Pressure of liquid given radius, help please

[harlow_barton]

A test tube filled with water is being spun around in an ultracentrifuge with angular velocity. The test tube is lying along a radius and the free surface of the water is at radius r(o).

Show that the pressure at radius r within the test tube is:

p = .5(p)(angular velocity)^2(r^(2) -r(o)^2)

where p is the density of the water. Ignore gravity and atmospheric pressure.


p = p - g(density)(height)


gravity or centripetal acceleration, a= r(angular velocity)^2

height or depth of water, h = r- r(o)

this only gets me to p= p + density*r*angular velocity^2(r-r(0))

I'm not sure where the rest comes from!

 

[Doc Al]

At any depth 'below' the surface, the pressure has to provide enough force to accelerate all the fluid 'above' it. Hint: Set up an integral.

 

[harlow_barton]

Doc Al, I'm not sure I understand what I should be taking the integral of. Could you explain further?

 

[Doc Al]

Doc Al, I'm not sure I understand what I should be taking the integral of. Could you explain further?

Write an expression for the net force on an infinitesimal slice (thickness dr) of the fluid in the tube; then integrate from r(0) to r to find the total force, and then the pressure, at any point along the tube.

 

[PJC01]

First, let's rearrange the equation for pressure: p = p + density*r*angular velocity^2(r-r(0))

We can rewrite this as p = p + density*r*angular velocity^2*r - density*r*angular velocity^2*r(o)

Next, we can factor out the density and angular velocity^2 terms: p = p + density*angular velocity^2*(r^2 - r(o)^2)

This is almost the same as the equation we are trying to prove, but we have an extra density term. To get rid of this, we can use the equation for density: p = m/V, where m is the mass of the water and V is the volume.

We can then substitute this into our equation for pressure: p = (m/V) + density*angular velocity^2*(r^2 - r(o)^2)

Next, we can use the equation for volume of a cylinder (since the test tube is in the shape of a cylinder): V = πr^2h, where h is the height or depth of the water.

Substituting this into our equation for pressure, we get: p = (m/πr^2h) + density*angular velocity^2*(r^2 - r(o)^2)

Now, we can use the equation for centripetal acceleration (a = r*angular velocity^2) to replace the angular velocity^2 term: p = (m/πr^2h) + density*a*r*(r^2 - r(o)^2)

Lastly, we can substitute in our expression for height (h = r - r(o)): p = (m/πr^2(r-r(o))) + density*a*r*(r^2 - r(o)^2)

Simplifying this further, we get: p = (m/πr(r-r(o))) + density*a*(r^2 - r(o)^2)

And finally, using the equation for mass (m = density*volume), we get: p = (.5density*r*angular velocity^2(r^2 - r(o)^2))

Which is the equation we were trying to prove! This shows that the pressure at a given radius within the test tube is directly proportional to the angular velocity squared and the difference in radii (r^2 - r(o)^2).