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

 Narcol2000 Aug22-08 11:10 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

$$dK = dm v^2$$

$$dm = \rho dV = \rho A(x) dx$$

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

$$A(x) = \pi r(x)^2$$

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

$$r(x) = r(1+x/l)$$

this gives

$$dm = \rho \pi r^2 (1+x/l)^2$$

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 11:12 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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