Deriving Kinetic Energy in a U-Tube with Varying Cross-Sectional Area

[RockenNS42]

1. Homework Statement
A U-tube has vertical arms of radii 'r' and '2r', connected by a horizontal tube of length 'l' whose radius increases linearly from r to 2r. The U-tube contains liquid up to a height 'h' in each arm. The liquid is set oscillating, and at a given instant the liquid in the narrower arm is a distance 'y' above the equilibrium level.

Show that the kinetic energy of a small slice of liquide in the horzional arm is given by

dk= (1/2)ρv² * (∏r²)dx/ (1+x/l)² (1)

where v is the velocity
ρ is denisty
l is the lengh from the centre of one arm to the centre of the other
x is the distance the slice is from the middle if the smaller arm

2. Homework Equations

k = 1/2mv²=1/2ρAlv² (2)

3. The Attempt at a Solution

So at first I seen the dK and dx and thought to intergrate, but I got a really messy problem so I figured that was it. SO i thought Id prove it bit by bit
i got that the l in (2) is the 1/(1+x/l)² from (1) but I am not sure how to go about proving this is the case...

 

[Spinnor]

The kinetic energy of the small slice is mv^2/2 where m = ρdV

dV = ∏R^2dx

And R = R(x)

Your expression,

dk= (1/2)ρv^2 * (∏r^2)dx/ (1+x/l)^2

Should read,

dk= (1/2)ρv^2 * (∏r^2)dx (1+x/l)^2

With no division symbol?

 

[RockenNS42]

it the text it has one

heres a pic of the page

 

[Spinnor]

Got it now I think. The velocity v is the velocity in the smaller vertical tube. You know that the product of the velocity of the fluid and the cross-sectional area at a point is constant?

Got to go to work, can help afterwards, hopefully someone else can fill in before then.

 

[RockenNS42]

Got it now I think. The velocity v is the velocity in the smaller vertical tube. You know that the product of the velocity of the fluid and the cross-sectional area at a point is constant?

Got to go to work, can help afterwards, hopefully
someone else can fill in before then.

That. Little was enough! I got it thanks :)