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Catie

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- Thread starter catie1981
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Catie

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There's a great little book by Taylor, "An Introduction to Error Analysis". It has a trainwreck on the cover.

A simple formula, derived in the book, describes how to calculate the total error in a measurement, if each of the different errors are independent from each other. Say you are calculating the kinetic energy: K = 1/2 m v^2. You independently measure the mass and velocity. You could take it a step further, if you measure the mass, distance, and elapsed time independently as well: K = 1/2 m (d/t)^2. In any case, the total error is given by:

[tex]\delta K = \sqrt{(\frac{\partial K}{\partial m} \delta m)^2+(\frac{\partial K}{\partial v} \delta v)^2}[/tex]

Does that help?

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jtbell

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I can say that the accuracy of the scale I used was only 0.1g, which could account for 2% of the error. However, I don't know how to show that it accounts for 2% of the error,

The most direct way to do this is to re-calculate your result (the angular velocity) after changing your measured value for the mass by 0.1g, while leaving all the other measured numbers constant. If this causes your result to change by 2%, then you would say that the 0.1g uncertainty in the mass contributes 2% to the uncertainty in the result.

To continue with this, suppose you also have to measure the radius of something, and the accuracy of your ruler is 0.1 mm. To find out how much this contributes to the uncertainty of your result, you would proceed similarly: re-calculate the result after changing the radius by 0.1 mm, while leaving all the other measured numbers at their original values. (Use the original value of the mass, not the changed one you used in the previous paragraph.) Let's suppose you find that this changes the result by 1%. Then you would say that the uncertainty in the radius contributes 1% to the uncertainty in the result.

In order to find the

[tex]\sqrt{(2)^2 + (1)^2} = 2.2[/tex]

(all of these are percentages)

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