Error Transmission (Moment of Inertia)

[Const@ntine]

Homework Statement

Compute the moment of intertia I_o of the two cylinders, per their axis of symmetry, and then, using Steiner's formula, the moment of inertia I_b, as per the axis of rotation.

Homework Equations

I_o = 1/2*m*R²
I_b = 2*(I_o + m*x²)

The Attempt at a Solution

Now, this is one of those huge exercises that you do at the Lab, which give you hours 3, plus work at home. So, this is just a part of it, but I'll type down what I've found already.

See, the problem here is computing the error. The formula for an f(x,y,z) fraction, is:

δf = √[(∂f/∂x*δx)² + (∂f/∂y*δy)²+ (∂f/∂z*δz)²]

So far so good, right? From the previous questions, I have (relevant to this exercise):

R +- δR = (2,10000 +- 0,00025)*10^-2 m
x +- δx = (5,00000 +- 0.00025)*10^-2 m
m +- δm = (435,150 +- 0.005)*10^-3 kg
I_o +- δI_o = (9,595 +- 0.002)*10^-5 kgm²

Plus, from the above formula: I_b = 2,36690115 * 10^-3 kgm²

From our professor, we know I_o should be in the range of 10^-5 and I_b in the rang of 10^-3, so that's all good. The problem I'm facing is when I try to find the error of I_b.

First of, I'm not sure if I should break down I_o into I_o = 1/2*m*R² and have x = m, y = R & z = x (going back to the general error transmission formula). At first I tried this, and then the next time I kept it as is, and went x = m, y = I_o & z = x. Still, the problem persists.

See, the SI units don't add up when I do the partial derivatives, apart from ∂I_b/∂m. And so, I can't add up the results of each derivation, and put them into the square root to get a result of [...] kgm².

So, what do I do now? In later questions, wehen we have to compute something else, certain of its quantities, we are instructed to use without their errors, as simple constants.

Any help is appreciated!

 

[gneill]

First of, I'm not sure if I should break down I_o into I_o = 1/2*m*R² and have x = m, y = R & z = x (going back to the general error transmission formula). At first I tried this, and then the next time I kept it as is, and went x = m, y = I_o & z = x. Still, the problem persists.

See, the SI units don't add up when I do the partial derivatives, apart from ∂I_b/∂m. And so, I can't add up the results of each derivation, and put them into the square root to get a result of [...] kgm².

So, what do I do now? In later questions, wehen we have to compute something else, certain of its quantities, we are instructed to use without their errors, as simple constants.

You should be able to go either way, expanding I_o or not. The units will work, you just have to realize that the units are just constants that multiply the value of the variable.

When you wrote the formula:

I_b = 2*(I_o + m*x²)

then I_o becomes a variable that has units "kg m²" When you take the partial derivative w.r.t. I_o the units associated with I_o are constants, and get carried along with the constant 2 in the derivative.

 

[Const@ntine]

You should be able to go either way, expanding I_o or not. The units will work, you just have to realize that the units are just constants that multiply the value of the variable.

When you wrote the formula:

I_b = 2*(I_o + m*x²)

then I_o becomes a variable that has units "kg m²" When you take the partial derivative w.r.t. I_o the units associated with I_o are constants, and get carried along with the constant 2 in the derivative.

I've actually never heard about that. Not knocking down or anything, but I've not seen it in any of my books, or had some professor tell it to us. In previous exercises, when it came to derivatives, anything that was x and nor x^y we just went ahead and wrote as 1, without its SI Units. And then the units worked out.

Okay, let's say, I break dowrn I_o into Io = 1/2*m*R². So, now I have:

ϑI_b/ϑm = R² + 2*x², both of which, after doing the necessary acts, have m² SI units. When multiplied by δm, which is in kg, I get the kgm² which I'm searching for.

Moving on, and carrying the SI Units:

ϑI_b/ϑR = 2*m*R + 2*m*x² || The problem here is, that, 2*m*R has kg*m SI Units, and 2*m*x² has kg*m² SI Units. When I multiply by δR, whose units are m, the units aren't the same, and I can't add them up. Same thing happens with ϑI_b/ϑx.

Now, if I keep I_o as is, and I keep the SI units like you said, I have:

ϑI_b/ϑI_o = 2*kg*m² + 2*m*x² || The SI Units now add up, but since I have to multiply by δI_o, whose SI Units are kg*m², then I get kg²*m⁴. Then I have to raise that to the second power, so I end up with kg⁴*m⁸, which makes the SI Units not add up again.

If I keep I_o, but without its SI Units, they again don't add up.

Generally, each derivative, when multiplied by its error, and then raised to the second power, should give me kg²*m⁴, since the end result will be under a square root, and in the end, will give me as a result, a value measured in the desirable kg*m² SI Units.

 

[gneill]

Moving on, and carrying the SI Units:

ϑIb/ϑR = 2*m*R + 2*m*x2 || The problem here is, that, 2*m*R has kg*m SI Units, and 2*m*x2 has kg*m2 SI Units. When I multiply by δR, whose units are m, the units aren't the same, and I can't add them up. Same thing happens with ϑIb/ϑx.

When you take a partial derivative, anything that isn't the variable you're differentiating with respect to is a constant. So here you're differentiating wither respect to R and so the term 2*m*x² becomes a constant that becomes zero upon the differentiation. Hence:

ϑIb/ϑR = 2*m*R

 

[Const@ntine]

When you take a partial derivative, anything that isn't the variable you're differentiating with respect to is a constant. So here you're differentiating wither respect to R and so the term 2*m*x² becomes a constant that becomes zero upon the differentiation. Hence:

ϑIb/ϑR = 2*m*R

Oh yeah, that's it. I haven't come across anything like this: f(x,y,z) = 5*x +7y*z in quite a while, so I got carried away and treated it like a case of f(x,y,z) = x*y*z. Yup, my bad... Well, that was embarassingly simple...

Thanks for the help though, I really appreciate it!