# Putting an equation into y = mx+b form

1. Aug 29, 2015

### amanda.ka

1. The problem statement, all variables and given/known data
I am doing an experiment where I have to test the validity of the equation: (m1-m2)g=(m1+m2+I/R^2)a. The lab instructions say to linearize this equation. What would be the "m" and "b" values?

2. Relevant equations
y=mx+b
(m1-m2)g=(m1+m2+I/R^2)a
3. The attempt at a solution
I reasoned that m1+m2 = slope and that the y-intercept will be zero, could someone please tell me if I did this correctly? Thank you!

2. Aug 29, 2015

### SammyS

Staff Emeritus
I suppose that you varied one or two of those parameters and perhaps measured another.

What was varied and what was measured?

Which of parameters are you trying to determine?

3. Aug 29, 2015

### amanda.ka

the weight difference (m1-m2)g was taken as the y variable and the acceleration (a) was taken as the x variable so I took these as varied and the slope and y-intercept to be the constant variables.

4. Aug 29, 2015

### SammyS

Staff Emeritus
Why did you leave $\ I/R^2 \ ?$ out of your slope?

5. Aug 29, 2015

### amanda.ka

Later in my lab manual it says to find the moment of inertia from slope so that is why I did not put "I" into my slope value.

6. Aug 29, 2015

### SammyS

Staff Emeritus
Leaving $\ I\$ out of the expression for slope would make it very hard (impossible) to then use the slope to determine $\ I\$ .

7. Aug 29, 2015

### amanda.ka

ok so my slope would be: m1+m2+I/R^2?
but then from a graphing software based on my data I got an actual value for the slope so how would I solve for I using that slope? Im just confused with how it all relates :(

8. Aug 29, 2015

### SammyS

Staff Emeritus
I'm assuming that when you varied m1 - m2 , you kept the sum, m1 + m2 constant. Is that right?

If so, then solve : slope = m1+m2+I/R2 for I.

9. Aug 29, 2015

### amanda.ka

Yes, (m1-m2)g was varied and I obtained the slope by plotting a graph of (m1-m2)g vs. a (acceleration).

10. Aug 29, 2015

### SammyS

Staff Emeritus
Yes, but was m1 +m2 maintained at the same value for all of the trials?

11. Aug 29, 2015

### amanda.ka

yes

12. Aug 29, 2015

### SammyS

Staff Emeritus
OK. then do the following:

13. Aug 29, 2015

### amanda.ka

14. Aug 29, 2015

### andrevdh

Tell us a bit about the apparatus.

15. Aug 29, 2015

### SammyS

Staff Emeritus
You're welcome.

Does your final result seem to make sense ?

16. Aug 29, 2015

### SammyS

Staff Emeritus
Likely it's Atwood's machine.

17. Aug 29, 2015

### amanda.ka

yes its an atwood machine (pulley) :) and my results make a lot more sense now, thanks again!