A rotating container injected with a liquid

[rbwang1225]

Homework Statement

A massive cylindrical container of inner radius $R$ is rotating freely with an initial angular velocity $w_0$. A liquid of density ρ is slowly injected into the container, until the container is fully filled except the center of the container. The angular velocity of the whole system reduces to $w$ after the injection. What is the container's moment of inertia, while the gravitational acceleration is $g$?

Homework Equations

The moment of inertia equation.

The Attempt at a Solution

I have no idea at the first place, could someone give me some advices?

 

[tiny-tim]

hi rbwang1225!

hint: use conservation of angular momentum (you know the density of the liquid is ρ) …

show us what you get ​

 

[rbwang1225]

After I tried to solve the problem by angular momentum conservation, I got stuck on the problem of calculation the moment of inertia of the liquid.
$Iω_0=I'ω$,
where $I$ is the moment of inertia of the containerm and $I'=I+I_{liquid}$.
I tried to calculate $I_{liquid}=∫r^2dm$, but had a trouble in the shape of the liquid.
$\tan\theta=\frac{w^2r}{g}$ The limit of $r$ is from the position $r$ on the curve line to $R$, but there is a $z$ dependence of the position r. I don't know how to get the relationship.
However, my way might be in the wrong direction.
Could you give me some ideas?

Sincerely.

 

[tiny-tim]

hi rbwang1225!

I tried to calculate $I_{liquid}=∫r^2dm$, but had a trouble in the shape of the liquid.
$\tan\theta=\frac{w^2r}{g}$ The limit of $r$ is from the position $r$ on the curve line to $R$, but there is a $z$ dependence of the position r. I don't know how to get the relationship.

you'd have to do it by integration, slicing the liquid (!) into cylindrical shells of thickness dr

however, i wouldn't bother …

the question doesn't tell you how tall the container is, so i reckon you're entitled to assume that the dip in the middle is too small to matter, and that the water is just a cylinder ​

(or is the diagram supposed to be showing the dip actually reaching the bottom of the container? in that case, yes you need to integrate )

 

[rbwang1225]

OK. Then suppose the liquid forms a cylinder, $I_{liquid}=\frac{ρVR^2}{2}$.
But the problem becomes we have no height of the cylinder, how could I overcome this?

Sincerely.