# Motion of the System (Pulley)

1. Oct 14, 2011

### Megzzy

1. The problem statement, all variables and given/known data
I'm currently doing a lab that focuses on a pulley and have got stuck on on part.

The goal here is to use the data from the Atwood’s pulley experiment to test the validity of (m1-m2)g=(m1+m2+1/R^2)a. In this equation, take the quantity (m1−m2)g as the y variable and the acceleration a as the x variable. This makes the equation linear. What does the slope and the y-intercept correspond to in this equation?

Regarding the y-intercept, it is expected to be zero according the equation. However, you will probably get a non-zero value for the y-intercept! What could be the reasons behind this deviation? Did we miss something while deriving the equation? Include a clear argument regarding this issue in your lab report.

2. Relevant equations
(m1-m2)g=(m1+m2+1/R^2)a
y=mx+b

3. The attempt at a solution
I honestly don't see what they are really asking in this situation. By looking at the equation as though it fits y=mx+b, the slope would be (m1+m2+1/R^2) however I am unsure as to what that is supposed to imply.

2. Oct 14, 2011

### ehild

What is R?

ehild

3. Oct 14, 2011

### Megzzy

R is the radius (2.25 cm).

4. Oct 14, 2011

### ehild

Your equation (m1-m2)g=(m1+m2+1/R^2)a is wrong. You can not add cm-2 to kg. How did you get that equation?

ehild

5. Oct 14, 2011

### Megzzy

6. Oct 15, 2011

### ehild

R is not radians, either. It is the radius of the pulley, but the correct formula is

(m1-m2)g=(m1+m2+I/R^2)a,

where I is the moment of inertia of the pulley.

This formula does not take friction into account. Because of friction, the system does not start to move when (m1-m2)g is less than the force of friction.

ehild

7. Oct 15, 2011

### Megzzy

Oh thank you! That makes much more sense. I'm still really confused about how it relates to the slope and y-intercept though.