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Homework Help: Bowling ball rolling, going up a ramp, and continues rolling.

  1. Feb 12, 2012 #1
    1. The problem statement, all variables and given/known data

    A bowling ball of mass 2.2kg is rolling along a track at 3.2m/s when it reaches a ramp angled at 55°, The ball rolls up the ramp to the top where the track levels out again, and the ball rolls at 0.3 m/s. How long is the ramp?

    mass = 2.2kg
    v1 = 3.2 m/s
    v2 = 0.3 m/s
    angle of the ramp = 55°
    Not sure about the friction so I'm going to assume it's frictionless.

    2. Relevant equations
    Fnet = ma
    Not sure

    3. The attempt at a solution

    Well from what we know I was able to find the acceleration going down the ramp was -8.2 m/s^2.

    I found this out because the only force acting on the ball while the ball was going up the ramp was the force fog ravity and I used the following equation.

    Fnet = ma
    Fx = ma
    -18 = 2.2a
    a = -8.2

    This is assming there is no friction and choosing up the ramp to be positive.

    I know I don't have much done but I'm lost as to what to do now. I was wondering if I could use the equation:

    v2^2 = v1^2 + 2ad

    I don't know if that would work since that would give us the displacement from the very beginning to the very end and not only the length of the ramp.
  2. jcsd
  3. Feb 12, 2012 #2
    You have the correct equation at hand with v2^2 = v1^2 + 2*a*d. You do not have the correct value for acceleration. Frictionless assumption is correct.

    Draw a free body diagram of the ball on the slope to assist in determining acceleration. Hint: Use trigonometry.
  4. Feb 12, 2012 #3
    I'm not able to find my mistake, wouldn't this be the free body diagram?

    I know this is a poor diagram since I did it on paint but am I missing something here

    http://sadpanda.us/images/840435-L0XGODW.jpg [Broken]
    Last edited by a moderator: May 5, 2017
  5. Feb 12, 2012 #4
    OK, I see that you are using 10 m/sec^2 for g. I used 9.81. So with your value of g, you have the correct acceleration.

    Now look at the equation you wrote: v2^2 = v1^2 + 2ad

    You have v2, you have v1, you have a. You seek d. So.........
  6. Feb 12, 2012 #5
    The only thing I'm not sure about is wouldn't the displacement be from where it started rolling to where it ended rolling? So it's not necessarily the length of the ramp.
  7. Feb 12, 2012 #6
    The ball has no acceleration on the flat surfaces so there is no change in velocity on those surfaces. The equation

    v2^2 = v1^2 + 2ad

    only applies to the distance (d) where there is an acceleration. It represents the change in velocity only over that distance (d) where (a) applies and (a) is constant.
  8. Feb 12, 2012 #7
    Ahh, I see, okay, thank you very much ^^
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