Motion of a Panel BRIEF ABSTRACT We seek to determine the position of the center of mass of a plate whose motion is governed by a constant velocity motion applied through a complex mechanism. The mathematics involves identifying the plane as complex coordinate system and using complex numbers to locate points on plane of motion. GENERAL INFORMATION FileName: Moving Full title: Motion of a Panel Developer: Jerry Fine, Department of Mechanical Engineering Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8353. Email: fine@nextwork.rose-hulman.edu. FAX: 812-877-3198. Contact: Brian J. Winkel, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8412. Email: winkel@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 The big picture A panel (seen here as an edge, the dark shaded region) moves with rollers at A and B in the vertical and horizontal slots shown. It is driven by an actuator which moves at a constant velocity of .025 m/sec. Find the position, velocity and acceleration of the center of mass of the panel, whose position is indicated in the figure. A suggested approach Mechanical engineers often use complex numbers in describing the motion of object which moves in parallel planes such as this panel. The plane of motion is taken to be the complex plane, and is superimposed on the plane of view above with the origin placed at the point of interest where the actuator is centered. The positions of different points on the moving body are represented by complex numbers in the polar form, i q r = R e = R ( cos q + i sin q ) The velocity and acceleration are obtained by differentiation with respect to time using the customary rules for complex numbers. Recall that the angle q is a function of time. The suggested approach for this problem is shown in the following figure. KEYWORDS Kinematics, position, velocity, acceleration, angular velocity, angular acceleration, complex numbers, differentiation, real part, imaginary part. TEACHER NOTES ISSUES RELATED TO THE PROBLEM This type of problem is called a "related rates" problem by teachers of calculus. There is a fairly easy solution: differentiate xA2 + yB2 = 1.62 twice. This will give equations which may be solved for the velocity and acceleration of point B. Then using similar triangles leads to the solution for velocity and acceleration at the mass center. To take full advantage of the problem, the student should try the complex vector approach which is more powerful. It also leads more easily to the velocity and position of the mass center. This problem is used to demonstrate a use of calculus and complex numbers with which the student is not already familiar. The science of kinematics (mathematical description of motion) is an important component of mechanics. The keys to solving the problem lie in the following realizations: 1. The addition of vectors to form a polygon. In this case the polygon is a right triangle. 2. The rules of differentiation. In particular the way complex exponentials differentiate. 3. Separation of the solution into two levels. A velocity analysis followed by and acceleration analysis. 4. That a complex number can be broken into real and imaginary parts that act like components of a vector. Prerequisites The student should be familiar with complex numbers, and in particular know the relationship between the cartesian form and the polar form. The student should know how to differentiate exponentials. Time allotment - time management This problem can be done in less than an hour. The velocity solution is simpler -- a progress check should verify the student has completed it in about 15 minutes. Expectations Some students might not care for the complex vector approach. It would then be better to let them use their own method to solve the problem. It might be advantageous to compare different approaches. But if the object of the presentation is to get them to think about the complex number representation then that should be emphasized. Future payoffs The student will be better prepared to study kinematics, while gaining familiarity with the use of complex numbers, an important skill in much advanced mathematics. Extensions If the student wishes, he or she might try to analyze a more complex mechanism, such as a slider crank or a four-bar. See the references. References and Sources Mechanisms and Dynamics of Machinery, by Hamilton Mabie and Charles Reinholtz, Fourth Edition, John Wiley and Sons, New York, ISBN 0-471-80237-9 POSSIBLE SOLUTION(S) Mathematica does not assume that a symbol such as x[t] or q[t] is real. This makes it a bit inconvenient , because we want to extensively use the real part (Re) and imaginary part (Im) functions. So what we do is simply define our own versions of these functions, telling Mathematica that symbols are real. A Mathematica Trick SetAttributes[realP,Listable]; SetAttributes[imagP,Listable]; realP[x_] := Thread[Re[x],Equal] /. Re[a_] -> a /. Im[b_] -> 0 imagP[x_] := Thread[Im[x],Equal] /. Re[a_] -> a /. Im[b_] -> 0 We have created the new functions, realP = Re and imagP = Im in a way such that they may be applied to equations as well as to single expressions. NOTE -- As defined these functions will work only when there is a symbol in the expression to be processed. In other words, realP[ a + b I] will work, but realP[ 3 + 4 I] won't. You will just have to use the conventional Re and Im when they apply. Data Here is a set of data describing the configuration of interest. data = { q[t] -> ( 180 - 38 ) Degree, xA'[t] -> -.025, xA''[t] -> 0 }//N; Complex Loop Equation for Position. A "loop" is a closed path including the mechanism of interest. Here is the basic equation defining position. We add the vectors defining the position of point A with respect to the origin, point B with respect to point A, and the origin with respect to point B. This concept is at the heart of kinematics. e1 = -I yB[t] + xA[t] + 1.6 Exp[ I q[t] ]==0 I q[t] 1.6 E + xA[t] - I yB[t] == 0 Here is the basic equation defining position at the moment of interest. e2=e1 /. data -1.26082 + 0.985058 I + xA[t] - I yB[t] == 0 Now we take real and imaginary parts. e3 = realP[e2] -1.26082 + xA[t] == 0 e4 = imagP[e2] 0.985058 - yB[t] == 0 Here is the position solution. solP = Solve[{e3,e4},{xA[t],yB[t]}] {{xA[t] -> 1.26082, yB[t] -> 0.985058}} Equation for Velocity Here we differentiate the complex loop equation for position. e5 = D[e1,t] I q[t] 1.6 I E q'[t] + xA'[t] - I yB'[t] == 0 Here is the basic equation defining velocity at the moment of interest. e6 = e5 /. data -0.025 + (-0.985058 - 1.26082 I) q'[t] - I yB'[t] == 0 Now we take real and imaginary parts. e7 = realP[e6] -0.025 - 0.985058 q'[t] == 0 e8 = imagP[e6] -1.26082 q'[t] - yB'[t] == 0 Here is our velocity solution. solV = Solve[ {e7,e8},{q'[t],yB'[t]}] {{q'[t] -> -0.0253792, yB'[t] -> 0.0319985}} Comments: 1. The term q'[t], given in radians per second, is negative. That means that the panel is rotating clockwise. This is associated with the convention that we measure angles counterclockwise positive in the complex plane. This term is called the ANGULAR VELOCITY. 2. The term yB' is the velocity of point B. Note that it is positive (upward) as we would expect. Equation for Acceleration Here we differentiate the complex loop equation for velocity. e9 = D[e5,t] I q[t] 2 I q[t] -1.6 E q'[t] + 1.6 I E q''[t] + xA''[t] - I yB''[t] == 0 Here we customize the equation to describe the configuration of interest. e10 = e9 /. data /. solV[[1]] 0.000812098 - 0.00063448 I + (-0.985058 - 1.26082 I) q''[t] - I yB''[t] == 0 We next take real and imaginary parts. e11 = realP[e10] 0.000812098 - 0.985058 q''[t] == 0 e12 = imagP[e10] -0.00063448 - 1.26082 q''[t] - yB''[t] == 0 Solution for accelerations solA = Solve[{e11,e12},{q''[t],yB''[t]}] {{q''[t] -> 0.000824416, yB''[t] -> -0.00167392}} Comments 1. The quantity q''[t] is called the angular acceleration. It is measured in radians per second/second. It uses the same sign convention (counterclockwise positive) used by angular velocity. 2. Not that the point B has an acceleration downward, though it is moving upward. This is in spite of the fact that the point A is moving at a constant velocity. Kinematics is INTERESTING in that it contains a number of counter-intuitive results. Position, Velocity and Acceleration of mass center. Here is position. rG = xA[t] + 0.8 Exp[ I q[t] ] I q[t] 0.8 E + xA[t] rG1 = rG /. data /. solP[[1]] 0.630409 + 0.492529 I or in cartesian coordinates. {Re[rG1], Im[rG1]} {0.630409, 0.492529} Here is velocity vG = D[rG,t] I q[t] 0.8 I E q'[t] + xA'[t] vG1 = vG /. data /. solV[[1]] -0.0125 + 0.0159993 I or in cartesian coordinates {Re[vG1], Im[vG1]} {-0.0125, 0.0159993} Here is acceleration aG = D[vG,t] I q[t] 2 I q[t] -0.8 E q'[t] + 0.8 I E q''[t] + xA''[t] aG1 = aG /. data /. solV[[1]] /. solA[[1]] 0. - 0.000836959 I {Re[aG1], Im[aG1]} {0., -0.000836959} Comments: 1. It is interesting to note that there is no x acceleration, but there is a y acceleration. 2. Again, the center of mass is accelerating downward, while it has an upward velocity. ISSUES IN SOLUTION Probably the main issue that the student needs to grasp is the setting up of the vector loops. There are two of them. The first involves the whole panel -- starting at the origin moving to A and then to B and then back to the origin. Each leg is a complex number or vector. The second expression involves getting to the mass center by moving first to A and then along the panel to the mass center. The student must realize that this is the same as the vector directly from the origin. Another solution issue is getting the computer algebra system to take real and imaginary parts of expressions containing symbols. A partial solution to this problem is offered. Unfortunately, this solution is system dependent. Maple, for example, would handle the situation differently.