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Momentum of water physics

  1. Sep 5, 2007 #1
    1. The problem statement, all variables and given/known data
    An inverted garbage can of weight W is suspended in air by water from a geyser.The water shoots up from the ground wit a speed v0 at a constant rate k. What is the maximum height the garbage can rride on due to the force on the water.What assumption have you made while solving this problem.
    2. Relevant equations
    dp/dt=F


    3. The attempt at a solution
    Assumption.e=1 so the force k wil apply on the can will be maximum as the change in p will be maximum.

    Then come the force equation of the body.

    2kv-w=ma
    here v is a function of height.
    v=sqrt(v0^2-2gh)

    Now what we have is a differentiol equation which we have to solve and finally we will have to equate dh/dt=0.

    My professor in class has taken the acceleration of the garbage can as zero and has got the answer os 15.3 when the answer is actually q7.

    Since i dont know how to slve this equation can someone actually help me out.

    According to the book the value of hmax should be approx17.

    Thanking you,
    Mehrzad.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 5, 2007 #2

    learningphysics

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    Is k given in kg/s ? Where does the 2kv come from... wouldn't it just be kv? Can you give the numbers for this problem...
     
  4. Sep 5, 2007 #3

    Doc Al

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    This sounds reasonable to me. After all, the garbage can is assumed to be suspended. So no need to solve any differential equations.

    Are you given values for W, k, and v0?
     
  5. Sep 5, 2007 #4
    In the given sum all the values that is k,W,m are variables. I need an equation relating the height between the max. height and these variables. in order to check the answer the following have been given.

    [tex]v_o=0m/s[/tex]
    W=10 kg
    k=.5kg/s
     
  6. Sep 5, 2007 #5
    I am sorry
    [tex]v_o=20m/s[/tex]
     
  7. Sep 5, 2007 #6
    v0=20m/s
     
  8. Sep 6, 2007 #7

    Doc Al

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    Using these values (W = 10 N, not 10 kg) and the same simplifying assumptions, I get the same answer as your professor. Do you see how we got that answer?
     
  9. Sep 6, 2007 #8
    I have an argument with what my sir did in vlass.

    What i am saying is that when the acceleration is equated to 0 then the answer which we get is 15.3.

    But note that when a body moves with constant velocit a=0 but still it is at a greater height and that is why(according to me) there is a difference.

    So therefore what i thought is right is to find h=f(t) and then equate dh/dt=0 which will indicate the instant where the garbage can does not move therefore it is the maximum height.

    For this we will have to solve the differential equation which i have no clue of how to solve.
    Here is the equation once again

    [tex]2k(v_o^2-2gh)^\frac{1}{2}-W=2\frac{d^2h}{dt^2}[/tex]

    From this we will have to evaluate [tex]\frac{dh}{dt}[/tex] and then finally equate that to 0.

    If there is anything wrong in my thinking then please correct me.
    Thanking you.
     
  10. Sep 6, 2007 #9

    learningphysics

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    Not sure I fully understand the problem, particularly the simplications that are being used... shouldn't the force a fluid exerts on an object be proportional to v^2 (with v being relative velocity between the fluid and the object?) Anyway, going with 2kv...

    Assuming that g = 10, W = 10N, hence m = 1kg, vo = 20m/s, k=0.5

    Shouldn't the equation be:

    2k(vfluid - vobject) - w = ma

    substituting in everything...

    [tex]v_{fluid} - \frac{dh}{dt} - 10 = \frac{d^2h}{dt^2}[/tex]

    Then using the equation for the speed of the fluid

    [tex]\sqrt{400-20h} - \frac{dh}{dt} - 10 = \frac{d^2h}{dt^2}[/tex]

    so the question is when dh/dt = 0 (this is when the height is a maximum), is d^2h/dt^2 also zero... if we assume that then we get out h = 15m.

    I tried to solve this another way using work energy...

    [tex]\int_0^h{2k(v_{fluid} - v_{object})dy} = mgh + \frac{1}{2}mv_{object}^2[/tex]

    [tex]\int_0^h{(\sqrt{400-20y} - v_{object})dy} = 10h + \frac{1}{2}v_{object}^2[/tex]

    Then taking the derivative of both sides with respect to h...

    [tex]\sqrt{400-20h} - v_{object} = 10 + v_{object} * \frac{dv_{object}}{dh}[/tex]

    substitute that vobject is 0 (note that vobject is just dh/dt)... then you get:

    [tex]\sqrt{400-20h} = 10[/tex], giving h = 15... so we see that when dh/dt = 0 h=15... and so substituting back into our previous equation we see that d^2h/dt^2 is also zero at this point...

    The fact that d^2h/dt^2 is also zero at the same time that dh/dt is 0 really wasn't obvious to me... I'm thinking the object rises up reaches some maximum speed then slows down while still rising and then comes to rest at the top...
     
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