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Finding a damping constant

  1. May 1, 2014 #1
    This is actually for a engineering course in modelling, and not in physics per se, but it seems to me to be fairly basic physics. Apologies in advance if it's out of place.

    1. The problem statement, all variables and given/known data
    A car with a mass of 1000 kg is held still on a slope with an inclination of 5.8 °, and then let go. Its speed v(t) is measured at intervals (t seconds): v(0) = 0 m/s, v(10) = 2.05 m/s, v(20) = 3.30 m/s, v(30) = 4.15 m/s, v(40) = 4.85 m/s, v(50) = 5.20 m/s, v(60) = 5.55 m/s. Find the effective damping constant [itex]b[/itex].


    2. Relevant equations
    Ideal damper: [itex]F_b (t) = bv(t)[/itex]


    3. The attempt at a solution
    Okay, let me preface by saying that the problem before this one was similar: a car with mass 1600 kg is sped up, and the gas pedal is let go at t=0; speed is measured once every ten seconds from t=0 to t=60 s. (Measure data is v(0) = 4.6 m/s, v(10) = 3.1 m/s, v(20) = 2.0 m/s, v(30) = 1.37 m/s, v(40) = 0.88 m/s, v(50) = 0.64 m/s, v(60) = 0.38 m/s.) That problem asked me to model the car as a mass with a damper, and to just approximate [itex]\frac{dv}{dt}[/itex] by drawing a curve of [itex]v(t)[/itex] and estimate the slopes at points in order to find the damping constant [itex]b[/itex].

    I solved that problem (at least I think I did) by saying that the resulting force on the mass is [itex]-bv(t)[/itex] which is equal in size to [itex]m \frac{dv}{dt}[/itex], so [itex]b = \frac{-m}{v(t)} \frac{dv}{dt}[/itex]; I got [itex]b[/itex] to between 60-68 kg/s, except for t=60s where it went up to 106 kg/s. So my answer there was that the damping constant [itex]b[/itex] was around 65 kg/s.

    Now! This previous problem implies that the problem which is the topic of this thread is to be solved in a similar manner. My thinking is that, if friction and drag can be rolled into this damper model, gravity will make the car accelerate until the damper force cancels it out: [itex]m \frac{dv}{dt} = mg_{\text{parallel}} - bv(t)[/itex] (where i have [itex]g_{\text{parallel}} = 9.82 \sin 5.8°[/itex]). My thinking was that I could just, as before, solve for [itex]b[/itex] and plug in values for [itex]v(t)[/itex] and approximations of [itex]\frac{dv}{dt}[/itex] without even having to solve a differential equation, but I only get wildly varying values of [itex]b[/itex]. I.e. if [itex]b(t) = \frac{mg_{\text{parallel}} - m \frac{dv}{dt}}{v(t)}[/itex], then b(10)=404, b(20)=267, b(30)=222, b(40)=195, b(50)=183, b(60)=171. Solving the differential equation to [itex]v(t) = \frac{mg_{\text{parallel}}}{b} (1-e^{-\frac{bt}{m}})[/itex], inputting known values and solving for [itex]b[/itex] (thank you, Wolfram Alpha) gave me values b=480, b=300, b=239, b=205, b=191, b=179. This damping constant doesn't look very constant to me.

    And that's where I hit the wall. Can someone tell me, what am I doing wrong here? Are my assumptions wrong? Am I misunderstanding what is requested? I suspect that the differential equation is lacking something, but I don't know what, and I can't seem to get any clues from the accompanying course text.
     
  2. jcsd
  3. May 1, 2014 #2

    ehild

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    Determine dv/dt for the intervals 0-10 s, 10-20 s, ...and also the average speeds in these intervals. Plot dv/dt in terms of the average speeds. Fit a straight line to the points, the tangent of the line should be -b/m. See attached picture.

    You can also notice that the constant terms is not g sin(5.8°). There should some other force than (b v) exist that opposes the acceleration of the car. That force - some kind of kinetic friction- is independent on the speed.
    The equation should be mdv/dt =mgsin(a)-f(friction)-bv.

    ehild
     

    Attached Files:

  4. May 1, 2014 #3
    Ooh! I get it now, rewriting the equation as a straight line, I wish that had come to me naturally... Thank you so much!
     
  5. May 1, 2014 #4

    ehild

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    You are welcome.

    ehild
     
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