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object-oriented programs: impedance & batons 872v0v0LXXrmaaXrmbbmgFigure 4.6 Left: The baton before it is thrown. “X” marks the CM. Center: The initial conditionsfor the baton as it is thrown. Right: The baton spinning in the air under the action of gravity.If we ignore air resistance, the only force acing on the baton is gravity, and it actsat the CM of the baton (Figure 4.6 right). Figure 4.7 shows a plot of the trajectory[x(t),y(t)] of the CM:(x cm ,y cm )=(V 0x t, V 0y t − 1 )2 gt2 , (v x,cm ,v y,cm )=(V 0x ,V 0y − gt) ,where the horizontal and vertical components of the initial velocity areV 0x = V 0 cos θ, V 0y = V 0 sin θ.Even though ω = constant, it is a constant about the CM, which itself travels alonga parabolic trajectory. Consequently, the motion of the baton’s ends may appearcomplicated to an observer on the ground (Figure 4.7 right). To describe the motionof the ends, we label one end of the baton a and the other end b (Figure 4.6 left).Then, for an angular orientation φ of the baton,φ(t)=ωt + φ 0 = ωt, (4.22)where we have taken the initial φ = φ 0 =0. Relative to the CM, the ends of thebaton are described by the polar coordinates( )( )L L(r a ,φ a )=2 ,φ(t) , (r b ,φ b )=2 ,φ(t)+π . (4.23)The ends of the baton are also described by the Cartesian coordinates(x ′ a,y a)= ′ L 2 [cos ωt, sin ωt] , (x′ b,y b)= ′ L [cos(ωt + π), sin(ωt + π)] .2−101COPYRIGHT 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 87