Classical Mechanics with Maxima by Todd Keene Timberlake, J. Wilson Mixon

By Todd Keene Timberlake, J. Wilson Mixon

This e-book courses undergraduate scholars within the use of Maxima―a machine algebra system―in fixing difficulties in classical mechanics. It capabilities good as a complement to a standard classical mechanics textbook. by way of difficulties which are too tough to unravel by way of hand, desktop algebra platforms that may practice symbolic mathematical manipulations are a beneficial device. Maxima is especially appealing in that it truly is open-source, multiple-platform software program that scholars can obtain and set up for free. classes realized and functions built utilizing Maxima are simply transferred to different, proprietary software program.

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The code below generates a numerical solution to the ODEs in Eq. 23 using the appropriate parameters and initial conditions. 1 s and find the solution from t D 0 to 3 s. 8$ vx0:float(v0*cos(theta))$ vy0:float(v0*sin(theta))$ 2 Some of the more sophisticated numerical ODE tools in Maxima, such as rkf45, use a variable step size. See Sect. 3. 1] )$ The list of output from rk has been stored in an object named data. This object is, in fact, a list that consists of embedded lists. To get an idea of what the output looks like, look at one of the elements (the fifth) of this list.

This object is, in fact, a list that consists of embedded lists. To get an idea of what the output looks like, look at one of the elements (the fifth) of this list. (%i) data[5]; (%o) Œ0:4; 2:693; 2:0795; 4:6921; 1:8126 As noted, this element is itself a list. The first element of this list is the value of the independent variable, t D 0:4 s. 1 s, and so on). The next four values in this list are the values for x, y, vx , and vy , respectively, at t D 0:4 s. 1 s to t D 3 s. Therefore, data should have a total of 31 elements, as we can verify using Maxima’s length command.

K. W. 1007/978-1-4939-3207-8_2 25 26 2 Newtonian Mechanics Fig. 2) Also, the sum of the torques about any point on the rod must be zero. For convenience we choose our origin at the point where the rod touches the wall, as this will eliminate both Fwx and Fwy from our torque equation (since they act at the origin and therefore do not exert any torque about that point). We can treat the weight of the rod as though it acts at the center of the rod, 1 m from the origin. The weight of the sign acts at the far end of the rod, 2 m from our chosen origin.

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