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Translational Speed of a Bowling Ball

  1. Dec 4, 2009 #1
    1. A bowling ball with a mass of 6.95 kg and a radius of 0.194 m starts from rest at a height of 2.21 m and rolls down a 49.2 degree slope. What is the translational speed of the ball when it leaves the incline?


    2. Unsure! Here's what I've got, though:
    ω = v/r
    mgh=1/2mv2+1/2Iω2
    And so: ω= √(2mgh/(mr2+I))
    I of a solid sphere = 2/3(m)(r2)


    3. The attempt at a solution
    I = 2/3(6.95)(.1942) = 0.174
    ω= √(2(6.95)(9.8)(2.21)/((6.95)(.1942+0.174)) = √(301.05)/(0.436) = 26.29
    v = ωr
    v = 26.20 X 0.194 = 5.1
    Alas, 5.1 is not the correct answer. I'm sure I need to factor in the angle somewhere, but I don't know how! Help, please!
     
  2. jcsd
  3. Dec 4, 2009 #2
    Are you sure the moment of inertia for a solid sphere is 2/3?
    I thought that was for a hollow sphere.
    Try 2/5?

    http://img80.imageshack.us/img80/5226/picture2hr.png [Broken]
     
    Last edited by a moderator: May 4, 2017
  4. Dec 4, 2009 #3
    Oh my goodness, has this been my problem the whole time?

    I tried it, it was! You are such a lifesaver, thank you so much! I couldn't for the life of me figure out what I was doing wrong! Thanks again!
     
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