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Is this right?, conservation of energy!

  1. Mar 4, 2012 #1
    1. The problem statement, all variables and given/known data

    The problem relates to a water tank see link
    http://www.phys.unsw.edu.au/~jw/1131/IntroLS.pdf
    and scoll down a little!


    2. Relevant equations



    3. The attempt at a solution
    so I think to go about this problem I need to use a conservation of energy approach.
    ie ke(i) + gpe(i)= ke(f) + gpe(f)

    so...
    initially the tap is not realesed thererore there is no kinetic energy at this time
    if I add the element (h) height, I can say the system at this a t=0 has gpe = (M + m)g(h)

    Hence, (M+m)g(h) = 1/2mv^2
    therefore, v^2 = (2(M+m)g(h))/m
    v= √ (2(M+m)g(h))/m (m/s)

    Is this approach correct.
    And I am assuming that the tank would keep on going in the -x direction as the surface is frictionless...correct?

    And can I introduce the element (h)?
    Or would I have to use calculus to find a solution?

    Any help is appreciated!
     
  2. jcsd
  3. Mar 4, 2012 #2
    what do you mean keep going in the -x direction? why do you think it was going in the -x direction in the first place?
     
  4. Mar 4, 2012 #3
    I would assume that since when the tap is released a force would be created in the x direction..ie, the water flowing to the end of the pipe before it is released.... so there would be an equal but opposite force in the -x direction..newtons third law.
    Is this wrong?
     
  5. Mar 4, 2012 #4
    EDIT: I have thought about this problem a bit and it seems to me that the tank should move to -x direction with a constant velocity(conservation of momentum) until the water hits the vertical part of the tap after which the tank would stop moving. Therefore the tank would be displaced a little to the left.
     
    Last edited: Mar 4, 2012
  6. Mar 4, 2012 #5
    Ok I think I agree
     
  7. Mar 4, 2012 #6

    gneill

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    Staff: Mentor

    When the tank is full the tank + water have a horizontal center of mass. Since the surface is frictionless and all horizontal forces are internal to the system, the center of mass should not change. Where's the center of mass after the water has run out?
     
  8. Mar 4, 2012 #7

    BruceW

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    Homework Helper

    This is a duplicate thread! I guess sg001 wanted more advice from other people. Which is fair enough I guess.
     
  9. Mar 4, 2012 #8
    can some explain why the final velocity is not zero?

    I have the equation vf = √((2mgh)/M) m/s
    I thought this eqn was for the water... perhaps it is for the tank?
     
  10. Mar 4, 2012 #9

    gneill

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    Staff: Mentor

    Do you know for a fact that it is nonzero?
     
  11. Mar 4, 2012 #10
    Yes
    according to my physics teacher, when I saaid
    vf=0 and distance travelled =0.
    he said no.
     
  12. Mar 4, 2012 #11

    gneill

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    Staff: Mentor

    Did he say "No" to both?
     
  13. Mar 4, 2012 #12
    no, he just said no in general.
    So are you saying that it did have a final vf ≠0
     
  14. Mar 4, 2012 #13

    gneill

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    Staff: Mentor

    Not necessarily :smile: It could have a nonzero displacement and vf = 0, for example.
     
  15. Mar 4, 2012 #14
    But the tank is on a frictionless surface so I would assume that once it had a velocity it would remain constant.

    And I dont see where the force would come from to make the object have a vf =0, once the object was in motion.
     
  16. Mar 4, 2012 #15

    gneill

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    Suppose that you stand on a frictionless surface and you have a heavy ball in your arms. Now you extend your arms out in front of you holding the ball. After your arms are extended you drop the ball straight down. Describe the motions and positions.
     
  17. Mar 4, 2012 #16
    what exactly do you mean by HORIZONTAL centre of mass??
    does this mean the mass is evenly distributed along the horizonatl axis of the tank(not including the pipe).

    If the centre of mass changes could this provide the fore need to make vf=0??
     
  18. Mar 4, 2012 #17
    This seems weird but I guess you would fall on your back, after the ball is released?

    Or even possibly land flat on your stomach if you were to quickly extend your arms out.?
     
  19. Mar 4, 2012 #18

    gneill

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    The center of mass will have some location (x,y). The tank cannot move in the y direction, but it can in the x direction.
    The center of mass (at least for those parts of the system that we consider to be free to move) cannot change. That's the whole point of center of mass and conservation of momentum.
     
  20. Mar 4, 2012 #19

    gneill

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    Okay, so you're sitting stably on a sled and can't fall over.
     
  21. Mar 4, 2012 #20
    To me it would seem that you would go towards the left because if you cant fall over you should at least have some "kick" to move you a little.
     
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