# Homework Help: Uniform acceleration: air track, data error?

1. Aug 19, 2010

### Archy

1. The problem statement, all variables and given/known data
: In case anyone didn't get a clear picture, in a nutshell the experiment involves us releasing a glider down an air track within a fixed distance x, where the time, t taken is then recorded down. Sorry for not mentioning this before if it's of any use.

To determine the acceleration due to gravity on an air track,
Note that distance, x is the ordinate and t^2 is the abscissa.

variables:
manipulated: distance, x(m)
responding: time (t^2)
constant: mass of the glider, angle of inclination of the air track

samples of the data obtained (respectively):
x = 0.25; 0.50; 0.75... ; 1.75
t^2= 7.0225; 13.3590; 21.8089... ; 49.2120

height, h= 0.29m; total distance of the track, d=1.9834m

2. Relevant equations
x= (at^2)/2, therefore a= 2m; m=slope
g= a/sin Ө;
sin Ө=0.29/1.9834= 0.1462

3. The attempt at a solution
Via linear least square fit, I got
slope, m= 0.03498 m/s^2

therefore, acceleration, a= 2m= 0.06996 m/s^2

therefore logically
g= (0.06996/0.1462) m/s^2
g= 0.4785 m/s^2 ...?

That doesn't make sense.
but if we reverse the equation with g= 9.80
then a= 9.80 (0.1462)= 1.43276 m/s^-2

However, the data derived acceleration doesn't come any closer to that value.

So it's either the experiment gone wrong or I miscalculated something somewhere; I have informed my other partners to ask whether they can calculate the final answer but I've received no reply yet.
I would appreciate it so much if someone can pin point a possible error?
Currently running on low battery now. Going to sleep and then go over the whole thing when I wake up.

Thanks.

Last edited: Aug 19, 2010
2. Aug 19, 2010

### kuruman

Linear least square fit to what? What did you plot against what?

3. Aug 19, 2010

### Archy

distance, x is the ordinate and t^2 is the abscissa.
So that'll be distance versus times squared?

4. Aug 19, 2010

### kuruman

OK, then what kinematic expression do you know that relates x and t2?

5. Aug 19, 2010

### Archy

x-x0=(v0t)+1/2(at2); initial distance, x0= 0, initial velocity, v0=0

then x= (1/2)(at2)?

Last edited: Aug 19, 2010
6. Aug 19, 2010

### kuruman

Good. So the measured slope is m = (1/2)a. How can you introduce g into the picture?

7. Aug 19, 2010

### Archy

Well, I guess since

a= g sin Ө

then
g sin Ө= 2m= a?

8. Aug 19, 2010

### kuruman

So g=2m/g sinӨ, and when you put in the numbers, you get an absurd value for g and I agree. There appears to be no error in the analysis, so the problem lies with the data or with the least-squares fit. Can you tabulate the original data with units?

9. Aug 19, 2010

### Archy

Yep.

[PLAIN]http://img683.imageshack.us/img683/3620/dataf.jpg [Broken]

 Oh and I used a java app to recalculate the LLSF and again the same answer.

Last edited by a moderator: May 4, 2017
10. Aug 19, 2010

### kuruman

When you say h = 0.2900 m, what is this a measurement of? From where to where?

11. Aug 19, 2010

### Archy

h is the maximum height of the inclined plane from the ground
while d being the length of the whole plane.

In terms of trigonometry it's expressed as sin Ө= 0.29/1.983

12. Aug 19, 2010

### kuruman

The plane has two ends. The high end of the plane is 0.29 m above ground, I can see that. How high above ground is the other (low) end of the plane?

13. Aug 19, 2010

### Archy

It's treated as a right angled triangle; So that'll be zero on the other side relative to the higher end, I believe.

14. Aug 19, 2010

### kuruman

You used a tilted air track. You measured 0.29 m from the ground to say the top of the high end of the track. You need to measure the distance from the ground to the corresponding top of the low end of the track, take the difference Δh and say sinθ = Δh/d. That's how you get a right triangle, otherwise you get a trapezoid. That's the problem with your analysis.

15. Aug 19, 2010

### Archy

Actually we used a wooden block to provide inclination? I'm afraid that I should have included that in the description, apologies; We just measured the height gain provided by the wooden block. Which was effectively 0.29 m.

edit: Assuming that h is the problem, altering h to be smaller as in Δh=(0.29m-h1) does allow for larger values of gexperiment but Δh has to be effectively very small for that to work; Δh has to be 0.01416 for gexperiment to be approximately 9.8 ms-2?

Last edited: Aug 19, 2010
16. Aug 19, 2010

### collinsmark

Something doesn't seem to be adding up.

So the track itself (stated as d = 1.9834 m) had a length a little longer than most people are tall (more-or-less)? So it was somewhere around a "person" sized track?

And the block of wood (stated as h = 0.290 m), was about as tall as a foot (more-or-less)?

The data would make more sense if either a) one end of the track was raised only about a centimeter or two above the other, or b) it was a much longer track (much longer than a person is tall).

If both h and d check out, the problem could also be in the units that time was measured. What mechanism did you use to measure time?

[Edit: As a matter of fact, things work out pretty well if the height of the block is 0.029 m (i.e. 2.9 cm) and the length of the track was 4 meters (a bit longer than twice the height of a person). Of course, there are an infinite amount of other length/height combination that would work too -- just not the one's you've measured. Are you sure you measured your d and h correctly?]

Last edited: Aug 19, 2010
17. Aug 19, 2010

### kuruman

I am familiar with the equipment Archy mentions. It is probably this
http://store.pasco.com/pascostore/showdetl.cfm?&DID=9&Product_ID=1619&Detail=1
(is this the track, Archy?)

The problem is not the length measurement, but the height difference between the ends. The track is set on a table top anf has a three point support. Two rubber feet about 1/3 of the way from one end and one adjustable (for leveling) foot about 1/3 of the way from the other end. The track is normally inclined by placing a block under the two-feet support, so the height of the block is irrelevant. What matters is the height difference between two matched points on the track separated by two meters. A convenient point would be the edge of the square that has the vertical as its diagonal. One measures the distance from the table top of that edge at the two ends of the track, then one finds the difference. Of course this method fails if the table top itself is not horizontal ...

Assuming a 2.0 m track (most air tracks have a scale pasted on them so there can be no mistake about its length), then the height difference ought to be given by

gsinθ =g Δh/2.0 m = 2*0.035 = 1.4 cm (one or two centimeters as collinsmark remarked.)

18. Aug 19, 2010

### Archy

The length of the track would be just that as mentioned by kuruman, since that is the exact track used in this experiment.

I didn't do the measuring for the apparatus personally, aside from handling the stopwatch and glider my partners did the rest, but I'll be matching up data with members from other groups to see how valid (or possible) it is. But Collinsmark is right on the part of the block being 0.029m, seeing that how a 29 cm block would make even less sense. That's just about right, it was conversion problem. Sorry on my behalf.

Collectively, there is a possibility if the Δh is half of what it is it would match the 9.8 benchmark quite properly. Which means that it is possible for the real value of Δh to be quite close to half of h;
0.029m/2=0.0145m, which will make gexperiment=9.5696 ms-2.
Which could mean the likeliest problem could be that in the measurement-calculations there.
I see what you mean now, kuruman does have a good point here as well, thank you.

I'll sync up some data with people from other groups to verify this error and maybe be on my way. Or back here.