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Angular Velocity of the Hubble Space Telescope

  1. May 15, 2009 #1
    1. The problem statement, all variables and given/known data

    The Hubble Space Telescope is poered by two solar panels as shown. The body of the telescope has a mass of 11 Mg and a radii of gyration kx = 1.64m and ky = 3.85m, whereas the solar panels can be considered as thin plates, eac having a mass of 54 kg. due to an internal drive, the panels are given an angular vlocity of .6j rad/s measured relative to the telescope. Determine the angular velocity of the telescope due to the rotation of the panels. Prior to roatin the panels, the telescope was originally traveling at vg = -400i + 250j + 175k m/s

    2. Relevant equations



    3. The attempt at a solution

    G--> m = 11,000 kg, kx = 1.64m, ky = 3.85m, vinitail = -400i + 250j + 175k m/s

    Panels--> 6m X 1.5m, m = 54 kg each, w = .6j rad/s

    mtotal = 11,000 + 2(54) = 11108 kg

    Ix = mkx2 = 11108(1.64)2 = 29876.0768

    Iy = mky2 = 11108(3.85)2 = 163794.1248

    Should i then use principle of impulse and momentum by doing for each axis.
     

    Attached Files:

  2. jcsd
  3. May 15, 2009 #2

    Doc Al

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    What's conserved?
     
  4. May 15, 2009 #3
    by that do u mean using V = Wy

    do i use that for each access,

    like for Vx = Wy = 11108(-400)

    because it is moving -400i
     
  5. May 15, 2009 #4

    Doc Al

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    :confused: Not sure what you are talking about.

    What quantity is conserved when the panels rotate?
     
  6. May 15, 2009 #5
    I'm confused, what do u mean by conserved

    w = .6j when the panels rotate, does this somehow apply, should i use w = v/r
     
  7. May 15, 2009 #6

    Doc Al

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    Conserved meaning "doesn't change", as in "conservation of energy". What's conserved in this case?

    You will certainly need the angular velocity of the plates, which is given as 0.6 rad/s with respect to the telescope. Which way will the telescope start to rotate?
     
  8. May 15, 2009 #7
    the weights will not change

    also the velocity in the k direction will remain

    the velociy in the j direction changes becuase of the w = .6j, but will it also change because of the radii of gyration

    will the velocity in the i direction change also because of the radii of gyration



    do u just add the .6j to the 250j to get velocity = 250.6 j m/s in terms of just j
     
  9. May 15, 2009 #8

    Doc Al

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    The linear velocity is not going to change.

    What's conserved? Big hint: It has something to do with angular velocity.
    No. You can't add an angular velocity (in rad/s) to a linear velocity (in m/s).
     
  10. May 15, 2009 #9
    Ok so the tele scope will travel continiously on the linear velocity vector

    the panels hoever will be rotating @ .6 rad/s so that the y axis is the axis of revolution

    "You will certainly need the angular velocity of the plates, which is given as 0.6 rad/s with respect to the telescope. Which way will the telescope start to rotate? "

    why would this cause the telescope to rotate. isnt the rotation independant
     
  11. May 15, 2009 #10

    Doc Al

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    The problem asks "Determine the angular velocity of the telescope due to the rotation of the panels." The rotation of the panels is caused by the telescope exerting a torque on them--thus the panels exert a torque back on the telescope, making it rotate.

    During all this, what's conserved?
     
  12. May 15, 2009 #11
    Ah i think i gotcha I need to use Conservation of Angular Momentum equation
    H1 = H2

    rmv = rmv

    my radius for the panel is 3m
    the mass is 54 kg but theres two of them so i think i should us 108kg
    v = .6 rad/s

    radius of the telescope is unknown
    mass is 11,000 kg
    v = -400i + 250j +175k m/s

    do i have to convert the velocities somehow
     
  13. May 15, 2009 #12

    Doc Al

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    Yes, you need to use conservation of angular momentum. (Usually an L is used to represent angular momentum.)

    "rmv" is the angular momentum of a point mass, not what you need here. The form you need here is L = Iω. You're given the rotational inertia of the telescope; you'll need to figure out the rotational inertia of the panels.
     
  14. May 15, 2009 #13
    ok i see now

    rotation inertia is k right

    L = Iw = .6I

    I = mk2 = 54k2 should i double the 54 since there are two of them

    L = .6(54k2 = 32.4k2

    is this the right track
     
  15. May 16, 2009 #14

    Doc Al

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    k is the radius of gyration; I = mk2.

    Careful: The rotational speed of the panels is not 0.6--that's its speed with respect to the telescope.

    Hint: Call the unknown rotational speed of the telescope ω. What would be the rotational speed of the panels in terms of ω? Use that to write the conservation of angular momentum.

    You aren't given the k for the panels, but you know their mass and dimensions so you can figure out their rotational inertia.
     
  16. May 17, 2009 #15
    I = mr2
    I = 108(32) = 972 I double the weight since there are two panels

    k=[tex]\sqrt{I/m}[/tex] = [tex]\sqrt{972/108}[/tex] = 3

    u used equation I = mr2 & k = [tex]\sqrt{I/m}[/tex], but doens that mean that r = k, so im assuming this is wrong
     
  17. May 17, 2009 #16

    Doc Al

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    The panels are rectangular plates. What formula would you use to calculate their rotational inertia? (Hint: Compare to a rod.)
     
  18. May 17, 2009 #17
    oooooooo

    I = 1/12ML2 = 1/12(108)(6)2 = 324

    I double the mass since theres two plates

    therefore k for the plates is

    k = [tex]\sqrt{324/108}[/tex] = 1.732

    so this is in the y directino so do i add this to the k for the telescope in the y direction
     
  19. May 18, 2009 #18

    Doc Al

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    Good.

    The only purpose for needing k would be to find I. But you just found I!

    No.

    This problem is the rotational analog to a problem like this: Say two kids were on ice. The boy shoves the girl so that the girl moves at speed X with respect to him. How fast is the boy moving after the shove?

    This is a similar problem but with rotation. The telescope turns the panels. But angular momentum of the system (telescope + panels) cannot change. It started at zero and will remain at zero. If the panels move clockwise, the telescope has to move counterclockwise.
     
  20. May 18, 2009 #19
    Ok im confused as to what to do now,

    do i plug it into the conservation of angular momentum
     
  21. May 18, 2009 #20

    Doc Al

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    Here's how conservation of angular momentum would apply to this problem:
    Itelescopeωt,i + Ipanelsωp,i = Itelescopeωt,f + Ipanelsωp,f
     
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