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Newtonian Mechanics - Ball Falling through Syrup

  1. Jan 27, 2013 #1
    1. The problem statement, all variables and given/known data
    A steel ball bearing falls through a viscous syrup. The equation of motion is
    m(dv/dt) = mg - (alpha)v where g = 9.8 m/s^2. The terminal velocity is 0.2 m/s. (a) Determine the value of alpha. (b) Determine the speed at t = 0.04 s. [initial velocity = 0; mass m = 12.6E-3 kg] (c) Determine the distance that the ball bearing will fall from t=0 to t=0.1s.



    2. Relevant equations
    ma = mg-(alpha)v


    3. The attempt at a solution
    OK, so for this problem, I would like confirmation on my work, not necessarily the solution to the question. For part (a) I have alpha = (mg)/v_t where v_t is the terminal velocity. Part (b) is where I am unsure of my work. Using differential equations, I obtain the equation for velocity to be v(t) = (1-e^((-g*t)/v_t))*v_t, which at t=0.04s gives me a velocity of 0.172 m/s. However, when I met with a group member to work on this, his velocity equation was v(t) = (gt)/(1+(gt)/v_t) which gave v(0.04s) = .132 m/s. The difference in our integration was at the step (dv/dt) = g - (v*alpha)/m. My partner merely switched the dt over and integrated dv = gdt - ((v*alpha)/m)dt, whereas I integrated dv/(1-((alpha*v)/(mg))) = gdt because I collected the velocity terms on the left had side. I wanted to know which was the correct way, if either?

    For part c, using my velocity equation, the position equation I obtained was
    x(t) = t*v_t + (v_t/g)*e^((-gt)/v_t) + (v_t)/g which from t = 0 to t = 0.1 s gave me a displacement of -0.0002m, which doesn't seem right. Again, I don't really want solutions so much as someone to check my work. Thanks in advance.
     
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  3. Jan 27, 2013 #2

    cepheid

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    My response to your friend is: HUH? You don't know what the function v(t) is. In fact, that is precisely what you are supposed to solve for. So how exactly do you expect to solve this equation just by directly integrating with respect to time, when v is in the integrand?


    You seem to have obtained the correct solution for v(t) (you'll notice that your v asympotitcally approaches vt), although I'm puzzled as to how you did it. As far as I can see, the pesky g term makes this equation non-separable, which means that you can't just collect like terms and then integrate. Once I had the equation in the form ##\dot{v} + (\alpha / m)v = g##, I used the method of integrating factors to turn the left hand side into an exact differential. THEN I was able to integrate both sides with respect to time. Note: the dot denotes a derivative w.r.t. time in physics.
     
  4. Jan 27, 2013 #3
    Ok, so using your equation (dv/dt) + (alpha/m)*v = g, I came to a new equation for v(t):
    v(t) = v_t - 1/e^((gt)/v_t)) Does this check out? It has been a year since I've taken differential equations, and I sadly haven't used it since, so I'm feeling very unsure of my work.

    Also, now for the final part, I integrated the equation for position, obtaining x(t) = (v_t)t + (v_t/g)*e^(-(gt)/v_t) - (v_t)/g, which gives me a negative displacement of -.00026 m. I can't help but feel that is incorrect. My work for this integration was using:
    dx = (v_t - 1/e^((tg)/v_t))dt = (v_t)dt - dt/e^((tg)/v_t). Is that even remotely correct?
     
    Last edited: Jan 27, 2013
  5. Jan 27, 2013 #4

    cepheid

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    Like I said before, your original solution for v(t) (that was of the form A[1-exp(-t/b)]) was correct. I just have no idea how you arrived at it. In any case, did you look up the Method of Integrating Factors?


    It's kind of hard to parse your equations, and I don't have time right now. Just integrate:

    x(t) = ∫v(t)dt

    and since v(t) is of the general form A[1 - exp(-t/b)], you get

    x(t) = ∫Adt - A∫exp(-t/b)dt

    = At + Abexp(-t/b) + C

    where you can fill in the constants A and b with what they are actually supposed to be, and
    figure out the constant of integration C yourself.
     
  6. Jan 27, 2013 #5

    haruspex

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    Doesn't look right. You should get something like vt(1-e-kt).
     
  7. Jan 27, 2013 #6

    cepheid

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    Which is the solution that he/she had in the OP, like I said above (twice).
     
  8. Jan 27, 2013 #7
    Well there we go. I am finished with this problem and confident in my answers. Thank you both for your help, and sorry for confusing the process, I misunderstood your (cepheid) comment on my initial v(t) equation, but now I can regroup with my partner and finalize this. Again, thank you.
     
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