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Angular momentum, moment of inertia etc.

  1. Dec 29, 2006 #1
    1. The problem statement, all variables and given/known data
    A space ship is located in a gravity free region of space. It consists of a large diameter, thin walled cylinder which is rotating freely. It is spinning at a speed such that the apparant gravity on the inner surface is the same as that on earth. The cylinder is of radius r and mass M.
    (a) discuss the minimum total work which had to be done to get the cylinder spinning.

    (b) radial spokes of negligible mass, connect the cylinder to the centre of rotation. An astronaught of mass m cllimbs a spoke to the centre. What will the fractional change in the appearant gravity on the surface of the cylinder?

    (c)If the astronaught climbs halfway up a spoke and lets go how far from the base of the spoke will he hit the cylinder, assuming the astronaught is point-like?

    2. Relevant equations
    omega.r=0 so that means all the normal equations simplify out nicely.

    3. The attempt at a solution
    I'm doing this question for revision, so since no answers are provided I'd be grateful if someone could check my answers for (a) and (b):

    for (a): I used acceleration as a function of angular velocity and kinetic energy as a function of moment of inertia and angular velocity to get:
    ke=0.25Mgr

    for (b) using conservation of angluar momentum:
    sorry for the mess, I couldn't get latex to come out right (I'll head off to read the introduction angain now!):
    a/a0 = (1+m/M)^2
    where a0 is original acceleration

    I'm not really sure how to start with part (c), so if someone could give me an idea how to start that would be nice.
    Thanks
     
    Last edited: Dec 29, 2006
  2. jcsd
  3. Dec 29, 2006 #2

    Hootenanny

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    I've not checked your answers to (a) and (b), but for question (c) you may wish to consider the astronauts tangential velocity...:wink:

    Edit: Just looking through your post, you may wish to reconsider you answer to (a), your almost there. What is the moment of inertia for a thin walled cylinder?
     
    Last edited: Dec 29, 2006
  4. Dec 29, 2006 #3
    edit: ignore this post, previously contained a wrong post by me.
     
    Last edited: Dec 29, 2006
  5. Dec 29, 2006 #4
    edit, just looked at edit by hootenanny, doh! thanks for pointing that out.
    so (a) should be 0.5mgr
    (b) is still (1+ m/M)^2

    (c) Since the astronaught's in free space they experience no acceleration, let w= angular velocity,
    v=wr/2
    distance he travels before hitting the rim using pythag:
    d=root(3)/2*r
    time=root(3)/w
    angle progressed by base of spoke= root(3)

    using more trig, the angle subtended by the astronaught to the centre relative to where he originally was is pi/3,
    therefore the arc length between them is (root(3)-pi/3)r and it's straight forward to find out the actual straight line length using more trig.

    Could someone please let me know if the above is along the right lines, thanks
     
    Last edited: Dec 29, 2006
  6. Dec 29, 2006 #5

    Hootenanny

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    No you don't. This is one and the same;
    Since the astronaut is travelling tangentally (in a straight line) he impacts the rim and an angular displacement of [itex]\theta=\sqrt{3}[/itex] from the base of the spoke. Do you follow?
     
  7. Dec 29, 2006 #6
    I don't sorry. I've attatched a diagram of what I thinks happening, hopefully this will make it clearer where I'm going wrong.
     

    Attached Files:

  8. Dec 29, 2006 #7

    Hootenanny

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    Spot on, looks good to me. I'm just waiting for your attachment to be approved by a mentor.
     
    Last edited: Dec 30, 2006
  9. Dec 29, 2006 #8
    Didn't realise attatchments had to be approved, it's really not a masterpiece anyway.
    My thinking was that he's releasing himself from halfway up a spoke, so his path forms a triangle, with hypotenuse r and a smaller side r/2, so the anglular displacement= arccos(0.5)= pi/3
    Thanks for your help.
     
  10. Dec 30, 2006 #9

    Hootenanny

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    Just looked at your attachment now. Yeah, your diagram and working looks good to me, I did it a slightly different way but arrived an the same answer (eventually):smile:. So your answer of [itex]S = (\sqrt{3}-\pi/3)r[/itex] is correct. Sorry for any confusion.

    I also forgot to welcome you to the Forums, and thank you for using the posting Template provided :smile:
     
    Last edited: Dec 30, 2006
  11. Dec 30, 2006 #10
    Thank you for your help.
     
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