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Homework Help: Kinetic Energy Ratio/Rotational Equilibrium

  1. Feb 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Two equal masses, each m, are resting at the ends of a uniform rod of length 2a and negligible mass. The system is in equilibrium about the center C of the rod. A piece of clay of mass m is dropped down on the mass at the right end, hits it with velocity v as shown below and sticks to it.

    The ratio of the kinetic energy Ef just after the collision to the kinetic energy Ei just before the collision, Ef/Ei, would be:

    (A) 1
    (B) 3/4
    (C) 2/3
    (D) 1/2
    (E) 1/3

    2. Relevant equations

    Not really sure.

    3. The attempt at a solution

    If the system is initially in equilibrium, then it's not moving, and I don't understand how it could have kinetic energy if it's not moving. I want to write: Ef/0, but that's obviously wrong.
     

    Attached Files:

  2. jcsd
  3. Feb 21, 2012 #2

    tiny-tim

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    Hi Victorzaroni! :smile:
    "The system" includes the clay, and that is moving! :wink:
     
  4. Feb 21, 2012 #3
    ooohhh. wow i feel dumb. that makes sense. so its the initial kinetic energy of the system is the kinetic energy of the clay.
     
  5. Feb 21, 2012 #4
    So Ei=(1/2)mv2. Now what? I know torque is involved somehow.
     
  6. Feb 21, 2012 #5

    tiny-tim

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    torque has nothing to do with it :confused:

    now find Ef (as a function of vf)
     
  7. Feb 21, 2012 #6
    Energy is a function of velocities (kinetic) and positions (potential), energy isn't a function of accelerations and torques
     
  8. Feb 22, 2012 #7
    Alright... So the new kinetic energy after the clay has stuck to the mass would be KE=(1/2)(M+M)r2? How do I account for the mass on the other end of the board?
     
  9. Feb 22, 2012 #8

    tiny-tim

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    add it on! :smile:

    (it'll have the same speed, won't it? :wink:)​
     
  10. Feb 22, 2012 #9
    Alright so I have: KEi=(1/2)Mv^2, and KEf=(1/2)(M+M+M)v^2. Dividing KEf by KEi yields (3/2)/(1/2), which is 3, which is not a choice? Dividing them the other way yields 1/3, choice E...
     
  11. Feb 22, 2012 #10

    tiny-tim

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    i'm lost :redface:

    what conservation equation are you using? :confused:
     
  12. Feb 22, 2012 #11
    I'm not using conservation, KEf/KEi is what the question is asking for and the other stuff is what you've been saying. lol
     
  13. Feb 23, 2012 #12

    tiny-tim

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    how are you going to find the relation between before and after without a conservation law? :confused:
     
  14. Feb 23, 2012 #13
    So how do i use a conservation law?
     
  15. Feb 23, 2012 #14

    tiny-tim

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    what do you think will be conserved?

    (and i'll repeat my previous question, since i'm still wondering …

    how were you intending to find the relation between before and after without a conservation law?)​
     
  16. Feb 23, 2012 #15
    The kinetic energy of the clay equals the rotational kinetic energy of the system?
     
  17. Feb 23, 2012 #16

    tiny-tim

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    no, this is a perfectly inelastic collision
    … so energy is not conserved
     
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