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Inverted garbage can and water hydrant

  1. Nov 19, 2012 #1

    WannabeNewton

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    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 at a constant rate [itex]\frac{\mathrm{d} m}{\mathrm{d} t}[/itex] with velocity [itex]v_{0}[/itex]. What is the maximum height at which the garbage can rides? The upwards direction is positive. Answer clue: if [itex]v_{0} = 20\frac{m}{s}, W = 98.1N, \frac{\mathrm{d} m}{\mathrm{d} t} = .5\frac{kg}{s}[/itex] then [itex]h_{max} \approx 17m[/itex].

    3. The attempt at a solution
    The net force on the garbage can is [itex]F = F_{I} - W[/itex] where [itex]F_{I}[/itex] is the impact force due to the stream of water. Consider a small amount of the stream of droplets, [itex]\Delta m[/itex] that, at time [itex]t = t_{0}[/itex], is just about to hit the can. We have that [itex]P_{0} = v_{0}\Delta m[/itex] and at [itex]t_{0} + \Delta t[/itex], the time instantly after the rebound\reflection of the small amount of droplets, [itex]P_{f} = -v_{0}\Delta m[/itex] so [itex]\frac{\Delta P}{\Delta t} = \frac{-v_{0}\Delta m - v_{0}\Delta m}{\Delta t} = \frac{-2v_{0}\Delta m}{\Delta t}[/itex]. Therefore, [itex]\lim_{t \rightarrow 0}\frac{\Delta P}{\Delta t} = \frac{\mathrm{d} P}{\mathrm{d} t} = -2v_{0}\frac{\mathrm{d} m}{\mathrm{d} t} = F_{I}[/itex] thus [itex]\frac{W}{g}a = -(W + 2v_{0}\frac{\mathrm{d} m}{\mathrm{d} t})[/itex] so [itex]a = \frac{-(W + 2v_{0}\frac{\mathrm{d} m}{\mathrm{d} t})}{\frac{W}{g}}[/itex]. At the maximum height, the can is instantaneously at rest so [itex]v_{0}^{2} = -2ah_{max}[/itex]. Then, [itex]h_{max} = \frac{Wv_{0}^{2}}{2g(W + 2v_{0}\frac{\mathrm{d} m}{\mathrm{d} t})}[/itex]. Now if I plug in the parameters from the answer clue then I do indeed get [itex]h_{max} \approx 17m[/itex] (16.93480102m to be exact when using g = 9.81m\s^2). My only problem is I had to use a major assumption when arriving at this solution: the water droplets would be rebounded\reflected perfectly so that the velocity of the small amount of droplets instantly after the rebound would be the same as the velocity instantly before hitting the can. If I drop this assumption then I really can't conceive of any way to find the new velocity using only the given information since I wont know how much of the velocity was lost to the can upon impact. If I keep the assumption, I can't exactly convince myself physically why the assumption is true. Any help would be appreciated, thanks!
     
  2. jcsd
  3. Nov 19, 2012 #2
    You are given the velocity of the water, is that the velocity at ground level? If so as the water jet rises it will lose velocity, have you taken that into account? Your assumption that the garbage can "reflects" the water with no lose in velocity is OK I think.
     
  4. Nov 19, 2012 #3

    WannabeNewton

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    The hydrant pumps out a constant steady jet of water at that velocity for all time so I think we can say the jet of water doesn't lose velocity as it rises.
     
  5. Nov 19, 2012 #4
    Any jet of water can only rise so high.

    Max height ≈ v^2/2g if shot straight up.

    As a water jet rises in our gravitational field it loses kinetic energy.

    Maybe others can help?
     
  6. Nov 19, 2012 #5

    WannabeNewton

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    Yeah I forgot that the droplets do indeed lose velocity as they go up as you stated. I can re-work the problem. Weird that the answer clue worked out for this solution though. Thanks again.
     
  7. Nov 20, 2012 #6

    WannabeNewton

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    Hey Spinnor, I figured out the solution and there wasn't much to change. I did find out that Kleppner had a mistake in his answer clue for this problem however which is why the answer came out right for the incorrect solution above haha! In the answer clue, dm \ dt should be 6kg\s not .5kg\s which incidentally wouldn't even be enough to raise the garbage can of that weight up that high (17 meters). So anyways I just did [itex]v^{2} = v_{0}^{2} - 2gy[/itex] for the trajectory of the water droplets. Then, [itex]v_{h} = (v_{0}^{2} - 2gh_{max})^{\frac{1}{2}}[/itex] so, keeping the same procedure for finding the impact force as before but using [itex]v_{h}[/itex], [itex]\frac{\mathrm{d} P}{\mathrm{d} t} = 2(v_{0}^{2} - 2gh_{max})^{\frac{1}{2}}\frac{\mathrm{d} m}{\mathrm{d} t} = W[/itex] at the max height so [itex]h_{max} = \frac{1}{2g}(v_{0}^{2} - (\frac{W}{2\frac{\mathrm{d} m}{\mathrm{d} t}})^{2})[/itex].
     
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