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Falling bodies and terminal velocity

  1. Dec 30, 2012 #1
    Is it correct to assume that two objects, with different masses, dropped from the same height at the same time will fall at the same rate until one of the objects reaches its terminal
    velocity? What are the applicable equation(s) used to calculate this?
     
  2. jcsd
  3. Dec 30, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi CatastrophicF! Welcome to PF! :smile:
    nooo …

    the equation of motion is ma = mg - f(v),

    where f(v) is the function giving the air resistance for that particular body at speed v :wink:

    (usually depends on size and shape but not on mass)
     
  4. Dec 30, 2012 #3
    Re: Welcome to PF!

    Thank you for the welcome and response

    ma = mg. I understand that equation is essentially telling me that in a vacuum, regardless of mass, two bodies will fall/accelerate at the same rate (on Earth 9.8m/s/s). I think I start to become a little confused once those bodies are falling through a fluid, in this case air. I read and watch videos of people telling me things like the acceleration of gravity is the same on all objects (I'm happy with that) and that intertia is the reason why if I held and dropped a basketball and a bowling ball they would hit the ground at the same time (I'm also happy with that). So my quest to self educate myself in the laws of physics begins with me thinking that mass is a inconsequential property of falling bodies buuuuttttt I also know about terminal velocity, in which mass is a variable. Sooooo my qunadary is as follows; if all bodies accelerate at 9.8m/s/s due to gravity and their terminal velocity is calculated by Cd, surface area and m... why would they not fall at the exact same rate up until the moment one of the objects acheived its terminal velocity?
     
    Last edited: Dec 30, 2012
  5. Dec 30, 2012 #4

    tiny-tim

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    Hi CatastrophicF! :smile:
    because Cd, surface area and m are different (and have different ratios) for different bodies :wink:

    (in particular, denser bodies of the same size will have the same air resistance, but more weight, and so will fall faster)

    btw, technically, nothing ever reaches terminal velocity! :biggrin:
     
  6. Dec 30, 2012 #5

    Chestermiller

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    Re: Welcome to PF!

    Look at Tiny Tim's equation in response #3. The drag force is independent of the mass m, so when air drag is present, the mass m no longer cancels from the equation.
     
  7. Dec 30, 2012 #6
    So technically speaking a heavier object...will...fall faster than a lighter one. For example if I dropped two bowling balls (same height/same time) that were the same size, one with a mass of 10lbs and one with a mass of 20lbs...the 20lbs bowling ball would, mathematically/physically although maybe not perceivably, fall faster and hit the ground first...?
     
    Last edited: Dec 30, 2012
  8. Dec 30, 2012 #7

    tiny-tim

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    yup! :biggrin:
     
  9. Dec 30, 2012 #8
    Thank you so much for helping me work that out. It's amazing that such a common assertion
    ("heavy and light objects fall at the same rate") is fundamentally incorrect. How did that happen?
     
  10. Dec 30, 2012 #9

    tiny-tim

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    galileo first demonstrated it …

    i don't know whether he pointed out that it only works exactly in a vacuum
     
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