Mass momentum of inertia in lab, error analysis

AI Thread Summary
The discussion focuses on a lab assignment involving the measurement of a ball's mass moment of inertia using an inclined plane. The participant is attempting to derive the moment of inertia from the conservation of mechanical energy and is grappling with error analysis, particularly how to account for the dependencies of the variable k on mass and radius. Suggestions include solving explicitly for moment of inertia to apply error propagation methods, which would yield multiple values to average, and considering the uncertainty in height measurements. The participant is also confused about how to incorporate the relationships between variables, particularly how velocity depends on height. Overall, the thread emphasizes the complexities of error analysis in experimental physics.
Regel
Messages
9
Reaction score
0

Homework Statement


Hi, I've a lab assingment, and the labwork must be planned beforehand, but I have some trouble figuring out some parts of my error analysis.

So, I'm supposed to measure (as in not use integrals to find out) the mass momentum of a ball. The plan is to place the ball on an inclined plane, and let it roll to the ground. I'm going to measure the mass, radius, the velocity just before the ball hits the ground, and the distance to ground from the beginning position.

This is done with 10 or so different starting distances (and consecutively with 10 different end velocities).

Homework Equations



First I derive the formula for my calculations from the principle of concervation of mechanical energy:
mgh &= \frac{1}{2}I_{cm} \omega _{cm}^2+\frac{1}{2}mv_{1}^2 \newline \ldots \Leftrightarrow mgh = \frac{I_{cm}+mr^2}{2r^2}v_{1}^2\\, where m is the mass, r radius, g is the acceleration due gravity, h is the initial position when the ball is at rest, v is the velocity just before the ground, and I_cm is the mass momentum of inertia.

Now, I have 10 different results, I use linear fit on a set of points [ m*g*h, v^2], so I get an equation like U(v^2) = kv^2+b. Now, computer gives me k with it's error limit (and b, which is just the systematic error).

k=\frac{I_{cm}+mr^2}{2r^2} \Leftrightarrow 2kr^2 = I_{cm}+mr^2 \Leftrightarrow I_{cm} = 2kr^2-mr^2

Now I have the equation for my mass moment of inertia.

The Attempt at a Solution


The problem is, the when using the propagation of uncertainty (http://en.wikipedia.org/wiki/Error_propagation):
\displaystyle\delta I_{cm} &amp;= \sqrt{\sum_{i}^n (\frac{\partial I_{cm}}{\partial x_i} \delta x_i)^2} = <br /> \sqrt{(\frac{\partial I_{cm}}{\partial k} \delta k)^2+(\frac{\partial I_{cm}}{\partial r} \delta r)^2+<br /> (\frac{\partial I_{cm}}{\partial m} \delta m)^2}<br /> \\ &amp;= \sqrt{4r^4\delta k^2+4r^2(2k-m)^2\delta r+r^4\delta m^2} = r\sqrt{4r^2\delta k^2+4(2k-m)^2\delta r+r^2\delta m^2 }.

k is not an independent variable, as it depends on mass m and radius r. So the big question is, how am I supposed to change the last equation, when taking into account that k depends on m and r?

As for my bad english, I apologize, please ask, if I didn't make myself clear. Thank you.
 
Last edited by a moderator:
Physics news on Phys.org
Welcome top PF.

Actually I might consider trying to use a different k,

k = k'm

such that you could write your equation as

I = mr2*(2k' - 1)

Then you might be able to use the RSS of the relative errors of m,2r,k'

(Note I think your r2 term should be carried as ∂r/r twice because it should be treated as r*r for taking the RSS.)
 
Still thinking about this, but the equation you have for propigation of uncertenty assumes you have an explicit function for moment of Inertia I as a function of all your variables. You don't have this. If you wanted to to this, you could solve for I and you would have both of your data values (h and v) on the left side of the equation. This, however would eliminate using the graphical method to find a slope. Each run would yield values for height and velocity to plug into an equation, and you would have ten different values for moment of inertia. You could simply average these.

For error analysis, don't forget to put some uncertenty in height into your equation too.

To more acurately measure height in the lab, I would consider measuring the distance up the ramp and using trig to find it's actual height. More accurate
 
Ok. I think I've got it. Basically two ways to do this error analysis.

1. Solve explicitly for moment of inertia and use your propigation of uncertenty you found from wikipedia. This eliminates the graphical method and leaves you with 10 values for moment of inertia to average in the end. With this model, you are GUESSING theoretical uncertenties for all variables in you equation for moment of inertia.

2. still thinking
 
flatmaster said:
Ok. I think I've got it. Basically two ways to do this error analysis.

1. Solve explicitly for moment of inertia and use your propigation of uncertenty you found from wikipedia. This eliminates the graphical method and leaves you with 10 values for moment of inertia to average in the end. With this model, you are GUESSING theoretical uncertenties for all variables in you equation for moment of inertia.

2. still thinking

If I solve explicitly for moment of inertia:
I=mr^2 \frac{2gh-v^2}{v^2}
(did I understand you correctly?)

Then mass moment of inertia depends on m,r,h and v. But v depends on h :O How to take this into consideration?


And lowlyPion's solusion was somewhat confusing: I =mr2*(2k' - 1), but k still depends on m and r, doesn't it?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top