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Rolling Motion, confusing with linearizing a graph?

  1. Feb 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Rolling Motion

    Theoretically, a=⅔g sin θ suggests a nonlinear relation between a and θ. Since a linear graph is a very convenient method of testing theoretical equations, it is a good idea to first linearize a=⅔g sin θ. A simple way to do this is to assign a as the y-variable and (g sin θ) as the x-variable. If a=⅔g sin θ is linearized in this way, what are the expected values of the slope and y-intercept of the linear graph?

    Plot a versus g sin θ. Perform a linear fit to the data, and determine the slope and the y-intercept. Compare the results with the expected values according to a=⅔g sin θ.


    2. Relevant equations
    a=⅔g sin θ


    3. The attempt at a solution
    I'm not exactly sure where to start! To be honest, I've never heard of 'linearizing' a graph before. Any insight into how to go about this would be greatly appreciated.

    PS I made my graph from my data of acceleration and g sin θ, and it is indeed very linear. I'm just not sure how to relate it or how to find the 'expected' values?
     
    Last edited: Feb 13, 2013
  2. jcsd
  3. Feb 13, 2013 #2

    TSny

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    You're plotting ##a## versus ##gsin\theta##. So, think of ##a## as the "##y##" variable and ##gsin\theta## as the "##x##" variable. What would the equation ##a = \frac{2}{3} g sin\theta## look like using the symbols ##x## and ##y##?
     
  4. Feb 13, 2013 #3
    y = 2/3 x ?
    So 2/3 is the slope? And since there's no b, this means that y = 0?
     
  5. Feb 13, 2013 #4

    TSny

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    Yes. Good. Note that y = (2/3) x is a linear equation. Thus, the word 'linearizing'.
     
  6. Feb 13, 2013 #5
    Well... that was a lot easier than expected. Thanks for your time!
     
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