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I have been asked to plot the relationship of $log(I)$ versus $log(R)$, where $I$ is the moment of inertia of an object and $R$ is the radius used to calculate $I$. What does the slope of this plot represent?

Also, what would the vertical axis intercept of this plot represent?

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$$ y = ax^k $$ taking logs we have $$ \log y = k \log x + \log a $$ if we relabel as $$ \bar{y} = k \bar{x} + c $$ we should see that the gradient of the last equation i.e. the $k$, maps to be the gradient in the log-log plot which in turn maps to being the exponent of the original equation.

So in short the gradient of the log-log determines if the original equation is a power law one, and if the gradient indeed does not change then we can assume that the exponent of the power law equation is the gradient.

The Intercept $c = \log a$ which is basically the coefficient of the power law.

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  • $\begingroup$ So, if $I$ is proportional to $R^2$, then the slope of the log-log plot should be 2? $\endgroup$
    – Alexander
    Nov 17, 2016 at 17:16
  • $\begingroup$ yes one should conclude that if you plot $\log I$ on the $y$-axis and $\log R$ on the other then yes the gradient will be $2$. $\endgroup$
    – Chinny84
    Nov 17, 2016 at 17:20
  • $\begingroup$ Thank you! Sorry if this is getting more off topic/is a stupid question, but would the slope of an $I$ versus $R^2$ plot represent the $k$ value in $I = kR^2$? $\endgroup$
    – Alexander
    Nov 17, 2016 at 17:27
  • $\begingroup$ Why would $I = k$ when $R=0$? Since they are multiplied, should $I$ not equal 0? Also, I did not mean to use the same variable. So, if $k$ does not function as the coefficient of the power law, would the slope of $I=kR^2$ equal $k$? $\endgroup$
    – Alexander
    Nov 17, 2016 at 17:35
  • $\begingroup$ Totally right - I was confusing another function. But remember the form of the log-log equation and the intercept is basically the log of the coefficient you multiply the function by. $\endgroup$
    – Chinny84
    Nov 17, 2016 at 17:39

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