# Setting up a graph for acc. of gravity

1. Nov 8, 2004

### MattF

Hi, I'm having a bit of a problem with an experiment I recently did. Basically I had two objects on a pulley (let's say masses M and m, M being the heavier object), at five different heights. I released the objects three times for each height. I then calculated the acceleration for each seperate time. Thus I made fifteen data points (fifteen accelerations for each time interval).

Now, here is the problem. I need to set up a graph with fifteen data points, and a slope of g . I already know the value of g(the slope) will NOT be 9.81 m/s^2. According to my data it will be roughly in the range of 8 m/s^2 (estimated). This is due to reaction time and other outside influences distorting the data, since it is not a perfect experiment in ideal conditions.

The y-axis must be acceleration, right? What will the x-axis be? I got a hint that it should be (M-m)/(M+m). This seems to be a valid method, because I get a reasonable value dividing acceleration by that. However, I only have those two masses, so if I graph my points I will get a vertical line. The masses do not change, thus the x-axis will only have one value. I obviously need a diagonal line in order to get a slope of g .

Basically my big question is, what values do I assign the x-axis in order to get a slope of g ? I have the height, time, final velocity, and acceleration. ANY help would be appreciated!!

2. Nov 8, 2004

### Integral

Staff Emeritus
If the slope of your line is supposed to be acceleration, then the plot axis must be velocity or displacement, and time. If you plot displacement vs time your graph will be parabolic. Average velocity vs time is the more likely quantity to plot. It would be easy to get (displacement/time) and will yield a linear plot with slope of g.

3. Nov 8, 2004

### MattF

I'm sorry but I think there's a little misunderstanding :). I already HAVE the acceleration taken from average velocity vs time. I did as you suggested and did v/t and d/t, but neither gives the right slope. I may have totally missed something, but I don't get it.

What I have are fifteen different accelerations of the pulley system, derived from three time intervals for each height. I have a, but I need gravitational acceleration. Acceleration of the pulley system needs to be on the y-axis, but what I'm confused about is the x-axis. What value goes on it? The linear plot will have a slope of g. The only thing that I know of that works for one data point is (M-m)/(M+m). For instance a=0.6227 m/s^2, M=16.31 g, and m=13.92 g. Thus g is about 7.876 m/s^2. However, I need values for the x-axis, and I only have those two masses.

I need further explanation :)

4. Nov 8, 2004

### MattF

I'll try and clear this up a bit more. The whole experiment is working with an Atwood device. I just realized that I totally messed up in one area. In this experiment I needed to variate the masses, which I forgot to do. Thus, I have the same two masses for each different height.

So, when I tried to graph the data, it failed because acceleration has to be a function of (M-m)/(M+m). A graph of the measured acceleration as a function of (M-m)/(M+m) yields the acceleration due to gravity as the slope. Basically the y-axis is acceleration and the x-axis is (M-m)/M+m), giving a slope of g which I'm trying to get. That's the answer! Does this make sense now? I hope this sounds logical.

Since I don't have more masses to do the x-axis on the graph, is it impossible to graph this? I've spent hours trying to find an alternative. Does anyone have an idea?

Integral, thanks for trying to help. I suppose I sounded a bit confusing, huh? :)