Center of Water In homework 1, you used dimensional analysis to investigate the mass of a fluid of non-uniform density, specifically a column of water with varying salinity. In this application, you return to the water column to determine its center of mass. Knowing this is critical, since many effects, especially rotational effects, depend on the where the center of mass of an object is. The position of the center of mass of an object is is the position of the object. We discsuss the center of mass in two different ways, one is where we have several discrete bodies, each with their own mass and position. In this case, the center of mass is just the weighted average of the positions of individual objects. What is a weighted average? It is an average where heavier things have more influence than lighter things, so if we had two objects, a heavy one M, at position X, and a lighter one M2 at position X2, their center of mass is Xem = Mix + M2 82 M1 + M2 This can be extended to multiple objects in a natural manner. If, however, we look at a system with a varying density like the water column in homework 1, we have a slightly different problem. Now, the geometric center of the water column has more mass under it than over it. The center of mass is the plane at the height where there is an equal amount of mass both above and below it, and that's where we would, for example, apply the force from the water's weight in a problem. So, need a way in which to find this position. This position is found in the same way In equilibrium the salinity of a water column of area 29 cm and height 85 cm tall reduces linearly from 94 9/kg (grams of salt per kilogram of water) at the bottom to 28 9/kg at the top. Assume that the density of water molecules remains constant throughout the water column, and measure from the bottom of the column. Part 3 of 3 This was all repeated from Lesson 1. Now, you'll find the center of mass of the cylinder. The center of mass is the weighted average of the position up the cylinder, where u is the first moment of the mass. This is found like the mass is, through an integral of the linear mass density, but this time multiplied by the position of the mass M = >(2) z dz (d) Find the ist moment of the fluid in the cylinder. M = 12230 g cm x k (e) Find the center of mass of the fluid in the cylinder. *cm = .6 cm xk Submit Skip Part 1 of 3 (a) What is the geometric center of the water column? Re = 42.5 cm| |42.5 cm k Part 2 of 3 (b) What is the linear mass density of the fluid? *(z) = (1000 kg/m x - 31.7 9/cm|+ (No Response) -0.0225 g/cm^2|z) k (c) Find the mass of the fluid in the cylinder. m = (No Response) 2620 gSolv ) . IL = 100occ , and I ; ky is same as I Therefore, Salinity of water column at the bottom is , L to = gyg _ 949 = 0.094/ g ) loooce and Salinity of water column at the top is, SH = 289 - 289 L 100OCC = 0.028 / 7 ) Also we are going to assume the density of water remains constant throughout the water column, and it is, HO = 1(2 ) Now By Straight line procedure to write a linear can, we can write Volume mass density" as a function of 'z' as, P( # ) = SHot Pot 13 H - So H = 1 + 0ogy + 0. 028 - 0.094) 2 85 = 1 . ogy + - 0. 066 85 = 1.ogy - (0.06 6 ) 85To convert this into linear mass density" we Should multiply, P(2) with area of base of Column, "A'. go , 7 ( 2 ) = 8 ( 2 ) . A = 1 1.094 - (0-256) 27 x 29 () 2 ( 2 ) = 31.726 - 0.022 2 O (d) The first moment of the fluid in the cylinder 18, H Mi = 2 (2 ) 2 12 85 cm = ( 31 . 726 - 0 . 022 2 ) 2 dz 85cm = 131 . 726 2 2 ) - 0.022 / 23 ) O ( 31 . 726 ) ( 85 ) 2 2 - ( 0 . 022 ) / 1 85 ) 3 = ( 85 ) 2 31. 726 _ 1. 87 Icm 2 3 =) M = 110106 . 59 gem(e ) To Calculate the center of mass of the fluid in the cylinder is given by, Xem = M m firstly, we need to calculate total mass of the fluid inside the cylinder, by using Eq" , 85 cm m = 1 ( 2 ) dz 85 cm ( 31 . 726 - 0 . 02 2 2 ) dz O 85 cm = 31 . 726 2 - 0.022 22 2 0 = [ 31. 726 ( 85 ) - 0.022 ( 85 ) 2 9 = 85 31 . 726 - (0 . 01 1 ) ( 85 ) | 9 M = 2617 . 235 J Therefore, by using Egh above,110106. 59 CM 2617. 235 - X cm = 42.07 cm So the center of mass of the fluid in the cylinder 18, 42.07 CM