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Wet Umbrella Problem

  1. Mar 25, 2016 #1
    • Thread moved from the technical forums, so no HH Template is shown.
    The problem: A wet umbrella is held upright and is twirled about the handle at a uniform rate of 21 revolution in 44 s. If the rim of the umbrella is a circle 1 m in diameter, and the height of the rim above the floor is 1.5 m, find how far the drops of water spun off the rim travel horizontally relative to the umbrella handle before they hit the floor?

    My attempt: My general set of goals for this particular problem were to first find the tangential velocity using the given values for revolutions per second, and then proceed with the tangential velocity as if it were a two-dimensional projectile problem.
    ω = Δθ/Δt, where 1 revolution is equal to 2π (I got 3.0 rad/s)
    Vt=rω, where r=.5m
    Vt=1.5m/s
    Then, I proceeded as aforementioned (two-dimensional projectile problem)
    I started with Δy=-1.5m and solved for t using Δy=Vi*t+.5at^2, with a=-9.8m/s^2 and Vi=0m/s (should be no initial vertical velocity, as far as I'm aware).
    I ended up getting t^2=.31s^2, and then t=.56s
    Then, I used that t-value and solved for Δx in the equation Δx=Vi*t+.5at^2, with Vi equal to the tangential velocity and a=0m/s^2.
    I ended up getting .84m (with intermediate roundings of values).
    Ultimately, my professor marked me off four points out of ten total points and gave the real answer as .97m. I've been looking at it for a while now and cannot seem to find my obvious error. Help?
     
  2. jcsd
  3. Mar 25, 2016 #2

    Nathanael

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    Homework Helper

    First thing is, we would probably expect some energy loss when escaping the surface tension, so what ever we calculate using this approach should be an upper bound on the true answer.

    Anyway your approach looks fine to me, except minor rounding errors (t = sqrt(3/g) is more like 0.55s).
    This makes the answer more like 0.83 ≈ (21pi/44)*sqrt(3/g).

    One thing I did notice is that 0.97m is the furthest radial distance from the axis of the handle: √(0.83^2+0.5^2) = 0.97m = your professors answer, so maybe you misread the question and it really asked for the radial distance from the center (as opposed to asking for the displacement from where it lost contact).
     
  4. Mar 25, 2016 #3

    mfb

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    2016 Award

    Staff: Mentor

    I would expect the same. The problem statement is poorly phrased. The relative motion is clearly the 84 cm (83 cm without rounding errors). The change in distance is 47 cm (the final distance of 97 cm minus the initial distance of 50 cm).
    The final horizontal distance to the handle is 97 cm.
     
  5. Mar 25, 2016 #4
    Thank you both for your replies! It helped.
     
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