# Introductory Lab Problem

1. Feb 1, 2010

### maherelharake

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

Given that the uncertainty in the angular velocity is δω = 0.48 rad/s and the uncertainty in x is δx = 0.37 cm, calculate the uncertainty in y (δy) for the following points. The angular velocity of the bowl is 16.3 rad/s and you may ignore the uncertainty in H.

I have no idea how to start this problem, because in the lab lecture, uncertainty was never mentioned. Please help.

2. Relevant equations
I have attached what we were given below.

3. The attempt at a solution
Nothing yet since I don't know how uncertainty factors into the given formulas.

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Last edited: Feb 1, 2010
2. Feb 1, 2010

### Delphi51

It is a little difficult to answer your question without knowing the level of your course. And not much info on the problem is given. But the formula for y is the main thing.

Beginners do error calcs by calculating the y with the "best" values (not counting the error) and then again with the best values plus their errors (or minus the errors if that gives a higher answer). Take the difference between the two answers to get the error in y.

The next level is to use formulas for the error when multiplying, dividing, etc. There are some formulas here:
http://en.wikipedia.org/wiki/Propagation_of_uncertainty

Most advanced - use partial derivatives to figure out a formula for the error. That is explained at the same link.

In all cases, you will need values for ω and x. You don't seem to have a value for x, so I don't see how you can do the problem.

3. Feb 1, 2010

### maherelharake

Sorry. I attached a value for x, later. It's x=5.72 cm. And the class is an intro course where calculus isn't used. I tried plugging in values, but immediately became confused.

4. Feb 1, 2010

### Delphi51

What do you get when you work out y = H + ω²x²/(2g) using the
x = 5.72 cm, ω = 16.3/s ? Of course you will have an H in the answer but it will cancel out when you take the difference between this y value and the second one you calculate using the values of ω and x increased by their error estimates.

5. Feb 1, 2010

### maherelharake

I got:
y=H+0.04435 for the first one without error estimates and
y=H+0.05328 for the equation with error estimates.

6. Feb 2, 2010

### Delphi51

Just subtract the two values to get the delta y you are looking for!

7. Feb 2, 2010

### maherelharake

I subtracted them and got .893 cm but that answer is incorrect. Any ideas?

8. Feb 2, 2010

### Delphi51

The trouble with error calcs is that they are not exact.
If you do the high - best you get a slightly different answer than if you do the best - low value. Or if you use the formula method you get yet another slightly different answer. By the formula, you get
δy = 2ωx²/(2g)*δω + 2ω²x/(2g)*δx = .00835 meters
Of course it is silly to include 3 digits in an error estimate; .01 would be more sensible. To get your course's answer, you really need to know the expected method and rounding practise.

9. Feb 2, 2010

### maherelharake

Thanks for all the responses. I went to my TA today and he helped me out. We had to use a formula that wasn't given. Thanks again!

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