# Centripetal force + uncertainty

1. Sep 6, 2011

### Wa1337

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
F= (mv2) / r

3. 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

2. Sep 6, 2011

3. Sep 6, 2011

### Wa1337

edit: nvm

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

Last edited: Sep 6, 2011
4. Sep 6, 2011

### AC130Nav

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.

5. Sep 6, 2011

### Wa1337

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

6. Sep 7, 2011

### AC130Nav

You were closer in your original post.

7. Sep 8, 2011

### AC130Nav

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."

8. Sep 8, 2011

### NewtonianAlch

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.