Uncertainity and Error Question

  • Thread starter Thread starter inno87
  • Start date Start date
  • Tags Tags
    Error
Click For Summary
SUMMARY

The discussion focuses on calculating the volume of a cylindrical cookie with a diameter of 5.0 ± 0.1 cm and a thickness of 1.00 ± 0.01 cm. The most likely volume is determined to be 19.634 ± 0.11 cm³, with a percent uncertainty of 0.5603%. The absolute uncertainty in the volume is confirmed as 0.11 cm³. Additionally, the user seeks clarification on how to adjust the diameter's uncertainty to achieve a volume uncertainty of ± 3%, indicating a need for further understanding of uncertainty propagation.

PREREQUISITES
  • Understanding of basic geometry, specifically the formula for the volume of a cylinder.
  • Knowledge of uncertainty propagation techniques in measurements.
  • Familiarity with statistical concepts related to normal distribution.
  • Ability to perform calculations involving percentages and absolute values.
NEXT STEPS
  • Study the principles of uncertainty propagation in measurements.
  • Learn how to calculate percent and absolute uncertainties in physical measurements.
  • Explore the effects of varying measurement uncertainties on calculated results.
  • Investigate statistical methods for analyzing normally distributed uncertainties.
USEFUL FOR

Students in physics or engineering, educators teaching measurement and uncertainty, and professionals involved in quality control or experimental design will benefit from this discussion.

inno87
Messages
6
Reaction score
0
I'm not sure if I am doing this right, can anyone help me out?

A cylindrical cookie has a diameter of 5.0 [tex]\pm[/tex] 0.1 cm, and a thickness of 1.00 [tex]\pm[/tex] 0.01 cm.

A. Assuming the uncertainities are normally distrubited, what is the most likely value of the volume of the cookie?

V=pi*(d/2)^2*h=pi*(1 [tex]\pm[/tex] .01 cm * [(5.0 [tex]\pm[/tex] .1 cm)/2]^2=pi*[1 [tex]\pm[/tex] .01 cm * (2.5 [tex]\pm[/tex] .05 cm)^2=pi*[1 [tex]\pm[/tex] .01 cm * 6.25 [tex]\pm[/tex] .1 cm^2]=pi* 6.25 [tex]\pm[/tex] .11 cm^3 = 19.634 [tex]\pm[/tex] .11 cm^3

Most likely volume is 19.634 cm^3

B. What is the percent uncertainty in the volume?
.11/19.634=.5603%

C. What is the absolute uncertainty in the volume?
.11 cm^3 (taken from question 1)

D. Assuming the thickness uncertainity remains [tex]\pm[/tex] .01 cm, to what value would the diameter's uncertainty (in cm) have to be reduced in order to make the uncertainty in the volume [tex]\pm[/tex] 3%?

I tried setting up something like (.01+D_unc)/(19.634)=.03 but that would mean you'd have to increase the uncertainty so it seems I may have done something wrong here!
 
Physics news on Phys.org
Anyone have an answer to this? I'm still very unstable about this uncertainty question.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
809
Replies
4
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
15
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
13
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 13 ·
Replies
13
Views
4K
Replies
9
Views
10K