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Energy of a Sphere Rolling

  1. Dec 11, 2009 #1
    1. The problem statement, all variables and given/known data
    A solid metal ball of radius 0.500m, and a mass of 1.50 kg, is found to be rolling down a sloped floor whose angle is 30.0° to the horizontal (assume no slipping). The ball has a final angular velocity of 2.00 rad/s. What is the total energy of the ball when it it is 2.00m from the bottom of the slope? (Assume there is no friction)

    2. Relevant equations
    Energy final = 1/2mv^2 + 1/2Iw^2 + mghf
    v= rw
    v= .5 (2) = 1m/s
    I = moment of inertia = 2/5(mr^2)

    3. The attempt at a solution
    Ef= 1/2(1.5kg)(1m/s^2)+ 1/2(2/5(1.5kg(.5m^2) + 1.5kg(9.8m/s^2)(2m)
    Ef= 0.75 + 0.075 + 29.4
    Ef= 30.225 Joules

    Does this seem right?
  2. jcsd
  3. Dec 11, 2009 #2


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    I suspect they mean a distance of 2m along the slope to the bottom
    what is the 'h' in the potential energy formula
  4. Dec 11, 2009 #3
    thank you, that makes the answer right it ends up being sin30 (2m)= 1 which makes my answer right thanks a lot.
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