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K of Rotation vs Rotational Momentum

  1. Nov 17, 2013 #1
    1. The problem statement, all variables and given/known data
    A space station has the form of a hoop of radius R = 15 m, with mass M = 1000 kg. Initially its center of mass is not moving, but it is spinning with angular speed ωi = 4 rad/s. A small package of mass m = 22 kg is thrown at high velocity by a spring-loaded gun at an angle θ = 25 ◦ toward a nearby spacecraft as shown. The package has a speed v = 380 m/s after launch. What is the space station’s rotational speed ωf after the launch? You may ignore the weight of the package in calculating the moment of inertia of the space station. Answer in units of rad/s


    2. Relevant equations
    Krot = 1/2 Iω2
    L = Iω
    τ = rfsinθ
    Ktrans = 1/2mv2
    I = mr2

    3. The attempt at a solution

    [STRIKE]I tried to use kinetic energy for this, and I got an answer that seems reasonable, but this is the torque/rotational momentum chapter so I don't want to submit an answer until I think it's right.

    I used the energy principle to solve this:

    Krot,i = Krot,f + Ktrans

    Since they provide us with an angle, I can't help but think torque is probably involved. There wasn't a way to find the force applied by the box on the wheel, so I decided torque wasn't really valid to use. Another point of contest is the fact that the station could gain some of the translational kinetic energy. I was thinking about it, and if the package was shot off at a 90 degree angle, all the energy would be transmitted to the translational kinetic energy of the station.

    Now that I think about it, it may be smart to apply only the y component towards reduction of the[/STRIKE] ... blah blah blah I stopped typing and took a second look.

    I used the y component of the kinetic energy of the block as reduction of the rotational kinetic energy of the station. The rest will be applied as translational kinetic energy. So pretty much:

    Krot,i = Krot,f + Ktrans,block sinθ

    Assuming I'm competent enough to use a calculator, some validation on whether or not this line of thought is proper would be nice. Thanks :)
     
    Last edited: Nov 17, 2013
  2. jcsd
  3. Nov 17, 2013 #2

    haruspex

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    as shown?
     
  4. Nov 17, 2013 #3
    You don't need the image to solve it.
     
  5. Nov 17, 2013 #4

    haruspex

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    No, but I will then need a clearer description. What is angle theta the angle between?
    - trajectory and radius?
    - trajectory and tangent?
    - trajectory and axis?
    - trajectory and plane?
     
  6. Nov 17, 2013 #5

    tiny-tim

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    Hi Lamebert! :smile:
    Sorry, but this is completely wrong …

    this is a collision (in reverse) …

    (mechanical) energy is not usually conserved in a collision, but momentum, and angular momentum, are :wink:

    (and anyway kinetic energy is a scalar, not a vector, so it doesn't have components)
    i agree :redface:
     
  7. Nov 17, 2013 #6
    I figured it out earlier today, but thanks. :tongue:
     
  8. Nov 17, 2013 #7

    haruspex

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    Actually I missed some possibilities. If the package is fired in the same plane as the space station (which isn't clear), not only aren't we told whether the given angle is to the radius or to the tangent, we don't know whether it's in the the forward or backward direction in relation to the spin. Nor indeed whether the 380 m/s is relative to the gun or to an inertial frame in which the axis of the space station is initially at rest.
    So how did you manage to deduce all those facts from the OP?:confused:
     
  9. Nov 18, 2013 #8

    tiny-tim

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    hi haruspex! :smile:
    but the method is the same either way!

    if the OP has used the right method, we can say "correct, assuming the direction is forward :smile:"

    and if the OP has used the wrong method (as obviously here … though he sorted it out himself later! :approve:), why does it matter? :wink:

    since we're only helping, and not answering, we don't (usually) need to know the minutiae of the question o:)
     
  10. Nov 18, 2013 #9

    haruspex

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    Not if it had been at 25 degrees to the plane or to the axis of the space station.
     
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