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86 chapter 4new features without “breaking” the old ones and in which you may be modifyingcode that you did not write.4.8 Trajectory of a Thrown Baton (Problem)We wish to describe the trajectory of a baton that spins as it travels through theair. On the left in Figure 4.6 the baton is shown as as two identical spheres joinedby a massless bar. Each sphere has mass m and radius r, with the centers of thespheres separated by a distance L. The baton is thrown with the initial velocity(Figure 4.6 center) corresponding to a rotation about the center of the lowersphere.Problem: Write an OOP program that computes the position and velocity of thebaton as a function of time. The program should1. plot the position of each end of the baton as a function of time;2. plot the translational kinetic energy, the rotational kinetic energy, and thepotential energy of the baton, all as functions of time;3. use several classes as building blocks so that you may change one buildingblock without affecting the rest of the program;4. (optional) then be extended to solve for the motion of a baton with anadditional lead weight at its center.4.8.1 Combined Translation and Rotation (Theory)Classical dynamics describes the motion of the baton as the motion of its centerof mass (CM) (marked with an “X” in Figure 4.6), plus a rotation about the CM.Because the translational and rotational motions are independent, each may bedetermined separately, and because we ignore air resistance, the angular velocityω about the CM is constant.The baton is thrown with an initial velocity (Figure 4.6 center). The simplest wayto view this is as a translation of the entire baton with a velocity V 0 and a rotationof angular velocity ω about the CM (Figure 4.6 right). To determine ω, we note thatthe tangential velocity due to rotation isv t = 1 ωL. (4.20)2For the direction of rotation as indicated in Figure 4.6, this tangential velocity isadded to the CM velocity at the top of the baton and is subtracted from the CMvelocity at the bottom. Because the total velocity equals 0 at the bottom and 2V 0 atthe top, we are able to solve for ω:12 ωL − V 0 =0 ⇒ V 0 = 1 2 ωL, ⇒ ω = 2V 0L . (4.21)−101COPYRIGHT 2008, PRINCET O N UNIVE R S I T Y P R E S SEVALUATION COPY ONLY. NOT FOR USE IN COURSES.ALLpup_06.04 — 2008/2/15 — Page 86