Determining the gravity using inclined planes


by mikkeljuhl
Tags: determining, gravity, inclined, planes
mikkeljuhl
mikkeljuhl is offline
#1
Apr24-12, 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?
Phys.Org News Partner Science news on Phys.org
Cougars' diverse diet helped them survive the Pleistocene mass extinction
Cyber risks can cause disruption on scale of 2008 crisis, study says
Mantis shrimp stronger than airplanes
berkeman
berkeman is offline
#2
Apr24-12, 11:16 AM
Mentor
berkeman's Avatar
P: 39,683
Quote Quote by mikkeljuhl View Post
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?
I don't see the ball's moment of inertia in your calculations...
mikkeljuhl
mikkeljuhl is offline
#3
Apr24-12, 11:24 AM
P: 8
Shouldn't it be possible to calculate it without that?
And how would you include that in the calculation?

berkeman
berkeman is offline
#4
Apr24-12, 11:26 AM
Mentor
berkeman's Avatar
P: 39,683

Determining the gravity using inclined planes


Quote Quote by mikkeljuhl View Post
Shouldn't it be possible to calculate it without that?
And how would you include that in the calculation?
As the ball rolls down the ramp, energy goes into its translational motion, and into its rotational motion. You need to account for both effects if you want to use the ball's motion to estimate the acceleration of gravity.
mikkeljuhl
mikkeljuhl is offline
#5
Apr24-12, 11:28 AM
P: 8
Of course in order to get immaculate results. But what if we do not take friction into account...
George Jones
George Jones is offline
#6
Apr24-12, 11:44 AM
Mentor
George Jones's Avatar
P: 6,044
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 velocity-time 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.
mikkeljuhl
mikkeljuhl is offline
#7
Apr24-12, 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?)
George Jones
George Jones is offline
#8
Apr24-12, 12:26 PM
Mentor
George Jones's Avatar
P: 6,044
Quote Quote by mikkeljuhl View Post
0.2839/sin(2.8), I get 5.81169576.
What are you trying to do? Are you trying to find a value for g?
mikkeljuhl
mikkeljuhl is offline
#9
Apr24-12, 12:47 PM
P: 8
Yep, am I completely off?
George Jones
George Jones is offline
#10
Apr24-12, 12:54 PM
Mentor
George Jones's Avatar
P: 6,044
Including the moment of inertia for a uniform sphere makes things somewhat better. But you haven't taken moment of inertia?
mikkeljuhl
mikkeljuhl is offline
#11
Apr24-12, 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.
George Jones
George Jones is offline
#12
Apr24-12, 01:09 PM
Mentor
George Jones's Avatar
P: 6,044
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].
mikkeljuhl
mikkeljuhl is offline
#13
Apr25-12, 12:49 AM
P: 8
Snip
mikkeljuhl
mikkeljuhl is offline
#14
Apr25-12, 01:08 AM
P: 8
Snip


Register to reply

Related Discussions
Inclined planes Classical Physics 7
Inclined planes . General Physics 8
Inclined Planes Introductory Physics Homework 3
inclined planes? General Physics 5
Inclined Planes Introductory Physics Homework 6