Instructor Hentzel Office Phone: 515-294-8141 E-mail: hentzel@iastate.edu Math Department Fax: 515-294-5454 http://www.math.iastate.edu/hentzel/class.166.08 Textbook: Calculus by Varberg, Purcell, Rigdon, ninth edition. Monday, January 28 5.6 Moments p313: 4,16,22,30 Main Idea: How things balance. Key Words: Center of Mass, Centroid, Mx, My Goal: Find the centroid and the center of mass and know the difference. -------------------------------------------------------- Previous assignment: p306:(2),(4),(12),(16) Page 306 Problem 2 A force of 6 pounds is required to keep a spring stretched 1/2 foot beyond its normal length. Find the value of the spring constant and the work done in stretching the spring 1/2 foot beyond its normal length. How much work is done in stretching the spring 2 feet? (2,24) /| / | / | / | / | (1/2,6) / | /| | / | | _____/__|_____|__________ 1/2 2 k = 6/(1/2) = 12. x=1/2 | 2 | x=1/2 Work = INT 12 x dx = 12 | x /2 | = 3/2 ft lbs x=0 | | x=0 x=2 | 2 | x=2 Work = INT 12 x dx = 12 | x /2 | = 24 ft lbs x=0 | | x=0 ------------------------------------------------------- Page 306 Problem 4 It requires 0.05 joule (newton-meter) of work to stretch a spring from a length of 8 centimeters to 9 centimeters and another 0.10 joule to stretch it from 9 centimeters to 10 centimeters. Evaluate the spring constant and find the natural length of the spring. newtons /| | / | | /* | (0.095, 10.0 ) | / | | /| | | / | | | /* | | (0.085, 5.0 ) | / | | | /| | | y-5.00 = 5/0.01(x-0.085) | / |.05|.10| ______|_______/__|___|___| y = 500 x - 42.50 + 5.00 c .08 .09 .10 meters y = 500 x - 37.50 The area's are in newton*meters. y=0 when x = 0.075 1/2 ( K(0.08-c) + K(0.09-c) )(0.01) = 0.05 1/2 ( K(0.09-c) + K(0.10-c) )(0.01) = 0.10 ( K(0.08-c) + K(0.09-c) )(0.01) = 0.10 ( K(0.09-c) + K(0.10-c) )(0.01) = 0.20 ( K(0.08-c) + K(0.09-c) ) = 10 ( K(0.09-c) + K(0.10-c) ) = 20 0.17 K - 2 c K = 10 0.19 K - 2 c K = 20 0.02 K = 10 K = 500. 85 -1000 c = 10 -1000 c = -75 c = 0.075 The spring constant is 500 newtons/meter. The natural length of the spring is 0.075 meter. ----------------------------------------------------------------- Now we take the problem to calculus land and work it so nobody actually knows what is really going on. i.e. c is in meters and is the natural length of the spring. x=0.09 INT k (x-c) dx = 0.05 x=0.08 x=0.10 INT k (x-c) dx = 0.10 x=0.09 | 2 | x=0.09 | k (x-c) /2 | = 0.05 | | x=0.08 | 2 | x=0.10 | k (x-c) /2 | = 0.10 | | x=0.09 2 2 1/2 k( (0.09-c) -(0.08-c) ) = 0.05 2 2 1/2 k( (0.10-c) -(0.09-c) ) = 0.10 k( c^2 - 0.18 c + 0.0081 - c^2 + 0.16 c - 0.0064) = 0.10 k( c^2 - 0.20 c + 0.0100 - c^2 + 0.18 c - 0.0081) = 0.20 k( -0.02 c + 0.0017) = 0.10 k( -0.02 c + 0.0019) = 0.20 Subtracting gives k( -0.0002) = -0.10 0.10 1000 k = ------- = ------ = 500 newtons 0.0002 2 Substituting k = 500 and solving for c gives: 500( -0.02 c + 0.0017) = 0.10 -10 c + 0.85 = 0.10 -10 c = -0.75 c = 0.075 meters --------------------------------------------------------- Page 306 Problem 12 _____________________ / ____________________ 5| | | | /-| |----------. / | | /' / | | / / | | / ' <------6------>/ / ` ' / . ./ . ./ 2 2 ` .___. ' x + y = 9 Pump water to 5 feet above the tank. Tank is 10 feet long. y= 0 INT 62.4 (5-y) 10 2 Sqrt[9-y^2] dy = 55340 y=-3 y= 0 2 1248 INT (5-y) Sqrt[9-y ] dy y=-3 y= 0 2 y= 0 2 1248 INT 5 Sqrt[9-y ] dy - 1248 INT y Sqrt[9-y ] dy y=-3 y=-3 2 3/2 | (9-y ) | y= 0 6240 Pi 9/4 - 1248 | ---------- | | (3/2)(-2) | y=-3 14040 Pi + 416*27 = 11232 + 14040 Pi ft lbs = 55340. ft-lbs ------------------------------------------------------- Integrate[624/10 (5-y) 10 2 Sqrt[9-y^2],{y,-3,0}] = 11232 + 14040 Pi ------------------------------------------------------ Page 306 Problem 16 A cylinder and piston whose cross-sectional area is one square inch, contains 16 cubic inches of gas under a pressure of 40 pounds per square inch. If the pressure and volume of the gas are related adiabatically (i.e,. without loss of heat) by 1.4 the law p v = c, how much work is done by the piston in compressing the gas to 2 cubic inches. 1.4 c = 40 16 1.4 _ _ 1.4 40 16 | 16 | p = ---------- = 40 | ---- | 1.4 |_ v _| v This will be the work the gas does expanding from 2 to 16. This will be the work necessary to compress it. Since the cross section is 1 sq inch, the volume is the same as x. _ _ 1.4 x= 16 | 16 | work = INT 40 | --- | dx x= 2 |_ x _| _ _ x= 16 | -0.4 | 1.4 | x | work = 40 16 | ----- | |_-0.4 _| x=2 1.4 _ _ 40 16 | -0.4 -0.4 | work = -------- | 16 - 2 | -0.4 |_ _| 5.6 -1.6 -0.4 work = -100 2 ( 2 - 2 ) 4.0 5.2 work = -100 ( 2 - 2 ) = 2075.83 inch*pounds This is the work done by the gas, so the work done by the piston is also 2075.83 inch-pounds. toobig = 2464.72 toosmall = 1769.55 error = -695.167 ---------------------------------------------------------------------- Get["font.math"]; p[x_] := 40 16^(1.4)/x^(1.4); a = Plot[p[x],{x,2,16},PlotStyle->{Thickness[0.005]}]; b = Plot[0,{x,0,16}]; c = ParametricPlot[ {0,y},{y,0,p[2]}]; A = Table[ {i,p[i]},{i,2,16}]; d = ListPlot[A,PlotStyle->{RGBColor[1,0,0],PointSize[0.01]}]; e = ParametricPlot[{2,y},{y,0,p[2]}]; B = { { 2,p[ 2]}, { 3,p[ 2]}, { 3,p[ 3]}, { 4,p[ 3]}, { 4,p[ 4]}, { 5,p[ 4]}, { 5,p[ 5]}, { 6,p[ 5]}, { 6,p[ 6]}, { 7,p[ 6]}, { 7,p[ 7]}, { 8,p[ 7]}, { 8,p[ 8]}, { 9,p[ 8]}, { 9,p[ 9]}, {10,p[ 9]}, {10,p[10]}, {11,p[10]}, {11,p[11]}, {12,p[11]}, {12,p[12]}, {13,p[12]}, {13,p[13]}, {14,p[13]}, {14,p[14]}, {15,p[14]}, {15,p[15]}, {16,p[15]}, {16,p[16]}}; f = ListPlot[B,PlotJoined->True,PlotStyle->{RGBColor[1,0,0],Thickness[0.005]}]; X = { { 2,p[ 3]}, { 3,p[ 3]}, { 3,p[ 4]}, { 4,p[ 4]}, { 4,p[ 5]}, { 5,p[ 5]}, { 5,p[ 6]}, { 6,p[ 6]}, { 6,p[ 7]}, { 7,p[ 7]}, { 7,p[ 8]}, { 8,p[ 8]}, { 8,p[ 9]}, { 9,p[ 9]}, { 9,p[10]}, {10,p[10]}, {10,p[11]}, {11,p[11]}, {11,p[12]}, {12,p[12]}, {12,p[13]}, {13,p[13]}, {13,p[14]}, {14,p[14]}, {14,p[15]}, {15,p[15]}, {15,p[16]}, {16,p[16]} }; g = ListPlot[X,PlotJoined->True,PlotStyle->{RGBColor[0,1,0],Thickness[0.005]}]; h = Show[a,b,c,d,e,f,g,PlotLabel->"Page 306 Problem 16, The pressure ", PlotRange->All,Axes->True]; Display["p16.ps",h]; toobig = Sum[ p[i],{i,2,15}]; toosmall= Sum[ p[i],{i,3,16}]; error = (p[16]-p[2]) 1; --------------------------------------------------------------- New Material Page 309 Example 1. Masses of 4, 2, 6, and 7 kilograms are located at points 0, 1, 2, and 4 respectively along the x-axis. Find the center of Mass. _ 4*0 + 2*1 + 6*2 + 7*4 42 x = ----------------------- = ---- 19 19 --------------------------------------------------------------- Page 310 Example 2. The density d(x) of a wire at a point x-centimeters 2 from one end is given by d(x) = 3x grams per centimeter. Find the center of mass of the piece between x=0 and x=10. x=10 3 | 4 |x=10 INT 3x dx | 3x /4 | 7500 ___ x=0 | |x=0 x = ------------ = ------------ = ------- = 7.5 cm x=10 2 | 3 |x=10 INT 3x dx | 3x /3 | 1000 x=0 | |x=0 --------------------------------------------------------------------- Page 310 Example 3. Five particles, having masses 1, 4, 2, 3 and 2 units are located at points (6,-1), (2,3), (-4,2), (-7,4) and (2,-2) respectively. Find the center of mass. 1{6,-1}+ 4 {2,3}+2 {-4,2}+ 3{-7,4}+ 2{2,-2} {-11, 23} --------------------------------------------- = ----------- 1+4+2+3+2 12 Page 311 Example 4. Find the centroid of the region bounded by the 3 curve y = x and y = Sqrt[x]. 3/2 4 x=1 3 | x x |x=1 Area = INT Sqrt[x]-x dx = | ----- - --- | = (2/3 - 1/4) = 5/12 x=0 | 3/2 4 |x=0 5/2 5 x=1 3 | x x |x=1 My = INT x(Sqrt[x]-x ) dx = | ---- - --- | = (2/5-1/5) = 1/5 x=0 | 5/2 5 |x=0 x=1 3 3 Mx = INT 1/2(Sqrt[x]+x )(Sqrt[x]-x ) dx x=0 x=1 6 = INT 1/2 ( x - x ) dx x=0 2 7 | x x | x=1 = 1/2 | --- - --- | = 1/2( 1/2 - 1/7) = 5/28 | 2 7 | x=0 _ x = My/A = (1/5)/(5/12) = 12/25 _ y = Mx/A = (5/28)/(5/12) = 12/28 = 3/7 --------------------------------------------------- Get["font.math"]; P1 = Plot[Sqrt[x],{x,0,1}]; P2 = Plot[x^3,{x,0,1}]; P3 = ListPlot[{{12/25,3/7}},PlotStyle->{RGBColor[1,0,0],PointSize[0.01]}]; P4 = Show[P1,P2,P3,PlotLabel->"P311 Ex 4 ", PlotRange->All,AspectRatio->Automatic]; Display["ex3.ps",P4]; Page 314 Problem 21. Divide the indicated region into rectangular pieces and assume that the moments Mx and My of the whole region can be found by adding the corresponding moments of the pieces. (-3,1)/14 | (-2, 1) | ( 2, 1) 0-------------------+-------------------0 | | | | | | | | | | | | ------|-------------------+---------0---------0---------- | | | ( 2, 0) | | | | | | | | | | | | 0-------------------|---------0 (-2,-1) | ( 1,-1) | | 8{0,0}-1{3/2,-1/2} (-3/2,1/2) ------------------- = ---------- = (-3/14,1/14) 7 7 ----------------------------------------------------- P1 = ListPlot[{{-2,1},{2,1},{2,0},{1,0},{1,-1},{-2,-1},{-2,1}}, PlotJoined->True,PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; P2 = ListPlot[{{-3/14,1/14}},PlotStyle->{PointSize[0.03]}]; P3 = Show[P1,P2, PlotLabel->"P314 p21",AspectRatio->Automatic]; Display["p21.ps",P3]; -------------------------------------------------------------------- Page 314 Problem 30. Find the centroid of a triangle. Consider the triangle T of Figure 21 _ (a) Show that y = h/3 (and thus that the centroid of a triangle is at the intersection of the medians). (b) Find the volume of the solid obtained when T is revolved around y = k, (Pappus's Theorem). y | | k|___________________ | | h|________ | /\ | / \ | / \ | / \ | /========\ dy | / T \ | / \ |/______________\_____________________x b Area = 1/2 bh y=h y=h 2 | 2 3 | y=h M = INT y ( (h-y)/h) b = b/h INT hy-y = b/h | hy /2 - y /3 | x y=0 y=0 | | y=0 3 3 2 = b/h( h /2 - h /3 ) = b h /6 2 Mx b h /6 ----- = --------- = h/3 Area b h /2