Consider a fluid rotating about a vertical axis. The pressure p at radius r from the axis is giv

Consider a fluid rotating about a vertical axis. The...

Key Concepts

Integration of Differential Equations

The integration of differential equations is a fundamental mathematical technique used to derive expressions for variables, such as pressure, from their gradients. In the context of fluid mechanics, integrating an equation like dp/dr = -??²r allows us to obtain the pressure as a function of position by accumulating the contributions from the differential changes over a spatial domain.

Pressure Gradient in Fluids

This concept refers to the spatial variation of pressure within a fluid. In fluid mechanics, the pressure gradient indicates how pressure changes from one point to another and is central to understanding forces and fluid movement, particularly when the fluid is in a state of equilibrium or subject to external forces.

Centrifugal Force in Rotating Systems

In a rotating frame of reference, the centrifugal force emerges as an apparent force that acts outward from the axis of rotation. This force plays a critical role in determining the pressure distribution within rotating fluids, as it creates a pressure gradient that counterbalances other forces such as gravity in conventional hydrostatics.

Transcript

00:01 Hey guys, so in this problem, we're given that row is the density of our liquid, omega is the angular velocity, and r is the distance from the vertical axis.

00:12 The first thing that's always good to do in any physics problem is to draw a couple diagrams.

00:17 And as you can see, i've done that here already, a side cross -section view and the top view.

00:22 As you can notice, there's this little tiny box here which represents a tiny area or a tiny part of the water.

00:32 With with dr.

00:36 Okay.

00:37 So let's do real quick some, let's start by doing some force analysis on this box.

00:48 So if you can imagine this as a particle rotating around in the cylinder, you know that there's got to be a centrifugal force pushing this particle outwards towards the wall.

01:04 And we're going to call this fc which is a centrifugal force that force acting along the left area of this box is going to be called p rather the pressure is going to be called p okay so now what we also know is that on the right side that there's going to be another force and pushing inwards so we'll call that f in in and the pressure here is also going to be p, same pressure, but we also have to account for any changes in pressure that might have occurred over this tiny little area, or this tiny little distance, dr.

01:55 So what we can do is take our change, our, let's see, a little expression for the change in pressure, which is change in pressure with respect to r, and multiply it by that tiny distance, d .r.

02:12 Okay.

02:14 So now that we have that, let's see.

02:17 What else do we know? well, based on our knowledge of the acceleration, we know that the acceleration towards the center of this part, towards the center of the cylinder acting on this particle is going to be equal to, hold on, a, let me move that up a little bit.

02:40 A is going to be equal to r.

02:46 Omega squared, okay.

02:53 What we can deal, given this little net force diagram we have going on here, and our angular acceleration, sorry, our linear acceleration is we can try plugging that into newton's second law of motion and see what we get.

03:17 For our forces, our net forces, we're going to have f -in, which is p, plus change in pressure over dr and dr minus p.

03:40 Okay.

03:41 Now you might be saying yourself, well, this is not a force, and you'd be right.

03:46 A force has to, a pressure, what we can use, or what we can do rather, is multiply this pressure by the areas here on either side, which we're going to call da.

04:02 And since we know that area times pressure gives us a force, which is exactly what we're going on, exactly what we got going on here, we know that we have our net force.

04:18 Okay.

04:20 We're going to equate that to the right side, but you're going to notice here that you need a mass.

04:28 Okay, well, what we have, what we have is row, which is our density of the liquid.

04:37 And in order to get a mass from that, we need to multiply that by a volume.

04:40 Okay? so let's start by that.

04:42 Oh, let's try to make that more distant.

04:45 Here's a row...

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