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424 Chapter 6: Applications of Definite Integrals Moments and Centers of Mass Many structures and mechanical systems behave as if their masses were concentrated at a single point, called the center of mass (Figure 6.29). It is important to know how to locate this point, and doing so is basically a mathematical enterprise. For the moment, we deal with one- and two-dimensional objects. Three-dimensional objects are best done with the multiple integrals of Chapter 15. Masses Along a Line We develop our mathematical model in stages. The first stage is to imagine masses and on a rigid x-axis supported by a fulcrum at the origin.m 3 m 1 , m 2 , 6.4 x m 1 Fulcrum at origin m 2 m 3 x 1 x 2 x 30 The resulting system might balance, or it might not. It depends on how large the masses are and how they are arranged. 4100 AWL/Thomas_ch06p396-465 9/1/04 12:55 PM Page 424 6.4 Moments and Centers of Mass 425 Each mass exerts a downward force (the weight of ) equal to the magni- tude of the mass times the acceleration of gravity. Each of these forces has a tendency to turn the axis about the origin, the way you turn a seesaw. This turning effect, called a torque, is measured by multiplying the force by the signed distance from the point of application to the origin. Masses to the left of the origin exert negative (coun- terclockwise) torque. Masses to the right of the origin exert positive (clockwise) torque. The sum of the torques measures the tendency of a system to rotate about the origin. This sum is called the system torque. (1) The system will balance if and only if its torque is zero. If we factor out the g in Equation (1), we see that the system torque is Thus, the torque is the product of the gravitational acceleration g, which is a feature of the en- vironment in which the system happens to reside, and the number which is a feature of the system itself, a constant that stays the same no matter where the sys- tem is placed. The number is called the moment of the system about the origin. It is the sum of the moments of the individual masses. (We shift to sigma notation here to allow for sums with more terms.) We usually want to know where to place the fulcrum to make the system balance, that is, at what point to place it to make the torques add to zero.x M 0 = Moment of system about origin = a m k x k . m 1 x 1 , m 2 x 2 , m 3 x 3 sm 1 x 1 + m 2 x 2 + m 3 x 3 d sm 1 x 1 + m 2 x 2 + m 3 x 3 d, g # sm 1 x 1 + m 2 x 2 + m 3 x 3 d "('''')''''* a feature of the environment a feature of the system System torque = m 1 gx 1 + m 2 gx 2 + m 3 gx 3 x k m k g m k m k gm k (a) (b) FIGURE 6.29 (a) The motion of this wrench gliding on ice seems haphazard until we notice that the wrench is simply turning about its center of mass as the center glides in a straight line. (b) The planets, asteroids, and comets of our solar system revolve about their collective center of mass. (It lies inside the sun.) x m 1 Special location for balance m 2 m 3 x 1 x 2 x 30 x 4100 AWL/Thomas_ch06p396-465 9/1/04 12:55 PM Page 425 The torque of each mass about the fulcrum in this special location is When we write the equation that says that the sum of these torques is zero, we get an equa- tion we can solve for Sum of the torques equals zero Constant Multiple Rule for Sums g divided out, distributed Difference Rule for Sums Rearranged, Constant Multiple Rule again Solved for This last equation tells us to find by dividing the system?s moment about the origin by the system?s total mass: The point is called the system?s center of mass. Wires and Thin Rods In many applications, we want to know the center of mass of a rod or a thin strip of metal. In cases like these where we can model the distribution of mass with a continu- ous function, the summation signs in our formulas become integrals in a manner we now describe. Imagine a long, thin strip lying along the x-axis from to and cut into small pieces of mass by a partition of the interval [a, b]. Choose to be any point in the kth subinterval of the partition. x k ¢m k x = bx = a x x = a m k x k a m k = system moment about origin system mass . x x x = a m k x k a m k . a m k x k = x a m k a m k x k - a xm k = 0 m k a sm k x k - xm k d = 0 g a sx k - xdm k = 0 a sx k - xdm k g = 0 x: = sx k - xdm k g. Torque of m k about x = a signed distance of m k from x ba downward force b 426 Chapter 6: Applications of Definite Integrals x a H9004m k x k b The kth piece is units long and lies approximately units from the origin. Now ob- serve three things. First, the strip?s center of mass is nearly the same as that of the system of point masses we would get by attaching each mass to the point : x L system moment system mass . x k ¢m k x x k ¢x k 4100 AWL/Thomas_ch06p396-465 9/1/04 12:55 PM Page 426 6.4 Moments and Centers of Mass 427 EXAMPLE 1 Strips and Rods of Constant Density Show that the center of mass of a straight, thin strip or rod of constant density lies halfway between its two ends. Solution We model the strip as a portion of the x-axis from to (Figure 6.30). Our goal is to show that the point halfway between a and b.x = sa + bd>2, x = bx = a Second, the moment of each piece of the strip about the origin is approximately so the system moment is approximately the sum of the : Third, if the density of the strip at is expressed in terms of mass per unit length and if is continuous, then is approximately equal to (mass per unit length times length): Combining these three observations gives (2) The sum in the last numerator in Equation (2) is a Riemann sum for the continuous func- tion over the closed interval [a, b]. The sum in the denominator is a Riemann sum for the function over this interval. We expect the approximations in Equation (2) to improve as the strip is partitioned more finely, and we are led to the equation This is the formula we use to find x. x = L b a xdsxd dx L b a dsxd dx . dsxd xdsxd x L system moment system mass L a x k ¢m k a ¢m k L a x k dsx k d ¢x k a dsx k d ¢x k . ¢m k L dsx k d ¢x k . dsx k d ¢x k ¢m k d dsx k d,x k System moment L a x k ¢m k . x k ¢m k x k ¢m k , Density A material?s density is its mass per unit volume. In practice, however, we tend to use units we can conveniently measure. For wires, rods, and narrow strips, we use mass per unit length. For flat sheets and plates, we use mass per unit area. Moment, Mass, and Center of Mass of a Thin Rod or Strip Along the x-Axis with Density Function (3a) (3b) (3c) Center of mass: x = M 0 M Mass: M = L b a dsxd dx Moment about the origin: M 0 = L b a xdsxd dx Dsxd x a b a H11001 b 2 c.m. H11005 FIGURE 6.30 The center of mass of a straight, thin rod or strip of constant density lies halfway between its ends (Example 1). 4100 AWL/Thomas_ch06p396-465 9/1/04 12:55 PM Page 427 The key is the density?s having a constant value. This enables us to regard the function in the integrals in Equation (3) as a constant (call it ), with the result that EXAMPLE 2 Variable-Density Rod The 10-m-long rod in Figure 6.31 thickens from left to right so that its density, instead of being constant, is Find the rod?s center of mass. Solution The rod?s moment about the origin (Equation 3a) is The rod?s mass (Equation 3b) is The center of mass (Equation 3c) is located at the point Masses Distributed over a Plane Region Suppose that we have a finite collection of masses located in the plane, with mass at the point (see Figure 6.32). The mass of the system is Each mass has a moment about each axis. Its moment about the x-axis is and its moment about the y-axis is The moments of the entire system about the two axes are The x-coordinate of the system?s center of mass is defined to be (4)x = M y M = a m k x k a m k . Moment about y-axis: M y = a m k x k . Moment about x-axis: M x = a m k y k , m k x k . m k y k ,m k System mass: M = a m k . sx k , y k d m k x = M 0 M = 250 3 # 1 15 = 50 9 L 5.56 m. M = L 10 0 dsxd dx = L 10 0 a1 + x 10 b dx = cx + x 2 20 d 0 10 = 10 + 5 = 15 kg. = c x 2 2 + x 3 30 d 0 10 = 50 + 100 3 = 250 3 kg # m. M 0 = L 10 0 xdsxd dx = L 10 0 xa1 + x 10 b dx = L 10 0 ax + x 2 10 b dx dsxd = 1 + sx>10d kg>m. = a + b 2 . x = M 0 M = d 2 sb 2 - a 2 d dsb - ad M = L b a d dx = d L b a dx = dCxD a b = dsb - ad M 0 = L b a dx dx = d L b a x dx = dc 1 2 x 2 d a b = d 2 sb 2 - a 2 d ddsxd 428 Chapter 6: Applications of Definite Integrals The ?s cancel in the formula for x. d The units of a moment are mass * length . 10 0 x (m) FIGURE 6.31 We can treat a rod of variable thickness as a rod of variable density (Example 2). x y 0 x k x k y k y k m k (x k , y k ) FIGURE 6.32 Each mass has a moment about each axis. m k 4100 AWL/Thomas_ch06p396-465 9/1/04 12:55 PM Page 428 6.4 Moments and Centers of Mass 429 With this choice of as in the one-dimensional case, the system balances about the line (Figure 6.33). The y-coordinate of the system?s center of mass is defined to be (5) With this choice of the system balances about the line as well. The torques ex- erted by the masses about the line cancel out. Thus, as far as balance is concerned, the system behaves as if all its mass were at the single point We call this point the system?s center of mass. Thin, Flat Plates In many applications, we need to find the center of mass of a thin, flat plate: a disk of alu- minum, say, or a triangular sheet of steel. In such cases, we assume the distribution of mass to be continuous, and the formulas we use to calculate and contain integrals in- stead of finite sums. The integrals arise in the following way. Imagine the plate occupying a region in the xy-plane, cut into thin strips parallel to one of the axes (in Figure 6.34, the y-axis). The center of mass of a typical strip is We treat the strip?s mass as if it were concentrated at The moment of the strip about the y-axis is then The moment of the strip about the x-axis is Equa- tions (4) and (5) then become As in the one-dimensional case, the sums are Riemann sums for integrals and approach these integrals as limiting values as the strips into which the plate is cut become narrower and narrower. We write these integrals symbolically as x = 1 x ' dm 1 dm and y = 1 y ' dm 1 dm . x = M y M = a x ' ¢m a ¢m , y = M x M = a y ' ¢m a ¢m . y ' ¢m.x ' ¢m. s x ' , y ' d.¢m s x ' , y ' d. yx sx, yd. y = y y = yy, y = M x M = a m k y k a m k . x = x x, Moments, Mass, and Center of Mass of a Thin Plate Covering a Region in the xy-Plane (6) Center of mass: x = M y M , y = M x M Mass: M = L dm Moment about the y-axis: M y = L x ' dm Moment about the x-axis: M x = L y ' dm To evaluate these integrals, we picture the plate in the coordinate plane and sketch a strip of mass parallel to one of the coordinates axes. We then express the strip?s mass dm and the co- ordinates of the strip?s center of mass in terms of x or y. Finally, we integrate and dm between limits of integration determined by the plate?s location in the plane. x ' dm, y ' dm, s x ' , y ' d x y 0 Balance line Balance line y H11005 y x H11005 x c.m. y x FIGURE 6.33 A two-dimensional array of masses balances on its center of mass. x y ~ x0 Strip c.m. ~ y ~ x ~ y Strip of mass H9004m ~~ (x, y) FIGURE 6.34 A plate cut into thin strips parallel to the y-axis. The moment exerted by a typical strip about each axis is the moment its mass would exert if concentrated at the strip?s center of mass s x ' , y ' d. ¢m 4100 AWL/Thomas_ch06p396-465 9/1/04 12:55 PM Page 429 EXAMPLE 3 Constant-Density Plate The triangular plate shown in Figure 6.35 has a constant density of Find (a) the plate?s moment about the y-axis. (b) the plate?s mass M. (c) the x-coordinate of the plate?s center of mass (c.m.). Solution Method 1: Vertical Strips (Figure 6.36) (a) The moment The typical vertical strip has The moment of the strip about the y-axis is The moment of the plate about the y-axis is therefore (b) The plate?s mass: (c) The x-coordinate of the plate?s center of mass: By a similar computation, we could find and Method 2: Horizontal Strips (Figure 6.37) (a) The moment The y-coordinate of the center of mass of a typical horizontal strip is y (see the figure), so The x-coordinate is the x-coordinate of the point halfway across the triangle. This makes it the average of y 2 (the strip?s left-hand x-value) and 1 (the strip?s right-hand x-value): x ' = sy>2d + 1 2 = y 4 + 1 2 = y + 2 4 . > y ' = y. M y : y = M x >M.M x x = M y M = 2 g # cm 3 g = 2 3 cm. M = L dm = L 1 0 6x dx = 3x 2 d 0 1 = 3 g. M y = L x ' dm = L 1 0 6x 2 dx = 2x 3 d 0 1 = 2 g # cm. x ' dm = x # 6x dx = 6x 2 dx. distance of c.m. from y-axis: x ' = x. mass: dm = d dA = 3 # 2x dx = 6x dx area: dA = 2x dx width: dx length: 2x center of mass sc.m.d: s x ' , y ' d = sx, xd M y : M y d = 3 g>cm 2 . 430 Chapter 6: Applications of Definite Integrals x (cm) y (cm) 0 2 1 (1, 2) y H11005 2x x H11005 1 y H11005 0 FIGURE 6.35 The plate in Example 3. x y 0 2 1 (1, 2) Units in centimeters Strip c.m. is halfway. x 2x dx y H11005 2x (x, 2x) ~~ (x, y) H11005 (x, x) FIGURE 6.36 Modeling the plate in Example 3 with vertical strips. x (cm) y (cm) 0 2 1 (1, 2) Strip c.m. is halfway. y dy ? ? ? ? ~~ (x, y) H11005 4 y H11001 2 , y ? ? ? ? 2 y , y 2 2 y 1 H11001 2 y x H11005 (1, y) 2 y 1 H11002 FIGURE 6.37 Modeling the plate in Example 3 with horizontal strips. 4100 AWL/Thomas_ch06p396-465 9/1/04 12:55 PM Page 430 6.4 Moments and Centers of Mass 431 We also have The moment of the strip about the y-axis is The moment of the plate about the y-axis is (b) The plate?s mass: (c) The x-coordinate of the plate?s center of mass: By a similar computation, we could find and If the distribution of mass in a thin, flat plate has an axis of symmetry, the center of mass will lie on this axis. If there are two axes of symmetry, the center of mass will lie at their intersection. These facts often help to simplify our work. EXAMPLE 4 Constant-Density Plate Find the center of mass of a thin plate of constant density covering the region bounded above by the parabola and below by the x-axis (Figure 6.38). Solution Since the plate is symmetric about the y-axis and its density is constant, the distribution of mass is symmetric about the y-axis and the center of mass lies on the y-axis. Thus, It remains to find A trial calculation with horizontal strips (Figure 6.38a) leads to an inconvenient inte- gration We therefore model the distribution of mass with vertical strips instead (Figure 6.38b). M x = L 4 0 2dy24 - y dy. y = M x >M.x = 0. y = 4 - x 2 d y.M x x = M y M = 2 g # cm 3 g = 2 3 cm. M = L dm = L 2 0 3 2 (2 - y) dy = 3 2 c2y - y 2 2 d 0 2 = 3 2 (4 - 2) = 3 g. M y = L x ' dm = L 2 0 3 8 s4 - y 2 d dy = 3 8 c4y - y 3 3 d 0 2 = 3 8 a 16 3 b = 2 g # cm. x ' dm = y + 2 4 # 3 # 2 - y 2 dy = 3 8 s4 - y 2 d dy. distance of c.m. to y-axis: x ' = y + 2 4 . mass: dm = d dA = 3 # 2 - y 2 dy area: dA = 2 - y 2 dy width: dy length: 1 - y 2 = 2 - y 2 How to Find a Plate?s Center of Mass 1. Picture the plate in the xy-plane. 2. Sketch a strip of mass parallel to one of the co- ordinate axes and find its dimensions. 3. Find the strip?s mass dm and center of mass 4. Integrate dm, dm, and dm to find and M. 5. Divide the moments by the mass to calculate and y. x M x , M y , x ' y ' s x ' , y ' d. 4100 AWL/Thomas_ch06p396-465 9/1/04 12:55 PM Page 431 The typical vertical strip has The moment of the strip about the x-axis is The moment of the plate about the x-axis is (7) The mass of the plate is (8) Therefore, The plate?s center of mass is the point EXAMPLE 5 Variable-Density Plate Find the center of mass of the plate in Example 4 if the density at the point (x, y) is twice the square of the distance from the point to the y-axis. d = 2x 2 , sx, yd = a0, 8 5 b. y = M x M = s256>15d d s32>3d d = 8 5 . M = L dm = L 2 -2 ds4 - x 2 d dx = 32 3 d. = d 2 L 2 -2 s16 - 8x 2 + x 4 d dx = 256 15 d. M x = L y ' dm = L 2 -2 d 2 s4 - x 2 d 2 dx y ' dm = 4 - x 2 2 # ds4 - x 2 d dx = d 2 s4 - x 2 d 2 dx. distance from c.m. to x-axis: y ' = 4 - x 2 2 . mass: dm = d dA = ds4 - x 2 d dx area: dA = s4 - x 2 d dx width: dx length: 4 - x 2 center of mass sc.m.d: s x ' , y ' d = ax, 4 - x 2 2 b 432 Chapter 6: Applications of Definite Integrals x y 0 4 (0, y) (a) ?2 2 c.m.dy x y 0 4 (b) ?2 2 dx x Center of mass y H11005 4 H11002 x 2 y H11005 4 H11002 x 2 ? ? ? ? ~~ (x, y) H11005 2 4 H11002 x 2 x, 2 y 4 H11002 x 2 2H208574 H11002 y FIGURE 6.38 Modeling the plate in Example 4 with (a) horizontal strips leads to an inconvenient integration, so we model with (b) vertical strips instead. 4100 AWL/Thomas_ch06p396-465 9/1/04 12:55 PM Page 432 6.4 Moments and Centers of Mass 433 Solution The mass distribution is still symmetric about the y-axis, so With Equations (7) and (8) become (7?) (8?) Therefore, The plate?s new center of mass is EXAMPLE 6 Constant-Density Wire Find the center of mass of a wire of constant density shaped like a semicircle of radius a. Solution We model the wire with the semicircle (Figure 6.39). The dis- tribution of mass is symmetric about the y-axis, so To find we imagine the wire divided into short segments. The typical segment (Figure 6.39a) has Hence, The center of mass lies on the axis of symmetry at the point about two-thirds of the way up from the origin (Figure 6.39b). Centroids When the density function is constant, it cancels out of the numerator and denominator of the formulas for and This happened in nearly every example in this section. As far as and were concerned, might as well have been 1. Thus, when the density is constant, the location of the center of mass is a feature of the geometry of the object and not of the material from which it is made. In such cases, engineers may call the center of mass the centroid of the shape, as in ?Find the centroid of a triangle or a solid cone.? To do so, just set equal to 1 and proceed to find and as before, by dividing moments by masses.yxd dyx y.x s0, 2a>pd, y = 1 y ' dm 1 dm = 1 p 0 a sin u # da du 1 p 0 da du = da 2 C-cos uD 0 p dap = 2 p a. distance of c.m. to x-axis: y ' = a sin u. mass: dm = d ds = da du length: ds = a du y,x = 0. y = 2a 2 - x 2 d sx, yd = a0, 8 7 b. y = M x M = 2048 105 # 15 256 = 8 7 . = L 2 -2 s8x 2 - 2x 4 d dx = 256 15 . M = L dm = L 2 -2 ds4 - x 2 d dx = L 2 -2 2x 2 s4 - x 2 d dx = L 2 -2 s16x 2 - 8x 4 + x 6 d dx = 2048 105 M x = L y ' dm = L 2 -2 d 2 s4 - x 2 d 2 dx = L 2 -2 x 2 s4 - x 2 d 2 dx d = 2x 2 , x = 0. Mass per unit length times length x y 0?aa (a) x y 0?aa a c.m. A typical small segment of wire has dm H11005 H9254 ds H11005 H9254adH9258. (a cosH9258, a sinH9258) dH9258 H9258 y H11005 H20857a 2 H11002 x 2 (b) ~~ (x, y) H11005 0, a 2 H9266 ? ? ? ? FIGURE 6.39 The semicircular wire in Example 6. (a) The dimensions and variables used in finding the center of mass. (b) The center of mass does not lie on the wire. 4100 AWL/Thomas_ch06p396-465 9/1/04 12:55 PM Page 433 Commercial_CD 4100 AWL/Thomas_ch06p396-465

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Author: Sree Ram M.

Created: 2009-12-06

Updated: 2009-12-06

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Created: 2009-12-06

Updated: 2009-12-06

File Size: 10 page(s)

Views: 13

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