# Rolling Motion, confusing with linearizing a graph?

1. Feb 13, 2013

### tariel

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. Feb 13, 2013

### TSny

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 $y$?

3. Feb 13, 2013

### tariel

y = 2/3 x ?
So 2/3 is the slope? And since there's no b, this means that y = 0?

4. Feb 13, 2013

### TSny

Yes. Good. Note that y = (2/3) x is a linear equation. Thus, the word 'linearizing'.

5. Feb 13, 2013

### tariel

Well... that was a lot easier than expected. Thanks for your time!