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Elastic problem

  1. Apr 20, 2015 #1
    1. The problem statement, all variables and given/known data
    5R1RjVK.png

    2. Relevant equations
    KEi = KEf

    3. The attempt at a solution

    since it is perfectly elastic, kinetic energy before = kinetic energy after

    (1/2) (m) (20m/s^2) = (1/2) (2m) (VBeamf^2) + (1/2) (m) (VBlockf^2)

    is this the right approach?
     
  2. jcsd
  3. Apr 20, 2015 #2

    ehild

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    Not quite. The beam can only rotate about the axis. What is its kinetic energy?
     
  4. Apr 20, 2015 #3
    (1/2) (I) (ω^2) ?
     
  5. Apr 20, 2015 #4

    ehild

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    Yes. And you need one more equation. What else is conserved?
     
  6. Apr 20, 2015 #5
    i have no idea, momentum?
     
  7. Apr 20, 2015 #6

    ehild

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    Momentum is conserved when translation is involved. But the beam can not translate. What can be conserved?
     
  8. Apr 20, 2015 #7
    I can't think of anything else besides momentum and kinetic energy. any hints?
     
  9. Apr 20, 2015 #8

    ehild

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    For rotation, you have quantities analogous to translational ones. For position, it is the angle of rotation. For velocity, it is angular velocity. For mass, it is moment of inertia. What do you have for momentum?
     
  10. Apr 20, 2015 #9
    angular momentum?

    but if the initial angular momentum is 0, then the final has to be 0 too? since it was at rest
     
  11. Apr 20, 2015 #10

    ehild

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    The angular momentum of the beam is zero, but the little mass has initial and final angular momentums with respect to the axis. How is the angular momentum defined for a point-like body, moving in the plane normal to an axis, travelling with velocity v, at distance D from an axis?
     
  12. Apr 20, 2015 #11
    L = mvr ?
     
  13. Apr 20, 2015 #12

    ehild

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    yes. What is r in this problem?
     
  14. Apr 20, 2015 #13
    distance traveled before the block hit the beam? I have no idea
     
  15. Apr 20, 2015 #14

    ehild

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    No, it is the distance of the line of track of the little mass from the axis of rotation. It is the same as the distance from the axis of the point where the small mass hits the beam .
     
  16. Apr 20, 2015 #15
    so it is just D/2?
     
  17. Apr 20, 2015 #16

    ehild

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    Yes. Now you can write the equation for conservation of angular momentum.
     
  18. Apr 20, 2015 #17
    i don't understand the use of this equation here.

    m vi D/2 = m vf D/2 ?
     
  19. Apr 20, 2015 #18

    ehild

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    The beam also has got angular momentum, you need to include it.
     
  20. Apr 20, 2015 #19
    m vi D/2 = (m vf D/2) + ( I ω)

    ?
     
  21. Apr 20, 2015 #20

    ehild

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    Yes, it is. You know the expression for the moment of inertia for the beam. And you have the equation for the energies.

    Eliminate ω and solve for vf.
     
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