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Circular motion ap physics lab

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  1. Oct 22, 2011 #1
    1. The problem statement, all variables and given/known data
    So i have to do a lab about circular motion
    we had to swing a rubber stopper around while holding a glass tube and when there was a mass hanging at the end of it
    "plot a graph so the slope of the straight line graph may be used to determine the mass of the stopper"
    we changed the mass at the bottom and varied the length of the string, but i don't know if this is a problem or not
    we were going to plot a centripetal force as a function of centripetal acceleration graph, and find the slope to find the mass of the stopper
    we measured time (t) in seconds, the length of the string (r) in meters, the mass hanging mass (m1) in kilograms, and we were supposed to calculate centripetal acceleration (ac) in meters per second squared, centripetal force, (Fc) in newtons, and velocity (v) in meters per second, and we had to find the mass of the stopper (m2) in kilograms, but we measured it for percent error
    (this is one of the trials
    m1 = .05 kg, t =.486 s, r =.295 m, Fc = .49N, ac = ?, v = ? m2 = ?

    2. Relevant equations
    Fc=m1g
    Fc=m2ac = m2v2/r
    C=2∏r
    v= C/t


    3. The attempt at a solution
    So, i tried to find the velocity and the acceleration and they just did not make any sense at all, so i'm wondering if anyone can help me figure out what i did wrong.
    C=2∏r
    C=(2)(∏)(.295m)
    C=1.854m
    v=C/t
    v=1.854m/.45s
    v=4.12m/s
    ac=v2/r
    ac[/SUP]=(4.12m/s)2/.295m
    ac=57.5m/s2
    which does not make any physical sense at all, HELP.
     
  2. jcsd
  3. Oct 22, 2011 #2

    Delphi51

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    Homework Helper

    Welcome to PF, Rebeka.
    This is a difficult experiment I have done many times with grade 11 classes. For starters it is difficult to measure the time for one turn. Typically, you measure the time for 20 turns and divide by 20. But it is hard to keep turning at the same speed that long. Your "t" is the time for one turn, isn't it?
    Secondly, there is a good deal of friction where the string turns at 90 degrees on the top of the glass tube. You must not expect near perfect results.
    Thirdly, the string "droops" at an angle below horizontal so there is a vertical component of force on the string that balances gravity in addition to the horizontal component that is the centripetal force. The force of gravity on the mass hanging below the glass tube is the combination of these two forces. It turns out that the results are very sensitive to the angle of droop and you must work very hard to measure that angle in order to successfully show that F = mv²/r to within experimental error. Of course the angle is different for every run. Oh - and you need to estimate the error in your measurements, especially of the angle because it has the greatest effect.

    It looks like you did not measure the angle of droop, so you must make note of it in your writeup and proceed anyway - not expecting good results. It would be impressive to your marker if you showed how the angle matters and come up with a good way to measure it. Any chance of repeating it? This is one experiment you really can do at home.

    The usual approach is to make a table of values with your measurements and calculations of F and v²/r. I always tried to keep the answer from my students, so they were supposed to try combinations other than v²/r, too. Anyway, you graph F vs v²/r, put error bars on if you have any error estimates, and then try to draw a straight line through the error bars. If you can, you have F = k*v²/r and perhaps you can show that k is somewhere near the mass. At least it has the right units.
     
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