Determining the gravity using inclined planesby mikkeljuhl Tags: determining, gravity, inclined, planes 

#1
Apr2412, 11:11 AM

P: 8

I have been using the formula a = g sin(theta) to process my data I am however pretty sure something is wrong. I doubt it is my data that is wrong, even though when I look at them they look weird in the sense that I would think that after one second the acceleration should have been doubled, whereas it hasn't. The data was measured using Vernier's Go Motion sensor. It set up a ramp (2m long) and then let a ball roll and got the following data:
Shouldn't the equation: a / sin(2.8) give me, when it is at one second, 9.8 m/s? At one second acceleration was(taking average at five trials) 0.3454 m/s^2 / sin(2.8) = 7.07065768m/s^2 And one of the closest (I didn't take average here, just looked for the highest number within my data) was at 1.70 s, which is: 0.466/ sin(2.8 degrees) = 9.540 m/s^2 Can you see what I have done wrong? 



#2
Apr2412, 11:16 AM

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#3
Apr2412, 11:24 AM

P: 8

Shouldn't it be possible to calculate it without that?
And how would you include that in the calculation? 



#4
Apr2412, 11:26 AM

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Determining the gravity using inclined planes 



#5
Apr2412, 11:28 AM

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Of course in order to get immaculate results. But what if we do not take friction into account...




#6
Apr2412, 11:44 AM

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Introducing the moment of inertia will multiply the right side of a =gsin(theta) by a constant, i.e., a = C g sin(theta), where moment of inertia determines C.
Your data looks very noisy at the start. If acceleration a = Cgsin(theat), and C and g and are all constants, then acceleration a should be, roughly, constant. The data looks not too bad near the end. To measure constant acceleration using motion sensors, it is often better to measure velocity, and to then find the line of best fit for a velocitytime graph. The slope of this line is the acceleration. Motion sensors measure position.To measure velocity, motion sensor (software) differentiates numerically. To get acceleration, the software numerically differentiates a second time. Numerical differentiation of data is a very noisy process, and a double differentiation should be avoided. 



#7
Apr2412, 12:10 PM

P: 8

I have done that now, graphed in a velocity vs. time graph.
I ended out with the equation: y = 0.2839x + 0.0971 y = velocity; a = acceleration, x = time; and let's pretend b is zero as that is the only plausible outcome. Is that correct? But this is where you lose me. Because (I don't have the motion of inertia, yet) if I put 0.2839 into the equation: 0.2839/sin(2.8), I get 5.81169576. This would be an awful lot of friction wouldn't it? And I am also in doubt whether I would have to time it by some friction (with slope/180 degrees, or something similar  or if that's what sin(theta) does, which I think it is, but I don't know if I am off here?) 



#8
Apr2412, 12:26 PM

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#9
Apr2412, 12:47 PM

P: 8

Yep, am I completely off?




#10
Apr2412, 12:54 PM

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Including the moment of inertia for a uniform sphere makes things somewhat better. But you haven't taken moment of inertia?




#11
Apr2412, 01:00 PM

P: 8

No, I have not. But I may be able to do it, in one of the following days, hopefully.
I would, however, like you to confirm that this is the right method to go about it, if you can do that? I just thought that it wouldn't be such a big uncertainty not to take moment of inertia into account. But I may be wrong. 



#12
Apr2412, 01:09 PM

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I have taught labs that use this method to find g (rolling steel balls). Results vary. The moment of inertia doe make a difference. The acceleration of an object rolling down an inclined plane is given by
[tex]a = \frac{g\sin\theta}{1+\frac{I}{MR^2}},[/tex] where [itex]I[/itex] is the momemt of inertia of the rolling object about the axis of rotation. For a uniform sphere, [itex]I=\frac{2}{5} MR^2[/itex]. 



#13
Apr2512, 12:49 AM

P: 8

Snip




#14
Apr2512, 01:08 AM

P: 8

Snip



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