1-Mar-2017: determination of g and some analyzing data

When we solve problems about motion, we always regard gravity to be a constant (9.8 m/s^2, or 9.81 m/s^2 in some case) in every situation. However, the real earth is not plat or a good-looking sphere. The real earth is like following.

(From GFZ, @GFZ_Potsdam)

From the image, we know that the hight is not the same in every place, which cause different gravity (from law of universal gravitation F = G • M1 • M2 / R^2). So, what is the gravity in Walnut? 

Moreover, after we grab all datas from different groups. We found we get different values about gravity. How is that happen, and which value should we use? 

The following lab will show the answer of all those questions. 

Part I:

Purpose: determine g.

At the beginning, Professor Wolf showed how to use the sparker and get a spark paper. 

Procedure:

1. Pull a piece of paper tape between the vertical wire and the vertical post of the device. Clip it with a weight to keep the paper "tight".
2. Turn the dial hooked up to the electromagnet up a bit. 
3. Hang the wooden cylinder with the metal ring around it on the electromagnet. 
4. Turn on the power on the sparker thing. 
5. Hold down the spark button on the spark box. 
6. Turn the electromagnet off so that the thing falls. 
7. Turn off power to the sparker thing. 
8. Tear off the paper strip.

After we get the spark paper, we picked a part of dots and measured the length between them. 

Then, we entered the datas into a Excel to make a graph on it. 

Then, we found out the function of distance is y = 4.7697x2 + 0.6901x + 0.0103

                         and the function of velocity is  y = 9.629x + 0.5247

Analyzing:

For any acceleration is constant velocity, there is function v = a*t + v0

Thus, the average velocity of any time interval will be (a*t1+ v0 + a*t2 + v0)/2 = a*(t1 + t2)/2 + v0

the middle of time interval is a*(t1 + t2)/2 + v0, which is the same as the average velocity. 

For the distance function: X = (a*t^2)/2 + v0*t + X0, 4.7697 = a/2, a = 9.54 m/s^2

It is only 0.12 m/s^2 away from average value, so it is accepted. 

For the velocity function: v = a*t + v0, a = 9.629 m/s^2

There is nearly no error from the average. It is a good data. 

Therefore, we found out our acceleration of gravity is 9.629 m/s^2

Part II:

Purpose: Analyzing the datas

Analyzing:

After all the groups done measurements, professor Wolf grabbed all the datas of gravity. And we found out that none of us get the exactly same result. 

There are various reasons (including professor gave us some fake value). One of the most important reasons is we will make some uncertainties, but as long as the error is acceptable, it will be fine. 

One way to value the error is using the function ∑i|xi-xavg|/N or √[∑i(xi-xavg)^2/N], which is called the standard deviation of the mean. 

As a result, we got 10.01 (cm) for the standard deviation of the mean, and it is acceptable. 

Analyzing:

Our value of g is the closest one to the average, and I think it is pretty good. For class, most datas are close to 9.6285 m/s^2, but there are two datas which are pretty far from the average value. 

I think most datas that are close to 9.6 m/s^2 are random errors, but those two far from average are systematic errors. 

In this lab, we got the real value of g in our classroom, but more importantly, we learned that there are some random errors that are hard to avoid. The best way to solve this problem is to do the lab over and over again, which shows the importance of comparing the value with other groups. What's more, we can check whether we make some mistake in the lab by checking the value acceptable or not.