How Does Rotation Affect Fluid Pressure and Surface Height in a Spinning Pan?

[moosemagoo]

A circular pan of liquid (density= rho) is centered on a horizontal turntable rotating with angular speed w. Its axis coincides with the rotation axis. Atmospheric pressure is Pa. R=10 cm
Find expressions for (a) the pressure at the bottom of the pan
and (b) the height of the liquid surface as a function of the distance r from the axis, given that the height at the center is h0.

I have been reading and rereading this problem, but am at a complete loss. I would really appreciate any help!

 

[OlderDan]

A circular pan of liquid (density= rho) is centered on a horizontal turntable rotating with angular speed w. Its axis coincides with the rotation axis. Atmospheric pressure is Pa. R=10 cm
Find expressions for (a) the pressure at the bottom of the pan
and (b) the height of the liquid surface as a function of the distance r from the axis, given that the height at the center is h0.

I have been reading and rereading this problem, but am at a complete loss. I would really appreciate any help!

I'm sure the problem is to be solved in the steady state condition where all of the liquid is circulating with the same angular velocity. Consider the liquid near the bottom
of the pan. Each little bit of mass of the liquid is in circular motion, which requires a centripetal force. Where does that force come from? How does that force depend on the distance from the axis of rotation? How can the necessary force be achieved?

 

[m2003]

Sure, I can help you with this problem. Let's break it down step by step.

First, let's define the variables we are given:

Density of liquid (ρ) = ?
Angular speed (ω) = ?
Atmospheric pressure (P_a) = ?
Radius of pan (R) = 10 cm
Height at center (h_0) = ?

We are asked to find expressions for the pressure at the bottom of the pan and the height of the liquid surface as a function of the distance r from the axis.

To find the pressure at the bottom of the pan, we need to use the hydrostatic equation, which states that the pressure at a certain depth in a fluid is equal to the atmospheric pressure plus the product of the density of the fluid, the acceleration due to gravity, and the depth.

So, for the bottom of the pan, the pressure (P) can be expressed as:

P = P_a + ρgh

Where g is the acceleration due to gravity and h is the height of the liquid at the bottom of the pan.

Next, we need to find an expression for the height of the liquid surface as a function of the distance r from the axis. To do this, we can use the fact that the liquid in the pan will experience centrifugal force due to the rotation of the turntable. This centrifugal force will cause the liquid to rise up the sides of the pan, creating a concave shape.

To find the height of the liquid surface, we can use the equation for centrifugal force:

F_c = mω^2r

Where m is the mass of the liquid, ω is the angular speed, and r is the distance from the axis.

We can also express the mass of the liquid as the product of its density and the volume of the liquid, which is given by the area of the circular pan (πR^2) multiplied by the height of the liquid (h):

m = ρπR^2h

Putting these together, we get:

F_c = ρπR^2hω^2r

This centrifugal force is balanced by the force of gravity acting on the liquid:

F_g = ρπR^2hg

Setting these two forces equal to each other, we can solve for the height of the liquid (h):

ρπR^2hω^2r = ρπR^2hg

h = r