Question #aa46e

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Answer 1 · 2017-02-07T23:58:59

$0.6 %$ $0.6%$

In the classical sense , the centripetal force $F$ $F$ can be measured as:

$F = \frac{m v^{2}}{r}$ $F = (m v^2)/r$

We can take logs and say that:

$ln F = ln (\frac{m v^{2}}{r})$ $ln F = ln ((m v^2)/r)$

$= ln m + {ln v}^{2} - ln r$ $= ln m + ln v^2 - ln r$

$= ln m + 2 ln v - ln r$ $= ln m + 2 ln v - ln r$

Using differentials:

$\frac{d F}{F} = \frac{d m}{m} + 2 \frac{d v}{v} - \frac{d r}{r}$ $(dF)/F = (dm)/m + 2 (dv)/v - (dr)/r$

$= 2 % + 2 \cdot 1.3 % - 4 % = 0.6 %$ $= 2% + 2 * 1.3% - 4% = 0.6%$

Answer 2 · 2017-02-08T00:29:35

$4.66 %$ $4.66%$

The formula for error propagation using standard deviations:

$(sigmaF)/|F|=sqrt(((sigmaa)/|a|)^2+((sigmab)/|b|)^2+((sigmac)/|c|)^2+...)$

Where a, b, c, ... are parameters used to determine the uncertainty of $F$ .

$(sigmaF)/|F|$ is basically a percentage in error if we multiply by 100%. Therefore,

Uncertainty in your centripetal force is $sqrt(2^2+1.3^2+4^2)=4.65725%$

Round it off to get $4.66%$ .

There is a more robust way of measuring uncertainties using calculus and it is my go to formula.

Uncertainty of a quantity, $sigmaQ=sqrt((sigmaa(delQ)/(dela))^2+(sigmab(delQ)/(delb))^2+(sigmac(delQ)/(delc))^2+...)$

$(sigmaa(delQ)/(dela))$ is basically $sigmaa$ also known as the uncertainty to be multiplied by the derivative of $Q$ with respect to parameter $a$ .

Cheers.