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A Sticky Problem!

  1. Mar 21, 2004 #1
    Hello Everybody!
    Er well i'm new to this forum so hi guys!
    I have a small problem which is made even worse by the fact i know very little about the subject. Ok the problem is about steel ball bearings falling through a liquid.

    The Situation:
    A wide cylindrical plastic tube 1.0 to 1.5 m long with a tight fitting bung at one end is filled with a viscous fluid. A steel ball bearing is dropped through the fluid and the time, t, is measured for various distances, s.

    I did not complete this experiment the results were given to me. The results i have are for three different fluids, sunflower oil, glycerol and syrup. For each liquid there are three sets of results which were taken at different temperatures. In each of the tables there are three results for small, medium and large (i am assuming that they are different sizes of ball bearing) and the results for the viscosity of the liquid in those conditions is given. Also there is a average result.

    Somehow i have to be able to deduce a displacement-time graph from this information. Any help? i found one equation which is:

    viscosity=( 2(p)ga^2 )/ 9v

    where p=difference in densities of fluid and ball, g=graity,v=velocity,a=radius of ball(unknown) but this leaves me as stuck as i was before.

    Any help would be gratefully appreciated!

    Cheers Guys

    P.S. Physics Rock!
  2. jcsd
  3. Mar 21, 2004 #2
    There are 3 forces on the ball - buoyancy, viscous retarding, and its weight.

    [tex]\Sigma F = ma = mg - F_{\mbox{buoyancy}} - F_{\mbox{viscous}}[/tex]

    [tex]r = \mbox{ball radius}[/tex]
    [tex]\rho = \mbox{ball density}[/tex]
    [tex]\rho ' = \mbox{fluid density}[/tex]
    [tex]\eta = \mbox{fluid viscosity}[/tex]
    [tex]m = \mbox{ball mass} = \frac{4}{3}\pi r^3\rho[/tex]

    Now substitute everything...

    [tex]\frac{4}{3}\pi r^3\rho a = \frac{4}{3}\pi r^3\rho g - \frac{4}{3}\pi r^3\rho 'g - 6\pi \eta rv[/tex]

    Reduce it a little and change sides:
    [tex]a = \frac{r^2\rho g - r^2\rho 'g - 8\eta v}{r^2\rho }[/tex]

    Now take the derivative of that by [tex]dt[/tex]:
    [tex]\frac{da}{dt} = -\frac{8\eta }{r^2\rho }\frac{dv}{dt} = -\frac{8\eta }{r^2\rho }a[/tex]

    You see something very special here. The derivative of the acceleration is proportional to the acceleration itself. The mathematical function that adheres to this is:
    [tex]f(x) = ae^{-kx}[/tex]
    [tex]f'(x) = -kae^{-kx}[/tex]
    (a and k being constant) Where a is the inital value of the function when x = 0.

    To fit the equation we got above for the acceleration into that function we need to find the inital acceleration, which is achieved when the ball first enters the liquid and its velocity is zero.
    [tex]a_{v = 0} = \frac{r^2\rho g - r^2\rho '}{r^2\rho } = \frac{\rho g - \rho 'g}{\rho } = g\frac{\rho - \rho '}{\rho }[/tex]
    Let's call that [tex]a_0[/tex].

    After all of this we find that the acceleration of the ball as a funciton of time is:
    [tex]a(t) = a_0e^{-kt}[/tex]
    [tex]k = \frac{8\eta }{r^2\rho }[/tex]

    So you have the acceleration, but you want the velocity of the ball. Well that's easy, since a(t) is the derivative of V(t), V(t) is the integral of a(t).
    [tex]V_t = \int a_tdt = \int (a_0e^{-kt})dt = \frac{a_0e^{-kt}}{-k} + C[/tex]

    So now you just need to find C. How? We say that at [tex]t = 0[/tex], the velocity of the ball is zero:
    [tex]V_{t = 0} = 0 = \frac{a_0}{-k} + C[/tex]
    [tex]C = \frac{a_0}{k}[/tex]

    [tex]V_t = \frac{a_0e^{-kt}}{-k} + \frac{a_0}{k}[/tex]

    Reorganizing a bit gives us the final formula for the ball's velocity:
    [tex]V_t = \frac{a_0}{k}(1 - e^{-kt})[/tex]

    *PHEW* :smile:
  4. Mar 21, 2004 #3
    Ehh, exactly what do you need to do? Just find the graph of X(t)? Because if "the time, t, is measured for various distances, x" and you have those results, just plug them into Excel and let it draw the graph for you!

    Oh my, I think I did all that work for nothing. :frown:
  5. Mar 21, 2004 #4

  6. Mar 21, 2004 #5
    Well, I'm thinking you also need to provide the formula of x(t), which you can find it by integrating V(t). So it wasn't all for nothing.

    [tex]x_t = \int V_tdt = \int \frac{a_0}{k}(1 - e^{-kt})dt = \frac{a_0}{k}\int (1 - e^{-kt})dt[/tex]
    [tex]x_t = \frac{a_0}{k}(t + \frac{e^{-kt}}{k} + C')[/tex]

    To find C' you say that at [tex]t = 0[/tex] the ball's displacement is 0:
    [tex]x_{t = 0} = \frac{a_0}{k}(0 + \frac{e^0}{k} + C') = 0[/tex]
    [tex]C' = -\frac{1}{k}[/tex]

    So finally:
    [tex]x_t = \frac{a_0}{k}(t + \frac{e^{-kt}}{k} - \frac{1}{k}) = \frac{a_0}{k^2}(kt + e^{-kt} - 1)[/tex]

    Can you analyze the ball's movement using this formula? :smile: Can you also tell how the ball's motion will change if [tex]0 < k < 1[/tex]?
    Last edited: Mar 21, 2004
  7. Mar 21, 2004 #6
    I really don't see why one would need to solve the whole differential equation for the starting phase of the ball's movement; in this experiment (with very viscous liquids) the "steady state" with constant velocity is reached after a few inches.

    So all you need is Stoke's law

    [tex] F=6 \pi \eta r v [/tex]

    The displacement-time graph would have to be all linear since as I said v=const.
  8. Mar 21, 2004 #7
    Wow hi Thank you for replying!

    The main problem i had was that i have no information about the ball at all apart from the fact it is steel and i no nothing about the dimensions of the tube. All i have is the viscosity. I will look into what you have given me and also i am writing a program that will be able to show this.

    My head hurts, lol

    Thank You Again

  9. Mar 21, 2004 #8
    wouldnt this also be true:




    a=[ 6(pi)rnv ]/m
    a=[ 6(pi)rnv ]/ p4/3(pi)r^3

    sorry for the crude typing, i dont have your fancy equation generators

  10. Mar 21, 2004 #9
    No, you're forgetting the weight of the ball.

    [tex]\Sigma F = F_s - mg = 0[/tex] (since you assume the velocity is constant)

    [tex]mg = 6\pi r\eta v[/tex]
  11. Mar 21, 2004 #10
    By the way, you can get this constant velocity kuengb is speaking of from this formula still:

    [tex]V_t = \frac{a_0}{k}(1 - e^{-kt})[/tex]

    As [tex]kt[/tex] tends to infinity, [tex]e^{-kt}[/tex] tends to 0. From the mathematics point of view, this would take forever but in our case it only takes a few moments. So the constant velocity, also known as the terminal velocity, is:
    [tex]V_t = \frac{a_0}{k} = \frac{gr^2}{8\eta }(\rho - \rho ')[/tex]

    (Of course, if you ignore the part where the ball develops this velocity, you don't have to go through all the trouble that I have...)
  12. Mar 21, 2004 #11
    What I wanted to say was: There is no need for going through all the calculations like you did (wow! That latex stuff would have KILLED me ) if you know that the difference between the "more exact" and the simpler model doesn't matter for the experiment. With the given data, I'm sure that already after maybe 0.1 sec the numeric difference between "exponential" and "constant model" will be far beyond measure accuracy.

    The best mathematical model for a physical problem is not always the most complicated.
  13. Mar 21, 2004 #12
    I don't disagree. :smile: BUT, if he's required to find the value of these three unknowns:
    [tex]r = \mbox{ball radius}[/tex]
    [tex]\rho = \mbox{ball density}[/tex]
    [tex]\rho ' = \mbox{fluid density}[/tex]

    Then he does need to develop the formulas like I did so eventually he will get this:
    [tex]x_t = gr^2\rho \frac{\rho - \rho '}{64\eta ^2}(\frac{8\eta }{r^2\rho }t + e^{-\frac{8\eta }{r^2\rho }t} - 1)[/tex]

    Then to find the three unknowns he would have to pick 3 convenient points and get 3 equations with 3 unknowns. If he skips everything and just assumes the velocity is constant at all times, he is ignoring the inital acceleration of the ball when it first enters the liquid. Therefore the density of the liquid doesn't play a factor in any of the calculations, and he won't be able to find it.
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