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An angular momentum problem

  1. Mar 8, 2009 #1
    1. The problem statement, all variables and given/known data

    A thin bar of length L and mass M, and a small blob of putty of mass m. The system is supported on a frictionless horizontal surface. The putty moves to the right with velocity v, strikes the bar at a distance d from the center of the bar, and sticks to the bar at the point of contact. Obtain expressions for the velocity of the system's center of mass and for the angular velocity of the system about its center of mass.

    2. Relevant equations

    L=I*omega and P=mv

    3. The attempt at a solution

    So since the question *** for the velocity of the center of mass, I think I need to find the moment of inertia by parallel axis theorem right ??

    so I= Icom + Mh^2 so Icom=(1/12)ML^2

    Thus, I= (1/12)ML^2 + (M+m)*d^2. Am I right so far ? Do I need to find the center of mass ?
  2. jcsd
  3. Mar 8, 2009 #2
    What is this d?
    Are you going to work the rest of the problem?
  4. Mar 8, 2009 #3
    d is the distance from center of the bar to the point of contact.

    I have not solved the whole thing yet but I think I can use angular momentum and linear momentum to solve the problem right ? but the point is I am not sure whether or not to find the new center of mass or not. Do I have to do so ?
  5. Mar 8, 2009 #4
    Would it hurt you to find it if eventually you found that you did not need it?
  6. Mar 8, 2009 #5
    Because I am struggling finding the new center of mass too so if I don't need it, it's better to focus on other parts of the question.
  7. Mar 8, 2009 #6
    Will the center of mass be: r= {(m*d)+ (M*L/2)} / (m+M)
  8. Mar 8, 2009 #7
    Am I right ?
  9. Mar 8, 2009 #8


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    Yes, I agree with that r (from the end nearest the center of mass).
    Was the bar sitting perpendicular to the path of the blob before the impact?
  10. Mar 8, 2009 #9
    Yeah, does it make a difference ?
  11. Mar 8, 2009 #10
    so I got the new center of mass. In order to get new moment of inertia I need to find h to substitute in I= Icom + Mh^2.

    How should I find h ? do I just use h= L/2 - r ?
  12. Mar 8, 2009 #11


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    Oops, I think we have an error in the center of mass. That "d" is from the L/2 center of the rod, so the center of mass ought to be at
    c = ML/2 + m(L/2-d) all divided by m+M.
    Last edited: Mar 8, 2009
  13. Mar 8, 2009 #12


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    In the parallel axis theorem, R is the distance between the center of mass and the axis of rotation. That would be R = L/2 - c, wouldn't it? So
    I = ML^2/12 + M(L/2-c)^2 + m(d-c)^2
    I fear this is getting a little too complicated for me! Hope someone else will help.
  14. Mar 9, 2009 #13
    In finding for new center of mass, I thought r is the position so r of m is d. Why should it be L/2 - d then ?
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