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### Transcript of Solucionario Raymond Serway - Capitulo 10

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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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_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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_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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_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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_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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_ atar = tan1

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]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 z-axis.For the rod along the y-axis, I = 13 mL2 from the table.For the rod parallel to the z-axis, the parallel-axistheorem givesI = 12 mr2 + m

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_L2 2 14 mL2xM mLLx xaxis of rotationChapter 10 Solutions 11 2000 by Harcourt College Publishers. All rights reserved.In the rod along the x-axis, 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 +

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]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

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_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

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_vR 2 = m1g

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_d2 + m2g

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_ d2 12

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_m1 + m2 + 12 M v2 = (m1 m2)g

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_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

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_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

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_v2R2 = v2

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_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. Free-body 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 work-energy theorem can take account of multiple objects more easily than Newton'ssecond law. Like your bratty cousins, the work-energy 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)

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_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 real-world 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

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_vr 2 + 12 mv2 = mghI v2r 2 = 2mgh mv2I = mr2

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_2ghv2 1 10.48 E = 12 I2 = 12

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_12 MR2

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_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 merry-go-round 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 =

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_32 g l sin rThen

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_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 =

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_103 revmin

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_1 min60.0 s =

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_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)

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_14.2 11.6 rR

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_103 kgm3 dr=

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_23 4(14.2 103) R55

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_23 4(11.6 103) R56 I = 83 (103) R5

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_14.25 11.66 M = dm= 0R4r2

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_14.2 11.6 rR 103 dr= 4 103

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_14.23 11.64 R3IMR2 = 83 (103) R5 ,_ 14.25 11.664 103R3R2 ,_ 14.23 11.64 = 23

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_.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=

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_12

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_12 (1.00 kg)(0.500 m) 2

]11

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_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

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_L2 , where I = 13 mL2 = 3g/ L (b) = I so that in the horizontal, mg

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_L2 = mL23 = 3g2L (c) ax = ar = r2 =

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_L2 2 = 3g2 ay = at = r =

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]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

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_Mm + 2M (sin cos ) 10.64 ( a ) E = 12

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_25 MR2 (2) E = 12 25 (5.98 1024)(6.37 106) 2

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_286400 2 = 2.57 1029 J (b)dEd t = dd t

]1112

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_25 MR2

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_2T2 = 15 MR2(2)2(2T3) dTd t = 15 MR2

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_2T 2

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_2T dTd t = (2.57 1029 J)

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_286400 s

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_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 parallel-axis theorem,I = MR2 + nmR23 + (M + nm)R2 = 2MR2 + 4 nmR23 *10.67 Every particle in the door could be slid straight down into a high-density 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 +

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_R22 = 58 MR2(5)Substituting (2), (3), (4) and (5) into (1) we findf = m

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_g 2yt2 R 58 MR22y(Rt2) = R

]11m

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_g 2yt2 54 Myt2 10.69 ( a ) While decelerating,f = I' = (20 000 kg m2)

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_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)

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_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

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_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

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_13 ml2 = 32

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