Physics Lab Question! Please Help!?

Question

Ok, I have this really simple lab but its just not working out and I dont know why. It seems complicated but its not. If you could read it out and help me it would be GREATLY appreciated. I dont get where Im going wrong. (See details if I run out of space.) The goal is to find the acceleration due to gravity. (9.8m/s^2)

We dropped a plum bob and as it dropped it marked in a piece of wax paper ever 1/30 seconds. We circled EVERY OTHER DOT. We then measured the distance from each circled dot to the next starting at dot one. (0 to dot two, dot two to dot four, dot four to dot six, etc. Aka, every other dot.) The time between each dot is 1/60 of a second. (its really 1/30, but we are measuring every other dot so its 1/60) My lab sheet tells me to make a displacement vs. time graph but Im not exactly sure why. Im trying to find acceleration which would be the slope of a Velocity/Time graph. So anyway, here are the displacements I got from measuring the dots mentioned before:

.56cm, 1.56cm, 2.51cm, 3.72cm, 4.73cm, 5.73cm, 6.92cm, 7.87cm, 9.03cm, 10.23cm, 11.21cm, 12.26cm, 13.34cm, 14.39cm, and 15.38cm. (15 dots were circled)

So those are my displacements. I converted them to meters for all calculations (since the goal is to find acceleration due to gravity which is in meters. I thought this was my problem at first, but it didnt help at all after I remembered to convert.)

So I went into excel and made my graph. For example I put .0056m and 2/60s together and .0156m and 4/60s together. I continued this for all my points. I got a really nice looking graph (sloped up to the right and the points were very even, etc.) I placed the treadline in there and got the R^2 value. The value was .9998. I dont know how to interpret that or what it means. I know the slope of my p/t graph is velocity, but does that mean my average velocity was .9998m/s^2

My lab sheet then told me to graph my velocities using an x-y scatter and find the slope of the velocity graphs for each data set. This kinda confused me because it it says graphS. But maybe it is refering to my other set of data points. Anyway, I calculated velocity by doing the change in displacement divided by the change in time. So for example I did:

(.1056m - .0056m)/((4/60s)-(2/60s)) = .168m/s

and

(.0251m - .0156m)/((6/60)-(4/60)) = .300m/s

I did this for all my data points. (For the first point I did (.0056m - 0m)/((2/60s)-(0s)) = .168m/s) I got 15 velocity quantities in all. I then made my velocity vs. time graph in excel. For example, I plotted:

.168m/s and 2/60s together and .3m/s and 4/60 together, etc. I continued this for all my velocity points. I then added a treadline and got the R^2 value. This value should be close tp 9.8m/s^2 since the slope of the velocity vs. time graph is acceleration. However, my R^2 value came out as .1335. I have NO idea why it isnt coming

out at 9.8 and any help would be greatly appreciated. Sorry I typed so much, but I wanted someone to be able to see where Im going wrong and not just guess what the problem could possibly be.

Also, for the other points we didnt circle (every other point starting at the first)... On this set of dots we didnt measure from dot to dot, but the distance from the first dot to the dot. So from dot one to dot three. Then from dot one to dot five, etc. It doesnt have any instructions on the lab sheet. All I know is he gave us the formula: a=2s/t^2 Im assuming s is the distance we measured but when I plug that in I get weird numbers. For example, we got 0cm, 1.2cm, 4.5cm, etc. So a=2(1.2cm)/(3/60)^2 (This is for the second dot) which is 960m/s^2 which makes NO sense.


So any help with why Im not getting 9.8m/s^2 and what Im supposed to do with this second set of dots would be GREAT. Im really sorry this was so long but Im so lost its ridiculous.

I meant to say it marks the paper 1/60s. So since we measure every other one its 1/30s.

Answer 1

I spent some time with your data and noticed the following:

First, your time interval is 2/30 s
or 1/15, not 1/60. This is a simple algebraic error, but doesn't account for the data having almost linear relationship to time.


Displacement should be d=1/2*a*t^2

I took the difference of each d and found the following sequence
100
95
121
101
100
119
95
116
120
98
105
108
105
99

This is not an accleration, it is almost sinusoidal.

So I thought about your experiment. Since it is a plum bob, was it swinging like a pendulum to mark the wax paper?

How was the experiment set up to make the marks?

Since the plumb bob was falling, it won't exactly follow the plumb bob motion, but do you see where I'm headed with this question?
j

Answer 2

Looking at your data I think the time interval is really 1/60 sec between each dot, or 1/30 sec between circled dots.

Using your numbers for the time interval (1/30 s between all dots, 1/15 s between circled dots) and your data, we can get an estimate of g by looking at the first and last pairs of data:

The velocity between dots 0&2 is
0.56cm / (1/15s) = 8.4 cm/s
Let this be the velocity at dot 1.

velcocity between the last two dots, 28&30, is
15.38cm / (1/15s) = 230.7 cm/s
We'll say this is the velocity at dot 29

The change in velocity from dot 1 to 29 is
(230.7 - 8.4) cm/s = 222.3 cm/s

The time difference is (29-1)*(1/30 s) = 0.933 s
(Here we are counting all dots from 1 to 29, so the time difference is 1/30 s per dot)

Acceleration is change in velocity / change in time, or
(222.3 cm/s) / 0.933 s = 238.2 cm/s^2

This converts to 2.382 m/s^2. You're about a factor of 4 off.
Most likely, the time increment is really 1/2 of what you thought it was, i.e. 1/60 sec between all dots, and 1/30 s between circled dots.

Using the different time increment, we'd get 9.527 m/s^2
which is more reasonable.

Hope this helps.

Answer 3

hi :-) As you develop the voltage the present additionally will develop inflicting the filament in the sunshine bulb to get warm, the present starts to even out because of the fact the filament has already replace into warm enought and might't get any warmer, the resistance additionally develop because of the fact once you utilize the formula V=IR to get R we use R= V/I and if the voltage remains increasing by using a great quantity however the present isn't then you quite are basically getting a much better resistance because of the fact the the present could greater or much less stay the comparable yet you're dividing a much better voltage with the comparable present day so that's greater effective, that's the comparable for the smaller present day, the resistance and voltage is extremely proportional because of the fact as you develop V, R has to develop to maintain the wonderful present day in the direction of the lighbulb, I=V/R if the voltage is increasing and the resistance is aswell then i does no longer replace because of the fact the two V and R are increasing :-) in case you wanna understand the rest e mail me ;-) i'm from the united kingdom aswell :-)