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Determining the value of acceleration due to gravity

  1. Nov 28, 2010 #1
    1. The problem statement, all variables and given/known data

    "The relation between the distanse S of the trolley as it moves on air track and the time t is given by S = Vt+(1/2)at^2, where V is the initial velocity of the trolley. The following set of data were taken for the distance S of a trolley on an air track apparatus, whose angle of inclination is 5 degrees, as a function of time t."

    Given data:
    S: 0 - 0.15 -0.388 - 0.600 - 0.930 - 1.37 - 2.45
    t: 0 - 0.975 - 1.48 - 2.04 - 2.46 - 3.03 - 3.95

    The questions:

    a-) Calculate t^2 for each case.
    b-) Plot a graph of S (y-axis) against t^2 (x-axis)
    c-) From the graph determine the magnitude of the acceleration of the trolley.
    d-) From the graph, determine the initial velocity of the trolley.

    From the above results, determine the experimental value of the acceleration due to gravity.

    2. Relevant equations

    s = vi*t + 0.5at^2, where s is the distance, vi is the initial velocity, t is the time, and a is the acceleration.

    3. The attempt at a solution

    Parts a and b are easy, as you just square t and plot the graph. As for part c, as I recall the slope of a distance against a time squared graph is equal to half the acceleration, and as such the acceleration is equal to 2 times the slope.
    As for d, the initial velocity can be determined by the formula given. According to my prof, any value of vi is OK (given the given data is used, obviously).

    What I can't get right is the last part, determining the experimental value of acceleration due to gravity. I tried a couple of times, but I keep getting a = 0.30 from the graph, which, when divided by sin of the angle (5 degrees), gives the value 3.44 (a = g*sin (theta)).

    What am I doing wrong? Is the acceleration determined in a different way? Does the experimental value really have that big of an error percentage?

  2. jcsd
  3. Nov 28, 2010 #2


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    Something to consider.. Does the problem state that this is on Earth? :biggrin:

    From your data alone, a=2*S/t^2

    a=2*0.15/0.975^2 ~ 0.315
    a=2*0.388/1.48^2 ~ 0.354
    a=2*0.6/2.04^2 ~ 0.288
    and so on..

    even if I take the average of these 3, a~0.319

    and a=g*sin(theta) yields g~3.66

    So maybe its not on Earth?
  4. Nov 28, 2010 #3
    This is a question on my physics lab report. We did a whole experiment on how to calculate Earth's gravitational acceleration using air tracks, light sensors, etc. As such, it is pretty obvious that the question is asking for earth's gravitational acceleration. And sadly, this is all the data we have.

  5. Nov 28, 2010 #4


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    Well then you have friction to account for, and maybe its not 5 degrees? If it was 2-3 degrees.. then it would make sense
  6. Nov 28, 2010 #5
    Since the trolley is moving on an air track, friction is almost negligible. The angle of inclination is is specifically stated as 5 degrees.

    Is the slope of a distance against time squared graph really equal to half of the acceleration?

  7. Nov 28, 2010 #6


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    [tex]S = \int v dt[/tex] [tex]= \int (v_{initial}+at) dt =[/tex]
    [tex]S_{initial} +[/tex] [tex]v_{initial}t +[/tex] [tex]\frac{at^{2}}{2}[/tex]

  8. Nov 28, 2010 #7


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    Well come to think of it, the problem does ask for Vinitial, so it can't be 0

    In that case, S= V*t + 0.5*a*t^2

    Which will give you a bunch of linear equations
    0 = V*0 + 0.5*a*0
    0.15 = V*0.975 + 0.5*a*(0.975)^2
    0.388 = V*1.48 + 0.5*a*(1.48)^2
    0.600 = V*2.04 + 0.5*a*(2.04)^2
    0.930 = V*2.46 + 0.5*a*(2.46)^2
    1.37 = V*3.03 + 0.5*a*(3.03)^2
    2.45 = V*3.95 + 0.5*a*(3.95)^2

    7.80125 a+3.95 V = 2.45
    4.59045 a+3.03 V = 1.37
    3.0258 a+2.46 V = 0.93
    2.0808 a+2.04 V = 0.6
    1.0952 a+1.48 V = 0.388
    0.475312 a+0.975 V = 0.15

    Which then reduces to nonlinear least squares problem

    [7.80125,3.95;4.59045,3.03;3.0258,2.46;2.0808,2.04;1.0952,1.48;0.475312,0.975]*[a;V] = [2.45; 1.37; 0.93; 0.6;0.388;0.15]

    b=[2.45; 1.37; 0.93; 0.6;0.388;0.15]

    x=[a;V]= inverse(transpose(A)*A)*transpose(A)*b

    [a;V] = [.3172542164;-0.01304253740]

    Vinitial = -0.013

    Either that or the more likely problem is that the incline is less than 5 degrees, and there is friction
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