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Centre of Mass and translational motion

  1. Oct 28, 2009 #1
    1. A 280kg flatcar 21 m long is moving with a speed of 7.0 m/s along horizontal frictionless rails. A 85kg worker starts walking from one end of the car to the other in the direction of motion, with speed 1.5 m/s with respect to the car. In the time it takes for him to reach the other end, how far has the flatcar moved?



    2. Relevant equations
    VCM= 1/mtotal X [tex]\sum[/tex] m1[tex]\Delta[/tex]r1/[tex]\Delta[/tex] t

    Where V= velocity t= time r= distance.
    I set r=0 where the worker is at the start of the flatcar and the flatcar is also at 0
    3. The attempt at a solution
    So I have this I found t for the worker to be 56.667 using V= r/t

    so then I have: VCM= 1/85+280 X [tex]\sum[/tex] 85 X[tex]\Delta[/tex]21 + 280 X r2/[tex]\Delta[/tex] t

    Im trying to solve for r2, the distance that the flatcar travels but I don't know what VCM is. Is it equal to the speed of the flatcar?
     
  2. jcsd
  3. Oct 28, 2009 #2

    Andrew Mason

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    How so? It takes him 14 seconds to walk to the other end. t = r/v = 21/1.5 = 14 seconds.

    The speed of the centre of mass is the speed of the rail car at the beginning. At the end, the speed of the rail car is the same as at the beginning. After the person has accelerated and is walking along the car and before he stops, the car is moving more slowly.

    The easiest way to do this is to figure out how far the centre of mass has moved (that is constant throughout) in 14 seconds and then determine how far the centre of mass of the car/person has moved relative to the car.

    AM
     
  4. Oct 28, 2009 #3

    kuruman

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    Assuming that the flatcar is not moving, can you calculate the distance by which it will move back if the man goes from one end to the other? If, yes, then add this negative distance to the positive distance the car travels in 56.667 s.
     
  5. Oct 29, 2009 #4
    Alright so I found r2(r2= 123.45)
    Then i plugged it into the RCM = [tex]\frac{1}{mtotal}[/tex] [tex]\sum[/tex][tex]mr[/tex]
    r2= 123.45
    (sorry for my formatting still getting the hang of it)
    came out with the answer 98.1.

    Im not getting the correct answer anything that I'm doing wrong?
     
  6. Oct 29, 2009 #5

    Andrew Mason

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    The total mass is 365 kg. The centre of mass of the car is in the middle. The centre of mass of the worker is where he is standing. Treat each as a point mass. Set the origin at the car's centre of mass. The worker is initially at +10.5 m. We are only concerned about one dimension here.

    So the centre of mass of the car/worker is a point, p, somewhere between 0 and +10.5m such that 85(10.5-p) + 280(0-p) = 0. Work that out.

    When the worker reaches the other end, the centre of mass of the car/worker is a point, p, somewhere between 0 and -10.5m such that 85(-10.5-p') + 280(0-p') = 0. Work that out. Careful with the signs because p' is negative.

    What is the distance between p and p'? That is how much the centre of mass has moved relative to the car. Now turn it around: How much has the CAR moved relative to the CM? Call this displacement [itex]\vec d_{car-cm}[/itex]

    Now let's look at the motion of the centre of mass relative to the track. Since the centre of mass is moving at 7 m/sec (and from conservation of momentum we know that this is constant throughout), the centre of mass has moved how far in 14 seconds? Call this displacement [itex]\vec d_{cm-t}[/itex]

    Add those two displacements to determine how far the car has moved relative to the track.

    AM
     
    Last edited: Oct 29, 2009
  7. Oct 29, 2009 #6
    Did it slightly different from your method but you really helped me to understand the concepts. Both of you actually Thank you. :) My answer was 93m.
     
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