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problem3 - PROBLEM 13.180 A 300-g collar A is released from...

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Unformatted text preview: PROBLEM 13.180 A 300-g collar A is released from rest, slides down a frictionless rod, and strikes a 900—g collar B which is at rest and supported by a spring of constant 500 N/m. Knowing that the coefficient of restitution between the two collars is 0.9, determine (a) the maximum distance collar A moves up the rod after impact, (17) the maximum distance collar B moves down the rod after impact. SOLUTION After impact Velocity of A just before impact, v0 v0 = ,/2g = 2(9.81m/52)(1.2 m)sin30° = (9.81)(1.2)(0.5) = 3.431m/s Conservation of momentum t4“. mAvo = vaB — mAvA: 0.3v0 = 0.9vB — 0.3vA (1) Restitution (vA + v3) = e(v0 + 0) = 0.9v0 Substituting for v3 from (2) in ( 1) 0.3v0 = 0.9(0.9v0 — vA) — 0.3vA, 1.2vA = 0.51v0 vA = 1.4582 m/s, v3 = 1.6297 m/s (a) A moves up the distance d where; %MAvfi = MAgdsin30°; %(1.4582 m/s)2 = (9.81m/s2)d(o.5) A), 61,, = 0.21675 m = 217 mm 4 “3 ID", ,‘f (D (b) Static deflection = x0, B moves down All)? @ / I Afi’x‘, Conservation of energy (1) to (2) dc . " “35" : Position (1) — spring deflected, x0 TI+V1=T2+V2; Tl=ém3v123$ T2=0 PROBLEM 13.180 CONTINUED VI = Ve + Vg = ékxé + mBgstin30° , , x + 1 V2 = Ve + Vg = [00 dBlocdx = 314d; + 2de0 + x3) 1 2 ékxg + mgdg Sin30° + EvaB = %k(d§ + Zde0 + x3) + 0 + 0 m; = vaé; 500d; = 0-9(1-6297)2,d3 = 0-0691 m dB = 69.1mm4 éll) 111111 T__ SOLUTION PROBLEM 13.185 A 9.13-kg sphere A of radius 90 mm moving with a velocity of magnitude v0 = 2 m/s strikes a 830-g sphere B of radius 40 mm which is hanging from an inextensible cord and is initially at rest. Sphere B swings to a maximum height h after the impact. Determine the range of values of h for values of the coefficient of restitution 6 between zero and one. Angle of impulse force from geometry of A and B 8 30mm somm‘:Zf 6 8 6 = cos—(fl) = 22.62° :zomm Momentum consideration Ball A: Ball B: Restitution M41539 + = vama- 0*“ C mmrfl’x TRAt FAt EC) -l- O :. 0—4 mBVizsz 23h (1) my -0?» Approach Separation v’ 0056 — v’ c036 + ' Sing 6: B (A)x (VA)y VA =V0=2m/s VA c059 e _ v33 — (v’A)x + (12:4)y tan0 _ v2; — (v;)x + (v'A)y(%) VA 2 PROBLEM 13.185 CONTINUED A: mAvA sin19 = mA(v:4)x(sint9) + mA(v'A)y(cosé?) 10 = 5(v;,)x +12(v;,)y (2) . __ I r A +B. mAvA — mA(vA)x + vaB 2 = (v'A)x + % (3) From the Equation for e I I 7 5 e = 0; VB — (VAX + (ml-13) = 0 (4) I I I 5 e=l; vB—(vA)x+(vA)y[Ti-]=2 (5) Simultaneous solution of Equations (2), (3) and (4) for e = 0 and Equations (2), (3) and (5) for e = 1 yeilds e = 0: (v'A)x = 1.8357 m/s, (my = 0.0685 m/s, v2; = 1.8072 m/s e =1: (V'A)x =1.6714m/s, (v2) = 0.1369 m/s, v53 = 3.6144m/s y , 2 h = (VB) = 0.1665 m, 0.6658m 166.5 5 h S 666mm‘ PROBLEM 15.6 When the power to an electric motor is turned on the motor reaches its rated speed of 2400 rpm in 4 s, and when the power is turned off the motor coasts to rest in 40 5. Assuming uniformly accelerated motion, determine the number of revolutions that the motor executes (a) in reaching its rated speed, (b) in coasting to rest. 2400 2 ml = 2400 rpm 2 (——6())(-fl = 807r rad/s, (00 = 0, ll: 45 (a) col = 010 + at = at, a = % 2 gig— = 207: rad/s2 .91: wot + lafi = 0 + l(207:)(42) =160mad = £5 = 80rev 2 2 27: 61 = 80 rev 4 (b) 601 = 807: rad/s, wz = 0, 1‘2 — t1 = 40 s (02 =wl+a(t2—tl), a: wz—w‘ 20—80” =—27Irad/s2 t2 — r1 40 92 — .91 = mtg _ q) + 50502 _ t1)2 = (807:)(40) + %(—27r)(40)2 = 1600” radians = 1600” = 800 rev 192 — 61 = 800 rev 4 2n PROBLEM 15.13 The bent rod ABCDE rotates about a line joining points A and E with a constant angular velocity of 12 rad/s. Knowing that the rotatiOn is clockwise as viewed from E, determine the velocity of comer C . 'l SOLUTION rA/E =—l6i+16j+8kin. [AE =\/162+162+32 =24in_ Angular velocity. (0 12 co=——r =— —l6i+l6'+8k [AE A/E 24( J ) co = —(8 rad/s)i + (8 rad/s)j + (4 rad/s)k r05 = —(16 in.)i + (6 in.)j Velocity ofC. i j k vC =coer/E = —8 8 4 = —24i—64j+80k —16 6 0 v6 = —(24.0 in./s)i — (64.0 in./s) j + (80.0in./s)k4 Angular Acceleration. (1 = 0 Acceleration of C. i j k aC=aer/E+o)xvc=0+ —8 8 4 —24 —64 80 896i + 544j + 704k ac = (896 in./sz)i + (544 in./sz) j + (704in./s2)k 4 PROBLEM 15.18 The belt shown moves over two pulleys without slipping. Pulley A starts from rest with a clockwise angular acceleration defined by the relation 05 = 120 — 0.002602, Where a is expressed in rad/s2 and a) is expressed in rad/s. Determine, after one-half revolution of pulley A, (a) the magnitude of the acceleration of point B on the belt, (b) the acceleration of point P on pulley C. SOLUTION ~1—rev = nradians, rA =160mm = 0.160 m, rC 2100mm = 0.100m 2 a=120_0,002w2 zwi‘lfl, Jim—2.2M?=dg d6 120 — 0.00%) 60000 — 0) Integrating and applying initial condition a) = 0 at 6 = 0 and noting that :9 = 7r radians at the final state, 1:; 500 male) 60000 2 = —2501n(60000 - wz : Ede = 7r " a) _ 2 —250[h1(60000 — 02) — 11160000] 2 —250 111%— = 7r — 02 = (BAH/250 60000 002 = 60000[1— (gr/25°] = 749.26 radz/sz a) = 27.373 rad/s) a = 120 — 0.002 of = 120 — (0.002)(749.26) = 118.50 rad/s) (a) Tangential velocity and acceleration of point B on the belt. v3 = v, = 73,60 = (0.160)(27.373) = 4.3797 m/s a3 = aA = rAa = (0.160)(118.50) = 18.96 m/s2 a3 = 18.96 m/s2 4 (b) Acceleration of point P on pulley C. p = 0.100 m C VP = v3 = 4.3797 m/s i _ 4.37972 0.100 = 191.82 m/szw PROBLEM 15.18 CONTINUED (aP), = a3 = 18.96 m/s2 1 up = (191.82)2 + (18.96)2 = 192.8 m/s2 18.96 , = 5.64° 191.82 fl tanfl= aP = 192.8 m/s2 7 5.64°4 ...
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