Solution Part 1 Because the helt does not slip, every point on the belt that is in contact with a pulley has the same velocity as the adjacent point on the pulley. Therefore, the speed of any point on the belt is v= Raog = Rycon @ so that Ra 0.10 wn = ~404 = 04 = 0.504 Rp 0.20 Differentiating with respect to time, we obtain for the angular acceleration of pulley B Oy = 0.504 = 0.5(—2.51) = —1.250 radis? Because ay = derg/dt, we have dig = ag dt, oF On= | wgdt= | =1.25¢dt = -0.625° + C, The initial condition, @p = 20 rad/s when 1 = 0, yields C; = 20 rad/s. Hence, the angular velocity of pulley B is wg = —0.6251? + 20 rad/s Answer Part 2 We let 0g be the angular position of a line in B measured from a fixed re line. Recalling that op = dOg/dt, we integrate dg = @pdt to obtain Oy = | opdi = fooese + 20)d = —0.20837° + 207 + C2Letting @; = 0 when t = 0, we have C; = 0, which gives 95 = —0.208315 + 207 rad The pulley comes to rest when wg = —0.6251? + 20 = 0, which yields 5.657 s. The corresponding angular position of the line in B is | ,-5.657 » = —0-2083(5.657)> + 20(5.657) = 112.0 rad Therefore, the angular displacement of pulley Bas it ts to a stop is AOp = On| Op\,_) = 112.00 = 112.0 rad Answer 1=5.657 5 Because the direction of rotation does not change, the total angle turned through by pulley B during the deceleration is also 112.0 rad. Part 3 Substituting Rg = 0.2 mand wp = —0.6251? + 20 rad/s into Eq. (a), the speed of point C (which is the same for all points on the belt) is ve = 0.2(—0.625¢? + 20) = -0.1257 +4 m/s (b) Answer Because the path of point C on the belt is a straight line, the acceleration of Cis ac = ie = —0.25¢ m/s? Answer We could obtain the same result by observing that a¢ is equal to the tangential component of acceleration of a point on the rim of pulley B (pulley A could also be used). Thus ac = Rgap = 0.2(-1. 25¢ m/s” Observe that the expression for vc in Eq. (b) is valid for the entire deceleration period (0 < ¢ < 5.657 s), whereas the answer for ag applies only for those times when Cis not in contact with either pulley. Points on the belt follow circular paths when they are in contact with a pulley.SOLUTION BA=-(00 mm)j+(240 mmk —-BA= 260 mm Ba _ 100+ 240 Pa 0 BA 260 1 ° 26 + coo} + 240K) An = 24f i 0005 240) Point: Foy, = (120 mami = (80 mu) ~ (120 mek Ve = OXtn ijk =| 0 -10 24 120-80 -120 = G12 mus)i + 2880 munis))-+ 01200 mm/s) y= G.l2 meh 42.88 m}j +(.200 mk By OX ty, + OX(WAEg,) =OF OXY, a, =0xv,, i jk =| 0 -0 24 3120 2880 1200. = 181120 mis? + 74880 mvs + 31200 mens?) +012 ms) € Sce Problem 15.14 for Aj and @ fy, Ome HOH 260) = (10 rad easmh!, ane =(65 als 24 rad/s 0005 +240 aah 240m a= “25 radis® + (GO rad/s" Point E: oj. = 12D muni ~ (80)j— (120k ijk -|o 0 24 120 80-120 120 mums + (2880 mins) + (1200 maw ve = (G12 mis) + (2.88 m/s\f + 1.200 mis) 8p POXFp, HOX(@XE | “OK, HOV i oiok boi ok ap-| 0-25 60 [+] 0-10 24 120-80 -120| | 3120 2880 1200 == (7800 ses! (7200 mods )-+ 3000 mals (81120 mans} + (74880 mes?) + 31200 mrs? (73820 munis" + (82080 munis" + 64200 mms) =-C733 n+ 82.1 Wve) + G42 0/6 <-¥ bc0890 +906 sine " Ylocose-y): y 74 2ybbsind tybasine tybd cso ae 020, J(b-[ese])so +04 0 + Cork)a (Mi) af 214 (0:28). 9 909 a (4p) 6) 8=%2, b/sinp =L/sinf, , >i (218 sin) « 21.248 yso YF caso Leow: 0.14 Cos +0. 1co521.24°= ee 2284m 670 0.14c0s8 ~0, 2264) sororaz294(aid) (-2? sink +0 0.09520) = -0.08555, O918 wn/s* (etow) y SOLUTION (VECTOR ANALYSIS) Kinematic Diagram. The kinematic diagrams for both AB and BC are shown in Fig. 16-31b. Here ac is vertical since C moves along a straight-line path. Acceleration Equation. Expressing each of the position vectors in Cartesian vector form tp = {0.25 sin 45% + 0.25 cos 45°j} ft = {-O.177i + O.177j} ft rca = {0.75 sin 13.6% + 0.75 cos 13.6%} ft = {0.177 + 0.7295} ft Crankshaft AB (rotation about a fixed axis): = Gap X ty ~ wary = (-20k) * (0.1771 + 0.177) — (10)?(—0.1771 + 0.177j) = {21.21i — 14.145} ft/s? Connecting Rod BC (general plane motion): Using the result for aj and noting that a¢ is in the vertical direction, we haveConnecting Rod BC (general plane motion): Using the result for ay and noting that ac isin the vertical direction, we have ag = ay + age X top ~ wcteyp 14.14) + (eck) (0.177 + 0.7299) - (243)*(0.177% + 0.7299) aj = U.2i~ 14.14] + 0.17 Taped ~ 01.729agci ~ 1.04% ~ 4.305 0= 2017 - 07290a¢ ¢ = O.1TIaye ~ 1845 Solving yields aac = 27.7 rad/s? ae = -135 fy/8 Ans. Body BC extended ®%p The distances to B and C from O, found from the triangle OBC, are gio = 50/tan 30° = 86.60 mm reio = 50/ sin 30° = 100 mmConsidering the motion of AB (its instant center is at A), we find that Va =Feya@an = 60(2) = 120 mm/s, directed as shown in Fig. (b). Analyzing the motion of BC (its instant center is at O) yields vp _ 120 conc = “8 = > = 1.386 rad/s 7 Tao 86. pc = 1.386 rad/s © Answer and Ye = re1o@nc = 100(1.386) = 138.6 mm/s be Vc = 138.6 mm/s Answer Because C is also a point on bar CD (ts instant center is at D), the angular velocity of bar CD is @cp = 1.733 rad/s O Answer