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Terminal Velocity of a Concrete Sphere

  1. Jul 19, 2007 #1
    1. The problem statement, all variables and given/known data

    I need to find the approximate terminal velocity of a baseball shaped concrete sphere and what distance it would take to reach that terminal velocity if dropped on earth under normal weather conditions.

    I have found that a baseball made of concrete would have a mass of approximately .5307 kg, radius of 1.46 inches and a cross sectional area of .004321 m^2

    I know that the acceleration due to gravity is 9.8 m/s^2
    and according to wikipedia the air density of dry air at sea level at 20 degress celsius is 1.2 kg/m^3

    Using wikipedia I found the equation for the terminal velocity.
    http://en.wikipedia.org/wiki/Terminal_velocity

    I am having trouble finding the approximate drag coefficient of the concrete sphere. Is there anyway to estimate this to get a reasonably close value, or must it be derived experimentally?
     
  2. jcsd
  3. Jul 19, 2007 #2
    I'm sure I've seen this question before and I'm sure drag coeffecients are worked out experimentally, ie in wind tunnels. You might be lucky and stumble across an object that is fairly close to a concrete sphere from which you could approximate as this is a fairly simple object, but AFAIK there is no good value without experiment, so what your looking for is for someone to have done the work before.

    My advice take a drag coefficient of something like a concrete baseball, then apply that, and mention that this is not the exact value, but should produce a fairly reliable result.

    EDIT: I found the drag coeffeicent for a rough sphere is 0.4. After a bit of rooting around.

    This will probably suffice?

    http://aerodyn.org/Drag/tables.html

    Table is here.
     
    Last edited: Jul 19, 2007
  4. Jul 19, 2007 #3
    Thanks a lot! I'm sure it will suffice.
     
  5. Jul 19, 2007 #4

    PhanthomJay

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    Yes, I agree 0.4 sounds good for the drag coefficient, that will get you the terminal velocity, but it is another matter to determine in what distance the ball will reach that velocity. Firstly, it is never reached, only approached, so you might want to consider the distance it takes to reach say 95% of terminal velocity. Secondly, you've got to be good at differential equations (at least better than me) in order to arrive at the solution, because acceleration will be variable due to the variable drag force which is a function of the square of the speed which is continually changing during decent.
     
  6. Jul 19, 2007 #5

    mgb_phys

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    Try a numerical solution first, just to give you an answer to check.
    Using a spreadsheet (or your favorite langauge) calculate the acceleration and speed every say 0.1sec and draw a graph of that.
     
  7. Jul 19, 2007 #6

    PhanthomJay

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  8. Jul 20, 2007 #7
    Thanks, that website is awesome. Can someone tell me what the exp means in the impact and velocity equations.

    It turns out the concrete sphere fell from a height low enough that it never reached terminal velocity. When I was first calculating the time and velocity at various heights I used the wrong equation but want to now if my logic made sense. I set the height equal to the integral of 9.8t from 0 to x, and solved for x.
    [tex] h = \int_{0}^{x} (9.8t)dt[/tex]
     
    Last edited: Jul 20, 2007
  9. Jul 20, 2007 #8

    mgb_phys

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    Exponent?
    [tex] e^{x}[/tex]
     
  10. Jul 20, 2007 #9

    PhanthomJay

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    Your integral (from 0 to t, not x) applies only to free fall with a constant acceleration of g=9.8m/s/s, no air resistance, and boils down to [tex] h =1/2gt^2 [/tex]for a body falling from rest. You cannot use that equation for the air resistance case, since the acceleration is not constant.
     
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