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Verifying gravity from our lab.

  1. Jan 30, 2009 #1
    1. The problem statement, all variables and given/known data
    We had our lab yesterday in engineering physics where we had a car with nearly no friction descending down an incline plane. The incline was at 16 degrees, and we used data studio to record a crazy amount of trials at various distances. The point of this is to calculate gravity.
    I should also add, that there was a time taken from when the car passed two gates.. so we have an average of distances vs times (10 sets of data).. thats what the slope is.

    2. Relevant equations and what i've tried

    Originally for the prelab i solved what i considered to be the correct general solution to find acceleration from the position equation of motion.

    I came up with -2( r sin 16) / t^2. This is obviously wrong because i'm not getting anywhere near gravity with this.

    I'm pretty sure i'm supposed to use the slope of a linear fit of the data points somehow into this, but for some reason i don't really understand how it relates to this equation. I thought the slope for the distance vs time^2 was supposed to be the acceleration.. guess i'm wrong.. anything to help me understand this and get me to the desired solution would be amazing.
    Last edited: Jan 30, 2009
  2. jcsd
  3. Jan 30, 2009 #2


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    Homework Helper

    It isn't clear what has been measured.
    Did you measure the time for the car to roll down a certain distance from an initial position where the speed was zero? (Was the first gate practically at the starting position?)

    If so, you could apply the formula d = 0.5at^2 and solve for the acceleration down the ramp. This would be less than 9.81 m/s^2 because of the angle of the ramp, so it would have to be divided by sin(16). Pretty much what your formula says.

    You could calculate the acceleration from each of your data sets and average them.

    A more sophisticated analysis would show the data on a graph so that the trend of the data can be seen and found with a best fit line. To do this, you must see
    d = 0.5at^2 as a linear equation corresponding to y = mx + b.
    What will you choose as the "y" to put on the vertical axis of the graph?
    What will you choose as the "x" to put on the horizontal axis?
    You want the slope or "m" to be the thing you want to find - acceleration.
  4. Jan 30, 2009 #3
    I'm sorry i was too vague. I measured distances between two gates with a photo lens. The time it took for the car to roll down from point a to point b was also measured. Then those are the graphs i did the fit on.

    I've been trying things for the last hour non stop.. so far i'm here.

    2(dx/dt^2) = g sin 16 dx/dt^2 = 1.373x+.112 x = .8900s x=t where t was the average time value of the ten trials. dx/dt^2 was the slope of the fit on d vs. t^2

    I get 9.8m/s^2 if i do this.. but i don't know if how i set it up is valid.

    y axis is distance x axis is time squared
  5. Jan 30, 2009 #4


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    Is the first gate at the top, or starting point, where the speed is zero?
    Do points a and b refer to the locations of these two gates?

    I have no way of understanding how you found 2(dx/dt^2) so I can't say if it is correct. Your derivation should begin with some commonly known formula. Is dx/dt a derivative?
  6. Jan 30, 2009 #5
    delta x= v naught *t +.5at^2
    initial velocity was indeed zero
    a=2(change of x/ chang of t^2)

    then change of x/ change of y=1.373(.8900)+.112 i let x = the average time it took on all the runs. thats the equation for the slope so i figured since the fit line was on the distance vs. time squared
    that is equal to acceleration.

    then i found g sin 16 = a from a free body diagram

    g sin 16 = 2(1.373(.8900)+.112)
    then i did 2(1.373(.8900)+.112)/ sin 16

    Sorry for being so crude i've never tried talking geek on a forum :). I'll learn how to do the real symbols soon. And learn to be more clear on what i'm trying to get across.
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