Centripetal force + uncertainty

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Homework Help Overview

The discussion revolves around calculating centripetal force using the formula F = (mv²) / r, while also considering the uncertainties in the measurements of mass, velocity, and radius. Participants are attempting to determine the correct force and its associated uncertainty based on given values and their uncertainties.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are discussing how to properly calculate the uncertainty in the centripetal force, with some attempting to apply percentage uncertainties to the measurements. There are questions about whether the final uncertainty value is correct and how to account for the squared velocity in the calculations.

Discussion Status

There is ongoing dialogue about the correct approach to calculating uncertainties. Some participants have provided insights on how to handle the squared term in the velocity, while others express confusion about the calculations and seek clarification. Multiple interpretations of the uncertainty calculations are being explored.

Contextual Notes

Some participants indicate that they are new to the topic, which may affect their understanding of the uncertainty calculations. There are references to external resources for clarification, and a mix of confidence levels in the provided answers is evident.

Wa1337
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Homework Statement


The centripetal force is given by F= (mv2) / r. If the mass is measured to be 2.8 +/- 0.1 kg, the velocity is 14 +/- 2 ms^-1, and the radius 8.0 +/- 0.2 m, find the force on the mass, including the uncertainty.


Homework Equations


F= (mv2) / r


The Attempt at a Solution


0.1/2.8 = .04
2/14 = .14
.2/8.0 = .025

add all up and i got .205

I did the equation and got 69 (2 sig figs) and finally got F = 69 +/- .21

Is that the proper uncertainty answer? If not please help.
 
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edit: nvm

final answer then: F = 270 +/- .21 ? (2 sig figs both)
 
Last edited:
Wa1337 said:
edit: nvm

final answer then: F = 270 +/- .21 ? (2 sig figs both)

I did not say the force was wrong but you changed it; moreover, you still don't have the uncertainty right as I understand it.

I guess I'm not helping.
 
I'm sorry I kind of don't get it cause I'm really new to this stuff but my final answer without changing the force is:

F = 69 +/- 14

is that better? thanks
 
Wa1337 said:
I'm sorry I kind of don't get it cause I'm really new to this stuff but my final answer without changing the force is:

F = 69 +/- 14

is that better? thanks

You were closer in your original post.
 
Wa1337 said:

Homework Statement



The Attempt at a Solution


0.1/2.8 = .04
2/14 = .14
.2/8.0 = .025

add all up and i got .205

I did the equation and got 69 (2 sig figs) and finally got F = 69 +/- .21

Is that the proper uncertainty answer? If not please help.

Why is no one helping with this? I'm about to give him the answer as I think it is--it's really not something I do day-to-day but IAW to site I posted previously.

We are surpassing Winston Chruchill adage here "The Americans and the British are a single people divided only by a common language."
 
Well he has to get the percentage uncertainties for each of the values.

So for example, for velocity it would 2/14*100 = 14.28% (quite a large uncertainty there)

Since it's squared, the uncertainty would be double: 28.57%

Do this for each of the variables, and add up the percentages. Then your final percentage of the calculated value for force should be your uncertainty in Newtons.
 

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