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Rolling Bowling Ball (Rolling Energy)

  1. Nov 11, 2009 #1
    1. The problem statement, all variables and given/known data
    A 20-cm-diameter, 6.0kg bowling ball rolls along a horizontal floor at 4.0 m/s. The ball then rolls up a 2.0-m-long ramp sloped at 30 degrees up from horizontal. Rolling friction is negligible. What is the ball's speed at the top of the ramp?


    2. Relevant equations
    K(final) + U(final) = K(initial) + U(initial)
    I = 2/5 * M * R^2
    omega = V/R


    3. The attempt at a solution
    The ball starts out with all kinetic energy so the above equation becomes k(final) + U(final) = k(initial). I set up the following equation:

    1/2*I*omega(final)^2 + 1/2*M*v(final)^2 + M * g * h = 1/2*I*omega(initial)^2 + 1/2*M*v(initial)^2

    After substituting in the values for I and omega, the radius and mass cancel out and I am left with:

    7/10*v(final)^2 = 7/10*v(initial)^2 - g * h

    After doing some simple trig on the slope I found that h = 1m, and solving for v(final) I end up with:

    v(final) = 1.41 m/s

    Since I don't know the answer to the problem and I get credit on an all or nothing basis, I wanted to come here and have someone verify the validity of my work before I hand it in.
     
  2. jcsd
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