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Loss of energy to rotation

  1. Dec 8, 2013 #1
    1. The problem statement, all variables and given/known data
    A merry-go-round is sitting in a playground. It is free to rotate, but is currently stationary. You can model it as a uniform disk of mass 220 kg and radius 100 cm (consider the metal poles to have a negligible mass compared to the merry-go-round). The poles near the edge are 86 cm from the center.

    Someone hits one of the poles with a 9 kg sledgehammer moving at 15 m/s in a direction tangent to the edge of the merry-go-round. The hammer is not moving after it hits the merry-go-round.

    How much energy is lost in this collision? (enter a positive number for the absolute value in Joules)

    2. Relevant equations
    conservation of momentum
    KE of rotation =.5mv2
    KE of linear motion = .5Iω2
    Lost of energy=KErot-KElinear

    3. The attempt at a solution

    I assume that hummer hit pole tangentially and collision was elastic.
    My way of thinking is that energy transfer of KE of linear motion to KE of rotation. Thus I need to find KElinear and KErot. Where KElinear is easy to find.

    To find KErot, first, apply conservation of momentum.
    mhvh=mdvd
    Linear momentum was transferred to rotational motion. Where vd is tangential velocity of disk. vd is velocity of hummer. Then, v=ωd (d-distance from center to metal pole)
    From these, I can find ω and put it into my KErot

    Than I can find my energy loss. Right?
     
    Last edited: Dec 8, 2013
  2. jcsd
  3. Dec 8, 2013 #2

    jedishrfu

    Staff: Mentor

    Minor nit but you wrote your relevant equations a bit off interchanging linear KE and rotational KE equations

    From what the problem said all the linear KE is transferred to the rotational KE so that means

    KE linear = KE rotation

    so you need to compute the I in the Ke rotation knowing that its a disk of a certain mass and diameter rotating about a certain axis.
     
  4. Dec 8, 2013 #3
    @ jedishrfu
    I don't think these KE are equal.

    I found, that KE of rot is less than KE linear. Maybe because the force is applied not at the rim but at some distance. That's why I got some energy loss.

    Is my approach to find omega right? omega is only thing missing in KE of rot.
     
  5. Dec 8, 2013 #4

    mfb

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    I moved your thread to introductory physics homework.

    Momentum can be transferred to the axis of rotation, it does not have to be conserved.
    If the collision would be elastic, no energy would be lost.
    That approach does not work. The problem statement directly says that energy is lost (and you can verify this with a calculation).


    Angular momentum is the correct approach.
     
  6. Dec 8, 2013 #5

    haruspex

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    Just to clarify mfb's remark...
    When the hammer struck the pole, the merry-go-round did not go sliding across the park. This is because there was an equal and opposite (parallel) reaction from the axle in the middle. That's why we can't use conservation of linear momentum here - we don't know the magnitude of that impulse. We can eliminate it from consideration by taking moments about the axle. It has no moment about that point.
     
  7. Dec 8, 2013 #6
    @ haruspex,
    if I consider torques about center of rotation, I then come to conclusion that angular momentum right before the collision equals angular momentum right after, right?
     
  8. Dec 8, 2013 #7

    mfb

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    Torques don't help (you do not have a collision time), but angular momentum before and after the collision is the same, right (if you define your axis properly).
     
  9. Dec 8, 2013 #8
    @mfb
    I considered that angular momenta before and after are conserved about center of rotation of the merry-go-round. I don't see any other axis helpful. What do you mean properly?
     
  10. Dec 8, 2013 #9

    haruspex

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    I believe that's the axis mfb had in mind. Perhaps "smartly" would have been clearer.
    But mfb's point is that the hammer supplied an impulsive angular momentum, i.e. a change to the angular momentum in an unknown short time. Hence the torque, as such, cannot be known, but is uninteresting.
    So, what is the total angular momentum of the system (merry-go-round+hammer) about the axis just before impact?
     
  11. Dec 8, 2013 #10
    Angular momentum of the system is just product of distance to the axis and moment of hammer before the impact.
     
  12. Dec 9, 2013 #11

    haruspex

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    Yes.
     
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