If a vertical cylinder of water (or any other liquid) rotates about its axis, as shown in
FIGURE CP8.72, the surface forms a smooth curve. Assuming that the water rotates as a unit (i.e., all the water rotates with the same angular velocity), show that the shape of the surface is a parabola described by the equation z = (ω^2 / 2g) r^2. Hint: Each particle of water on the surface is subject to only two forces: gravity and the normal force due to the water underneath it. The normal force, as always, acts perpendicular to the surface.

Question

If a vertical cylinder of water (or any other liquid) rotates about its axis, as shown in 
FIGURE CP8.72, the surface forms a smooth curve. Assuming that the water rotates as a unit (i.e., all the water rotates with the same angular velocity), show that the shape of the surface is a parabola described by the equation z = (ω^2 / 2g) r^2. Hint: Each particle of water on the surface is subject to only two forces: gravity and the normal force due to the water underneath it. The normal force, as always, acts perpendicular to the surface.

Accepted Answer

Hi, everyone. In this practice problem, we are being asked to find the equation for a curve, we will have a container filled with a fiscus fluid rotating at a constant angular velocity resulting in a fluid surface that follows a parabolic curve. We're being asked to find the equation for this curve in terms of Z omega and R where Z is the height of the fluid surface omega is the angular velocity. R is the distance from the center of rotation to the point of the fluid surface. And we want to also remember that the fluid particles on the surface experience only two different forces which are gravity and also the normal forces exerted by the fluid underneath it. The options given for the equation for the curve are for A Z equals omega squared divided by two G multi multiplied by R B Z equals two omega squared divided by G multiplied by R squared. C Z equals omega divided by G multiplied by R over two or half of R and D. Finally Z equals omega squared divided by two G multiplied by R squared. So the way we want to tackle this is to by first drawing the system that we have in the problem statement. So we'll have our container right here. So the container will be filled with the fiscus fluid. And because of that, it will also have a fluid surface that follows a parabolic curve when it is uh spin or rotated with a constant angular velocity. So this is going to be the parabolic surface and the container itself is going to be uh rotated based on this axis of rotation right here, just like. So we wanna look at a particle of the fluid right here, which is I'm gonna name this point A which is essentially just a particle of fluid. And in this case, we wanna look at point A through two different AX accesses. The first one is going to be through the R axis which is going to be just the radius or the R of the container itself. And the second one is going to be the Z axis which is pretty much the height going vertically just like. So also, so next, we want to actually uh zoom into point a uh closer so that we will get a better look at the forces acting upon point A or the the specific particle of fluid. So first, I'm gonna draw the two different axis. The vertical one is going to be our Z axis and the horizontal one is going to be our R axis and then the point in the intersection is going to be point A. So this is going to be our particle of fluid of interest just like. So and then according to the problem statement, there will be only two different forces. The first one is gravity, which is of course, it's going to be pointing downwards just like. So, so this is going to be F G and the second one is going to be the normal force. And remember, so point A is going to have the slope of the uh fluid surface in a diagonal angle just like. So if you're looking at it from this point of perspective, so this is going to be the fluid surface. And because of that, then the normal force is going to be perpendicular to this fluid surface. So it will be going to this direction just like. So I'm going to indicate that it is going to perpendicular with this right angle right here. And this is going to be our F and or our normal force, the angle between the R axis and R F N, we will indicate that with the, and we will indicate that the angle between the fluid surface and our axis with just like. So, so next, what we wanna do is to do the projection for the uh normal force. First, we want to project it onto the C axis by using or utilizing theta. So that will actually be F N multiplied by sign of theta because it is on the other side of the known angle. And then on the R axis, we will also do the projection just like. So and that will essentially be F N multiplied by cosine of data because it is on two or it is on the same side as the known angle just like. So, so that will be um the free body diagram of our system. And in this case, we know that the curve shown in the figure is going to be the surface of the fluid and point A on the surface of the fluid is considered as a particle of fluid with a mass of M. So we want to apply Newton's second law of motion at this particle both in the radical or in the R and also in the Z direction so that we can develop our equation of interest. So let's start with actually applying the Newton second law onto the radial direction. So the net force acting on the particle radially is actually what's keeping it rotating about its axis, the net force is going to actually equal to the center pedal force. So that will be Newton Second law. I'm gonna indicate that this with R D which is meaning it's gonna be in the radial direction. So according to the new second law, Sigma F R in the radial direction will equals to M multiplied by a in the radial direction which in this case is going to be ac or the center acceleration M F R, the only force acting upon the R direction is actually just going to be the cental force which in this case is going to be the F N multiplied by cosine of theta. So in this case, Sigma F R equals F C which going to equals F N multiplied by cosine of theta, which is the only force acting upon the R or the radial direction or the radial axis. So then our equation will then become um F N multiplied by cosine of data equals to M multiplied by AC. And we want to recall that the cental acceleration will actually equals to V squared divided by R and want to substitute that into our equation right here. So then our equation will then become F N multiplied by cosine of the equals M multiplied by V squared divided by R. And you want to rearrange this so that we can get an equation just for cosine of data which will help us in eventually actually finding our equation in the end. So cosine theta equals then M V squared divided by F and multiplied by R just like so awesome. So I'm gonna name this the first equation that we have. And then the second equation will also uh involve Newton's second law. But in this case, it is going to be in the Z direction. So I'm gonna write down Z there for the Z direction. So uh on or on the Z direction, the two different forces acting are F N multiplied by sine of data. And also F G. So we want to actually utilize that Sigma F Z well, that equals to M multiplied by A or in this case because in the Z direction the um particle outlet is not moving at all, it is constant. So what we wanna do is we want to actually equals this to zero or equals the ma equals to zero because the acceleration is zero. So I'm gonna just cross this out equals to zero. So that Sigma F Z is that equals to zero. OK. Ziga F C as we have uh announced previously is only F N multiplied by sine of data minus F G. So we wanna write that down F N multiplied by sign of data minus F G equals to zero. And we wanna rearrange this so that we get an equation for sign off data which will then be F G divided by F N F G is just the normal M multiplied by G divided by F N. And we wanna find the equation for sin of data because we want to actually um combine them combine that with our first equation that we have developed previously. Awesome. So the second equation is sin theta equals M G divided by F N. And now what we wanna do next is to simplify equation two and equation one by actually dividing equation two by equation one. So we want to divide two by one. So that will be sign of theta divided by co sign of theta equals M G divided by F N. All of that is going to then be divided by M V squared divided by F N multiplied by R. So we can actually uh simplify this in order for us to get, we can cross out the MS, we can cross out the F MS and that will leave us with tangent of theater. Well, then equals two G divided by V squared of that multiplied by R. So next, you wanna recall that V or the tangential velocity is given by R multiplied by omega. So we want to substitute that into our tangent equation so that we can then get tangent of data equals G multiplied by R divided by R omega squared, which will then can be simplified to G divided by R omega squared just like so awesome. So this will then be our third equation that we have, which is tangent of data equals G divided by R omega squared. For a Parabola, we want to recall that Z well, actually equals to a multiplied by R squared. And what we want to utilize next is we want to find an equation uh that will relate our tangent of theta here so that we can substitute that back into our third equation and then remove the tangent theta. So for a Parabola Z will equals to A R squared. We wanna take the derivative or we want to differentiate this equation so that we can get the slope of the curve at point A. So we wanna differentiate this equation or take the derivative with respect to R. So D Z divided by D R from this A R squared equation will come out to be two A R and that will essentially equals to the slope of the curve at point A. And if you look at our free body diagram previously, the slope of the curve at point A, which is this point right here is going to equal to the tangent of five. And then we want to use the relationship between five data and our degrees right angle in order for us to get tangent of data included into our equation. So D Z divided by D R equals two A R equals the slope of the curve at point A equals tangent of P five. And we wanna recall that theta plus five actually equals degrees. And then that will actually mean Phi equals to 90 degrees minus theta. So tangent of five will equals to the tangent of degrees minus theta. And according to the trigonometry principles, we can then change the tangent of 90 degrees minus data into one divided by tangent of data. So the right side will then be one divided by a tangent of data. And the left side will still remains to be two A R. And we want to actually rearrange this so that we get an equation for tangent of data, which will then equals to one divided by two A R. And that will essentially be our fourth equation just like. So, so now looking into our third and also our fourth equation, we can sort of equal them into one another and get a simplified final equation with removing our tangents. So first, according to the third equation tangent of theta will equals to G divided by R omega squared. And according to our fourth equation tangent of data will equals to one divided by two A R. We have R S on both sides. So we can cross that out and then we can simplify and rearrange this equation so that we get an equation for A which will then equals to omega squared divided by two G. As discussed before uh Parabola will have the equation of Z equals to A R squared. And we wanna imply that again, C equals to A R squared and D A will can, can then be actually substituted from our equation that we have developed so that we can get Z equals to omega squared R squared divided by two G rearranging this so that we can get an equation according to one of our answer choices. Then we can get open parenthesis, Omega squared divided by two G, close close parenthesis. All of that multiplied by R squared. So that will essentially be the final equation that we get for the curvature of the parabolic curve for this particular practice problem. And that will actually correspond to option D in our answer choices. So option D is actually going to be the answer to this practice problem. And that will be it for this practice video. If you guys still have any sort of confusion on this one, please make sure to check out our other lesson videos. But other than that C equals omega squared divided by two G of that multiplied by R squared will actually be the answer to this practice video. Thank you. Mhm.

Verified Solution

Key Concepts

Centrifugal Force

Hydrostatic Pressure

Equilibrium of Forces