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(III) A cylindrical bucket of liquid (density $\rho )$ is rotated about its symmetry axis, which is vertical. If the angular velocity is $\omega,$ show that the pressure at a distance $r$ from therotation axis is$$P=P_{0}+\frac{1}{2} \rho \omega^{2} r^{2}$$where $P_{0}$ is the pressure at $r=0$

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$P=P_{0}+\frac{1}{2} \rho \omega^{2} r^{2}$

Physics 101 Mechanics

Chapter 13

Fluids

Fluid Mechanics

Cornell University

Rutgers, The State University of New Jersey

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Lectures

03:45

In physics, a fluid is a substance that continually deforms (flows) under an applied shear stress. Fluids are a subset of the phases of matter and include liquids, gases, plasmas and, to some extent, plastic solids.

09:49

A fluid is a substance that continually deforms (flows) under an applied shear stress. Fluids are a subset of the phases of matter and include liquids, gases and plasmas. Fluids display properties such as flow, pressure, and tension, which can be described with a fluid model. For example, liquids form a surface which exerts a force on other objects in contact with it, and is the basis for the forces of capillarity and cohesion. Fluids are a continuum (or "continuous" in some sense) which means that they cannot be strictly separated into separate pieces. However, there are theoretical limits to the divisibility of fluids. Fluids are in contrast to solids, which are able to sustain a shear stress with no tendency to continue deforming.

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A circular pan of liquid w…

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A liquid in a container is…

Hey, guys. So in this problem, we're given that rose. The density of our liquid omega is the angular velocity and or the distance from the vertical axis. The first thing that's always good to do in any physics volume, it's deductible diagrams. And as you can see, I've done that here already. Side cross section view on top of you. As you notice, there's this little tiny box here which represents a tiny area, a tiny part of the water with with D or Okay, so let's do real quick. Some Let's start by doing some force analysis on this box. Okay, 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 the pushing this particle outwards towards the wall, and we're gonna call this FC centrifugal force. That force acting along the left area of this box is gonna be called P. Rather, the pressure is gonna be called p. Okay, so, no, but we also know is that on the right side that there's gonna be another force and pushing inwards. So we'll call that f ine, uh, in okay. And the pressure here is also gonna be p same pressure. But we also have to account for any changes in pressure that might have occurred over this tiny little area of this tiny little distance D r. So we can d'oh is take art change. Let's see little expression for the change in pressure, which is changing pressure with respect to our and multiply it by that tiny distance. Okay, so now that we have that, that's what else do we know? Well, based on our knowledge of a big celebration, we know that Duke's elevation towards the center off this part, uh, towards the center of the cylinder, acting on this particle is gonna be equal to hold on. Hey, let me look that up a little bit. A gonna be equal to our Omega squared quick. Okay? No, What? We can deal, given this little net force diagram we have going on here and I angular acceleration. I'm sorry. I linear ex elevation is we can try plugging that into Newton's second Law of Motion and see what we get. So, for our forces are net forces, we're gonna have f in which is he? Plus change in pressure over e. R. We are minus P. Okay, now, you might be saying yourself. Well, this is not a force. And you'd be right. A force has toe a pressure. What we can use what we can do. Rather is multiply this pressure by the areas here on I decide which we're gonna call D A. And since we know that Oh, since we know that area tons, pressure gives us a force, which is exactly we're going on exactly what we got going on here. We know that, um, we have our net force. Okay, We're gonna quit that two the right side. But you're gonna notice here that you need a mass. Okay, well, what we have well, we have is rope, which is our density of liquid. And in order to get a mass from that, we need to multiply that by a boy. Okay, so let's start, like, try to make that more. Here's a row, okay? And we need to multiply that by 1,000,000,000. Okay, So what would the volume of this little box? Well, let's see. We know that it's with is gonna be d r, and we can get a new area which we use on this side also. Yeah. Okay, so now we have a special to mass. And now all we need to do is plug in acceleration, which is our Omega. Quick. All right, so simplifying this whole equation, what we're left with is change in pressure is equal to that seems Let me make that a little bit better equal to row, which is air density. Times are omega squared. Okay, Now what we want to do is to find an expression pressure we got to do is integrate both sides. And the way we're gonna do that is on this side, that straw that in right now we're gonna be integrating from our initial pressure, which, based on the problem we know is gonna be key zero. Okay, to some toe, a final pressure at distance or which is going to be P. Okay. And now we look at the right side trying the intubation, and we're gonna be going from Let's just say some reference point where the radius zero that would be at the center. 10 two. Whatever radius we're looking for are and we also need real quick deal right there. Just so that we know we're integrating with respect to Ah, the variable or Okay, So carrying that out on the left side, we're gonna get Yeah, final pressure, P minus P zero. The initial pressure, the inside right there. And that's gonna be equal to, well, side all we really have to look at. These are just Constance. Ah, that is Rosa Constant and Omega Square. It's constant to imagine they're not there. Integrate, though. And what we're going to get is one. Uh huh. Row City row, Omega Squared square. That's just some basic integration. Okay, now we're almost done. The very last thing we have to do when I remember that the problem was asking for a form of pressure and we got that real easy. All we have to do is moved the initial pressure over to the other side of the equation. And what we get is pressure. That's some distance or is equal that the pressure zero plus one over two bro Omega squared. I squared. Okay. And that's all

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