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Center of Mass Motion Question

  1. Nov 10, 2011 #1
    1. The problem statement, all variables and given/known data
    A person with mass m1 = 69 kg stands at the left end of a uniform beam with mass m2 = 108 kg and a length L = 2.5 m. Another person with mass m3 = 67 kg stands on the far right end of the beam and holds a medicine ball with mass m4 = 9 kg (assume that the medicine ball is at the far right end of the beam as well). Let the origin of our coordinate system be the left end of the original position of the beam as shown in the drawing. Assume there is no friction between the beam and floor.

    3)What is the new x-position of the person at the left end of the beam? (How far did the beam move when the ball was throw from person to person?

    4)To return the medicine ball to the other person, both people walk to the center of the beam. At what x-position do they end up?

    2. Relevant equations
    [tex]\frac{1}{M_{total}} \sum_{i=1}^{n} m_{i}r_{i} [/tex]


    3. The attempt at a solution
    I answered the first 2 parts of the question that have to do with finding the center of mass and I am fine with any sort of easy "find the center of mass of this situation", but I cannot figure out these sort of relative motion questions to do with the distance something is moving from the center of mass.

    I understand that since there are no external forces, the center of mass remains at the same place, I just don't know how to quantify what the motion of ball being thrown from one side to the other would do and I have trouble with relative motion questions.

    The question is just for practice and I want to figure it out. Any help would be greatly appreciated.
     
  2. jcsd
  3. Nov 11, 2011 #2

    Andrew Mason

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    Don't worry about the motion. All that matters is the change in location of the ball in 3) and of the people and the ball in 4).

    For 3), determine the location of the centre of mass of the beam + people + ball in the frame of reference of the beam before and after the ball changes position. You know that the location of the centre of mass relative to the floor does not change, so that will tell you how far the beam has to move due to the change in position of the ball.

    Do the same for 4).

    AM
     
  4. Nov 11, 2011 #3
    Yes, I've got that far but how is it that "it tells me how far the beam has to move due to the change in the position of the ball" I conceptually understand everything you mentioned and I've calculated everything I just can't put it together.
     
  5. Nov 11, 2011 #4
    So I've finally figured out question 3 but now I'm lost on 4. Any ideas?
     
  6. Nov 11, 2011 #5

    Andrew Mason

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    If everyone moves to the centre of the beam, where is the centre of mass of the system (beam + people + ball) in the frame of reference of the beam? Where was the centre of mass of the system before in the beam frame of reference? Does the centre of mass of the system move in the reference frame of the floor? So how far does the beam move relative to the floor so that the cm of the system is in the same place relative to the floor?

    AM
     
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