What is the torque exerted by the malfunctioning thruster on the Jupiter 6?

  • Thread starter Thread starter Zyrn
  • Start date Start date
  • Tags Tags
    Thruster Torque
Click For Summary
SUMMARY

The torque exerted by the malfunctioning thruster on the Jupiter 6 is calculated to be 8.445 x 1011 Nm in the j-direction. The spaceship has a radius of 5000 m, a height of 200 m, and a mass of 8.00 x 108 kg. The thruster, located 30 degrees from the x-axis, exerts a force of 78i + 0j - 150k mega-Newtons for 2 seconds. The moment of inertia is determined using the formula I = (1/2)MR2.

PREREQUISITES
  • Understanding of torque calculation using τ = R × F
  • Familiarity with moment of inertia, specifically I = (1/2)MR2
  • Knowledge of angular motion equations, including Δθ = ωt + (1/2)αt2
  • Basic vector analysis for force components in three dimensions
NEXT STEPS
  • Study the principles of angular momentum and its conservation
  • Learn about the dynamics of rotating bodies in space
  • Explore advanced torque applications in engineering contexts
  • Investigate the effects of variable forces on rotational motion
USEFUL FOR

Physics students, aerospace engineers, and anyone interested in the dynamics of rotating spacecraft and the effects of malfunctioning propulsion systems.

Zyrn
Messages
4
Reaction score
0

Homework Statement



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

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 into 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.
 
Last edited by a moderator:
Physics news on Phys.org
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.
 
Yeah. That was it. Can't believe I didn't realize that sooner.
τ=8.445*10^11 Nm j
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 30 ·
2
Replies
30
Views
4K
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
8K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
13K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K