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Translational Velocity Problem (Sphere)

  1. May 15, 2012 #1
    1. The problem statement, all variables and given/known data
    "A bowling ball encounters a .760m verticle rise on the way back to the ball rack. Ignore frictional losses and assume the mass of ball is distributed uniformly. Translational speed of ball =3.5 m/s at the bottom of the rise. Find translational speed at the top."

    -Ball is going up a vertical hill (IMPOSSIBRU?!?!?)
    acceleration is -9.8m/s^2
    translational speed = 3.5 m/s at BOTTOM of vertical hill (initial translational speed)
    hill is .76m high

    2. Relevant equations
    KE=PE
    mgh=.5mv^2


    3. The attempt at a solution
    -m's cancel in equations-
    .5(9.8m/s^2)(.760m)=.5(v)^2
    v^2 = 14.xxx
    v>3.5(3.5 is initial velocity), this is impossible because the translation velocity is supposed to decrease when ball rolls up hill, not increase.
     
  2. jcsd
  3. May 15, 2012 #2

    gneill

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    Staff: Mentor

    What does the initial ".5" represent in the expression indicated above?

    The vertical rise only provides a decrease in kinetic energy as KE is traded for increased PE.
    Find the initial KE and the final KE, and thus final velocity.
     
  4. May 15, 2012 #3
    The .5 is in the formula for Kinetic Energy (KE), which is .5mv^2.
     
  5. May 15, 2012 #4

    gneill

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    Staff: Mentor

    Why is it multiplying a potential energy term on the left hand side?
     
  6. May 16, 2012 #5
    Because when the ball reaches the top, it will have gravitational potential energy.
     
  7. May 16, 2012 #6

    gneill

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    Staff: Mentor

    Sure, but what is that 0.5 doing there? It's not the mass of the ball, it wasn't given. It's not the change in height, because that's 0.760m and is already there to the right of the gravitational acceleration (g = 9.8 m/s2). The kinetic energy change is on the right side of the equals sign, and it has the expected 0.5.
     
  8. May 17, 2012 #7
    The 0.5 is part of the universal equation to find Kinetic energy.

    Google "Kinetic energy" > Wikipedia entry > "Kinetic energy of rigid bodies" and you will see the equation KE= 0.5mv^2.
     
  9. May 17, 2012 #8

    gneill

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    Staff: Mentor

    The kinetic energy is on the right hand side of your equation. The left hand side is the potential energy (it involves the mass, the acceleration due to gravity, and the change in height).

    Your procedure should be to find the change in potential energy due to the change in elevation, and then apply that change in energy to the initial kinetic energy to find the final velocity.
     
  10. May 17, 2012 #9
    My bad. Thanks!
    Hold on... I used
    KE(initial) + PE(initial) = KE(final) + PE (final)
    The initial PE is 0, because the initial height/elevation is 0.
    so:
    .5m(3.5m/s)^2 = m(9.8)(.760) + .5m(v)^2.
    The m's (masses) cancel, but I got
    6.125 = 7.448 + .5(v)^2.
    This must be wrong as I would get a negative velocity. I tried making "g" into (-9.8m/s^2) but still no luck. This is weird,,,
     
    Last edited: May 17, 2012
  11. May 17, 2012 #10
    Maybe the final velocity is negative?????
    But then there can't be a square root of a negative number. Help!
     
  12. May 17, 2012 #11

    gneill

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    Staff: Mentor

    Yup, it looks as though the ball hasn't sufficient energy to make it to the top of the rise.
     
  13. May 17, 2012 #12
    Wait.. I found out my error! I also had to take in the rotational energy of the ball into account :D

    /marked as solved
     
    Last edited: May 17, 2012
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