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hw9 - £50 I 16100111111 “ll“ PROBLEM 17.11 The double...

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Unformatted text preview: £50 I 16100111111 “ll“ . PROBLEM 17.11 The double pulley shown has a mass of 14 kg and a centroidal radius of gyration of 165 mm. Cylinder A and block B are attached to cords that are wrapped on the pulleys as shown. The coefficient of kinetic friction between block B and the surface is 0.25. Knowing that the system is released from rest in the position shown, determine (a) the velocity of cylinder A as it strikes the ground, (1)) the total distance that block B moves before coming to rest. 25!) mm SOLUTION Let VA = speed of block A, VB 2 speed of block B, a) = angular speed of pulley. Kinematics. VA = rAa) = 0.250a) v3 = rBa) = 0.15060 SA = rAB = 0.2506 s3 = r36 = 0.1506 (a) Cylinder A falls to ground. SA = 0.900 m s3 = M(0.900) = 0.540 m 0.250 Work ofweightA: UHZ = mAgsA = (11.5)(9.81)(0.900) = 101.534 J Normal contact force acting on block B: N = mg = (9)(9.81) = 88.29 N Friction force on block B: Ff = ,ukN = (0.25)(88.29) = 22.0725 N Work offriction force: UH2 = —FfsB = —(22.0725)(0.540) = —11.919 J Total work: UH2 =101.534 — 11.919 = 89.615 J . . 1 2 1 2 1 2 K1net1c energy: T1 = 0; T2 = —mAvA + —Ia) + —vaB 2 2 2 T2 = lmArja)2 + lmckza)2 + lmgrgw2 2 2 2 = %[(11.5)(0.250)2 + (14)(0.165)2 + (9)(0.150)2]a)2 = 0.6512502 Principle of Work and Energy. T1 + UH2 = T: 0 + 89.615 = 0.6512w2 a) = 11.7309 rad/s Velocity of cylinderA: VA = (0.250)(11.7309) vA =2.93m/s§< PROBLEM 17.11 CONTINUED (b) Block B comes to rest. For blockB and pulley C. T3 = élwz +ém3v123; T4 = 0 : 1 2 2 1 2 2 1 2 2 2 T3 = 3ka a) +EmBrBa) = 5 (14)(0.165) +(9)(0.150) ](11.7309) = 40.159 J Work of friction force: U3<)4 = —Ffsj9 = —22.07255}; Principle of Work and Energy. T3 + U3_,4 2 T4: 40.159 — 22.0725s;g = o s}, = 1.819 In Total distance for block B. d = s3 + SE: d =0.540+1.819 SOLUTION PROBLEM 17.15 A slender 6-kg rod can rotate in a vertical plane about a pivot at B. A spring of constant K = 600 N/m and an unstretched length of 225 mm is attached to the rod as shown. Knowing that the rod is released from rest in the position shown, determine its angular velocity after it has rotated through 90°. Position I . BG = 0.270 m CD = 0.5252 + 0.1802 = 0.555 m x1 = CD — 10 = 0.555 — 0.225 = 0.330 m 12 = %(600)(0.330)2 = 32.67 J h1 = BG = 0.270 m (Vg)1 = Wh1 = mgh = (6)(9.81)(—O.270) = —15.892 J Kinetic energy: 7] = 0 Position 2. CD = 0.525 — 0.180 = 0.345 m Spring: x2 = CD — [0 x2 = 0.345 — 0.225 = 0.120m 1 )2 = 5(600)(0.120)2 = 432 J Gravity: Kinetic energy: = 1—12—(6)(0.900)2 = 0.405 kg-m2 172 = (30% = 0.27002 T2 =%m1722 + gig); = %(6)(0.270a)2)2 + %(0.405)a)22 = 0.4212a;22 Conservation of Energy. T1 + VI = T2 + V2: 0 + 32.67 — 15.892 = 0.4212022 + 0 + 4.32 (02 = 5.44 002 = 5.44 rad/s) 4 SOLUTION PROBLEM 17.29 The mechanism shown is one of two identical mechanisms attached to the two sides of a ZOO-lb uniform rectangular door. Edge ABC of the door is guided by wheels of negligible mass that roll in horizontal and vertical tracks. A spring of constant k = 40 lb/ft is attached to wheel B. Knowing that the door is released from rest in the position 0 = 30° with the spring unstretched, determine the velocity of wheel A just as the door reaches the vertical position. Kinematics. Locate the instantaneous center at point 1. VA = (5 sin 6) a) Moment of inertia. T = LEOOY = 51.7601b-sZ-ft 12 32.2 Kinetic energy. T = lfl(5cosl9)2 02 + l(51.760)a)2 2 32.2 2 = (77.640cos2 0 + 25.880)a)2 Potential energy. Gravity. Datum at levelA. Vg = Wh = (200)(—55in6) - ~1000sin€ ft-lb Two springs. Ve = 2(é—ke2j = k(5$in(9 ~ Ssin30°)2 = 25k(sin6 — 0.5)2 Position 1. (9 = 30° = —500 ft~1b Position 2. 0 = 90° 0) (02 T2 = (77.640cos2 90° + 25.880)a)2 = 25.88002 ‘,~\_ PROBLE V2 = M 17.29 CONTINUED (V )2 + (Vg)2 = (25)(40)(sin90° — 0.5)2 — 500sin90° 8 —750 ft-lb Conservation of energy. T1 + Vl = T2 + V2: Kinematics. 0 — 500 = 25.880w2 — 750 a) = 3.1080 rad/s VA = (55in90°)(3.1080) VA = 15.54 ft/s m4 PROBLEM 17.62 A drum of 4—in. radius is attached to a disk of 8-in. radius. The disk and drum have a combined weight of 10 lb and a combined radius of gyration of 6 in. A cord is attached to the drum at A and pulled with a constant force 1’ of magnitude 5 lb. Knowing that the disk rolls without sliding and that its initial angular velocity is 10 rad/s clockwise, determine (a) the angular velocity of the disk after 5 s, (b) the corresponding total impulse of the friction force exerted on the disk at B. SOLUTION V = VG = rBa) ‘2 Moment of inertia. I = Wk g Kinetics. Syst Momenta1 + Syst Ext Imp1 _,2 = Syst Momenta2 x. . — W — - / moments about B. [(01 + Ever + WtrB srnfl — Pt(rB — rA) = 7002 + [7er g £0? + r§)(a)2 — ml) 2 [WrB sinfl — P(rB - rA)]t g PROBLEM 17.62 CONTINUED WrB sinfl — P(rB — rA) (02 = w1+ W(/?2 + r13) gt W=lOlb, rA =4in.=%ft, rB =8in.=§ft i=6in.=%ft P=51b t=55 ml 2 —10rad/s ,8 = 30° (10)(§)sin30° — (5)(§ — 4) ———2———2—(32.2)(5) mm) + (4) ] (1.66667)(32.2)(5) 6.9444 602 : —10+ = —lO + = 28.64 rad/s (02 = 28.6 rad/s ‘34 +/ components parallel to incline: E17] + Wtsinfl — Pt — det = 14:72 g g 1th = (Wsinfl — 10%ng _ 7,) = (Wsinfl - P)t = (105in30° — 5)(5) — = —8.001b-s det = 8.001b-s 730% /\ PROBLEM 17.68 A sphere of radius r and mass m is placed on a horizontal floor with no linear velocity but with a clockwise angular velocity wo. Denoting by ,uk the coefficient of kinetic friction between the sphere and the floor, determine (a) the time z‘1 at which the sphere will start rolling without sliding, (b) the linear and angular velocities of the sphere at time t1. SOLUTION Moment of inertia. Solid sphere. I— = gmr2 rim KN / 1w: \\ ‘i‘ 1; + -- 2 — / _ G * 2 m4}; "N "'1’” x”; . , “we!" Ft N’L Syst Momenta1 Syst Ext Imp1 —>2 Syst Momenta2 +Ty components: Ntl — th = 0 N = W = mg (l) “31.x components: Ft1 = mv2 (2) +3 moments about G: lwo — Ftlr = 7602 (3) Since F = ,ukN = ykmg, Equation (2) gives ykmgt1 = mv2 or v2 2 ,uk gt1 (4) Using the value for l in Equation (3), Emrza)0 — ,ukmgtlr = Zera)2 or (02 = (00 — i’ukgtl (5) 5 5 2 r (a) Time II at which sliding stops. From kinematics, v2 = r0) 5 2 m; #kgt1 = “00 — ‘flkgfi t1 = —_‘£ ‘ 2 7 #kg (b) Linear and angular velocities. . 2 From Equation (4), v2 = ,ukggfl v2 = —ra)0 --> { 7 #kg 7 From Equation (6), wz = 2 = Zwo 002 = 12— 0) 4 r 7 7 /"\ PROBLEM 17.81 A 1.8-kg collar A and a 0.7-kg collar B can slide without friction on a 7 frame, consisting of the horizontal rod 0E and the vertical rod CD, which is free to rotate about its vertical axis of symmetry. The two collars are connected by a cord running over a pulley that is attached to the frame at 0. At the instant shown, the velocity VA of collar A has a magnitude of 2.1 m/s and a stop prevents collar B from moving. The stop is suddenly removed and collar A moves toward E. As it reaches a distance of 0.12 m from 0, the magnitude of its velocity is observed to be 2.5 m/s. Determine at that instant the magnitude of the angular velocity of the frame and the moment of inertia of the frame and pulley system about CD. ‘ __| 1— ——1 SOLUTION Components of velocity of collar A. vi, = (VA )3 + (VA )2 (1) Constraint of rod 0E. (VA )0 = rAa) (2) Constraint of cable AB. ArA = AyB, (VA )r = v3 (3) Position 1. (NA) = 0.1m, [(vAH1 = 0, (VA)1 = 2.1m/s 2 From Equation (1), (2.1)2 = 0 + [(vA )01 [(VA)6:L = 2.1m/s A From Equation (2), (2.1) = 0.1601 (01 = 21 rad/s From Equation (3), v3 = 0 Potential energy. Take position 1 as datum. V1 = 0 (4) Angular momentum. (H0)1 = 1021 + mA [(VA)r]1(rA)1: (H0)1 = [(21) + (1.8)(2.1)(0.1) (H0)1 = 211 + 0.378 (5) . . 1 2 1 2 1 2 Kinetic energy. 71 = —Ia)1 + —mAvA + —vaB: 2 2 2 1 2 l 2 T1 = E[(21) + E(1.8)(2.1) T1 = 220.51 + 3.969 (6) Position 2. (rA)2 2 0.12m, (VA)2 = 2.5 m/s 0) = wz From Equation (2), [(VA )012 = 0.12602 . 2 2 2 2 2 2 From Equation (1), [(VA),:|2 = (VA)2 — [(vA)0]2 = (2.5) — (0.12) (02 : 6.25 — 0.014460% L A A‘ PROBLEM 17.81 CONTINUED From Equation (3), v; = 6.25 — 0.0144022 Change in radial position. ArA = (rA)2 — (0)1 = 0.02 m From Equation (3), AyB : 0.02 m Potential energy. V2 = mBg(AyB) = (0.7)(9.81)(0.02) V2 = 0.137341 (7) Angular momentum. (H0)2 = 1602 + mA [(VAMZ (rA)2: (H0)2 = [(02 + (1.8)(0.12a)2)(0.12) (H0)2 = (1 + 0.02592)a)2 (8) Kinetic energy. T2 = %1w§ + émAvj + évag: T2 2%103 + %(1.8)(2.5)2 + %(0.7)(6.25 — 0.0144022) 12 = (0.51 — 0.00504)a)22 + 7.8125 (9) Conservation of angular momentum. (HO)1 = (H0)2: 211 + 0.378 = (I + 0.02592)a)2 Solving for (02, (02 = % = % (10) + . Conservation of energy. T1 + VI = T2 + V2: 220.51 + 3.969 = (0.51 — 0.00504)a)22 + 7.8125 + 0.13734 2 220.51 — (0.51 — 0.00504)% — 3.98084 = 0 220.51132 — 0.51N2 + 0.00504N2 — 3.98084D2 = 0 220.51(12 + 0.051841 + 0.0006718464) — 051(44112 + 15.8761 + 0.142884) + 0.00504(44112 + 15.8761 + 0.142884) — (3.98084)(12 + 0.051841 + 0.0006718464) = 0 013 +1.7345212 — 0.049651671 — 0.001954378 = 0 Solving the quadratic equation for I, _ 0.04965167 i 0.126590 1 = 0.050804 and —0.022179 3.46904 PROBLEM 17.81 CONTINUED Reject the negative root. _ (21)(0.050804) + 0.378 From Equation (10), Q2 — 0 050804 + 0 02592 a) = 18.83 rad/s { I = 0.0508 kg-m2 4 PROBLEM 17.98 The slender rod AB of length L forms an angle fl with the vertical axis as it strikes the frictionless surface shown with a vertical velocity V1 and no angular velocity. Assuming that the impact is perfectly elastic, derive an expression for the angular velocity of the rod immediately after the impact. SOLUTION Moment of inertia. Perfectly elastic impact. Kinetics. M Syst Momenta1 Syst Ext Imp1 _,2 Syst Momentaz _+> horizontal components: 0 + 0 = m17x Kinematics. v0 2 VA + vGM [Vy l] = [v1 l l+[(vA)x w» ]+[3w‘§ ,6] Velocity components T : )moments about A: mvl 55in fl + 0 = —mVy gsin fl + fa) mvl 55infl = m[§a)sinfl — vljésinfl + émLza) (émL2 + imL2 sin2 ,ij = mvlL sinfl 12' “—81% £4 l+3s1n flL SOLUTION Moment of inertia. 3 Wall}? Syst Momenta1 Sphere A: +T components: +T linear components: 3:. linear components: PROBLEM 17.109 The 9-kg rigid body BD consists of two identical 60-mm-radius spheres and the rod which connects them and has a centroidal radius of gyration of 250 mm. The body is at rest on a horizontal frictionless surface when it is struck by the 3-kg sphere A which has a radius of 60 mm and is moving as shown with a velocity VI of magnitude 4 m/s. Assuming a perfectly plastic impact, determine immediately after the impact (a) the angular velocity of body BD, (b) the velocity of point G. ll + Syst Ext Imsz = 0+0=mA(vA) (VA) =0 y y Let (VA )x be named vA and note that Q = -— = —. Then, 4;) moments about B: Kinematics. a): 0+0=mA(vA)y+mVy Vy=0 mAvI +0 2 mA(vA)x + mix 17x = Inmihl —(vA)x] 3 1 m 9 3 __ 1 Vx=-3‘(V1‘VA) 0 + 0 = I_a) — (bsin60°)mi7x mb sin 60° _ b sin 60° —————v — (vI m]? X‘ 31:2 W) VB 2 VG +vB/G = [Vx M ]+[ba) TI 30°] - 2 (v3)x = l(v1 — VA) + bcos30°$ 3 (V1 — VA) Impact condition. For perfectly plastic impact, (a) (b) PROBLEM 17.109 CONTINUED 1 3b2 (v3)x = §[l + Ejbfi — VA) _ 1 H M _ 3 (4)(0.250)2 (v3)x — VA = e(v1— 0) 0.69333(v1— VA) — VA = ev1 (v1— VA) = 0.69333(v] — VA) 1.69333vA = (0.69333 — e)v1 e = 0 1.69333vA = 0.69333v1 VA = 0.40945v1 v1 — VA = 0.59055v1 = (0.59055)(4) = 2.3622 m/s (0.300)sin 60° w = (3)(0.250)2 (2.3622) 1 _ =— 2.3622 v 3( > <| co = 3.27 rad/s )4 = 0.787 m/s —~ 4 ...
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