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Diluted Gravity

  1. Sep 17, 2005 #1
    Hey,

    I'm new here but I have a physics question with my homework! hehe

    We're working with dilution in gravity and we did a lab where we were to roll a marble down a ramp and record such variables as time and distance in order to afterwards calculate the acceleration and gravity force.

    Basically, the problem I am having is that my calculated gravity for one of the exercices is 11.0 m/s^2 ! And I am wondering if this is usual considering that regular gravity is only 9.81 m/s^2. So, if the latter were diluted, wouldn't that decrease the amount?

    The formulas we were told to use are a=2d/t^2 and g=a(d/h)

    Please help me in any way possible! =) By the way, nice forum!

    PS
     
  2. jcsd
  3. Sep 17, 2005 #2

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    Could you give us the data from your trial, and how you used it to arrive at the answer you got?
     
  4. Sep 17, 2005 #3
    11.0 m/s^2 is incorrect when considering dilation of gravity. This error is most liekly due to human error in the recording of results e.g. time, distance, height. To improve experimental data experiments should always be repeated at least 3 times.
     
  5. Sep 17, 2005 #4
    Hi, thanks for the replies! I indeed repeated the lab three times in order to achieve more accurate results. So I suppose that would lead to the problem being in my calculations.

    My times were:

    3.08 s, 3.32 s and 3.33 s (average of 3.24)

    The height of the ramp's heightened edge was 0.0580 m and the length of the ramp was 1.83 m.

    Well, since I needed acceleration in the formula for gravity, I started with finding that...

    a=(2d)/t^2

    a=2(1.83m)/(3.24s)^2

    a= 3.66m/10.5s^2

    a=0.349 m/s^2

    Then, I needed to find gravity with the formula...

    g=a(d/h)

    g=0.349m/s^2(1.83m/0.0580m)

    g=0.349m.s^2(31.6)

    g=11.0 m/s^2

    Well, that's about it... see anything wrong in there? I'm usually rather good in physics, I finished with 100% last year, hehe, I'm just really tired and am not seeing anything wrong in my calculations. Then again, perhaps the human measurements were just really off somehow.

    I'd appreciate any help!
     
  6. Sep 18, 2005 #5
    Anyone have an idea?

    This is due soon and I want to have it right! =D

    Thanks
     
  7. Sep 18, 2005 #6

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    Look's just like margin of error in your experimental results. You did nothing wrong with your math, you've just got slightly funky data. But they weren't off by that much, your third time listed gives a 'correct' answer within 6%.
     
  8. Sep 18, 2005 #7
    Perhaps, my teammate holding the stopwatch had mistaken the 13.xx s for what he thought was 3.xx s I have replaced them to 13 for experimental sake and the results make more sense. Anyways, this lab won't be worth much, at least it's the data and not my calculations! Thanks a lot for the help! =D
     
  9. Sep 18, 2005 #8

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    Whoa! your answer is not that far off! Maybe you need to rethink what answer you should expect to get for g. Ask yourself, What does g represent?
     
  10. Sep 19, 2005 #9
    Is g in this formula supposed to be equal to a value less than 9.81? I would think so because this gravity is being diluted by the ramp's presence. (i.e. it is not in free fall). So I would think that g represents the amount of acceleration due to gravity that is still present despite the dilution. Am I wrong?
     
  11. Sep 19, 2005 #10

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    "a" is the diluted value. 'g' is a constant for the surface of the Earth. I'm not trying to be difficult in my replies, I'm just hoping that if I nudge you right, you'll have an "AHA" moment And figure it out for yourself.
     
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