What is the uncertainty in y for given angular velocity and position values?

  • Thread starter Thread starter maherelharake
  • Start date Start date
  • Tags Tags
    Introductory Lab
Click For Summary

Homework Help Overview

The discussion revolves around calculating the uncertainty in a variable \( y \) based on given uncertainties in angular velocity and position values. The context is an introductory physics course where uncertainty propagation is being explored, particularly in relation to a formula involving angular motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various methods for calculating uncertainty, including using best values and error estimates, and mention the need for a specific formula for \( y \). Some express confusion about how to incorporate uncertainties into their calculations.

Discussion Status

Several participants have provided insights into different approaches for calculating uncertainty, including basic error propagation techniques and more advanced methods involving partial derivatives. There is acknowledgment of the need for specific values and formulas, and some participants have shared their attempts at calculations, although no consensus on the correct method has been reached.

Contextual Notes

Participants note the absence of certain values and the lack of clarity regarding the expected methods for error calculations in the context of their course. There are indications that the problem setup may not have been fully communicated in the initial post.

maherelharake
Messages
261
Reaction score
0

Homework Statement



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.



Homework Equations


I have attached what we were given below.


The Attempt at a Solution


Nothing yet since I don't know how uncertainty factors into the given formulas.
 

Attachments

  • Formula.jpg
    Formula.jpg
    4.9 KB · Views: 396
  • 1.jpg
    1.jpg
    1.8 KB · Views: 417
Last edited:
Physics news on Phys.org
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.
 
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.
 
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.
 
I got:
y=H+0.04435 for the first one without error estimates and
y=H+0.05328 for the equation with error estimates.
 
Just subtract the two values to get the delta y you are looking for!
 
I subtracted them and got .893 cm but that answer is incorrect. Any ideas?
 
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.
 
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!
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
18K
Replies
4
Views
3K
  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K