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7.6 Moments and Center of Mass in This Section We Want to find a Point on Which a Thin Plate of Any Given Shape Balances Horizontally As in Figure 7.6.1

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7.6 Moments and Center of Mass in This Section We Want to find a Point on Which a Thin Plate of Any Given Shape Balances Horizontally As in Figure 7.6.1 Arkansas Tech University MATH 2924: Calculus II Dr. Marcel B. Finan 7.6 Moments and Center of Mass In this section we want to find a point on which a thin plate of any given shape balances horizontally as in Figure 7.6.1. Figure 7.6.1 The center of mass is the so-called \balancing point" of an object (or sys- tem.) For example, when two children are sitting on a seesaw, the point at which the seesaw balances, i.e. becomes horizontal is the center of mass of the seesaw. Discrete Point Masses: One Dimensional Case Consider again the example of two children of mass m1 and m2 sitting on each side of a seesaw. It can be shown experimentally that the center of mass is a point P on the seesaw such that m1d1 = m2d2 where d1 and d2 are the distances from m1 and m2 to P respectively. See Figure 7.6.2. In order to generalize this concept, we introduce an x−axis with points m1 and m2 located at points with coordinates x1 and x2 with x1 < x2: Figure 7.6.2 1 Since P is the balancing point, we must have m1(x − x1) = m2(x2 − x): Solving for x we find m x + m x x = 1 1 2 2 : m1 + m2 The product mixi is called the moment of mi about the origin. The above result can be extended to a system with many points as follows: The center of mass of a system of n point-masses m1; m2; ··· ; mn located at x1; x2; ··· ; xn along the x−axis is given by the formula n X mixi i=1 Mo x = n = X m mi i=1 n X where the sum Mo = mixi is called the moment of the system about i=1 Pn the origin and m = i=1 mn is the total mass. Example 7.6.1 Point masses mi are located on the x−axis as shown in Figure 7.6.3. Find the moment M of the system about the origin and the center of mass x: Figure 7.6.3 Solution. The moment of the system about the origin is Mo = 12(−3) + 15(2) + 20(8) = 154: The center of mass is 154 154 x = = 12 + 15 + 20 47 2 Discrete System: Two dimensional case The concept of center of mass can be applied to two dimensional objects as well. The determination of the center of mass in two dimensions is done in a similar manner. If a mass m is located at a point (x; y) then we define the moment of m about the x−axis to be the product my and the moment of m about the y−axis to be the product mx: Let (x; y) be the center of mass. The procedure of finding formulas for x and y is the same as the one dimensional case. Add up the masses times their x−locations then divide by total mass to get x: Next, add up the masses times their y−locations then divide by total mass to get y: Hence the two formulas: n n X X ximi yimi My i=1 Mx i=1 x = m = n and y = m = n X X mi mi i=1 i=1 n X where Mx = yimi is the moment of the system about the x−axis i=1 n X and My = ximi is the moment of the system about the y−axis. i=1 Since mx = My and my = Mx; the center of mass (x; y) is the point where a single particle of mass m would have the same moments as the system. Example 7.6.2 Point masses mi are located at the points Pi: Find the moment Mx and My and the center of mass of the system: mi Pi 4 (2; −3) 2 (−3; 1) 4 (3; 5) 3 Solution. We have Mx =4(−3) + 2(1) + 4(5) = 10 My =4(2) + 2(−3) + 4(3) = 14 m =4 + 2 + 4 = 10 14 x = = 1:4 10 10 y = = 1 10 Continuous System:One Dimensional Case Next we consider a continuous system. Suppose that we have an object lying on the x−axis between x = a and x = b: At point x, suppose that the object has mass density (mass per unit length) of δ(x): To calculate the center of mass, we divide the object into n pieces, each of length ∆x: On each piece, the density is nearly constant, so the mass of the ith piece is mi ≈ δ(xi)∆x: The center of mass is then n n X X mixi xiδ(xi)∆x i=1 i=1 x = n ≈ n : X X mi δ(xi)∆x i=1 i=1 Letting n ! 1 we obtain R b xδ(x)dx x = a : R b a δ(x)dx Example 7.6.3 Find the center of mass of a 2-meter rod lying on the x−axis with its left end at the origin if its density is δ(x) = 15x2 kg=m: Solution. The total mass is Z 2 2 3 2 m = 15x dx = 5x 0 = 40 kg: 0 4 The center of mass is R 2 15x3dx 15 x4 2 x = 0 = = 1:5 m 40 40 4 0 Two Dimensional System: Continuous case In the continuous case, we consider a thin plate that occupies a region in the plane as shown in Figure 7.6.4. We assume the plate has uniform density ρ. Figure 7.6.4 Divide the interval [a; b] into n subintervals with endpoints xi = a + i∆x b−a ∗ xi−1+xi and length ∆x = n : Let xi = xi = 2 : Then the center of mass of 1 the rectangle Ri is Ci(xi; 2 f(xi)): The mass of this rectangle is mi = ρf(xi)∆x: Thus, My(Ri) = ρf(xi)∆xxi and [f(x )]2 M (R ) = ρ i ∆x: x i 2 We define n X Z b My = lim ρf(xi)xi∆x = ρxf(x)dx: n!1 i=1 a Likewise, n Z b X 1 2 1 2 Mx = lim ρ [f(xi)] ∆x = ρ[f(x)] dx: n!1 2 2 i=1 a Thus, R b ρxf(x)dx R b xf(x)dx x = a = a R b R b ρ a f(x)dx a f(x)dx 5 and R b ρ 1 [f(x)]2dx R b 1 [f(x)]2dx y = a 2 = a 2 : R b R b ρ a f(x)dx a f(x)dx Example 7.6.4 Find the center of mass of a semicircular plate of radius r: Figure 7.6.5 Solution. Due to symmetry, the center of mass must lie on the y− axis so that x = 0: Now, p R r 1 [ r2 − x2]2dx y = −r 2 πr2 2 3 r 2 2 x = 2 r x − πr 3 −r 4r = 3π If the thin plates occupies a region bounded by an upper curve f(x) and a lower curve g(x) then the center of mass is given by R b ρx[f(x) − g(x)]dx R b x[f(x) − g(x)]dx x = a = a R b R b ρ a [f(x) − g(x)]dx a [f(x) − g(x)]dx and R b ρ 1 [f(x)2 − g(x)2]dx R b 1 [f(x)2 − g(x)2]dx y = a 2 = a 2 : R b R b ρ a [f(x) − g(x)]dx a [f(x) − g(x)]dx 6 Example 7.6.5 Find the center of mass of the region bounded by the line y = x and the parabola y = x2: Assume ρ = 1: Figure 7.6.6 Solution. We have Z 1 1 m = (x − x2)dx = 0 6 Z 1 1 x =6 x(x − x2)dx = 0 2 Z 1 1 2 y =6 (x2 − x4)dx = 0 2 5 Remark 7.6.1 The center of mass of a body need not be within the body itself; the center of mass of a ring or a hollow cylinder of uniform density is located in the enclosed space, not in the object itself. Remark 7.6.2 Whenever a density is not mentioned in a problem we will assume that the density is 1. In this case, we will use the word centroid to stand for the center of mass. Example 7.6.6 Find the centroid of the region shown in Figure 7.6.7 7 Figure 7.6.7 Solution. Creating the rectangular system shown is Figure 7.6.8, we can locate the centroid of the three rectangles at (1=2; 3=2); (5=2; 1=2); and (5; 1): Figure 7.6.8 Using these points, we can find the centroid of the region (1=2)(3) + (5=2)(3) + (5)(4) x = = 2:9 3 + 3 + 4 (3=2)(3) + (1=2)(3) + (1)(4) y = = 1 3 + 3 + 4 Pappus Centroid Theorem Let R be a plane region with area A and centroid C. Let L be a line in the plane that does not interest the interior of R: Let r be the distance of C to the line L: The solid of revolution obtained by rotating R aroud the line L has volume given by the formula V = 2πrA: 8 Note that 2πr is the distance traveled by the centroid as the region is re- volved about the line.
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