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What is the drag coefficient of the cue ball

  1. Jul 21, 2009 #1
    1. The problem statement, all variables and given/known data

    hile playing pool on a cruise ship, you accidentally hit the cue ball over the railing and it falls into the ocean below. If the cue ball (a sphere of diameter = 57.15 mm, m = 160 g) reaches a terminal velocity of 1.55 m/s in the water (), what is the drag coefficient of the cue ball in water?

    2. Relevant equations

    Cd= Fd/ 1/2 pv^2A
    Cd= coef of drag
    Fd= drag force
    p = mass density of the fluid
    V= speed
    A = area of the sphere


    3. The attempt at a solution

    Ok from the above equation i am able to determine all of the items except Drag force

    I got the area from pi * d^2 for the sphere (pi*55.15^2= 10260.82)
    adn the rest you can fill in

    I am stuck trying to figure out the drag force
     
  2. jcsd
  3. Jul 21, 2009 #2

    mgb_phys

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    Re: Drag

    At terminal velocity the gravity force downwards is equal to the drag force upward.
     
  4. Jul 21, 2009 #3
    Re: Drag

    even when in water?
     
  5. Jul 21, 2009 #4
    Re: Drag

    so M * G ---) 1.568 = drag force?
     
  6. Jul 21, 2009 #5
    Re: Drag

    air is less dense than water is why i ask
     
  7. Jul 21, 2009 #6
    Re: Drag

    also converting the Area to m 10.26m is that correct?
     
  8. Jul 21, 2009 #7
    Re: Drag

    if my numbers are correct i arrange the problem

    1.56 / (1/2 * 1000*1.55^2 * 10.26)

    doing the math Coefficient of drag = 1.26e-4?
     
  9. Jul 21, 2009 #8
    Re: Drag

    i feel like im making a mistake in converting the reference area since the object is falling
     
  10. Jul 21, 2009 #9

    mgb_phys

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    Re: Drag

    It's the cross sectional area, so pi r^2 is correct, it would be safest to work in metres
     
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