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Falling air resistance, determining g

  1. Jun 7, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi, I'm trying to determine g (very precisely) using a falling ball in non-vacuum. I'm using three laser beams (along the z-axis), they're detecting when the ball enters the beam and leaves it again. This will give me six different times. To only problem is that we don't know where the ball crosses the laser (probably not in the middle).
    I think we have to know when the middle of the ball crosses the laser, and that is proving to be quite hard due to the complexity of the formula below. (If the speed would have been constant, the middle of the ball would cross the laser at t=(t1+t2)/2) but due to the acceleration of the ball that isn't the case).
    Also the exact moment the ball is droped isn't known. So the first time you'll know is the moment the ball enters the first laser(so t1=0).

    2. Relevant equations
    Falling object experiancing air resistance:
    [tex] \frac{dv(t)}{dt} = g - q v(t)^2 [/tex]
    [tex] v(t) = \frac{g} {\sqrt{g q}} tanh(\sqrt{g q} (t+c_1)) [/tex]

    3. The attempt at a solution
    I think the solution is to solve a system of equations using 8 know values (5 times (t1=0 so everything is relative to t1) and the locations of the 3 lasers). Eventually I'm trying to find g.
    Last edited: Jun 7, 2012
  2. jcsd
  3. Jun 7, 2012 #2
    Can't you:

    1) Just take the times as when the lower ball edge breaks the beam, and not worry about the middle

    2) It sounds like you drop the ball from above the lasers, and if you don't know the time you don't know the initial velocity. In that case the (fallen) distance would be
    d=Vinitial*t + 0.5*g*t^2
    After hitting the second laser you will have a d and a t. You'll have one equation and two unknowns, Vinitial and g. Hmm.
    After hitting the second laser you then have d2 and t2, so that makes 2 equations and two unknowns. You should be able to plug in and solve. To get g the most precise, probably rearrange to Vinitial subtracts out.
  4. Jun 8, 2012 #3
    I think you're probably trying to split hairs. You need to dope things out a little first. How much can the velocity of the ball possibly change during the time it passes through one of the lasers? It couldn't be more than if the ball were in vacuum. So look at the vacuum problem first, and calculate the times that the top, bottom, and center of the ball would pass through the laser. Then see how much of an error you would be making if you took the time for the center of the ball to pass as being equal to the average of the top and bottom. In the case where there is air drag, the error would be even smaller.

    Also, what makes you think that the air drag is proportional to the square of the velocity. This is only true if the drag coefficient is a constant. The drag coefficient is a function of the Reynolds Number. Look up the correlation for the drag coefficient as a function of Reynolds Number for air flow past a sphere. This will give you an idea of how the drag force actually varies with speed of the ball.

    Last edited: Jun 8, 2012
  5. Jun 11, 2012 #4
    Thanks for the help! I forgot the thing about the Reynolds Number! Do you think it makes a lot of difference?

    I worked out a solution with the help of Mathematica:

    [itex]t_{a1}[/itex] and [itex]t_{a2}[/itex] are known, [itex]t_a[/itex], the moment the center passes is unknown. However, I think (correct me if I'm wrong please) solving the below equation for [itex]t_a[/itex] works:
    Solving the integral gives:
    [itex]-\frac{Log(Cosh(\sqrt{g q}(c+t_{a1}))Sech(\sqrt{g q}(c+t_{a})))}{q}=-\frac{Log(Cosh(\sqrt{g q}(c+t_{a}))Sech(\sqrt{g q}(c+t_{a2})))}{q}[/itex]
    Solving the above equation for [itex]t_a[/itex] gives four solutions two are complex and one other doesn't satisfy [itex]t_{a1}<t_a<t_{a2}[/itex] to one left is:
    [itex]t_{a}=\frac{-c g q+\sqrt{g q} ArcSech(\sqrt{Sech(\sqrt{g q}(c+t_{a1}))Sech(\sqrt{g q}(c+t_{a2}))})}{g q}[/itex]

    Furthermore, the total distance traveled is r(t)=∫v(t)dt =[itex] \frac{Log(Cosh(\sqrt{g q}(c+t)))}{q}+C_1[/itex]

    Now the last step is to solve the system (Ra,b,c are the positions of the lasers):
    [itex]r(t_a)=R_a \\
    r(t_b)=R_b \\
    I haven't managed to solve the above system algebraically but Mathematica can do it numerically using FindRoot. If anybody has the time to check this or solve the system it would be awesome, but I understand if you don't :)
  6. Jun 11, 2012 #5
    Depending on the Reynolds Number, the drag force can vary from being linear in the velocity to being quadratic. For very small spheres like fog droplets, the drag is linear. For large spheres, the behavior approaches quadratic. To be sure, you can calculate the Reynolds Number, and then check the drag coefficient graph to see what regime you are in. If the drag coefficient is approximately constant over the range of Reynolds Numbers of interest, then the drag force vs velocity behavior is quadratic. If the drag coefficient is decreasing nearly linearly with Reynolds Number, then the drag force vs velocity behavior is linear. You owe it to yourself to check.

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