A volume and center of mass BRIEF ABSTRACT Boring a small cylindrical hole into another solid cylinder to determine a set center of mass and volume is the interest here. GENERAL INFORMATION FileName: HOLEBORE Full title: We determine the volume of a circular bore in a hollow cylinder. Last Revision Date: 27 May 1996. Developer: Brian J. Winkel, Department of Mathematical Sciences, United States Military Academy, West Point NY 10996 USA. Phone: 914-938-3200. Email: ab3646@usma2.usma.edu. FAX: 914-938-2409. Contact: Brian J. Winkel, Department of Mathematical Sciences, United States Military Academy, West Point NY 10996 USA. Phone: 914-938-3200. Email: ab3646@usma2.usma.edu. FAX: 914-938-2409. Aaron D. Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8151. Email: Aaron.Klebanoff@Rose-Hulman.Edu. FAX: 812-877-3198. Support: The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849: Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering and the Arvin Foundation of Columbus IN. STATEMENT OF PROBLEM We have a 16 cm long piston case whose outer shell is metal bounded by an outer circle of radius R = 10 cm and an inner circle of radius r = 8 cm (r < R). We wish to bore a circular hole of radius s cm (s < r < R) through the cylinder's inner and outer shell. The hole will be a circular shaft. And we require that the central axis of the hole be perpendicular to and go through the central axis of the piston. (1) Find the volume of the bored out material if s = .5; s = 1; s = 2; s = 8. (2) How big must s be so that we bore out a hole of size 100 cm^3? (3) Construct a circular bore hole with its center going through the central axis as above so that we change the z coordinate of the center of mass from z = 8 to some value around z = 8.3 cm? I.e., vary the height and radius of the bore hole until you get a z-coordinate for the center of mass which is quite near z = 8.3 cm. KEYWORDS Visualization, volume, center of mass, function of two variables, double integration, parametric equations, TEACHER NOTES ISSUES RELATED TO THE PROBLEM Prerequisites Functions of two variables, double integrals, cylinder description, volume. Time allotment - time management This problem can be set up in 30 minutes by students who have had some three dimensional geometry and who can describe cylinders in three space. However, the numerical work to obtain some of these results may take some computation time. Expectations Visualization is the problem here as we attempt to "See" what is really going on in the statement. Sketching is the order of the day and even model building might be appropriate. We would expect the students to be able to visualize and describe mathematically the regions of interest. Future payoffs Students will be able to compute integrals of visualized regions in three space. Extensions For question (3) ask the following question, "In this case how does this hole change the x and y coordinates of the center of mass?" Suppose we move the bore "off center." What will be the change in material bored out? Can we get material bored out as a function of distance of center of bore hole from the central axis of the piston case? And now suppose we make circular bore of radius s, with the central axis of the hole being at an angle of q to and going through the central axis of the central axis of the piston. WE have not attempted this and presume it could be VERY tough. References and Sources POSSIBLE SOLUTION(S) Part (1) We enter the dimensions of the piston. r = 8; R = 10; We consider the piston placed with its central axis on the z-axis and its inner surface to be the cylinder x^2 + y^2 = 8^2 and the outer surface to be the cylinder x^2 + y^2 = 10^2 - both with z values running from z = 0 to z = 16 cm. The bore hole will then be given by the cylinder x^2 + z^2 = s^2. Actually this last circle bore is computed on the assumption that the vertical piston goes through the xy-plane, say from z= -8 to z = 8. But we keep the original assumption, z = 0 to z = 16 cm, for later purposes and note that these early computaitons are unaffected. We present a double integral over the xz-region bounded by the circle x^2 + z^2 = s^2 from y = Sqrt[r^2 - x^2] up to y = Sqrt[R^2 - x^2]. We do this in two parts - first the inner integral which we will formally integrate. inner[x_,s_] = Expand[Integrate[ Sqrt[R^2 - x^2] - Sqrt[r^2 - x^2], {z,-Sqrt[s^2-x^2],Sqrt[s^2-x^2]}]] 2 2 2 2 -2 Sqrt[r - x ] Sqrt[s - x ] + 2 2 2 2 2 Sqrt[R - x ] Sqrt[s - x ] And then the outer integral which we shall have to do numerically. vol[s_] := NIntegrate[inner[x,s],{x,-s,s}] Then our desired volume for s = 0.5 cm is offered by vol(.5). vol[.5] 1.57141 We also find the other bore hole volumes. {vol[1],vol[2],vol[8]} {6.29305, 25.2929, 467.873} Part (2) Indeed we plot the volume of the bore hole for values of s = 0 to s = 8 cm. This plot takes quite a bit of time to plot - a few minutes. Times are on a Pentium 90MHz machine. Timing[Plot[vol[s],{s,0,8}]] {51.024 Second, -Graphics-} Timing[Plot[vol[s],{s,3.75,4}]] {32.516 Second, -Graphics-} And estimating from this plot we see that if s = 3.95 then volume of the bored out material is about 100 cm^3. Part (3) We enter the dimensions of the piston. r = 8; R = 10; We consider the piston placed with its central axis on the z-axis and its inner surface to be the cylinder x^2 + y^2 = 8^2 and the outer surface to be the cylinder x^2 + y^2 = 10^2 - both with z values running from z = 0 to z = 16 cm. The bore hole will then be given by the cylinder x^2 +( z - a)^2 = s^2. We present a double integral over the xz-region bounded by the circle x^2 +( z - a)^2 = s^2 from y = Sqrt[r^2 - x^2] up to y = Sqrt[R^2 - x^2]. We do this in two parts - first the inner integral which will formally integrate. inner[x_,s_,a_] = Expand[Integrate[ Sqrt[R^2 - x^2] - Sqrt[r^2 - x^2], {z,-Sqrt[s^2-x^2] + a,Sqrt[s^2-x^2] + a}]] 2 2 2 -2 Sqrt[64 - x ] Sqrt[s - x ] + 2 2 2 2 Sqrt[100 - x ] Sqrt[s - x ] And then the outer integral which we shall have to do numerically. vol[s_,a_] := NIntegrate[inner[x,s,a],{x,-s,s}] Then our desired volume for s = 0.5 cm at z = a = 3 cm is offered by vol(.5, 3). vol[.5,3] 1.57141 We note this is the same as in Part (1). This is not unexpected as the shift value of a does not affect the obtained volume. We now compute the z coordinate of the center of mass by adding the elements form the shell of the piston and subtracting the contribution from the bore hole - both to the moments in the numerator and to the mass in the denominator of the expression for the z coordinate of the center of mass, i.e. zcm(a, s). zcm[a_,s_] := (Integrate[ z (Pi R^2 - Pi r^2),{z,0, 16}]- a vol[s,a])/ (Integrate[ (Pi R^2 - Pi r^2),{z,0, 16}] - vol[s,a])//N We check result for a hole of radius "0" drilled with "center" at z = 4 cm and then at z = 6 cm where the z coordinate of the center of mass should "fall" at the midpoint value 16/2 = 8. zcm[4,0] 8. zcm[6,0] 8. We note thqat if we bore a hole, say of radius 0.5 cm, centered at z = 8 cm we expect the center of mass to remain at z = 8 cm as well. zcm[8,.5] 8. Suppose we put a s = 1 cm radius bore hole with center at z = 6 cm. Then the center of mass should move up and indeed it does move to z = 8.00698 cm. zcm[6,1] 8.00698 Suppose we put a s = 2 cm radius bore hole with center at z = 6 cm. Then the center of mass should move up even more and indeed it does move to z = 8.02835 cm > 8.00698 cm. zcm[6,2] 8.02835 We attempt to get the z coordinate of the center of mass up to 8.3 cm. zcm[2,2] 8.08505 zcm[3,3] 8.16374 zcm[4,4] 8.24207 zcm[4,3] 8.13099 zcm[5,5] 8.29919 zcm[6,6] 8.30855 zcm[6,4] 8.12103 zcm[7,7] 8.23146 zcm[6,2] 8.02835 zcm[8,8] 8. We now attempt a plot of the z coordinate of the center of mass function as a function of the two variables a and s. This may (and does!) take a long time - perhaps 10 minutes!! Timing[Plot3D[zcm[a,s],{a,2,14},{s,0,2}]] {550.02 Second, -SurfaceGraphics-} ISSUES IN SOLUTION We do not get a closed form solution for this problem and hence we have to estimate several results. This can be disconcerting to students, but it is the real world.