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Linear momentum in a pipe with a 90 degree curve

  1. May 23, 2010 #1

    This is my first thread related to an idea I have been working on and I would really appreciate some help. My education goes no further than high school, so I am doing my best.

    I apologize if I am doing this completely incorrectly. I am truly trying to do my best.

    To understand direction in this 2D model, I will use compass directions of N, E, S, W.

    We have a 1 kg steel ball that is inside a 10 kg pipe in space. The 1 kg mass is traveling at 10 m per s (ms) in a straight line going East inside the pipe that is at rest in comparison to the 1 kg mass. There is no resitance inside the pipe between the steel ball and the pipe.

    The pipe has a 90 degree turn in it, in the shape of a 1/4 circle that causes the 1 kg mass to go from an East direction to a South direction. To describe the corner it has an inside diameter of 1 m with the circumference of the 1/4 circle being equal to 1.57 m. (measured from center of pipe)

    When the 1 kg mass starts the corner it is going 10 ms but when it finishes it turn of 1.57 m it will be going 6.9 ms, therefor looses 31% of it's momentum due to the corner.

    I need to understand how much momentum is transferred to the pipe, and in which direction??

    The way I approached the concept is...

    We have two forces that will push the pipe both North as well as East. As the 1 kg ball goes South it will push the pipe in equivalence North, however when it goes East it will push East.

    I have been doing some experimenting on my air hockey board to make these assumptions.

    1) Ep = East Momentum of 1 kg
    Ep = mv
    Ep = 1 kg x 10 ms
    Ep = 10 p

    2) Sp = South Momentum of 1 kg (required to push the 1 kg mass south 1 m)
    Sp = mv

    a) Need to determine velocity requried
    t = d/v
    0.157 s = 1.57 m / 10 ms (starting velocity 1 kg)
    0.228 s = 1.57 m / 6.9 ms (end of 1/4 circle 1 kg)
    0.192 s Total of both times divided by 2

    v = d/t
    v = 1 m / 0.192 s (1 m based on radius of 1/4 circle turn)
    v = 5.2 ms

    b) Sp = mv
    Sp = 1 kg x 5.2 ms
    Sp = 5.2 p

    3) 1 kg Ball 31 % loss on 90 curve = Momentum (Ep & Sp) x 31%

    Ep = 10 p x 31%
    Ep = 3.1 p

    Sp = 5.2 p x 31%
    Sp = 1.61 p


    The momentum Sp will be Np on the pipe, so Np + Ep will equal the direction of the pipe.

    Am I close?

    Last edited: May 24, 2010
  2. jcsd
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