much rather we were given an assignment to complete as at least then you can receive constructive feedback as opposed to simply being told the answer is wrong.

I am happy to be wrong as long as I can understand why. I am sure I have just made a simple mistake, like use the wrong equation or miscalculate something, but with no guidance I can't fix it.

Thank you.

 

[Charles Link]

It looks almost completely correct to me, but I found one arithmetic mistake: $\Omega =1.16 E-11$. (You missed the exponent by one unit). At least I think that's one place that needs a correction... editing..that would make the answer 2.92 E-11 if everything else is correct, which I think it may be.

 

[Cutter Ketch]

Well, I'm not sure if you can call it helpful, but I don't see anything wrong with this. I am 99.9% sure this is the right approach, and I checked almost all of the math. (I could only approximate multiplying by pi in the last step. doing it in my head while lying in bed.)

 

[Cutter Ketch]

It looks almost completely correct to me, but I found one arithmetic mistake: $\Omega =1.16 E-11$. (You missed the exponent by one unit). At least I think that's one place that needs a correction... editing..that would make the answer 2.92 E-11 if everything else is correct, which I think it may be.

I suck. What hurts more than having said I didn't see anything wrong is saying so AFTER you already found this mistake. Jeez.

 

[Charles Link]

I suck. What hurts more than having said I didn't see anything wrong is saying so AFTER you already found this mistake. Jeez.

That tends to re-enforce my belief that this might be the OP's only error. Looking forward to hearing from the OP.

 

[Physics Dad]

Thank you all for the replies, turns out there was a mistake in the system and my answer of 2.92*10¹² was correct after all. Had this verified by 3 lecturers at uni today.

You can all sleep easy now.

Thanks again.

 

[Charles Link]

Thank you all for the replies, turns out there was a mistake in the system and my answer of 2.92*10¹² was correct after all. Had this verified by 3 lecturers at uni today.

You can all sleep easy now.

Thanks again.

Very good, except when you said you had the correct answer after all, I was pretty sure there must be a second error and I found it: $dt=(5)(60)(60)=18000$, instead of 1800. Glad you got the right answer, but looks like you had offsetting errors.