Finding Propagation of Uncertainty?

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Homework Help Overview

The discussion revolves around the propagation of uncertainty in the calculation of density based on measurements of mass, length, and diameter of a cylinder. Participants are exploring how to correctly apply uncertainty propagation formulas in this context.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the uncertainty in density using a specific formula but is uncertain about the coefficients used in the calculation. Some participants question the correctness of the approach and suggest that the original poster may have overlooked multiplying by the calculated density value.

Discussion Status

Participants are actively discussing the correct method for calculating uncertainty, with some providing insights into the formula's structure and the significance of the coefficients. There is acknowledgment that the original poster's approach was close, but adjustments were necessary for the online grading system to accept the answer.

Contextual Notes

There is mention of specific values and measurements, as well as a reference to an online grading system that has specific requirements for the format of the answer. The original poster expresses confusion regarding the derivation of certain coefficients in the uncertainty formula.

Monkey618
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You measure the mass of the cylinder to be m = 584.9 +- 0.5 grams, and you measure the length of the cylinder to be L = 18.195 +- 0.003 cm. Just like in the lab you performed, you now measure the diameter in eight different places and obtain the following results.
Diameter (cm)
2.125
2.090
2.065
2.240
2.110
2.100
2.080
2.240

This gives an average of 2.131 +- 0.0695

This makes the density = 9.01 g/cm^3 +- propagation of uncertainty

Trying to calculate this, I have: sqrt( ((1*0.5) / 584.9)^2 + ((-2 * 0,0695) / 2.131)^2 + ((-1 * 0.003) / 18.195)^2 ) = 0.0652

However the online grading system say's that I'm wrong. So where have I gone wrong with the uncertainty of the density. All of the other values have been graded and marked correct, so where did I mess up with the uncertainty.

I'm not really clear where the "1", "-2", or "-1" came from in the formula above, I was basing it on my notes from class.
 
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Monkey618 said:
You measure the mass of the cylinder to be m = 584.9 +- 0.5 grams, and you measure the length of the cylinder to be L = 18.195 +- 0.003 cm. Just like in the lab you performed, you now measure the diameter in eight different places and obtain the following results.
Diameter (cm)
2.125
2.090
2.065
2.240
2.110
2.100
2.080
2.240

This gives an average of 2.131 +- 0.0695

This makes the density = 9.01 g/cm^3 +- propagation of uncertainty

Trying to calculate this, I have: sqrt( ((1*0.5) / 584.9)^2 + ((-2 * 0,0695) / 2.131)^2 + ((-1 * 0.003) / 18.195)^2 ) = 0.0652

However the online grading system say's that I'm wrong. So where have I gone wrong with the uncertainty of the density. All of the other values have been graded and marked correct, so where did I mess up with the uncertainty.
It looks like you forgot to multiply the "sqrt" by the calculated value of the function being evaluated (the density value).
I'm not really clear where the "1", "-2", or "-1" came from in the formula above, I was basing it on my notes from class.
They are values that depend upon the exponent of the variable in the function. Write out the function being evaluated in one line (promote the variables in the denominator to the numerator and adjust exponents accordingly):

$$f(m,d,L) = \frac{4}{\pi}M^1 d^{-2} L^{-1}$$

You can pick out the values as the exponents of the variables. Thus, for example, the value "-2" is associated with the diameter variable d.
 
You calculated the relative uncertainty, and both the formula and the result look good. Similar to gneill, I think the online grading system wants the absolute uncertainty.
 
I've been meaning to get back here to say Thank You! Multiplying by the density was exactly what I needed to do for WebAssign (the online grading system) to accept the answer.
 

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