In this screencast we are going to look at rigid body rotation and look at the example
of what occurs when you take a fluid and say a cylindrical vessel, and spin that vessel
very rapidly.
So here we have a 10 cm high by 7 cm diameter mug that is filled to a height of 5 cm with
water and is sealed to the center of a plate.
The question is how fast can the plate be spun without loosing water out of the mug?
So let's take a look at what is happening here.
So here we have our mug, it is filled up halfway with water.
And it is attached to a plate and we start rotating that plate.
And we do so until a point at which this water is going to come out of the mug.
But how does that occur?
Well as we start spinning this what happens is that the centripetal acceleration starts
pulling the fluid to the sides.
And so if this were our free surface without any rotation, with some rotation we will start
getting a parabolic shape occurring for our fluid.
With more centripetal acceleration that will become more pronounced until some point this
free surface will reach the top of the mug and any further spinning will cause the water
to come out.
How do we set up a problem like this?
Let's start with the following equation for rigid body rotation where the height of our
free body surface is equal to the rotational velocity squared times the radius squared
over 2 times the acceleration due to gravity plus some constant.
In this case we are going to use h0.
Now h0 is going to be the distance at the center from the free surface to the bottom.
And again h is dependent on r.
So let's start with writing our volume of the water that we start with.
And we know that the volume for a cylinder is pi times the radius squared times our height.
So using large capital letters to signify the actual radius of our mug and the actual
height of the water we start with.
Now if we were to write this for any radius r, it would just be pi*r^2*h.
So a differential volume would look like 2*pi*r*h*dr.
Now this would be a way to calculate our volume for our spinning mug or a stagnant mug for
that case where we have some profile that we could use for h.
And we just so happen to have that.
So if we write our differential volume in terms of our parabola, we have omega squared,
r squared, 2g, plus h0.
So we would evaluate this to determine our volume of the fluid when the mug is spinning.
This would be equal to 2*pi*the integral of 0 to the outer radius of the mug, big R, times
omega squared, r cubed, over 2g, plus h0 times r.
So now we would evaluate this integral from 0 to R. And we would get the following simplified
form.
Now this following has to be constant if we haven't lost any water.
So this has to be equal to the original amount of water that we had which was pi large R^2
times large H. So what do we do with this?
If we take a look at our diagram, H is the original height of the water.
And h0 is the height to the free surface.
Now since we start halfway full in this mug then the bottom of the parabola, in this case
the height h0, is zero, that means that the edges of our free surface at the outer radius
of our mug, will be at the top of our mug.
So this will give us our rotational speed that we could achieve without losing any water.
Any higher speeds would end up causing that h0 to actually be negative to drop below the
surface of the bottom of the mug.
So if we look for H minus h0, and we know that this is 5 cm, we could set that equal
to the following.
And we could simplify it to take the following form.
Since we know the radius of our mug as 3.5 cm we make sure that our units are consistence
and use acceleration of gravity as 981 cm/s^2.
This allows us to solve for omega and we should get for omega, 40 radians per second, which
is approximately equal to 6.
4 revolutions per second.
So that is the maximum rational velocity that we could achieve without loosing any fluid
from the mug.
Now it turns out that compared to the original free surface when the container is not rotating
at all, that surface actually dips down at the center an amount that is equal to the
amount the surface rises at the rim.
So what I mean is that this difference between the original free surface and now our h0 that
we used before, when that becomes 5 cm, it reaches the bottom it must raise 5 cm at the
outer edge and therefore we get our 10 cm.
So you may also see a form of the equation where delta h is writing as omega squared
times big R^2 over 2g where delta h is equal to two times that change in distance that
we had before.
Hopefully this gives you an idea on performing these kinds of calculations for riding body
rotation.