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88 chapter 4qFigure 4.7 Left: The trajectory (x(t), y(t)) followed by the baton’s CM. Right: The appletJParabola.java showing the entire baton as its CM follows a parabola.The baton’s ends, as seen by a stationary observer, have the vector sum of theposition of the CM plus the position relative to the CM:(x a ,y a )=[V 0x t + L 2 cos(ωt), V 0yt − 1 2 gt2 + L ]2 sin(ωt) , (4.24)(x b ,y b )=[V 0x t + L 2 cos(ωt + π), V 0yt − 1 2 gt2 + L ]2 sin(ωt + π) .If L a and L b are the distances of m a and m b from CM, thenL a =m bm a + m b, L b = m am a + m b, ⇒ m a L a = m b L b . (4.25)The moment of inertia of the masses (ignoring the bar connecting them) isI masses = m a L 2 a + m b L 2 b. (4.26)If the bar connecting the masses is uniform with mass m and length L, then it hasa moment of inertia about its CM ofI bar = 1 12 mL2 . (4.27)Because the CM of the bar is at the same location as the CM of the masses, the totalmoment of inertia for the system is just the sum of the two:I tot = I masses + I bar . (4.28)−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 88