Calculating the acceleration of gravity using a ticker tape.

AI Thread Summary
The discussion focuses on calculating the acceleration due to gravity using data from a ticker tape experiment. Participants highlight the challenge of obtaining an accurate average acceleration from individual measurements of distance and time. The initial calculations yield an incorrect value for gravity, prompting suggestions to use the formula g = 2s/t² for more precise results. A recommendation is made to graph the position as a function of time squared to determine the slope, which should yield a more reliable estimate of gravity. The conversation emphasizes the importance of understanding the distinction between average and instantaneous acceleration in the context of the experiment.
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Homework Statement



It was an experiment we did at school. We placed a tape through a vibrator that leaves a black dot on a tape for every 1/100 s. We dropped the tape and registered many dots. We have narrowed the experiment down to 10 dots. So we chose a dot, and registered the length between the dot before it, and after it to calculate the velocity at that dot. We did this to a dot 10 dots further down to find the velocity at that dot, so we could use the difference in distance / difference in time. This is our measurements:

L_A = 0.036 m
L_B = 0.044 m

We "know" the time it takes for the vibrator to leave 10 dots is 10*1/100= 1/10=0.1, therefore the distance between the three dots is 0.02s. We get

(0.044/0.02)-(0.036/0.02) = 0.4

0.4/0.1= 4m/s^2 = g.

Thats very wrong!
Any suggestions?
 
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This way of calculating gives you average velocity. use s=0.5*g*t^2 that is g=2s/t^2.
 
Heimisson said:
This way of calculating gives you average velocity. use s=0.5*g*t^2 that is g=2s/t^2.

Hello,

Thank you for replying.

I still can not figure out how that formula will help me. That would give me the acceleration at one given point, right? And not the "avarage" acceleration for all the points, because that's what I am after.

Because the formula gives me: 0.8/0.1^2 = 80m/s^2, which obviously is way too high.

Please do correct me if I am wrong.
 
No this should be more like 0.08m from the data you are showing.
s is not velocity but position.

There are several different ways of getting some numerical value from a collection of data. What would be most professional is to make a graph of s as a function of t^2 (which should be linear), the slope of the best line through the points should give you 0.5*g. However your teacher should have thought you the proper way to get a average value from the data.

You can do this in a several different ways. If the instruction don't give you some specific way to do this, just do what you think is best.
 
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