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Ball goes down a frictionless ramp

  1. Dec 7, 2009 #1
    1. The problem statement, all variables and given/known data

    A hollow ball goes down a frictionless ramp. The ball starts at a height of 5 m. The ball has a radius of 4 cm and a mass of 0.5 kg.

    (a) What is the final velocity of the ball at the ball at the bottom of the ramp if it just slides down the ramp to a final height of 0 m. (Assume no rolling)

    (b) If the work done (by gravity on the ball) was used to rotate the sphere instead, what would the angular velocity of the ball be?



    2. Relevant equations

    I have no clue. I went through the section in my textbook that seems to relate to this practice test problem but they none of them seems to be anything I could use.


    3. The attempt at a solution

    Again, being I can't find any formulas or equations to work with. I don't know where to start with the values given.

    Thanks
     
    Last edited: Dec 7, 2009
  2. jcsd
  3. Dec 7, 2009 #2
    Hi there,

    Since your questions asked for the "work done", I presume that you are studying the chapter dealing with energy. Remember that a frictionless problem boils down to the mechanical energy conservation principle.

    Cheers
     
  4. Dec 7, 2009 #3
    I see. That definitely rang a bell.

    So simply, I could just break the problem down with K(i) + U(i) = K(f) + U(f) and in turn that would substitute to be (1/2)MVi^2 + mgyi = (1/2)MVf^2 + mgyf

    Also, for part b, would I just use for angular velocity w = sqrt(g/r)?
     
  5. Dec 7, 2009 #4
    No; I think what you need to use for b is conservation of energy where there are 2 forms of kinetic energy, rotational and translational:

    mgy=1/2mv^2+1/2(I)w^2 So you need to compute I for a hollow ball (I=2/5mr^2).

    Now altho it may look hopeless, one eqn with 2 unknowns it is not. The translational velocity (linear) and rotational or angular velocity are related how? Use this to substitute v with w (angular velocity).
     
  6. Dec 7, 2009 #5

    Doc Al

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    Staff: Mentor

    The work done by gravity--and thus the final kinetic energy--in both cases in the same. In the first, all the energy is translational KE; in the second, rotational KE. What's the formula for rotational KE?

    I interpret part (b) a bit differently than denverdoc did. I don't think it's asking for the rotational speed of a ball rolling down the hill (which is a more interesting problem). (If that is what it's asking, then it's very poorly worded!)
     
  7. Dec 7, 2009 #6
    Doc, I see your point. I thought it was ambiguous and decided on the more "interesting" interpretation. Your likely right as it does say "convert."
     
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