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-   -   [Something descriptive here will yield more responses] (http://www.physicsforums.com/showthread.php?t=251548)

Narcol2000 Aug22-08 10:10 PM

[Something descriptive here will yield more responses]
 
I'm trying to solve the question b) in this linkhttp://i36.tinypic.com/2h3wmtx.jpg

My reasoning is

[tex]
dK = dm v^2
[/tex]

[tex]
dm = \rho dV = \rho A(x) dx
[/tex]

where A(x) is a function of how the cross sectional area varies along the tube from left to right.

[tex]
A(x) = \pi r(x)^2
[/tex]

where r(x) is how the radius of the cross sectional area varies and is given by

[tex]
r(x) = r(1+x/l)
[/tex]

this gives

[tex]
dm = \rho \pi r^2 (1+x/l)^2
[/tex]

the answer given gives the (1+x/l)^2 term in the denominator but the method i have used will give it in the numerator, i don't see the flaw in what i have done though unfortunately.

Mårten Aug23-08 10:12 AM

Re: [Something descriptive here will yield more responses]
 
Hi Narcol2000!

This is a kind of question that I think is better posted in the Homework area (see link below). And don't forget to give the post a title! :smile:

http://www.physicsforums.com/forumdisplay.php?f=152


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