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Moment of Inertia and Collisions

  1. May 17, 2007 #1
    1. The problem statement, all variables and given/known data

    A uniform symmetric ellipsoid (Mass M) has a large semi axis c and small semi axis a. A particle of mass m<<M is moving along a straight line parallel to the x-axis. Its y-coordinate is a/2 and its z-coordinate it c/2. After an inelastic collision, it sticks to the ellipsoid and then it (The ellipsoid) starts to move and rotate. One can assume the moment of inertia of the composite system equals to that of the ellipsoid.

    Find its linear velocity after collision
    Find the angular momentum of the system
    Find the precession of the ellipsoid c-axis around direction of the angular momentum

    2. Relevant equations

    Moment of Inertia about z-axis is I=(2/5)*M*a^2
    Moment of Inertia about other axes is I=(1/5)*M*(a^2+c^2)
    KE=0.5*I*w^2 where w is angular velocity
    L= I*w

    3. The attempt at a solution

    I suspect that you need the linear velocity to determine both the angular momentum and the precession. But my linear velocity cancels out when I equate KE=0.5*(m+M)*v^2=0.5*(0.4Ma^2)*(v/a)^2 [v/a=w]

    Any help would be nice!!! Thanks a lot

    Regards,
    The Keck
     
  2. jcsd
  3. May 18, 2007 #2
    By conservation of linear momentum, mv(i)+M(0)=(m+M)v'. Here v' is velocity vector of the center of mass of the system.

    By conservation of angular momentum, m(v i)x(a/2 j+ c/2 k)=Iw, where I=I cm about the x axis. From this you can find the angular velocity of the system. [tex]w=\frac{mv}{I_{cm}}(\frac{a}{2}k-\frac{c}{2}j)[/tex].

    I dont know what you mean by precision but if you explain it, I may be able to help you.
     
  4. May 22, 2007 #3
    So that would mean you actually need the initial velocity of the particle m, cause the question seems to indicate you don't need it. (I a bit confused about what you are doing...are you finding w and v independently and then use the fact that v=a*w or something???)

    I'll check up on that the precession means, cause personally I'm not sure on it myself.

    Thanks a lot

    Regards,
    The Keck
     
  5. May 23, 2007 #4
    Okay...I checked up on my lecturer who wrote this question, and he said that he had forgotten to put in an initial velocity v(i). But I'm still not sure on what you are doing with the angular velocity. Are you saying that the angular momentum is mv(i) x (a/2j + c/2k)???

    Regards,
    The Keck
     
  6. May 23, 2007 #5
    Okay...I checked up on my lecturer who wrote this question, and he said that he had forgotten to put in an initial velocity v(i). But I'm still not sure on what you are doing with the angular velocity. Are you saying that the angular momentum is mv(i) x (a/2j + c/2k)???

    Regards,
    The Keck
     
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