Solucionario Raymond Serway  Capitulo 10

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2000 by Harcourt College Publishers. All rights reserved.Chapter 10 Solutions10.1 ( a ) = it = 12.0 rad/ s3.00 s = 4.00 rad/ s2 (b) = it + 12 t2 = 12 (4.00 rad/ s2)(3.00 s) 2 = 18.0 rad 10.2 ( a ) = 2 r ad365 days 1 day24 h 1 h3600 s = 1.99 107 rad/ s (b) = 2 r ad27.3 days 1 day24 h 1 h3600 s = 2.65 106 rad/ s *10.3 i = 2000 rad/ s = 80.0 rad/ s2( a ) = i + t = [2000 (80.0)(10.0)] = 1200 rad/ s (b) 0 = i + tt = i = 200080.0 = 25.0 s 10.4 ( a ) Let h and m be the angular speeds of the hour hand and minute hand, so thath = 2 rad12 h = 6 rad/ h and m = 2 rad/ hThen if h and m are the angular positions of the hour hand and minute hand with respectto the 12 o'clock position, we haveh = ht and m = mtFor the two hands to coincide, we need m = h + 2 n, where n is a positive integer.Therefore, we may write mt ht = 2 n, ortn = 2nm h = 2n2 6 = 12n11 h 2 Chapter 10 Solutions 2000 by Harcourt College Publishers. All rights reserved.Construct the following table:n t n (h) time (h:min:s)0 0.00 12:00:001 1.09 1:05:272 2.18 2:10:553 3.27 3:16:224 4.36 4:21:495 5.45 5:27:166 6.55 6:32:447 7.64 7:38:118 8.73 8:43:389 9.82 9:49:0510 10.91 10:54:33(b) Let s and s be the angular position and angular speed of the second hand, thens = 2 rad/ min = 120 rad/ h and s = stFor all three hands to coincide, we need s = m + 2k (k is any positive integer) at any ofthe times given above. That is, we needstn mtn = 2k, ork = s m2 tn = 1182 1211 n = (3)(4)(59)n11 to be an integer. This is possible only for n = 0 or 11. Therefore, all three hands coincideonly when straight up at 12 o' clock .10.5 i =
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_100 rev1.00 min
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_1.00 min60.0 s
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_2 r ad1.00 rev = 103 rad/ s, f = 0( a ) t = f i = 0 10/ 32.00 s = 5.24 s (b) = t =
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_f + i2 t =
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_106 rad/ s
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_106 s = 27.4 rad *10.6 i = 3600 rev/ min = 3.77 102 rad/ s = 50.0 rev = 3.14 102 rad and f = 0 2f = 2i + 20 = (3.77 102 rad/ s)2 + 2(3.14 102 rad) = 2.26 102 rad/ s2 Chapter 10 Solutions 3 2000 by Harcourt College Publishers. All rights reserved.10.7 ( a ) t = 0 = 5.00 rad t = 0 = dd t t = 0 = 10.0 + 4.00tt = 0 = 10.0 rad/ s t = 0 = dd t t = 0 = 4.00 rad/ s2 (b) t = 3.00 s = 5.00 + 30.0 + 18.0 = 53.0 rad t = 3.00 s = dd t t = 3.00s = 10.0 + 4.00tt = 3.00 s = 22.0 rad/ s t = 3.00 s = dd t t = 3.00s = 4.00 rad/ s2 *10.8 = 5.00 rev/ s = 10.0 rad/ sWe will break the motion into two stages: (1) an acceleration period and (2) a decelerationperiod.While speeding up,1 = t = 0 + 10.0 rad/ s2 (8.00 s) = 40.0 radWhile slowing down,2 = t = 10.0 rad/ s + 02 (12.0 s) = 60.0 radSo, otal = 1 + 2 = 100 rad = 50.0 rev 10.9 i = it + 12 t 2 and = i + tare two equations in two unknowns i and .i = t i = ( t)t + 12 t 2 = t 12 t237.0 rev
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_2 r ad1 rev = (98.0 rad/ s)(3.00 s) 12 (3.00 s)2232 rad = 294 rad (4.50 s2)4 Chapter 10 Solutions 2000 by Harcourt College Publishers. All rights reserved. = 61.5 rad4.50 s2 = 13.7 rad/ s2 Chapter 10 Solutions 5 2000 by Harcourt College Publishers. All rights reserved.*10.10 ( a ) = t = 1 rev1 d ay = 2 r ad86400 s = 7.27 105 rad/ s (b) t = = 1077.27 105 rad/ s
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_2 r ad360 = 2.57 104 s (428 min) *10.11 Estimate the tires radius at 0.250 m and miles driven as 10,000 per year. = sr = 1.00 104 mi0.250 m
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_1609 m1 mi = 6.44 107 radyr = 6.44 107 radyr
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_1 rev2 r ad = 1.02 107 revyr or ~107 revyr *10.12 Main Rotor:v = r = (3.80 m)
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_450 revmin
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_2 r ad1 rev
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_1 min60 s = 179 m/ s v =
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_179 ms
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_vsound343 m/ s = 0.522vsound Tail Rotor:v = r = (0.510 m)
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_4138 revmin
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_2 r ad1 rev
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_1 min60 s = 221 m/ s v =
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_221 ms
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_vsound343 m/ s = 0.644vsound 10.13 ( a ) v = r; = vr = 45.0 m/ s250 m = 0.180 rad/ s (b) ar = v2r = (45.0 m/ s)2250 m = 8.10 m/ s2 toward the center of track 10.14 v = 36.0 kmh
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_1 h3600 s
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_103 m1 km = 10.0 m/ s = vr = 10.0 m/ s0.250 m = 40.0 rad/ s 6 Chapter 10 Solutions 2000 by Harcourt College Publishers. All rights reserved.10.15 Given r = 1.00 m, = 4.00 rad/ s2, i = 0, and i = 57.3 = 1.00 rad( a ) = i + t = 0 + tAt t = 2.00 s, = (4.00 rad / s2)(2.00 s) = 8.00 rad/ s (b) v = r = (1.00 m)(8.00 rad/ s) = 8.00 m/ s ar = r 2 = (1.00 m)(8.00 rad/ s)2 = 64.0 m/ s2 at = r = (1.00 m)(4.00 rad/ s2) = 4.00 m/ s2 The magnitude of the total acceleration is:a = a2r + a2t = (64.0 m/ s2)2 + (4.00 m/ s2) 2 = 64.1 m/ s2The direction of the total acceleration vector makes an angle with respect to the radiusto point P: = tan1
,
_ atar = tan1
,
_4.0064.0 = 3.58(c) = i + it + 12 t2 = (1.00 rad) + 12 (4.00 rad/ s2)(2.00 s) 2 = 9.00 rad 10.16 ( a ) = vr = 25.0 m/ s1.00 m = 25.0 rad/ s (b) 2f = 2i + 2() = 2f 2i2() = (25.0 rad/ s)2 02[(1.25 rev)(2 rad/ rev)] = 39.8 rad/ s2 (c) t = = 25.0 rad/ s39.8 rad/ s2 = 0.628 s 10.17 ( a ) s = v t = (11.0 m/ s)(9.00 s) = 99.0 m = sr = 99.0 m0.290 m = 341 rad = 54.3 rev (b) = vr = 22.0 m/ s0.290 m = 75.9 rad/ s = 12.1 rev/ s Chapter 10 Solutions 7 2000 by Harcourt College Publishers. All rights reserved.10.18 KA + UA = Kp + Up 6.00 9.80 5.00 = 12 (6.00) v2p + 6.00 9.80 2.00v2p = 58.8 m2/ s2Radial acceleration at P,ar = v2pR = 29.4 m/ s2 Tangential acceleration at P,at = g = 9.80 m/ s2 10.19 ( a ) = 2f = 2 radrev 120060.0 rev/ s = 126 rad/ s (b) v = r = (126 rad/ s)(3.00 102 m) = 3.77 m/ s (c) ar = 2r = (126)2(8.00 102) = 1260 m/ s2 = 1.26 km/ s2 (d ) s = r = t r = (126 rad/ s)(2.00 s)(8.00 102 m) = 20.1 m 10.20 Just before it starts to skid,Fr = marf = mv2r = sn = smgs = v2rg = 2rg = ( 2 2i)rg = 2rg = 2atg s = 2(1.70 m/ s2)(/ 2)9.80 m/ s2 = 0.545 *10.21 ( a ) x = r cos = (3.00 m) cos (9.00 rad) = (3.00 m) cos 516 = 2.73 my = r sin = (3.00 m) sin (9.00 rad) = (3.00 m) sin 516 = 1.24 mr = xi + yj = (2.73i + 1.24j) m (b) 516 360 = 156. This is between 90.0 and 180, so the object is in the second quadrant .The vector r makes an angle of 156 with the positive xaxis or 24.3 with the negativexaxis.garP8 Chapter 10 Solutions 2000 by Harcourt College Publishers. All rights reserved.(d ) The direction of motion (i.e., the direction of thevelocity vector is at 156 + 90.0 = 246 from thepositive x axis. The direction of the accelerationvector is at 156 + 180 = 336 from the positive x axis.(c) v = [(4.50 m/ s) cos 246]i + [(4.50 m/ s) sin 246]j= (1.85i 4.10j) m/ s (e) a = v2r = (4.50 m/ s)23.00 m = 6.75 m/ s2 directed toward the center or at 336a = (6.75 m/ s2)(i cos 336 + j sin 336) = (6.15i 2.78j) m/ s2 ( f ) F = ma = (4.00 kg)[(6.15i 2.78j) m/ s2] = (24.6i 11.1j) N *10.22 When completely rewound, the tape is a hollow cylinder with a difference between the innerand outer radii of ~1 cm. Let N represent the number of revolutions through which the drivingspindle turns in 30 minutes (and hence the number of layers of tape on the spool). We candetermine N from:N = 2 = (t)2 = (1 rad/ s)(30 min)(60 s/ min)2 rad/ rev = 286 revThen, thickness ~ 1 cmN ~102 cm 10.23 m1 = 4.00 kg, r1 = y1 = 3.00 m; m2 = 2.00 kg, r2 = y2 = 2.00 m;m3 = 3.00 kg, r3 = y3 = 4.00 m; = 2.00 rad/ s about the xaxis( a ) Ix = m1r 21 + m2r 22 + m1r 22 = (4.00)(3.00)2 + (2.00)(2.00)2 + (3.00)(4.00)2Ix = 92.0 kg m2 KR = 12 Ix2 = 12 (92.0)(2.00) 2 = 184 J (b) v1 = r1 = (3.00)(2.00) = 6.00 m/ s v2 = r2 = (2.00)(2.00) = 4.00 m/ s v3 = r3 = (4.00)(2.00) = 8.00 m/ s 156mxvaxOy = 3.00 m 4.00 kg3.00 kg2.00 kgyy = 2.00 my = 4.00 mChapter 10 Solutions 9 2000 by Harcourt College Publishers. All rights reserved.K1 = 12 m1v21 = 12 (4.00)(6.00) 2 = 72.0 JK2 = 12 m2v22 = 12 (2.00)(4.00) 2 = 16.0 JK3 = 12 m3v23 = 12 (3.00)(8.00) 2 = 96.0 JK = K1 + K2 + K3 = 72.0 + 16.0 + 96.0 = 184 J = 12 Ix210.24 v = 38.0 m/ s = 125 rad/ sRATIO = 12 I212 mv2 = 12 ,_ 25 mr2 212 mv2 RATIO = 25 (3.80 102)2 (125)2(38.0)2 = 1160 10.25 ( a ) I = j mj r2j In this case,r1 = r2 = r3 = r4r = (3.00 m)2 + (2.00 m) 2 = 13.0 mI = [ 13.0 m]2 [3.00 + 2.00 + 2.00 + 4.00]kg= 143 kg m2 (b) KR = 12 I2 = 12 (143 kg m2)(6.00 rad/ s) 2= 2.57 103 J 10.26 The moment of inertia of a thin rod about an axis through one end is I = 13 ML2. The totalrotational kinetic energy is given asKR = 12 Ih 2h + 12 Im 2m , wit hIh = mhL2h3 = (60.0 kg)(2.70 m)23 = 146 kg m2, andIm = mmL2m3 = (100 kg)(4.50 m)23 = 675 kg m2x (m)y (m)01241 32.00 kg 3.00 kg2.00 kg 4.00 kg10 Chapter 10 Solutions 2000 by Harcourt College Publishers. All rights reserved.In addition, h = (2 rad)(12 h) 1 h3600 s = 1.45 104 rad/ s, whilem = (2 rad)(1 h ) 1 h3600 s = 1.75 103 rad/ s. Therefore,KR = 12 (146)(1.45 104) 2 + 12 (675)(1.75 103) 2 = 1.04 103 J 10.27 I = Mx2 + m(L x)2d Idx = 2Mx 2m(L x) = 0 (for an extremum)x = mLM + m d2Idx2 = 2m + 2M; therefore I is minimum whenthe axis of rotation passes through x= mLM + m which is also the center of mass of the system. The moment of inertia about an axis passingthrough x isICM = M
]11mLM + m 2 + m
]111 mM + m 2 L2= MmM + m L2 = L2where = MmM + m 10.28 We assume the rods are thin, with radius much lessthan L. Call the junction of the rods the origin ofcoordinates, and the axis of rotation the zaxis.For the rod along the yaxis, I = 13 mL2 from the table.For the rod parallel to the zaxis, the parallelaxistheorem givesI = 12 mr2 + m
,
_L2 2 14 mL2xM mLLx xaxis of rotationChapter 10 Solutions 11 2000 by Harcourt College Publishers. All rights reserved.In the rod along the xaxis, the bit of material between x and x + dx has mass (m/ L)dx and is atdistance r = x2 + (L/ 2)2 from the axis of rotation. The total rotational inertia is:Itotal = 13 mL2 + 14 mL2 + L/ 2L/ 2(x2 + L2/ 4)(m/ L)dx = 712 mL2 +
,
_mL x33L/ 2L/ 2 + mL4 xL/ 2L/ 2 = 712 mL2 + mL212 + mL24 = 11 mL212 *10.29 Treat the tire as consisting of three parts. The two sidewalls are each treated as a hollowcylinder of inner radius 16.5 cm, outer radius 30.5 cm, and height 0.635 cm. The tread region istreated as a hollow cylinder of inner radius 30.5 cm, outer radius 33.0 cm, and height 20.0 cm.Use I = 12 m(R21 + R22 ) for the moment of inertia of a hollow cylinder.Sid ewall:m = [(0.305 m)2 (0.165 m)2] (6.35 103 m)(1.10 103 kg/ m3) = 1.44 kgIside = 12 (1.44 kg) [(0.165 m)2 + (0.305 m)2] = 8.68 102 kg m2Tread:m = [(0.330 m)2 (0.305 m)2] (0.200 m)(1.10 103 kg/ m3) = 11.0 kgItread = 12 (11.0 kg) [(0.330 m)2 + (0.305 m)2] = 1.11 kg m2Entire Tire:Itotal = 2Iside + Itread = 2(8.68 102 kg m2) + 1.11 kg m2 = 1.28 kg m2 10.30 ( a ) I = ICM + MD2 = 12 MR2 + MR2 = 32 MR2 (b) I = ICM + MD2 = 25 MR2 + MR2 = 75 MR2 12 Chapter 10 Solutions 2000 by Harcourt College Publishers. All rights reserved.*10.31 Model your body as a cylinder of mass 60.0 kg and circumference 75.0 cm. Then its radius is0.750 m2 = 0.120 mand its moment of inertia is12 MR2 = 12 (60.0 kg)(0.120 m) 2 = 0.432 kg m2 ~ 100 kg m2 = 1 kg m2 10.32 = 0 = mg(3r) Tr2T Mg sin 45.0 = 0T = Mg sin 45.02 = 1500 kg(g) sin 45.02 = (530)(9.80) Nm = T3g = 530g3g = 177 kg r3r1500 kgm = 45Chapter 10 Solutions 13 2000 by Harcourt College Publishers. All rights reserved.10.33 = (0.100 m)(12.0 N) (0.250 m)(9.00 N) (0.250 m)(10.0N )= 3.55 N m The thirtydegree angle is unnecessary information.Goal Solution G: By simply examining the magnitudes of the forces and their respective lever arms, it appearsthat the wheel will rotate clockwise, and the net torque appears to be about 5 Nm.O: To find the net torque, we simply add the individual torques, remembering to apply theconvention that a torque producing clockwise rotation is negative and a counterclockwisetorque is positive.A: = Fd = (12.0 N)(0.100 m) (10.0 N)(0.250 m) (9.00 N)(0.250 m) = 3.55 N mThe minus sign means perpendicularly into the plane of the paper, or it means clockwise.L: The resulting torque has a reasonable magnitude and produces clockwise rotation as expected.Note that the 30 angle was not required for the solution since each force acted perpendicularto its lever arm. The 10N force is to the right, but its torque is negative that is, clockwise,just like the torque of the downward 9N force.10.34 Resolve the 100 N force into componentsperpendicular to and parallel to the rod, asFpar = (100 N) cos 57.0 = 54.5 NandFperp = (100 N) sin 57.0 = 83.9 NTorque of Fpar = 0 since its line of action passesthrough the pivot point.Torque of Fperp is = (83.9 N)(2.00 m) = 168 N m (clockwise) 10.0 N30.0aOb12.0 N9.00 N100 N2.00 m20.020.037.014 Chapter 10 Solutions 2000 by Harcourt College Publishers. All rights reserved.*10.35 The normal force exerted by the ground on each wheel isn = mg4 = (1500 kg)(9.80 m/ s2)4 = 3680 NThe torque of friction can be as large asmax = fmaxr = (sn)r = (0.800)(3680 N)(0.300 m) = 882 N m The torque of the axle on the wheel can be equally as large as the light wheel starts to turnwithout slipping.*10.36 We calculated the maximum torque that can be applied without skidding in Problem 35 to be882 N m. This same torque is to be applied by the frictional force, f, between the brake padand the rotor for this wheel. Since the wheel is slipping against the brake pad, we use thecoefficient of kinetic friction to calculate the normal force. = fr = (kn)r, so n = kr = 882 N m(0.500)(0.220 m) = 8.02 103 N = 8.02 kN 10.37 m = 0.750 kg F = 0.800 N( a ) = rF = (30.0 m)(0.800 N) = 24.0 N m (b) = I = rFmr2 = 24.0(0.750)(30.0)2 = 0.0356 rad/ s2 (c) aT = r = (0.0356)(30.0) = 1.07 m/ s2 *10.38 = 36.0 N m = I f = i + t10.0 rad/ s = 0 + (6.00 s) = 10.006.00 rad/ s2 = 1.67 rad/ s2( a ) I = = 36.0 N m1.67 rad/ s2 = 21.6 kg m2 (b) f = i + t0 = 10.0 + (60.0) = 0.167 rad/ s2 = I = (21.6 kg m)(0.167 rad/ s2) = 3.60 N m Chapter 10 Solutions 15 2000 by Harcourt College Publishers. All rights reserved.(c) Number of revolutions = i + it + 12 t2During first 6.00 s = 12 (1.67)(6.00) 2 = 30.1 radDuring next 60.0 s = 10.0(60.0) 12 (0.167)(60.0) 2 = 299 radtotal = (329 rad) rev2 r ad = 52.4 rev 10.39 For m1: Fy = may +n m1g = 0n = m1g = 19.6 Nfk = kn = 7.06 NFx = max 7.06 N + T1 = (2.00 kg)a (1)For the pulley = IT1 R + T2 R = 12 MR2
,
_aR T1 + T2 = 12 (10.0 kg) a T1 + T2 = (5 .00 kg)a (2)T1 T1T2n2fkfkT2 nm2gMgnm1gm2m1T2T1I, Rt16 Chapter 10 Solutions 2000 by Harcourt College Publishers. All rights reserved.For m2: +n m2g cos = 0n = 6.00 kg(9.80 m/ s2) cos 30.0 = 50.9 Nfk = kn = 18.3 N18.3 N T2 + m2 g sin = m2a18.3 N T2 + 29.4 N = (6.00 kg)a (3)( a ) Add equations (1) (2) and (3):7.06 N 18.3 N + 29.4 N = (13.0 kg)aa = 4.01 N13.0 kg = 0.309 m/ s2 (b) T1 = 2.00 kg (0.309 m/ s2) + 7.06 N = 7.67 N T2 = 7.67 N + 5.00 kg(0.309 m/ s2) = 9.22 N 10.40 I = 12 mR2 = 12 (100 kg)(0.500 m) 2 = 12.5 kg m2i = 50.0 rev/ min = 5.24 rad/ s = f it = 0 5.24 rad/ s6.00 s = 0.873 rad/ s2 = I = (12.5 kg m2)(0.873 rad/ s2) = 10.9 N mThe magnitude of the torque is given by fR = 10.9 N m, where f is the force of friction.Therefore, f = 10.9 N m0.500 m , andf = kn yields k = fn = 21.8 N70.0 N = 0.312 10.41 I = MR2 = 1.80 kg(0.320 m)2 = 0.184 kg m2 = I( a ) Fa(4.50 102 m) 120 N(0.320 m) = 0.184 kg m2(4.50 rad/ s2)Fa = (0.829 N m + 38.4 N m)4.50 102 m = 872 N (b) Fb(2.80 102 m) 38.4 N m = 0.829 N mFb = 1.40 kN 0.500 mf nn 70.0 NChapter 10 Solutions 17 2000 by Harcourt College Publishers. All rights reserved.10.42 We assume the rod is thin. For the compound objectI = 13 MrodL2 +
]1125 MballR2 + MballD2 I = 13 1.20 kg (0.240 m)2 + 25 20.0 kg(4.00 102 m)2 + 20.0 kg(0.280 m)2I = 1.60 kg m2( a ) Kf + Uf = Ki + Ui + E12 I2 + 0 = 0 + Mrodg(L/ 2) + Mballg(L + R) + 012 (1.60 kg m2) 2 = 1.20 kg(9.80 m/ s2)(0.120 m) + 20.0 kg(9.80 m/ s2)(0.280 m)12 (1.60 kg m2) 2 = 56.3 J (b) = 8.38 rad/ s (c) v = r = (0.280 m)8.38 rad/ s = 2.35 m/ s (d) v2 = v2i + 2a(y yi)v = 0 + 2(9.80 m/ s2)(0.280 m) = 2.34 m/ sThe speed it attains in swinging is greater by 2.35/ 2.34 = 1.00140 times 10.43 Choose the zero gravitational potential energy at the level where the masses pass.Kf + Ugf = Ki + Ugi + E12 m1v2 + 12 m2v2 + 12 I2 = 0 + m1gh1i + m2gh2i + 012 (15.0 + 10.0) v2 + 12
]1112 (3.00)R2
,
_vR 2 = 15.0(9.80)(1.50) + 10.0(9.80)(1.50)12 (26.5 kg) v2 = 73.5 J v = 2.36 m/ s 18 Chapter 10 Solutions 2000 by Harcourt College Publishers. All rights reserved.10.44 Choose the zero gravitational potential energy at the level where the masses pass.Kf + Ugf = Ki + Ugi + E12 m1v2 + 12 m2v2 + 12 I2 = 0 + m1gh1i + m2gh2i + 012 (m1 + m2) v2 + 12
]1112 MR2
,
_vR 2 = m1g
,
_d2 + m2g
,
_ d2 12
,
_m1 + m2 + 12 M v2 = (m1 m2)g
,
_d2 v = (m1 m2)gdm1 + m2 + 12 M 10.45 ( a ) 50.0 T = 50.09.80 aTR = I = I aR I = 12 MR2 = 0.0938 kg m250.0 T = 5.10
,
_TR2I T = 11.4 N a = 50.0 11.45.10 = 7.57 m/ s2 v = 2a(yi 0) = 9.53 m/ s 3 kgFgn T0.250 mChapter 10 Solutions 19 2000 by Harcourt College Publishers. All rights reserved.(b) Use conservation of energy:(K + U)i = (K + U)fmgh = 12 mv2 + 12 I22mgh = mv2 + I
,
_v2R2 = v2
,
_m + IR2 v = 2mghm + IR2 = 2(50.0 N)(6.00 m)5.10 kg + 0.0938(0.250)2 = 9.53 m/ s Goal Solution G: Since the rotational inertia of the reel will slow the fall of the weight, we should expect thedownward acceleration to be less than g. If the reel did not rotate, the tension in the stringwould be equal to the weight of the object; and if the reel disappeared, the tension would bezero. Therefore, T < mg for the given problem. With similar reasoning, the final speed mustbe less than if the weight were to fall freely: vf < 2gy 11 m/ sO: We can find the acceleration and tension using the rotational form of Newtons second law.The final speed can be found from the kinematics equation stated above and from conservationof energy. Freebody diagrams will greatly assist in analyzing the forces.A: ( a ) Use = I to find T and a.First find I for the reel, which we assume to be a uniform disk:I = 12 MR2 = 12 3.00 kg (0.250 m)2 = 0.0938 kg m2The forces on the reel are shown, including a normal force exerted by its axle. From thediagram, we can see that the tension is the only unbalanced force causing the reel tor ot at e.T50.0 N6.00 m20 Chapter 10 Solutions 2000 by Harcourt College Publishers. All rights reserved. = I becomesn(0) + Fg(0) + T(0.250 m) = (0.0938 kg m2)(a/ 0.250 m) (1)where we have applied at = r to the point of contact between string and reel.The falling weight has massm = Fgg = 50.0 N9.80 m/ s2 = 5.10 kgFor this mass, Fy = may becomes+T 50.0 N = (5.10 kg)(a) (2)Note that since we have defined upwards to be positive, the minus sign shows that itsacceleration is downward. We now have our two equations in the unknowns T and a forthe two linked objects. Substituting T from equation (2) into equation (1), we have:[50.0 N (5.10 kg)a](0.250 m) = 0.0938 kg m2 a0.250 m 12.5 N m (1.28 kg m)a = (0.375 kg m)a12.5 N m = a(1.65 kg m) or a = 7.57 m/ s2and T = 50.0 N 5.10 kg(7.57 m/ s2) = 11.4 NFor the motion of the weight,v2f = v2i + 2a(yf yi) = 02 + 2(7.57 m/ s2)(6.00 m)vf = 9.53 m/ s(b) The workenergy theorem can take account of multiple objects more easily than Newton'ssecond law. Like your bratty cousins, the workenergy theorem keeps growing betweenvisits. Now it reads:(K1 + K2,rot + Ug1 + Ug2)i = (K1 + K2,rot + Ug1 + Ug2)f0 + 0 + m1gy1i + 0 = 12 m1v21f + 12 I222f + 0 + 0Now note that = vr as the string unwinds from the reel. Making substitutions:50.0 N(6.00 m) = 12 (5.10 kg) v2f + 12 (0.0938 kg m2)
,
_vf0.250 m 2300 N m = 12 (5.10 kg) v2f + 12 (1.50 kg) v2f vf = 2(300 N m)6.60 kg = 9.53 m/ sL: As we should expect, both methods give the same final speed for the falling object. Theacceleration is less than g, and the tension is less than the objects weight as we predicted.Now that we understand the effect of the reels moment of inertia, this problem solution couldbe applied to solve other realworld pulley systems with masses that should not be ignored.Chapter 10 Solutions 21 2000 by Harcourt College Publishers. All rights reserved.10.46 = 12 I 2(25.0 N m)(15.0 2) = 12 (0.130 kg m2) 2 = 190 rad/ s = 30.3 rev/ s 10.47 From conservation of energy,12 I
,
_vr 2 + 12 mv2 = mghI v2r 2 = 2mgh mv2I = mr2
,
_2ghv2 1 10.48 E = 12 I2 = 12
,
_12 MR2
,
_3000 260.0 2 = 6.17 106 JP = Et = 1.00 104 J/ st = EP = 6.17 106 J1.00 104 J/ s = 617 s = 10.3 min 10.49 ( a ) Find the velocity of the CM(K + U)i = (K + U)f0 + mgR = 12 I2 = 2mgRI = 2mgR32 mR2 vCM = R 4g3R = 2 Rg3 (b) vL = 2vCM = 4 Rg3 (c) vCM = 2mgR2m = Rg PivotRg22 Chapter 10 Solutions 2000 by Harcourt College Publishers. All rights reserved.*10.50 The moment of inertia of the cylinder isI = 12 mr2 = 12 (81.6 kg)(1.50 m) 2 = 91.8 kg m2and the angular acceleration of the merrygoround is found as = I = (Fr)I = (50.0 N)(1.50 m)(91.8 kg m2) = 0.817 rad/ s2At t = 3.00 s, we find the angular velocity = i + t = 0 + (0.817 rad/ s2)(3.00 s) = 2.45 rad/ sand K = 12 I 2 = 12 (91.8 kg m2)(2.45 rad/ s) 2 = 276 J 10.51 mg l2 sin = 13 ml2 = 32 g l sin at =
,
_32 g l sin rThen
,
_32 g l r > g sin for r > 23 l About 1/ 3 the length of the chimney will have a tangential acceleration greater thang sin .*10.52 The resistive force on each ball is R = DAv2. Here v = r, where r is the radius of each ballspath. The resistive torque on each ball is = rR, so the total resistive torque on the three ballsystem is total = 3rR. The power required to maintain a constant rotation rate isP = total = 3rR. This required power may be written asP = total = 3r [DA(r)2] = (3r3DA3)With =
,
_2 r ad1 rev
,
_103 revmin
,
_1 min60.0 s =
,
_100030.0 rad/ s,P = 3(0.100 m)3(0.600)(4.00 104 m2)(1000/ 30.0 s)3or P = (0.827 m5/ s3) where is the density of the resisting medium.atgg sin Chapter 10 Solutions 23 2000 by Harcourt College Publishers. All rights reserved.( a ) In air, = 1.20 kg/ m3, andP = (0.827 m5/ s3)(1.20 kg/ m3) = 0.992 N m/ s = 0.992 W (b) In water, = 1000 kg/ m3 and P = 827 W 10.53 ( a ) I = 12 MR2 = 12 (2.00 kg)(7.00 102 m) 2 = 4.90 103 kg m2 = I = 0.6004.90 103 = 122 rad/ s2 = t t = = 1200 ,_ 260122 = 1.03 s (b) = 12 t2 = 12 (122 rad/ s)(1.03 s) 2 = 64.7 rad = 10.3 rev 10.54 For a spherical shell 23 dm r2 = 23 [(4r2dr)]r2I dI = 23 (4r2) r2(r)drI = 0R 23 (4r4)
,
_14.2 11.6 rR
,
_103 kgm3 dr=
,
_23 4(14.2 103) R55
,
_23 4(11.6 103) R56 I = 83 (103) R5
,
_14.25 11.66 M = dm= 0R4r2
,
_14.2 11.6 rR 103 dr= 4 103
,
_14.23 11.64 R3IMR2 = 83 (103) R5 ,_ 14.25 11.664 103R3R2 ,_ 14.23 11.64 = 23
,
_.9071.83 = 0.33024 Chapter 10 Solutions 2000 by Harcourt College Publishers. All rights reserved.I = 0.330MR2 Chapter 10 Solutions 25 2000 by Harcourt College Publishers. All rights reserved.10.55 ( a ) W = K = 12 I2 12 I2i = 12 I(2 2i ) where I = 12 mR2=
,
_12
,
_12 (1.00 kg)(0.500 m) 2
]11
,
_8.00 r ads2 0 = 4.00 J (b) t = 0 = ra = (8.00 rad/ s)(0.500 m)2.50 m/ s2 = 1.60 s (c) = i + it + 12 t2; i = 0; i = 0 = 12 t2 = 12
,
_2.50 m/ s20.500 m (1.60 s) 2 = 6.40 rads = r = (0.500 m)(6.40 rad) = 3.20 m < 4.00 m Yes 10.56 ( a ) I = 12 mr2 = 12 (200 kg)(0.300 m) 2 = 9.00 kg m2 (b) = (1000 rev/ min)(1 min/ 60 s)(2 rad)/ 1 rev = 105 rad/ sW = K = 12 I2 = 12 (9.00)(104.7) 2 = 49.3 kJ (c) f = 500 rev/ min 1 min/ 60 s(2 rad/ 1 rev) = 52.4 rad/ sKf = 12 I2 = 12.3 kJW = K = 12.3 kJ 49.3 kJ = 37.0 kJ *10.57 = 10.0 rad/ s2 (5.00 rad/ s3)t = d/ dt65.0d = 0t[10.0 5.00t]dt = 10.0t 2.50t2 = 65.0 rad/ s = dd t = 65.0 rad/ s (10.0 rad/ s2)t (2.50 rad/ s3)t2( a ) At t = 3.00 s, = 65.0 rad/ s (10.0 rad/ s2)(3.00 s) (2.50 rad/ s3)(9.00 s2) = 12.5 rad/ s 26 Chapter 10 Solutions 2000 by Harcourt College Publishers. All rights reserved.(b) 0d = 0t dt = ot[65.0 rad/ s (10.0 rad/ s2)t (2.50 rad/ s3)t2]dt = (65.0 rad/ s)t (5.00 rad/ s2)t2 (0.833 rad/ s3)t3At t = 3.00 s, = (65.0 rad/ s)(3.00 s) (5.00 rad/ s2)9.00 s2 (0.833 rad/ s3)27.0 s3 = 128 rad 10.58 ( a ) MK2= MR22 , K = R2 (b) MK2 = ML212 , K = L 36 (c) MK2 = 25 MR2, K = R 25 10.59 ( a ) Since only conservative forces act, E = 0, soKf + Uf = Ki + Ui12 I2 + 0 = 0 + mg
,
_L2 , where I = 13 mL2 = 3g/ L (b) = I so that in the horizontal, mg
,
_L2 = mL23 = 3g2L (c) ax = ar = r2 =
,
_L2 2 = 3g2 ay = at = r =
,
_L2 = 3g4 (d ) Using Newton's second law , we have Rx = max = 32 mg Ry mg = may or Ry = 14 mg xyLpivotChapter 10 Solutions 27 2000 by Harcourt College Publishers. All rights reserved.10.60 The first drop has a velocity leaving the wheel given by 12 mv2i = mgh1, sov1 = 2gh1 = 2(9.80 m/ s2)(0.540 m) = 3.25 m/ sThe second drop has a velocity given byv2 = 2gh2 = 2(9.80 m/ s2)(0.510 m) = 3.16 m/ sFrom = vr , we find 1 = v1r = 3.25 m/ s0.381 m = 8.53 rad/ s and 2 = v2r = 3.16 m/ s0.381 m = 8.29 rad/ sor = 22 212 = (8.29 rad/ s)2 (8.53 rad/ s)24 = 0.322 rad/ s2 10.61 At the instant it comes off the wheel, the first drop has a velocity v1, directed upward. Themagnitude of this velocity is found fromKi + Ugi = Kf + Ugf12 mv21 + 0 = 0 + mgh1 or v1 = 2gh1 and the angular velocity of the wheel at the instant the first drop leaves is1 = v1R = 2gh1R2 Similarly for the second drop: v2 = 2gh2 and 2 = v2R = 2gh2R2 The angular acceleration of the wheel is then = 22 212 = 2gh2/ R2 2gh1/ R22(2) = g(h2 h1)2R2 10.62 Work done = Fs = (5.57 N)(0.800 m) = 4.46 Jand Work = K = 12 I2f 12 I2i (The last term is zero because the top starts from rest.)Thus, 4.46 J = 12 (4.00 104 kg m2) 2f FAA28 Chapter 10 Solutions 2000 by Harcourt College Publishers. All rights reserved.and from this, f = 149 rad/ s Chapter 10 Solutions 29 2000 by Harcourt College Publishers. All rights reserved.10.63 Kf = 12 Mv2f + 12 I2f : Uf = Mghf = 0; Ki = 12 Mv2i + 12 I2i = 0Ui = (Mgh)i: f = N = Mg cos ; = vr ; h = d sin and I = 12 mr2 ( a ) E = Ef Ei or fd = Kf + Uf Ki Uifd = 12 Mv2f + 12 I2f Mgh(Mg cos )d = 12 Mv2 + (mr2/ 2)(v2/ r2)/ 2 Mgd sin 12
]11M + m2 v2 = Mgd sin (Mg cos )d orv2 = 2Mgd (sin cos )(m/ 2) + M vd =
]114gd M(m + 2M) (sin cos ) 1/ 2(b) v2 = v2i 2as, v2d = 2ada = v2d2d = 2g
,
_Mm + 2M (sin cos ) 10.64 ( a ) E = 12
,
_25 MR2 (2) E = 12 25 (5.98 1024)(6.37 106) 2
,
_286400 2 = 2.57 1029 J (b)dEd t = dd t
]1112
,
_25 MR2
,
_2T2 = 15 MR2(2)2(2T3) dTd t = 15 MR2
,
_2T 2
,
_2T dTd t = (2.57 1029 J)
,
_286400 s
,
_10 106 s3.16 107 s (86400 s/ day) dEd t = 1.63 1017 J/ day 30 Chapter 10 Solutions 2000 by Harcourt College Publishers. All rights reserved. 10.65 = tt = = (31.0/ 360) rev900 rev/ 60 s = 0.00574 sv = 0.800 m0.00574 s = 139 m/ s = 31vd 10.66 ( a ) Each spoke counts as a thin rod pivoted at one end.I = MR2 + n mR23 (b) By the parallelaxis theorem,I = MR2 + nmR23 + (M + nm)R2 = 2MR2 + 4 nmR23 *10.67 Every particle in the door could be slid straight down into a highdensity rod across its bottom,without changing the particles distance from the rotation axis of the door. Thus, a rod0.870 m long with mass 23.0 kg, pivoted about one end, has the same rotational inertia as thedoor:I = 13 ML2 = 13 (23.0 kg)(0.870 m) 2 = 5.80 kg m2 The height of the door is unnecessary data.Chapter 10 Solutions 31 2000 by Harcourt College Publishers. All rights reserved.10.68 f will oppose the torque causing the motion: = I = TR f f = TR I (1)Now find T, I and in given or known terms and substitute intoequation (1)Fy = T mg = ma then T = m(g a) (2)also y = vit + at22 a = 2yt2 (3)and = aR = 2yRt2 (4)I = 12 M
]11R2 +
,
_R22 = 58 MR2(5)Substituting (2), (3), (4) and (5) into (1) we findf = m
,
_g 2yt2 R 58 MR22y(Rt2) = R
]11m
,
_g 2yt2 54 Myt2 10.69 ( a ) While decelerating,f = I' = (20 000 kg m2)
,
_2.00 rev/ min (2 rad/ rev)(1 min/ 60 s)10.0 s f = 419 N mWhile accelerating, = I or f = I(/ t) = 419 N m + (20 000 kg m2)
,
_10.00 rev/ min (2 rad/ rev)(1 min/ 60 s)12.0 s = 2.16 103 N m (b) P = f = (419 N m)
]1110.0 revmin
,
_2 r ad1 rev
,
_1 min60 s = 439 W ( 0.6 hp) MmR/ 2R/ 2y32 Chapter 10 Solutions 2000 by Harcourt College Publishers. All rights reserved.10.70 ( a ) W = K + UW = Kf Ki + Uf Ui0 = 12 mv2 + 12 I2 mgd sin 12 kd 2 12 2 (I + mR2) = mgd sin + 12 kd 2 = 2mgd sin + kd2I + mR2 (b) = 2(0.500 kg)(9.80 m/ s2)(0.200 m)(sin 37.0) + (50.0 N/ m)(0.200 m)21.00 kg m2 + (0.500 kg)(0.300 m)2 = 1.18 + 2.001.05 = 3.04 = 1.74 rad/ s 10.71 ( a ) m2g T2 = m2aT2 = m2(g a) = (20.0 kg)(9.80 2.00)m/ s2 = 156 N T1 = m1g sin 37.0 + m1aT1 = (15.0 kg)(9.80 sin 37.0 + 2.00)m/ s2 = 118 N (b) (T2 T1)R = I = I
,
_aR I = (T2 T1)R2a = (156 118)N(0.250 m)22.00 m/ s2 = 1.17 kg m2 Goal Solution G: In earlier problems, we assumed that the tension in a string was the same on either side of apulley. Here we see that the moment of inertia changes that assumption, but we should stillexpect the tensions to be similar in magnitude (about the weight of each mass ~150 N), andT2 > T1 for the pulley to rotate clockwise as shown.If we knew the mass of the pulley, we could calculate its moment of inertia, but since we onlyknow the acceleration, it is difficult to estimate I. We at least know that I must have unitsof kgm2, and a 50cm disk probably has a mass less than 10 kg, so I is probably less than0.3 kgm2.O: For each block, we know its mass and acceleration, so we can use Newtons second law to findthe net force, and from it the tension. The difference in the two tensions causes the pulley torotate, so this net torque and the resulting angular acceleration can be used to find thepulleys moment of inertia.mRk37.015.0 kgT1m120.0 kgT22.00 m/ s2m2Chapter 10 Solutions 33 2000 by Harcourt College Publishers. All rights reserved.A: ( a ) Apply F = ma to each block to find each string tension.The forces acting on the 15kg block are its weight, the normal support force from theincline, and T1. Taking the positive x axis as directed up the incline, Fx = max yields:(m1g)x + T1 = m1(+a)(15.0 kg)(9.80 m/ s2) sin 37 + T1 = (15.0 kg)(2.00 m/ s2)T1 = 118 NSimilarly for the counterweight, we have Fy = may, or T2 m2g = m2(a)T2 (20.0 kg)(9.80 m/ s2) = (20.0 kg)(2.00 m/ s2)So, T2 = 156 N(b) Now for the pulley, = r(T2 T1) = I. We may choose to call clockwise positive. Theangular acceleration is = ar = 2.00 m/ s20.250 m = 8.00 rad/ s2 = I or (0.250 m)(156 N 118 N) = I(8.00 rad/ s2)I = 9.38 N m8.00 rad/ s2 = 1.17 kg m2L: The tensions are close to the weight of each mass and T2 > T1 as expected. However, themoment of inertia for the pulley is about 4 times greater than expected. Unless we made amistake in solving this problem, our result means that the pulley has a mass of 37.4 kg (about80 lb), which means that the pulley is probably made of a dense material, like steel. This iscertainly not a problem where the mass of the pulley can be ignored since the pulley hasmore mass than the combination of the two blocks!10.72 For the board just starting to move, = Img l2 cos =
,
_13 ml2 = 32
,
_gl cos Rmg34 Chapter 10 Solutions 2000 by Harcourt College Publishers. All rights reserved.The tangential acceleration of the end isat = l = 32 g cos Its vertical component is ay = at cos = 32 g cos2 If this is greater than g, the board will pull ahead of the ball in falling:( a )32 g cos2 g cos2 23 so cos 23 and 35.3 (b) When = 35.3 ( cos2 = 2/ 3), the cup will land underneath the releasepoint of theball ifrc = l cos = l cos2 cos = 2l3 cos (c) When l = 1.00 m, and = 35.3rc = 2(1.00 m)3 2/ 3 = 0.816 mwhich is (1.00 m 0.816 m) = 0.184 m from the moving end 10.73 At t = 0, = 3.50 rad/ s = 0e0. Thus, 0 = 3.50 rad/ sAt t = 9.30 s, = 2.00 rad/ s = 0e(9.30 s), yielding = 6.02 102 s1( a ) = dd t = d(0et)d t = 0()etAt t = 3.00 s, = (3.50 rad/ s)(6.02 102 s1)e3.00(6.02 102) = 0.176 rad/ s2 (b) = 0t0et dt = 0 [et 1] = 0 [1 et]At t = 2.50 s, = 3.50 rad/ s(6.02 102)1/ s [1 e(6.02 102)(2.50)] = 8.12 rad = 1.29 rev Chapter 10 Solutions 35 2000 by Harcourt College Publishers. All rights reserved.(c) As t , 0 (1 e) = 3.50 rad/ s6.02 102 s1 = 58.2 rad = 9.26 rev 36 Chapter 10 Solutions 2000 by Harcourt College Publishers. All rights reserved.10.74 Consider the total weight of each hand to act at the center of gravity (midpoint) of thathand. Then the total torque (taking CCW as positive) of these hands about the center of theclock is given by = mhg
,
_Lh2 sin h mmg
,
_Lm2 sin m = g2 (mhLh sin h + mmLm sin m) If we take t = 0 at 12 o'clock, then the angular positions of the hands at time t areh = ht, where h = 6 rad/ h and m = mt, where m = 2 rad/ hTherefore, =
,
_4.90 ms2 [(60.0 kg)(2.70 m) sin (t/ 6) + (100 kg)(4.50 m) sin 2t]or = 794 N m[sin(t/ 6) + 2.78 sin 2t], where t is in hours.( a ) ( i ) At 3:00, t = 3.00 h, so = 794 N m [sin(/ 2) + 2.78 sin 6] = 794 N m ( i i ) At 5:15, t = 5 h + 1560 h = 5.25 h, and substitution gives: = 2510 N m ( i i i ) At 6:00, = 0 N m (i v ) At 8:20, = 1160 N m (v) At 9:45, = 2940 N m (b) The total torque is zero at those times whensin(t/ 6) + 2.78 sin 2t = 0We proceed numerically, to find 0, 0.5152955, ..., corresponding to the times12:00:00 12:30:55 12:58:19 1:32:31 1:57:012:33:25 2:56:29 3:33:22 3:56:55 4:32:244:58:14 5:30:52 6:00:00 6:29:08 7:01:467:27:36 8:03:05 8:26:38 9:03:31 9:26:3510:02:59 10:27:29 11:01:41 11:29:05