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Acceleration of water jet

  1. Jul 31, 2016 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
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    I am stuck on C and D and E, any help would be appreciated!

    The speed of the water (which is calculated in B) is 4.43 m/s

    For C I think it might be law of conservation of momentum or use the formula P=(1/2)*rho*A*v^3 with P=FV and possibly F=ma???

    I think D is just STUVA (s=ut+(1/2)at^2) where u=0 (I think). S=0 + (1/2)a4^2...
     
    Last edited: Jul 31, 2016
  2. jcsd
  3. Jul 31, 2016 #2

    billy_joule

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    For c) you can find the thrust force more directly by T = m_ dot * v
    Your approach for d) will work.
     
  4. Jul 31, 2016 #3

    rude man

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    For (c) consider that the c.g. with respect to ground does not move.
    So after 1 sec. how far does the vehicle move to the left? How far after 2 sec.?
    Then you see why it accelerates and what the acceleration must be.

    (d) is totally elementary after you have (c)
     
  5. Jul 31, 2016 #4
    Mike: In your course, are you learning how to do macroscopic momentum balances on control volumes?
     
  6. Aug 1, 2016 #5
    @Chestermiller - It's just an introductory course so its nothing that complicated yet

    Thanks for all the feedback, I'll try the question again!
     
  7. Aug 1, 2016 #6

    rude man

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    All you need is conservation of linear momentum.
     
  8. Aug 4, 2016 #7

    rude man

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    Just to tease, since as usual the OP has long dropped out - what is your v?
     
  9. Aug 4, 2016 #8

    billy_joule

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    The water velocity found via the Bernoulli equation applied between the water surface and the outlet (also used to find the mass flow rate).
     
  10. Aug 4, 2016 #9

    rude man

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    Well I would dispute that! :frown: . Were it so, the c.g. of the system would move in time with respect to the ground! Which it can't of course.
     
  11. Aug 4, 2016 #10

    billy_joule

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    It'll find the correct instantaneous acceleration when velocity is zero and water has just begun to flow (and so c.g. hasn't moved ..yet), will it not?
    Which is all we need for (d.
    I agree that we'd need to apply a smarter approach if (d wasn't simplified.
     
  12. Aug 4, 2016 #11

    rude man

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    You can't compute the initial accelereation from data at t=0 only, since v is undefined until there is motion. You have to go to a small but finite amount of time. In (d) you assume the acceleration stays constant for 4 sec since that's what the problem is telling you to do.

    The c.g. can't move ever, at least not until the water hits the ground, at which point some momentum is transferred to the ground (earth).
     
  13. Aug 5, 2016 #12

    billy_joule

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    I meant vwater, which we have from b).

    Yes, right. I should've said before the cart moves, not the overall CG.

    So this is the method I had in mind, could you point out the error?
    Thrust due to water:
    vwater = √(2gh)
    T = m_dotwater * vwater = ρAvwater2 = ρA(√(2gh))2 = 2ρAgh

    Accelration of cart at t = 0
    a = F/mcart = T/mcart = 2ρAgh / mcart = 0.0036 m/s2
     
  14. Aug 5, 2016 #13

    rude man

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    The issue I have with what you wrote is this:
    dm/dt = ρAvwater = ρA√(2gh) as you wrote. Agreed.

    But the effective v of the water in the conservation of momentum equation is half of what you wrote. That's because you have to reference the c.g. of the water, not its front. And the speed of the c.g. of the jet as it evolves is only half the speed of the front.
    Thus vwater = √(gh/2) in the momentum equation which is mcart dvcart/dt = vwater dmwater/dt.

    This is easy to see if you think of the evolving system: in order for the c.g. of the system not to move, the distance from the starting point of the c.g. of the vehicle times its mass must always equal the distance from the same point to the c.g. of the water times its mass.

    Make sense?
     
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