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Angular + Linear Velocity problem

  1. Nov 29, 2011 #1
    I kno this is pretty simple and the answer is probably staring me in the face, but i'm lost for some reason. Our teacher gave back our tests so i have the answer, but I'm not sure how to get it. I havent had a chance to speak to my teacher yet, but i intend to tomorrow if no one helps me first.

    1. The problem statement, all variables and given/known data

    A bowling ball [(mass = 7 kg)(radius = .50m)] rolls without slipping down a 3m high ramp. Starting from rest, find the velocity at the bottom of the ramp.


    2. Relevant equations

    Bowling ball inertia = I = (2/5)(M)(R)^2 = (0.7)

    3. The attempt at a solution

    I used conservation of angular momentum

    (1/2)(I-initial)(omega-initial)[itex]^{2}[/itex] + mg(h-initial) = (1/2)(I-final)(omega-final)[itex]^{2}[/itex] + mg(h-final)

    after calculating i got omega-final = 24 (actually 24.25, but he instructed us to round to the nearest whole number)
    i converted that to v = 12

    which is wrong

    i was supposed to set it up as:

    (1/2)(I-initial)(omega-initial)[itex]^{2}[/itex] + mg(h-initial) = (1/2)(I-final)(omega-final)[itex]^{2}[/itex] + mg(h-final) + (1/2)m(v)[itex]^{2}[/itex]

    and v = 6.5 is the answer

    i was under the impression we were supposed to find the angular velocity at the bottom of the ramp and convert to linear velocity

    i'm not sure how he got v = 6.5
     
    Last edited: Nov 29, 2011
  2. jcsd
  3. Nov 29, 2011 #2

    Doc Al

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    Staff: Mentor

    In your solution, you ignored the translational motion of the ball. The ball ends up with both rotational speed and translational speed. Hint: How are those two speeds related for rolling without slipping?
     
  4. Nov 29, 2011 #3
    are you referring to v = rω?
     
  5. Nov 29, 2011 #4

    Doc Al

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    Staff: Mentor

    Yes.
     
  6. Nov 29, 2011 #5
    so would it look like:

    (1/2)(I-initial)(omega-initial)[itex]^{2}[/itex] + mg(h-initial) = (1/2)(I-final)(omega-final)[itex]^{2}[/itex] + mg(h-final) + (1/2)m(r)[itex]^{2}[/itex](omega-final)[itex]^{2}[/itex]

    just worked it out and got the right answer
    i actually set it up like that last night and worked it out but just now realized i made a simple error
    hate when that happens

    thanks doc!
     
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