7.6 The Use of Definite Integrals in Physics and Engineering


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1 Arknss Tech University MATH 94: Clculus II Dr. Mrcel B. Finn 7.6 The Use of Definite Integrls in Physics nd Engineering It hs been shown how clculus cn be pplied to find solutions to geometric problems such s problems concerned with computing re, volume, surfce re. nd rc length. In this section clculus is used to solve problems tht rise from Physics nd engineering. The Concept of Work The work done by constnt force, F, in moving n object distnce, d, is equl to the product of the force nd the distnce moved. Tht is, W = F d. The SI (interntionl) unit of work is the joule (J), which is the work done by force of one Newton (N) pushing body long one meter (m). Thus, 1 joule = 1 Nm. In the British system, unit work is the footpound. Since 1N =.489 lb(1 lb = 4.45 N) nd 1m = ft (1 ft =.35 m), we hve 1J = ft lb(1 ft lb = 1.36 J). Now, in most cses the pplied force is not constnt, but vries over the stright line of motion. For exmple, suppose tht the force, F (x), cting on prticle s it moves long the stright line from to b vries continuously. In order to find the totl work done by the force we divide the intervl [, b] into n smll equl subintervls [x i 1, x i ], ech of length x, so tht the chnge in F is smll long ech subintervl, i.e., pproximtely constnt. Then the work done by the force in moving the body from x i 1 to x i is pproximtely: W i F (x i ) x where x i 1 x i x i. Hence, the totl work is W F (x i ) x. As n the Riemnn sum t the right converges to the following integrl: W = F (x)dx. Remrk In physics, the kinetic energy of n object is the energy which it possesses due to its motion. It is defined s the work needed to ccelerte body 1
2 of given mss from rest to its stted velocity. Hving gined this energy during its ccelertion, the body mintins this kinetic energy unless its speed chnges. The sme mount of work is done by the body in decelerting from its current speed to stte of rest. Exmple Consider spring on the x xis so tht its right end is t when the spring is t its rest position. According to Hooke s Lw, the force needed to stretch the spring from to x is proportionl to x, i.e., F (x) = kx where k is clled the spring constnt. See Figure Figure Find the work done in stretching the spring length of. The work needed to stretch the spring from to is given by the integrl W = kxd k Exmple 7.6. (Work Done Filling (or Emptying) Tnk) A tnk in the shpe of right circulr cone of height 1 m nd rdius 4 m is inserted into the ground with its vertex pointing down nd its top t ground level. If the tnk is filled with wter (density ρ = 1kg/m 3 ) to depth of 8 m, how much work is performed in pumping ll the wter in the tnk to ground level? Wht chnges if the wter is pumped to height of m bove ground level? Set up coordinte system s shown in Figure 7.6..
3 Figure 7.6. Consider lyer of distnce x i from the bse of the cone nd with thickness x. The volume of such circulr lyer is Using similr tringles we find tht V i πr i x. r i 4 = 1 x i 1 nd consequently r i = 5 (1 x i ). Thus, V i 4π 5 (1 x i ) x. Hence its mss is m i = 1 4π 5 (1 x i ) 16π(1 x i ) x. The force required to rise this lyer is f i = m i g = 9.8[16π(1 x i ) x] = 1568π(1 x i ) x. The work done to rise it to the top of the tnk is W i 1568π(1 x i ) x i x. Adding the works done to rise these slices we obtin the totl work done to empty the tnk: W = π(1 x) xd J. 3
4 Now if the wter is pumped to height of m bove ground level then W = π(1 x) (x + )d J Force nd Pressure Pressure is the force per unit re cting on n object. The pressure is exerted eqully in ll directions nd it increses in depth. Consider thin horizontl plte of re A squre meters submerged in liquid t given depth d below the surfce. The volume of the liquid directly bove the plte is V = Ad nd its mss is m = ρad where ρ is the density of the liquid. Thus, the hydrosttic force exerted by the liquid on the plte is F = mg = ρgad where g is the ccelertion due to grvity. Hence, the pressure P on the plte is defined by P = F A = ρgd. The SI unit for pressure is clled Pscl. Thus, 1 P = 1 Newton per squre meter. Exmple Consider dm for storing wter s shown in Figure Set up nd clculte definite integrl giving the totl hydrosttic force on the dm if wter level is 4 m from the top of the dm. The density of wter is ρ = 1 kg/m 3. Figure We divide the dm into horizontl strips in which the pressure is lmost constnt. Let s find the re of the i th strip which is pproximtely rectngle 4
5 of height x nd width w i = (15 + ). From similr tringles, we hve 1 = 16 x i = = 8.5x i. Thus, A i (3.5x i ) (46 x i ) x. The pressure exerted on this strip of the dm is given by P i 1gx i. The hydrosttic force cting on this strip is F i = P i A i 1gx i (46 x i ) x. Thus, the totl force is F = 16 1gx(46 x)d N Moments nd Center of Mss In this section we wnt to find point on which thin plte of ny given shpe blnces horizontlly s in Figure Figure The center of mss is the soclled blncing point of n object (or system.) For exmple, when two children re sitting on seesw, the point t which the seesw blnces, i.e. becomes horizontl is the center of mss of the seesw. 5
6 Discrete Point Msses: One Dimensionl Cse Consider gin the exmple of two children of mss m 1 nd m sitting on ech side of seesw. It cn be shown experimentlly tht the center of mss is point P on the seesw such tht m 1 d 1 = m d where d 1 nd d re the distnces from m 1 nd m to P respectively. See Figure In order to generlize this concept, we introduce n x xis with points m 1 nd m locted t points with coordintes x 1 nd x. Figure Since P is the blncing point, we must hve Solving for x we find m 1 (x x 1 ) = m (x x). m 1x 1 + m x. m 1 + m The product m i x i is clled the moment of m i bout the origin. The bove result cn be extended to system with mny points s follows: The center of mss of system of n pointmsses m 1, m,, m n locted t x 1, x,, x n long the x xis is given by the formul m i x i The sum M = origin. m i m i x i is clled the moment of the system bout the 6
7 Exmple Point msses m i re locted on the x xis s shown in Figure Find the moment M of the system bout the origin nd the center of mss x. Figure The moment of the system bout the origin is M = 1( 3) + 15() + (8) = 154. The center of mss is = Discrete System: Two dimensionl cse The concept of center of mss cn be pplied to two dimensionl objects s well. The determintion of the center of mss in two dimensions is done in similr mnner. If mss m is locted t point (x, y) then we define the moment of m bout the x xis to be the product my nd the moment of m bout the y xis to be the product mx. Let (x, y) be the center of mss. The procedure of finding formuls for x nd y is the sme s the one dimensionl cse. Add up the msses times their x loctions then divide by totl mss to get x. Next, dd up the msses times their y loctions then divide by totl mss to get y. Hence the two formuls: x i m i y i m i where M nd M y = My m = m i nd y = Mx m = y i m i is the moment of the system bout the x xis x i m i is the moment of the system bout the y xis. Since m M y nd my = M x, the center of mss (x, y) is the point where single prticle of mss m would hve the sme moments s the system. 7 m i
8 Exmple Point msses m i re locted t the points P i. Find the moment M x nd M y nd the center of mss of the system: We hve m i P i 4 (, 3) ( 3, 1) 4 (3, 5) M 4( 3) + (1) + 4(5) = 1 M y =4() + ( 3) + 4(3) = 14 m = = = 1.4 y = 1 1 = 1 Continuous System:One Dimensionl Cse Next we consider continuous system. Suppose tht we hve n object lying on the x xis between nd b. At point x, suppose tht the object hs mss density (mss per unit length) of δ(x). To clculte the center of mss, we divide the object into n pieces, ech of length x. On ech piece, the density is nerly constnt, so the mss of the i th piece is m i δ(x i ) x. The center of mss is then m i x i Letting n we obtin m i x i δ(x i ) x. δ(x i ) x xδ(x)dx δ(x)dx. 8
9 Exmple Find the center of mss of meter rod lying on the x xis with its left end t the origin if its density is δ(x) = 15x kg/m. The totl mss is The center of mss is M = 15x d 5x 3 = 4 kg. 15x3 dx = [ x 4 4 ] = 1.5 m. Two Dimensionl System: Continuous cse In the continuous cse, we consider thin plte tht occupies region in the plne s shown in Figure We ssume the plte hs uniform density ρ. Figure Divide the intervl [, b] into n subintervls with endpoints x i = + i x nd length b n. Let x i = x i = x i 1+x i. Then the center of mss of the rectngle R i is C i (x i, 1 f(x i)). The mss of this rectngle is m i = ρf(x i ) x. Thus, nd we define M y = lim n M y (R i ) = ρf(x i ) xx i ρf(x i )x i ρxf(x)dx. 9
10 Likewise, Thus, nd M lim n ρ 1 [f(x i)] ρxf(x)dx ρ f(x)d xf(x)dx b f(x)dx y = ρ 1 [f(x)] dx ρ f(x)d 1 ρ[f(x)] dx. 1 [f(x)] dx f(x)dx. Exmple Find the center of mss of semicirculr plte of rdius r. Figure Due to symmetry, the center of mss must lie on the y xis so tht. Now, y = r r = πr = 4r 3π 1 [ r x ] dx πr [r x x3 3 Exmple Find the center of mss of the region bounded by the line y = x nd the ] r r 1
11 prbol y = x. Figure We hve A = 6 y = (x x )d 1 6 x(x x )d 1 1 (x x 4 )d 5 Remrk 7.6. The center of mss of body need not be within the body itself; the center of mss of ring or hollow cylinder of uniform density is locted in the enclosed spce, not in the object itself. 11
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