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Angular momentum: Sphere and 2 points

  1. Apr 25, 2010 #1
    1. The problem statement, all variables and given/known data
    A uniform sphere of mass [tex]m[/tex]and radius [tex]R[/tex] rotates around the vertical axis with angular speed [tex]\omega[/tex]. Two particles of mass [tex]m/2[/tex] are brought close to the sphere at diametrically opposite points, at an angle [tex]\theta[/tex] with the vertical. The masses, which are initially at rest, abruptly stick to the sphere. What angle does the resulting [tex]\omega[/tex] make with the vertical.


    2. Relevant equations
    [tex]\Delta L = r\times \int Fdt[/tex]
    Equations for principal moments


    3. The attempt at a solution
    I'm very confused about how to start this problem. The book I'm using only does motion after an impulsive blow and frequency of motion due to a torque types of problems. I believe this one is an impulsive blow problem. Angular momentum of the system changes when the two masses stick to the sphere. But I'm confused where to go from here. I don't know how to calculate the angular momentum of the system when the two masses stick, and I don't know how to find the change in the axis of rotation. Thanks for your help guys.
     
    Last edited: Apr 25, 2010
  2. jcsd
  3. Apr 25, 2010 #2

    diazona

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    Remember that angular momentum is conserved... so calculate the initial angular momentum, then calculate the final angular momentum, then set them equal to each other and work with the resulting equation.
     
  4. Apr 25, 2010 #3
    Well I know the initial angular momentum is [tex]2/5 mR^{2} \omega[/tex], but I don't know how to calculate it for the final angular momentum.
     
  5. Apr 25, 2010 #4

    diazona

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    OK, think about what quantities you would need in order to write a formula for the final angular momentum.
     
  6. Apr 25, 2010 #5
    Can I calculate it with respect to the initial axis of rotation? So the angular momentum of each point mass would be [tex]mR^{2}\omega sin\theta[/tex] added to [tex]2/5mR^{2}\omega[/tex], where [tex]\omega[/tex] is the new angular velocity vector?
     
  7. Apr 25, 2010 #6
    Oh and [tex]\theta[/tex] is the given angle plus the angle of the new angular velocity vector with respect to the vertical.
     
  8. Apr 25, 2010 #7

    diazona

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    I might be missing something, but I don't think that would work. Here's my reasoning: you know that the direction of angular momentum stays the same. Thus, in order for the direction of the angular velocity to change, the moment of inertia needs to be a tensor with off-diagonal terms. And in that case, I don't think you can get away without calculating the full tensor. (It's not that complicated in this case)

    By the way, I would strongly suggest using different letters for the angular velocity and angle of inclination after the masses are stuck on. Like, [itex]\omega'[/itex] and [itex]\theta'[/itex].
     
  9. Apr 25, 2010 #8
    Ohhhhhhh ok that makes much more sense. So then the tensor is not the one for the principal axes? I'm confused how to find tensors for objects like this, I've only calculated them for shapes like squares and spheres and rectangles.
     
  10. Apr 25, 2010 #9
    The only thing I can think of to do is multiply by a rotation matrix where the angle of the rotation matrix is the angle the new angular velocity vector makes with the vertical. But I still don't know how to include the two masses that are stuck on the sphere into the tensor.
     
  11. Apr 25, 2010 #10
    Also, is only the direction of the angular velocity changing, or is it both the direction and magnitude?
     
  12. Apr 25, 2010 #11

    diazona

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    Well of course there is a set of principal axes for the sphere with the two point masses. Actually there are multiple sets... I imagine it would be any set of axes in which one of them goes through the two spheres.

    Now, remember that a tensor is kind of like a vector in the sense that how you write it down depends on the coordinate system you pick. If you choose a coordinate system that corresponds to the principal axes, and write down the tensor in that coordinate system, it will only have diagonal terms. That's the definition of the principal axes, in fact. In this case, I think one of the principal axes would have to pass through the two small particles. (Think about that, and see if it makes sense or if maybe I got it wrong)

    However, knowing the moment of inertia tensor in the "principal axis coordinates" doesn't do you much good in this case because that coordinate system isn't stationary in space - it keeps rotating around with the object. (There are ways to account for that but that would be an unnecessary complication) It'll be much easier if you pick one coordinate system that is fixed in space, and compute everything in that coordinate system. In fact, you have a coordinate system defined by the problem already: it tells you that the sphere is initially rotating around the vertical axis, so I'd pick the direction of the initial angular velocity to be the z axis. (And you can then identify the direction of the initial angular momentum, right?)

    As for actually computing the moment of inertia tensor in that coordinate system, I'd go to the formula found on Wikipedia,
    http://en.wikipedia.org/wiki/Moment_of_inertia
    (The same formula should also be found in your textbook or whatever reference you have available) Scroll down to the section where it talks about the tensor. There is a formula for computing the moment of inertia of a collection of point particles, which is exactly what you have here - you have the two particles coming in at the angle. There is also the sphere, of course, but you don't need to recompute the moment of inertia of that; just add it in at the end to get the total moment of inertia.
     
  13. Apr 25, 2010 #12

    diazona

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    I don't know, I suppose you'll figure that out when you do the problem :wink:
     
  14. Apr 25, 2010 #13
    Ok so you're saying I can use superposition for the moment of inertia of the system? That's what I was mostly confused about. I have the formula for computing it for a collection of particles, I just wasn't sure if I could use that when the masses stuck to the sphere.
     
  15. Apr 25, 2010 #14

    diazona

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    Yes, you can always use superposition for moments of inertia.

    The moment of inertia of a system with multiple pieces is always equal to the sum of the moments of inertia of each of the individual pieces.

    (All the moments of inertia have to be computed with respect to the same center of rotation, though. You can't add moments of inertia computed around different centers of rotation.)
     
  16. Apr 25, 2010 #15
    Ok one more question and then I should be able to pretty much finish this problem. I'm finding the coordinates of the point masses with respect to the CM of the of sphere, but I feel like they fall into a two dimensional plane. For example, I believe one of the masses has coordinates [tex](Rsin\theta,0,Rcos\theta)[/tex] and the other one is the negative of this because it's diametrically opposite. I feel like I may be visualizing the problem incorrectly, but am I correct in assuming this?
     
  17. Apr 25, 2010 #16
    By the way, thank you so much for all of your help, I really appreciate it.
     
  18. Apr 25, 2010 #17
    Actually I think what I wrote is wrong. I have to calculate the tensor for the collection of particles with respect to the new axis of rotation don't I, so theta has to be replaced by [tex]\theta + \theta'[/tex] where [tex]\theta '[/tex] is the angle the new angular velocity vector makes with the z-axis.
     
  19. Apr 26, 2010 #18

    diazona

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    Yes you are (correct). Remember that the coordinate system you should use is the one which makes the problem as easy as possible. That means, as I said, that it should have its z-axis pointing in the direction of the initial angular velocity, but you still have some freedom to choose the direction of the x- and y-axes. It makes sense to choose them such that one of those coordinates (y in your case) is zero.
    No you don't. Everything should be calculated with respect to the same coordinate system. (Well... technically yes you could switch to a new coordinate system aligned with the new axis of rotation. That would make this problem far more complicated than it needs to be.)
     
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