Calculating Density of Rock from Mars w/ Uncertainty

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Homework Help Overview

The problem involves calculating the density of a rock sample retrieved from Mars, given its mass and dimensions, while also determining the associated uncertainty in the density calculation. The subject area includes concepts of density, volume of a cylinder, and propagation of uncertainty in measurements.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss methods for calculating the density and its uncertainty, with some questioning the accuracy of the volume uncertainty calculations. There are attempts to apply error propagation techniques, and participants express confusion over their results.

Discussion Status

The discussion is ongoing, with participants providing guidance on how to approach the uncertainty calculations. There are multiple interpretations of the error propagation methods being explored, and some participants are revisiting their calculations to identify potential mistakes.

Contextual Notes

Participants note that they have not been taught certain methods of error propagation, which may contribute to the confusion in their calculations. There is also mention of differing results among classmates, indicating a variety of approaches being taken.

shade585
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Homework Statement



the mass of a particular sample of rock retrived from Mars is m=772.2+or-0.2g. The sample is in the shape of cylinder of height h=10+or-0.1 cm and base radius r=2.5 +or-0.1cm. calculate the density of the rock sample in units of kg/m^3. You must state your result in the correct units with its associated uncertainty. The volume of a cylinder is V=pir^2h.

Homework Equations


p=m/v



The Attempt at a Solution



I can get the density but I cannot get the correct uncertainty.

The density I get is

(772.2g/196.35cm^3)((1 kg/m^3)/10^-3g/cm^3)=3932.77 kg/m^3

I believe the uncertainty is supposed to be +or-400 but I end up with four.

(delta p)/(3.9g/cm^3) = .2/772.2 + .14/196.35

delta p = .004 g/cm^3

.004 g/cm^3 ((1kg/m^3)/(10^-3 g/cm^3) = 4 kg/m^3

Im new to this propagation of errors stuff. Using sig figs other people in my class got either 4000 +or- 400(or +or- 100) I can't remember.
 
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The uncertainty for your volume seems off... can you show how you got it?
 
Start with the formula for the density...Take the derivatives with respect to the variables etc... try it and we will hepl along.
 
I tried to find the uncertainty for volume again. I am probably calculating the error wrong.

(delta v/196.35) = 2pi(.1/2.5)+(.1/10)

delta v = 51 cm^3

(delta p)/(3.9g/cm^3) = .2/772.2 + 51/196.35 = 1.014 g/cm^3


1.014 g/cm^3 ((1kg/m^3)/(10^-3 g/cm^3) = 1014 kg/m^3

this seems way to high. I was given an online manuel on how to calculate propagation of error but my calculations always seem off.

To Dr Transport I do not believe have never seen of been taught that method.
 
shade585 said:
I tried to find the uncertainty for volume again. I am probably calculating the error wrong.

(delta v/196.35) = 2pi(.1/2.5)+(.1/10)

I think this is the mistake:

looking at the quantity without pi, call it x = r^2*h:

(delta x/62.25) = 2(.1/2.5)+(.1/10) = 0.09

delta x = 0.09*62.5 = 5.625

So the error in pi*r^2*h is 3.14*deltax = 17.67

so delta v = 17.67
 

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