[Something descriptive here will yield more responses]

1. Narcol2000

26
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.

2. Mårten

127
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!

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