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A marble that is rolling without slipping approaches a hill

  1. Jun 9, 2015 #1
    1. The problem statement, all variables and given/known data
    A marble that is rolling without slipping approaches a hill at 8.5 m/s. How high vertically will the marble go under these circumstances:

    If the hill is rough enough to prevent any slipping?
    If the hill is perfectly smooth?

    Why does the marble rise to different heights when it had the same initial kinetic energy in both cases?
    2. Relevant equations
    ME=KE+PE
    KE=1/2mv^2+mgh
    PE=mgh

    3. The attempt at a solution
    KE(initial)=1/2(8.5)^2+9.8(0)=144.5
    I have no idea how to solve for final or even if thats what to do next.
    PE(initial)=0
    I'm not sure how to solve final without knowing the height for sure.
    Like, I said earlier I don't even know if this is the correct start to solve this.
     
  2. jcsd
  3. Jun 9, 2015 #2

    ehild

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    The marble rolls, that means it rotates about its centre. You have to take the rotational energy into account in addition to the translational kinetic energy.
     
  4. Jun 10, 2015 #3

    haruspex

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    No, try again.
    What's missing on the right?
    The height is what you are trying to find. What is conserved?
     
  5. Jun 10, 2015 #4
    In the first case, all of its initial KE (linear KE + rotating KE), ends up as PE
    In the second case. only the linear KE gets converted, and the marble is still rotating at the original rate when it stops up the hill
     
  6. Jun 10, 2015 #5

    haruspex

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    Dean, please give the OP a bit more chance to get there with hints.
     
  7. Jun 10, 2015 #6
    my apologies
     
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