A test-tube filled with a liquid and was swing around

[rbwang1225]

Homework Statement

A small cylindrical test-tube of inner radius R is initially filled with a liquid up to height $h_0$. The tube is connected by a long rope of length $L$ ($L»h_0$) and swinging horizontally with a constant angular velocity $ω$. There is a tiny round hole of radius $r$ ($r « R$) at the bottom of the test-tube and the liquid is spraying out from the hole. What's the remaining height of the liquid as a function of time?

Homework Equations

Continuity equation and circular motion equations.

The Attempt at a Solution

$v\pi r^2=wR\pi R^2$
$v=\frac{R^3ω}{r^2},$
where v is the speed of the liquid just outside the hole.
The small cube moving out the hole is $dx\pi r^2$ which is equal to the lose of the liquid in the tube $dh \pi R^2.$
$v=\frac{dx}{dt}$
Then the remaining height of the liquid as a function of time is $h(t)=h_0-∫dh=h_0-∫\frac{r^2}{R^2}dx=h_0-∫_0^t\frac{r^2}{R^2}\frac{R^3ω}{r^2}dt=h_0-Rωt$
I know I might miss something, could anyone point out for me.

Regards.

 

[Spinnor]

This is late and may be of no use, if so sorry.

Put a block of mass m in the bottom of a mass-less bucket. Attach a rope to the bucket as is done above with the test-tube and swing in a horizontal circle.

The force on the block is F = mv^2/r = mω^2r

The pressure of the block on the bottom of the bucket due to the block is,

F/Area of the block

With this you should be able to find the pressure at the bottom of the test tube? Now equate (the actual liquid velocity will be less but this is a good first approximation),

pressure*volume =

energy = ρ*volume*(velocity of liquid exiting hole)^2/2

Hope this helps (and is right!).