Torque from a thruster

  • Thread starter Zyrn
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Homework Statement



http://spiff.rit.edu/webwork2_course_files/phys216/tmp/gif/set24-prob2-./giant_saucer.gif [Broken]

The Jupiter 6 is a giant spaceship in the shape of a big, flat disk: it has radius R=5000 m, height h=200 m, and mass M=8.00E+08 kg. As it flies through space, it rotates around its center with a period of P=200 seconds to generate artificial gravity. The rotation looks counter-clockwise as seen from above the ship (if you can't tell from the diagram).

One day, one of the outboard thruster rockets malfuctions. It is located at the point marked with a big black dot on the diagram, on the outer rim of the ship, an angular distance theta=30 degrees away from the x-axis. The thruster exerts a force with components 78i+0j−150k mega-Newtons for a duration of t=2 seconds before the technicians can disable it.

What is the torque exerted around the center of the ship by the thruster?

Homework Equations



I=(1/2)MR^2
τ=R×F
Δθ=ωt+(1/2)αt^2
α=τ/I

The Attempt at a Solution



R=5000m
M=8*10^8kg
ω=(2π)/(200s)
F=(78i+0j−150k)MN

I=(1/2)(8*10^8kg)(5000m)^2=1*10^16kgm^2
R_x=Rcosθ
R_z=Rsinθ
τ=(150MN(5000m)cosθ+78MN(5000m)sinθ)j (using matrix)

α=(150MN(5000m)cosθ+78MN(5000m)sinθ)/(1*10^16kgm^2)
θ=((2π)/(200s))2s+(1/2)((150MN(5000m)cosθ+78MN(5000m)sinθ)/(1*10^16kgm^2))(2s)^2

I'm not entirely sure what to do at this point. I tried graphing to get θ, then plugging it in to the equation for τ multiple ways. Integrating from .5236 (which came from the 30°) to .5236-θ, taking the difference of the value of τ when plugging in .5236 and .5236-θ, just plugging in θ. I'm stuck of what to do and any input would help.
 
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Answers and Replies

  • #2
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I might have just figured out how I was going about it incorrectly. Rather than having the thruster force always in the same direction, it would be changing since it's attached to the ship. I'm going to try it that way.
 
  • #3
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Yeah. That was it. Can't believe I didn't realize that sooner.
τ=8.445*10^11 Nm j
 

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