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Terminal Speed and Height of Object

  1. Sep 30, 2008 #1
    1. The problem statement, all variables and given/known data
    (a) Calculate the terminal speed of a wooden sphere(density .830 g/cm^3)
    falling through the air if its radius is 8.00cm and its drag coefficient is .500.
    (b) From what height would a freely falling object reach this speed in the
    absence of air resistance?


    2. Relevant equations
    p =m/v
    m=pv
    Vt = sqrrt 2mg/DpA


    3. The attempt at a solution
    p = .830 g/cm3
    r = 8.00 cm = .080m
    A = pie(r2) = .020m2
    D = .500

    Volume of Sphere = 4/3 pie(r3) = .335m3

    I have the Velocity equation ready to go but I don't think that I'm converting
    p correctly. I get .830 g/cm3 = 8.3x10 -7 g/m3

    I'm not sure how to go to kg/m3, which is what I believe that I need to find
    speed since g = m/s2
    Once I get p to be correct I can get m and p and put them into the Vt equation.
    I think I'm confused on the conversion.
    Thanks,
    Kevin
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 30, 2008 #2

    PhanthomJay

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    The conversion factor is 1g/cm^3 = 1000kg/m^3. The density of the wooden sphere may be used to calculate the mass of the sphere, but that is not the correct density to use in your equation for V_term.
     
  4. Sep 30, 2008 #3
    ok so .830 g/cm3^3 = 830kg/m^3
    so
    m=pv = (830kg/m^3)(.335m^3) = 278kg
    Is that right?
    Thanks,
    Kevin
     
  5. Sep 30, 2008 #4
    Ooops I get Volume of sphere to be .0021m^3
    So m=pv = 830kg/m^3(.0021m^3) = 1.74kg

    So I get Vt to be 4.11m/s
    Can anyone help confirm this.
    Thanks,
    Kevin
     
  6. Sep 30, 2008 #5

    PhanthomJay

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    The density looks good, but the volume is (4/3)(pi)(r^3) =4/3(pi)(.08m)^3= .002m^3, no? Now after you get the density and volume of the wood sphere correct (it's not that easy with all those metric decimals), calculate the mass of the sphere. But i want to be sure you understand that when you calculate V-term, the density to use in the denominator of that equation is NOT the density of the wood, it is the density of the ____?
     
  7. Sep 30, 2008 #6
    I still get .0021m^3 for volume. Maybe I don't understand that.
    In the Vt don't I have to find the mass of the sphere and use that?
    Kevin
     
  8. Sep 30, 2008 #7

    PhanthomJay

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    I said .002 which is just a round off of .0021; your number is OK. But you're missing the point. When you calculate the terminal velocity using your relevant equation for V_t = sqrt2mg/DpA, the number to use for p is the density of the air, not the wood.
     
  9. Sep 30, 2008 #8
    That does make sense. Sorry I misunderstood.
    So would I use 1.2kgm^3 or do I have to calculate it from this
    problem?
    Kevin
     
  10. Sep 30, 2008 #9
    As the density of air
     
  11. Oct 1, 2008 #10

    PhanthomJay

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    You use the 1.2kg/m^3 which you have looked up and correctly identified as the approximate density of air near earth's surface. Now plug in all the numbers to solve for the terminal velocity.
     
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