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CHAPTER EIGHT SOLUTIONS CHAPTER EIGHT SOLUTIONS 8.12 Taking torques about the left end of the scaffold, we have T1(0) - (700 N)(1 m) - (200 N)(1.5 m) + T2(3 m) = 0. From which, 1 Then, from ΣFy = 0, we have T1 + T2 - 700 N - 200 N = 0. Since T2 = 333 N, we find (a) x (a) (b) 200 N 3m V 8.16 2 2m Tx = T⋅ cos60° = 0.5 T 3m (c) T ___________________________ See the diagram below: Ty = T⋅sin60° = 0.866T 700 N H (b) 1m 1.5 m T1 = 567 N. 8.14 700 N T T2 = 333 N. 200 N 80 N If x = 1 m, then Στleft end = (-700 N)(1 m) - (200 N)(3 m) - (80 N)(6 m) + (0.866T)(6 m). Equating this to zero gives: T = 342 N. From ΣFx = 0, H = 0.5 T = 171 N. From ΣFy = 0, V = 980 N - 0.866T = 683 n. If T = 900 N Στleft end = (-700 N)(x) - (200 N)(3 m) - (80 N)(6 m) + (779.4 N)(6 m). Equating this to zero and solving for x gives: x = 5.13 m. From Στ = 0 about the pivot shown, we have (Tsin30°)d - (200 N)d = 0 where d is the length of the beam. The forces H, V, and Tx produce Pivot no torque about the pivot point. 30° Solving for T, We find: T= 400 N. H Tx From ΣFx = 0, we have d H - Tcos30° = 0. V T = 400 N, so H = 346 N (to right). From ΣFy = 0, V + Tsin30° - 200 N = 0 , which yields V = 0. 104 T y 200 N CHAPTER EIGHT SOLUTIONS 8.12 We choose the pivot point at the left end, and use Στ = 0. -(700 N)(0.5 m) - (294 N)(1 m) + (T1sin40°)(2 m) = 0, which yields T1 = 501 N. T 700 N 2 Now use ΣFx = 0. T1cos40° - T3 = 0. 0.5 m With T1 = 501 N, this gives T3 = 384 N. T3 1m 1m Finally, ΣFy = 0 gives W = 294 N T2 + T1sin40°- 994 N = 0. From which, T2 = 672 N. T1 T 1y 40° 200 N T1x 8.13 F = Tension force in Deltoid Muscle, t F = Force exerted on arm by shoulder joint s F = F sin12° ty t Arm 0.08 m F sx θ Ftx = F cos12° t Point O (Shoulder Joint) 0.29 m W F sy Fsy ΣτO = Ftsin 12°(0.08 m) -(41.5 N)(0.29 m) = 0. From which Ft = 724 N (Tension in deltoid muscle) ΣFy = 0 gives - Fsy + Ftsin 12° - 41.5 N = 0, yielding, ΣFx = 0 gives Fsx = Ftcos 12° and Fsx = 709 N. Therefore, and F sx Fs Fsy = 109 N. (Fsx)2 + (Fsy)2 = 717 N, Fsy tanθ = = 0.1539, or θ = 8.75°. Fsx Fs = 8.14 Using the fact that the scaffold is in translational equilibrium, we have ΣFy = TL - 750 N - 345 N - 500 N - 1000 N + TR = 2595 N, or TL + TR = 2595 N (1) where TL and TR are the tensions in the left and right cables, respectively. Since the scaffold is also in rotational equilibrium, we may write, Στleft end = -(750 N)(1 m) - (345 N)(1.5 m) - (500 N)(2 m) - (1000 N)(2.5 m) + (TR)(3 m) = 0, or TR = 1589 N. Then equation (1) gives TL = 1006 N. 8.15 Choosing the pivot point at the point O shown, Στ = 0 becomes (50 N)(7.5 cm) + T(0) - R(3.5 cm) = 0. Thus, R = 107 N. R C = 50 N 3.5 cm T 7.5 cm Now, apply ΣFy = 0 to obtain -50 N + T - 107 N = 0 and T = 157 N. O 105 Pivot CHAPTER EIGHT SOLUTIONS Pivot d 5 Ty Tx d 2 Cx d2 V Cy C Fy V Fx F 8.16 Let us first resolve all forces into components parallel to and perpendicular to the leg, as shown. Use Στ = 0 about the pivot indicated. Ty(d/5) - Cy(d/2) - Fy(d) = 0, where d is the length of the lower leg. Cy = C sinV = (30 N)sin40° = 19.3 N, and Fy = FsinV = (12.5 N)sin40° = 8.03 N. Thus, Ty = 88.5 N, but Ty = T sin25°. So, T = 209 N. 8.17 Στ)point o = 0 gives (-1200 N) L 3L 3L cos 65°+(Tcos25°)( sin65°)+(Tsin25°)( cos65°)-(2000 N)(Lcos65°). 2 4 4 From which, T = 1465 N. From ΣFx = 0, H = Tcos 25° = 1328 N (toward right). From ΣFy = 0, V = 3200 N - Tsin25°= 2581 N (upward). Tx = Tcos25° Ty = Tsin25° L 3L/4 Ty Tx 2000 N L/2 8.18 1200 N We call the tension in the cord at the left end of the sign, T1 and the 65° H tension in the cord near the middle of the sign, T2, and we choose our Point O pivot point at V the point where T1 is attached. T1 T2 2 Στpivot = 0 = (-W)(0.5 m) + T2(0.75 m) = 0, so, T2 = W 3 0.75 m From ΣFy = 0, T1 + T2 - W = 0. Substituting the expression for T2 and solving, we find 1 T1 = W. 3 0.5 m W 106 CHAPTER EIGHT SOLUTIONS 8.19 Use ΣFy = 0: or F1 - 500 N - 800 N = 0, F1 = 1300 N. Now, apply ΣFx = 0: f - F2 = 0,or f = F2. The lever arm for the 500 N force is (7.5 m)cos60° = 3.75 m The lever arm for the 800 N force is dcos60° = d/2 . For F2, the lever arm is (15 m)sin60° = 13 m. F2 (1) (15 m)sin60° (2) 800 N F1 pivot 500 N 60° f (7.5m)cos60° d cos60° Using Στ = 0 with pivot pt. at base of the ladder, -(500 N)(3.75 m) - (800 N)(d/2) + F2(13 m) = 0 (3) (a) When d = 4 m, Equation (3) gives: F2 = 267 N Equation (2) then gives f = 267 N, and F1 = 1300 N from above. (b) If d = 9 m: Equation (3) gives F2 = 421 N, Equation (2) yields f = 421 N, and Equation (1) gives F1 = 1300 N. If the ladder is ready to slip, µs = f/F1 = (421 N)/1300 N = 0.324. 8.20 Use ΣFy = 0: F1 - 200 N - 800 N = 0. F1 Thus, F1 = 1000 N. The friction force, f, at the base of the ladder is f = µ sF 1 (8 m)sin50° when the ladder is on the verge of slipping. Thus, f= F 1 0.6(1000 N) = 600 N. 800 N Now use ΣFx = 0: f - F2 = 0, or F2 = f = 600 N. pivot 200 N 50° f Finally, use Στ = 0 with the pivot pt. at the base of the ladder. The lever arm for the 200 N force is (4mcos50°) = (4 m)cos50° 2.57 m. d cos50° The lever arm for the 800 N force is dcos50° = 0.643d, where d is the distance from the base of the ladder up to the position of the person. Finally, the lever arm for the force F2 is (8 m) sin50° = 6.13 m. We have: -(200 N)(2.57 m) - (800 N)(.643d) +(600 N)(6.13 m) = 0 giving, d = 6.2 m. 8.21 Total torque about the elbow gives -(0.33 m)(2 kg)(9.8 m/s2) + FBcos75°(0.08 m) = 0, or 8.22 First realize that the normal force N supports the entire weight of the body. Thus, N = w = 700 N. Also, observe the diagram at right and see that α = 0 since T is perpendicular to the end of the rod in the model. From the conditions for equilibrium, we have ΣFx = 0 = -Tsinθ + Rsin15° = 0, or Tsinθ = Rsin15° (1) Also, ΣFy = 0 = N - Rcos15° + Tcosθ = 0, or (with N = 700 N), Tcosθ = Rcos15° - 700 N. (2) FB = 312 N. T 5 cm 1 N = W = 700 N Στpivot = 0 ⇒ - N(18 cm)cosθ + (7 cm)T = 0, or with N = 700 N, T = (1800 N)cosθ. (3) (1800 N)sinθcosθ Substitute (3) into (1) to obtain R= . (4) sin15 107 7 cm 2 8 cm α=θ θ R 15° Choose Pivot here CHAPTER EIGHT SOLUTIONS Now substitute (3) and (4) into equation (2) to find (1800 N)sinθcosθcos15° (1800 N)cos2θ = - 700 N, which reduces to sin15° cos2θ + 0.3889 = 3.732sinθcosθ. Square both sides of the equation to obtain cos4θ = 2 2 2 0.7778cos θ + 0.15123 = 13.9282sin θcos θ. Using the identity sin2θ = 1 - cos2θ, this reduces to cos4θ - 0.8809cos2θ + 0.01013 = 0 (a quadratic equation in cos2θ). 2 The quadratic formula gives cos θ = 0.86926, or cos2θ = 0.0117 From these, we find that cosθ = ± 0.93234 or cosθ = ± 0.108 which gives θ = 21.2°, θ = 158.8°, θ = 83.8°, and θ = 96.2°. Of these solutions, θ = 21.2° is the only one that is physically possible. Then (with θ = 21.2°), equation (3) yields T = 1678 N, and equation (4) gives R= 2345 N. 8.23 8.24 8.25 Considering the torques about the point at the bottom of the bracket yields: F(0.06 m) = 0, so F = 66.7 N. (0.05 m)(80 N) - ⎛1.1x10-10 m⎞⎟ = 0.55 X 10-10 m from the center of mass, we have I = Σmr2, Since each atom is ⎜ 2 ⎝ ⎠ or I = 2 x (2.32 X 10-26 kg)(0.55 X 10-10 m)2 = 1.4 X 10-46 kg m2. Set L = 4 m, r = 0.5 m, M = 0.5 kg, m = 5 kg. Using the +α free-body diagram of the mass, we obtain (taking down as I positive), r ΣFy = mg - T = ma. (1) +a The torque on the spool gives 1 Support T Στcenter = Tr = Iα = Mr2α, or Force 2 1 T= Mrα. (2) 2 Since the string does not slip on the spool, rα = a, and equation (2) becomes, T= T M W = Mg 1 Ma. 2 Putting this into equation (1), gives, 1 mg (5 kg)(9.8 m/s2) mg - Ma = ma, or a = = = 9.33 m/s2. 2 M (5 kg + 0.25 kg) m+ 2 1 Since the mass starts from rest, L = voyt + at2, gives 2 2(4 m) = 0.926 s as the time for the mass to fall 4 m. 9.33 m/s2 a 9.33 m/s2 Also, α = = = 18.66 rad/s2, so at t - 0.926 s, r 0.5 m ω = ω0 + αt = 0 + (18.66 rad/s)(0.926 s) = 17.3 rad/s. t= 8.26 (a) (b) (c) 2L = a I = Σmiri2 First, apply the above equation about the x axis. We have, Ix = (3 kg)(9 m2)+(2 kg)(9 m2)+(2 kg)(9 m2)+(4kg)(9 m2) = 99.0 kg m2. About the y axis, we have Iy = (3 kg)(4 m2) + (2 kg)(4 m2) + (2 kg)(4 m2) + (4kg)(4 m2) = 44.0 kg m2. The distance, r, (from an axis through O and perpendicular to the page) out to each of the masses is found from the pythagorean theorem: r = 108 (2 m)2 + (3 m)2 = 13 m2 , CHAPTER EIGHT SOLUTIONS and the moment of inertia is I0 = (3 kg)(r2) + (2 kg)(r2) + (2 kg)(r2) + (4kg)(r2), or = (11 kg)(13 m2) = 143 kg m2. 8.27 We use τ = Iα . τx = Ixα = (99 kg m2)(1.5 rad/s2) = 149 N m τy = Iyα = (44 kg m2)(1.5 rad/s2) = 66.0 N m τ0 = I0α = (143 kg m2)(1.5 rad/s2) = 215 N m 109 CHAPTER EIGHT SOLUTIONS 8.28 We have already found (problem 8.26) Iy = 44.0 kg m2, and we know τ (20 N m) τ = Iα. Thus, α= = = 0.455 rad/s2, and I (44 kg m2) ω = ω0 + αt = 0 + (0.455 rad/s2)(3 s) = 1.36 rad/s. 8.29 Let θ = 30°, ω0 = 60.0 rad/s and R = 0.03/2 = 0.015 m. The forces acting on the coin parallel to the incline give ΣFparallel = - f - mgsinθ = ma = mRα, (1) where f is the friction force between the coin and incline. The net torque about the coin's center of mass is 1 1 Στcenter = fR = Iα = MR2α, or f = MRα. 2 2 1 Equation (1) then becomes MRα- mgsinθ = mRα, yielding 2 2gsinθ 2(9.8 m/s2)sin30° α=== -217.8 rad/s2. 3R (3)(0.015 m) Therefore a = rα = (0.015 m)(-217 rad/s2) = - 3.27 m/s2. 2 2 Then v = v 0 + 2as, with v0 = rω0 = (0.015 m)(60 rad/s) = 0.9 m/s and v = 0, gives: 0 = (0.9 m/s)2 + 2(-3.27 )s. This gives s = 0.124 m as the distance the coin goes up the incline. 0.25 = 0.125 m, and L = 4 m. A study of the forces parallel to the incline yields, 2 ΣF|| = mgsinθ - f = ma (1) where f is the friction force between sphere and incline. The net torque about the center of the sphere gives 2mR2 ⎛ a ⎞ a Στcenter = fR = Iα = , (α = , sphere rolls without slipping) 3 ⎝R⎠ R 2ma so that f = . Therefore equation (1) becomes, 3 2ma 3gsinθ mgsinθ = ma, yielding a = = 2.94 m/s2. 3 5 If the sphere starts from rest, then its center of mass moves a distance 1 2L 8 L = at2, which gives t = = = 1.7 s as the required time. 2 a 2.94 8.30 Let θ = 30°, R = 8.31 (a) (b) 1 1 mr2 = (0.85 kg)(4 X 10-2 m) 2 = 6.8 X 10-4 kg m2. 2 2 τnet = Iα = (6.8 X 10-4 kg m2) (66 rad/s2) = 4.49 X 10-2 kg m2. The torque exerted by the fish = Fr, so the net torque is τnet = Fr - 1.3 Nm = 4.49 X 10-2 kg m2. 1.345 Nm From which, F= = 34 N. 4 X 10-2 m 1 1 θ = ω0t + αt2 = 0 + (66 rad/s2)(0.5 s) 2 = 8.25 rad, and 2 2 -2 s = rθ = (4 X 10 m)(8.25 rad) = 0.33 m = 33 cm. I= 110 CHAPTER EIGHT SOLUTIONS 8.32 8.33 1 1 mr2 = (100 kg)(0.5 m) 2 = 12.5 kg m2, 2 2 ω0 = 50 rev/min = 5.24 rad/s ω - ω0 0 - 5.24 rad/s α= = = - 0.873 rad/s2 t 6s τ = Iα = (12.5 kg m2)(-0.873 rad/s2) = -10.9 Nm Also, the magnitude of the torque is given by fr = 10.9 N m, where f is the force of friction. 10.9 Nm Therefore, f= = 21.8 N, and 0.5 m f 21.8 N f = µkN yields µk = = = 0.31. N 70 N I= We first calculate the moment of inertia as 1 1 I = MR2 = (150 kg)(1.5 m) 2 = 168.8 kg m2. 2 2 Then we note that ωf = 0.5 rev/s = π rad/s, and calculate α as, ω = ω0 + αt or π rad/s = 0 + α (2 s). Thus, Then τ = Iα becomes Fr = Iα, or F = 8.34 8.35 α = π/2 rad/s2. Iα (169 kg m2)(π/2 rad/s2) = = 177 N. r 1.5 m The moment of inertia of the pulley is 1 1 I= MR2 = (5 kg)(0.6 m) 2 = 0.90 kg m2. 2 2 First we apply Newton's second law to the falling bucket of mass 3 kg (weight 29.4 N). We have 29.4 N - T = (3 kg)a. (1) Now apply τ = Iα to the pulley. We have a I a 0.90 kg m2 TR = Iα = I ⎛ ⎞ . Thus, T = 2 = a. (2) ⎝R⎠ R (.6 m)2 We can solve (1) and (2) simultaneously to find a. We find, (a) a = 5.35 m/s2 downward. 1 (b) To find how far it drops, we use y = v0t + at2. 2 1 y = 0 + (5.35 m/s2)(4 s) 2 = 42.8 m 2 a 5.35 m/s2 (c) α = = = 8.91 rad/s2. R 0.6 m The resultant torque is given by (120 N)(0.811 m) - (100 N)(0.811 m) = 16.2 N m 1 1 The moment of inertia is: I = mr2 = (2.1 kg)(0.811 m) 2 = 0.691 kg m2. 2 2 τ 16.2 Nm Then, τ = Iα gives α= = = 24 rad/s2. I 0.691 kg m2 111 R T T a 3 kg 29.4 N CHAPTER EIGHT SOLUTIONS 8.36 The free body diagram is shown at the right. The resultant torque about the center of the pulley is (1.14 X 104 N)r - T1r = (79.8 kg m4)α T1 - 9800 N = (1000 kg)a (2) Since there is no slipping, we also have a = rα = (0.762 m)α. Solving (1), (2), and (3) simultaneously, we find a = 1.41 m/s2. 8.37 8.38 (3) (a) Newton's second law for the cylinder is mg - T = ma . (1) 1 2 (b) τ = Iα becomes: Tr = mr α. (2) 2 (c) Solve (2) for T and substitute α = a/r. We find: mr2a ma T= 2 = . (3) 2 2r Finally, substitute (3) into (1) to find, ma 2 mg = ma which gives a = g. 2 3 (a) (b) r (1) From Newton's second law, applied in the vertical direction, we have r = 0.762 m α T R 1 4 T2 =1.14 x10 N T1 1000 kg a 9800 N T r a mg We first determine the angular acceleration due to the applied torque and the frictional torque by writing τapplied + τfriction = Iα1 with τapplied + τfriction = 36 N m. Now in six seconds, ω changes from 0 to 10 rad/s. Using ω = ω0 + α1t, we have 10 rad/s = 0 + α1(6 s), giving α1 = 1.67 rad/s2. Then, 36 N m= (1.67 rad/s2)I, or I = 21.6 kg m2. When the applied torque is removed, we have τfriction = Iα2. Using ω = ω0 + α2t we have 0 = 10 rad/s + α2(60 s), so α2 = -0.167 rad/s2. Therefore, (c) τfriction = (21.6 kg m2)(-0.167 rad/s2) = - 3.6 N m, and ⎥τfriction⎥ = 3.6 N. The angular displacement during the first 6 s is 0 + 10 rad/s θ1 = ω1ave(Δτ) = (6 s) = 30 rad, and that during the last 60 s is θ2 = 2 10 rad/s ω2ave(Δτ)2 = (60 s) = 300 rad. 2+0 θ = θ1 + θ2 = 30 rad + 300 rad = 330 rad = 52.5 rev. 112 CHAPTER EIGHT SOLUTIONS 8.39 The initial angular velocity of the wheels is zero, and the final angular velocity is v 50.0 m/s ωf = = = 40.0 rad/s. r 1.25 m ωf - ω0 40.0 rad/s - 0 Thus, α = = = 83.3 rad/s2. t 0.48 s τcenter of a wheel = fr, so τ = Iα gives Iα (110 kg m2)(83.3 rad/s2) = = 7.33 X 103 N, and r 1.25 m f 7.33 X 103 N µk = = = 0.524. N 1.4 X 104 N f= 8.40 We begin by using v = Rω, where R is the radius of the sphere. 1 2 ω2 2MR2ω2 1 Then, KErot = Iω2 =( MR2) = = Mv2. 2 5 2 10 5 The total kinetic energy is given by 1 1 7Mv2 KEtot = KErot + KEcm = Mv2 + Mv2 = . 5 2 10 KErot Then, = (1/5)/(7/10) = 2/7. KEtot 8.41 The moment of inertia of a long thin rod about an axis through one end is I = 1 ML2. The total 3 rotational kinetic energy is given as 1 1 KErot = I ω2 + I ω2 , with 2 h h 2 m m m hL 2h (60 kg)(2.7 m)2 Ih = = = 145.8 kg m2, and 3 3 m m L 2m (100 kg)(4.5 m)2 Im = = = 675 kg m2. 3 3 (2π rad) 1 h In addition, ωh = = 1.454 X 10-4 rad/s, while (12 h) 3600 s (2π rad) 1 h ωm = = 1.745 X 10-3 rad/s. Therefore, (1 h) 3600 s 1 1 KErot = (145.8)(1.454 X 10-4) 2 + (675)(1.745 X 10-3 ) 2 = 1.03 X 10-3 J. 2 2 8.42 (a) (b) 8.43 I = Σmr2 = (4 kg)(3 m)2 + (2 kg)(2 m)2 + (3 kg)(4 m)2 = 92.0 kg m2 1 1 KE = Iω2 = (92.0 kg m2)(2 rad/s) 2 = 184 J. 2 2 v4 = r4ω = (3 m)(2 rad/s) = 6 m/s, v2 = r2ω = (2 m)(2 rad/s) = 4 m/s, v3 = r3ω = (4 m)(2 rad/s) = 8 m/s. 1 1 1 1 KE = Σ mv2 = (4 kg)(6 m/s) 2 + (2 kg)(4 m/s) 2 + (3 kg)(8 m/s) 2 2 2 2 2 KE = 184 J. Using W = ΔKE = KEf - KEi = 2W 2(3000 J) we get I = 2 = ω (200 rad/s)2 1 Iω2 - 0, 2 = 0.15 kg m2 . 113 CHAPTER EIGHT SOLUTIONS 8.44 The moment of inertia of the cylinder is 1 1 I = mr2 = (81.6 kg)(1.5 m) 2 = 91.8 kg m2, and 2 2 the angular acceleration of the merry-go-round is found as α = τ/I = (F r)/I = (50 N)(1.5 m)/(91.8 kg m2) = 0.817 rad/s2. At t = 3 s, we find the angular velocity , ω = ω0 + αt, to be ω = 0 + (0.817 rad/s2)(3 s) = 2.45 rad/s, and 1 1 KE = Iω2 = (91.8 kg m2)(2.45 rad/s) 2 = 280 J. 2 2 1 1 mv2cm = Iω2, gives 2 2 2 1 1 2mR 1 1 mv2cm = ω2 = m(Rω)2 = mv2t , where vt is the tangential speed of a point 2 2 5 5 5 on the equator of the ball. Cancelling the mass gives v2cm 2 vcm = , or = 0.63. 2 5 vt v t 8.45 Setting KEtrans = KErot , or 8.46 (a) (b) (c) 8.47 1 1 mv2 = (10 kg)(10 m/s) 2 = 500 J. 2 2 1 1 1 v2 1 KErot = Iω2 = ( mr2)( 2 ) = (10 kg)(10 m/s) 2 = 250 J. 2 2 2 4 r KEtotal = KEtrans + KErot = 750 J. KEtrans = Let I be the moment of inertia of the object about the point of rotation. Then τ = Iα so that τ constant implies, α is constant. Then ω22 = ω12 + 2α(θ2 - θ1), or ω22 - ω12 = 2αΔθ 1 1 1 Also, W = ΔKE = Iω22 Iω21 = I(ω22 - ω21) = Iα(Δθ) = τ(Δθ) 2 2 2 8.48 The forces on a rolling object on an incline of slope angle θ are the force of friction, f, the normal force, N, and its weight, mg. We apply Newton's second law with the x axis along the plane, to find, mgsinθ - f = ma. (1) +Y N Now use, τ = Iα , or fr = Iα = I(a/r). Therefore, f = (I/r2)a. (2) Substitute into (1) for the frictional force f from (2) and solve the resulting equation for a. f gsinθ a= . I +X (1 + 2) mr θ This is the linear acceleration of the center of mass of the rolling W = mg 2 2 object. For a solid sphere, I = mr2 or (I/mr2) = . 5 5 From this, we see that a = gsinθ/1.4. I 1 Using the same approach for a solid cylinder, = , and a = gsinθ/1.5 2 2 mr Finally, for a hollow ring, (I/mr2) =, and a = gsinθ/2. Thus, we find asphere > acylinder > aring, so the sphere wins and the ring comes in last. 8.49 (a) The angular velocity of the flywheel is 5000 rev/min = 524 rad/s, and its moment of inertia 1 1 is, I = mr2 = (500 kg)(2 m) 2 = 103 kg m2. 2 2 114 CHAPTER EIGHT SOLUTIONS (b) Therefore, the stored energy is 1 1 KE = Iω2 = (103 kg m2)(524 rad/s) 2 = 1.37 X 108 J. 2 2 A 10 hp motor is equivalent to 7460 W. Therefore, if energy is used at the rate of 7460 J/s, the stored energy will last for t = E/P = 1.37 X 108 J/7.46 X 103 J/s = 1.84 X 104 s = 5.10 h. 8.50 Work done = Fs = (5.57 N)(0.8 m) = 4.46 J, and 1 1 1 Work = ΔKE = Iωf2 Iω02 = Iωf2. (ω0 = 0 since top starts from rest.) 2 2 2 1 Thus, 4.46 J = (4 X 10-4 kg m2) ωf2 and from this, ωf = 149 rad/s. 2 8.51 I= 2 2 mr2 = (5.98 X 1024 kg)(6.38 X 106 m) 6 = 9.74 X 1037 kg m2 5 5 2π rad The intial period = 24 h = 86,400 s, and ω0 = = 7.27 X 10-5 rad/s. 86400 s 1 Therefore, the initial kinetic energy is KEi = Iω02 = 1.57 X 1029 J. 2 2π rad The final period = 24 h + 1 min = 86,460 s, ωf = = 7.267 X 10-5 rad/s. 86460 s 1 1 The energy available = ΔKE = Iω02 Iωf2. 2 2 ΔKE = 3.57 X 1026 J, and time = Substituting the appropriate values gives, 3.57 X 1026 J = = 1.8 X 106 y 2 X 1020 J/y 8.52 energy available rate of energy use Let us first consider conservation of energy, with the zero level for potential energy at the base of the ramp. We have 1 1 1 1 mvi2 + Iωi2 + mgyi = mvf2 + Iωf2 + mgyf, or 2 2 2 2 1 1 mvi2 + Iωi2 + 0 = 0 + 0 + (mg)(s)sin20°, where s is the distance the hoop rolls up the 2 2 ramp. For a hoop, I = mr2. Thus, Iωi2 = mr2ωi2 = m(rωi)2 = mvi2. Therefore, our equation for conservation of energy becomes 1 1 mvi2 + mvi2 = (mg)(s)sin20°, or, solving for s, we have 2 2 s = vi2/gsin20°. Since vi = rωi = (3 m)(3 rad/s) = 9 m/s, we find s = (9 m/s)2/(9.8 m/s2)sin20° = 24 m. 8.53 (a) (b) The angular velocity of the earth on its axis is ω = 2π rad/day = 7.27 X 10-5 rad/s. We assume the earth is a uniform sphere of radius R and calculate its moment of inertia as 2 2 I = mR2 = (5.98 X 1024 kg)(6.38 X 106 m) 2 = 9.74 X 1037 kg m2. 5 5 The angular momentum is found as L = Iω = (9.71 X 1037 kg m2)(7.27 X 10-5 rad/s) = 7.08 X 1033 J s. For the orbiting earth, ω = 2π rad/yr = 1.99 X 10-7 rad/s, and the angular momentum is given by L = Iω = mr2ω = (5.98 X 1024 kg)(1.49 X 1011 m)2 (1.99 X 10-7 rad/s), L = 2.66 X 1040 J s. 115 CHAPTER EIGHT SOLUTIONS 8.54 The total angular momentum about the center point is given by m hL 2h (60 kg)(2.7 m)2 L = Ihω h + Im ω m with Ih = = = 145. 8 kg m2, and Im = 3 3 m m L 2m (100 kg)(4.5 m)2 = = 675 kg m2. 3 3 (2π rad) 1 h In addition, ωh = = 1.454 X 10-4 rad/s, while (12 h) 3600 s (2π rad) 1 h ωm = = 1.745 X 10-3 rad/s. Thus, (1 h) 3600 s L = (145.8 kg m2)(1.454 X 10-4 rad rad ) + (675 kg m2)(1.745 X 10-3 ), or s s L = 1.2 kg m2/s. 8.55 Let M and m be the mass of the cylinder and putty, respectively, and let R be the radius of the cylinder. Then the total angular momentum before the putty hits the cylinder is MR2 (10 kg)(1 m)2(7 rad/s) Lbefore = I0ω0 = ω0 = = 35 kg m2/s. 2 2 After the putty hits the cylinder, we have ⎛MR2 + mr2⎞⎟ ω Lafter = Iω + mr2(ω) = ⎜ ⎝ 2 ⎠ 2 ⎛(10 kg)(1 m) + (0.25 kg)(.9 m)2⎞⎟ ω = (5.20 kg m2)(ω). = ⎜ 2 ⎝ ⎠ Using Lbefore = Lafter , we have (5.20 kg m2)ω = 35 kg m2/s, so ω = 6.73 rad/s. 8.56 We use conservation of angular momentum, with Ii = I, and If = 0.9I I iω i I Thus, Ifωf = Iiωi gives ωf = = ω = 1.111ωi. 0.9I i ωf KEf - KEi⎞ The % change in kinetic energy = ⎛⎜ 100%. Substituting appropriate values, gives ⎝ KEi ⎟⎠ %change in KE = 11.1%. The kinetic energy has increased because she has had to do work to pull her arms inward. 8.57 First compute some needed constants: mass of people = mp = (5.5 x 109)(70 kg) = 3.85 X 1011 kg I(due to people) = Ip = mpR2E = (3.85 x 1011 kg)(6.37 x 106 m)2 = 1.56 x 1025 kg m2 . I for Earth alone: 2 2 IE = MER2E = (5.98 X 1024 kg)(6.37 X 106 m) 2 = 9.71 X 1037 kg m2. 5 5 2π rad 1 day Original angular speed of Earth: ω0 = = 7.27 X 10-5 rad/s. 1 day 86400 s The original angular momentum of the system (Earth plus people) is L0 = IEω0 + Ipω0 = (IE + Ip)ω0 The tangential speed of a point on the equator of the Earth is v = REω so when the people start running, their tangential speed is v' = v + 2.5 m/s = REω + 2.5 m/s, and their angular speed is v' 2.5 m/s ω' = =ω+ , where ω is the new angular speed of the Earth. RE RE The angular momentum of the system is now given by 116 CHAPTER EIGHT SOLUTIONS Lf = Learth + Lpeople = IEω + Ipω' = IEω + Ip(ω + 2.5 m/s ),or RE Ip (2.5 m/s) . RE Conserving angular momentum(Lf = L0) gives: Ip (2.5 m/s)Ip ω ω = ω0 (2.5 m/s) , or =1. RE(IE + Ip) ω0 ω0RE(IE + Ip) Lf = (IE + Ip)ω + ω (2.5)(1.56 X 1025) =1 , ω0 (7.27 X 10-5)(6.37 X 106)(9.71 X 1037 + 1.56 X 1025) ω = 1 - 8.67 X 10-16. Thus, the ratio of the length of the new day to the old day is ω0 T 2π ω0 1 1 = = = ≈ 1 + 8.67 X 10-16, and T0 ω 2π ω 1 - 8.67 X 10-16 ω0 the length of the new day is T ≈ T0(1 + 8.67 X 10-16) T = T0 + T0(8.67 X 10-16) = 24 h + (86,400 s)(8.67 X 10-16), or T ≈ 24 h + 7.5 X 10-11 s. Therefore, the day is lengthened by approximately 7.5 X 10-11 s. 8.58 The total angular momentum is given by Itotal = Iweights + Istudent = 2(mr2) + (3 kg m2) Before: r = 1 m. Thus, Ii = 2(3 kg)(1 m)2 + (3 kg m2) = 9 kg m2. After: r = 0.3 m. Thus, If = 2(3 kg)(0.3 m)2 + (3 kg m2) = 3.54 kg m2. We now use conservation of angular momentum Ifωf = Iiωi, or Ii 9 (a) ωf = ωi = (0.75 rad/s) = 1.9 rad/s. If 3.54 1 1 (b) KEi = Iiωi2 = (9 kg m2)(0.75 rad/s) 2 = 2.5 J, 2 2 1 1 KEf = Ifωf2 = (3.54 kg m2)(1.91 rad/s) 2 = 6.4 J. 2 2 8.59 Using conservation of angular momentum, we have Lapogee = Lperihelion, or (mr2a)ωa = (mr2p)ωp. Thus, va vp (mr2a) = (mr2p) , giving ra rp rp 0.59 AU rava = rpvp , or va = v = (54 km/s) = 0.91 km/s. ra p 35 AU 117 CHAPTER EIGHT SOLUTIONS 8.60 (a) The moment of inertia of the system is given by 1 1 I = Iman + Iwheel = mr2 + MR2 = (80 kg)r2 + (25 kg)(2 m) 2. 2 2 When the man is at a distance of r = 2 m from the axis, the equation above gives Ii = 370 kg m2 , and when the man moves to a point 1 m from the center (r = 1 m), the moment of inertia becomes, If = 130 kg m2 . The initial angular velocity of the system is ωi = 0.2 rev/s = 1.26 rad/s, and we find the final angular velocity via conservation of angular momentum, as Ii 370 ωf = ωi = (1.26 rad/s) = 3.58 rad/s. If 130 1 1 (b) The change in kinetic energy is KEf - KEi = Ifωf2 - Iiωi2 , or 2 2 1 1 ΔΚΕ = (130 kg m2)(3.58 rad/s) 2 - (370 kg m2)(1.26 rad/s) 2 = 540 J. 2 2 This difference results from work done by the man on the system as he walks inward. 8.61 I0 = mr02 = (0.12 kg)(0.4 m)2 = 1.92 X 10-2 kg m2 If = mrf2 = (0.12 kg)(0.25 m)2 = 7.5 X 10-3 kg m2, and v0 I0 0.8 m/s ω0 = = = 2 rad/s. Now, use conservation of angular momentum: ωf = ω = r0 0.4 m If 0 1.92 X 10-2 kg m2 (2 rad/s) = 5.12 rad/s. 7.5 X 10-3 kg m2 1 1 The work done = ΔKE = I ω 2 - I ω 2. Substituting the appropriate values found earlier, we 2 f f 2 0 0 have: work done = 5.99 X 10-2 J. 8.62 (a) The table turns opposite to the way the woman walks, so its angular momentum cancels that of the woman. From conservation of angular momentum, we have Lf = Li = 0, so Lf = Iwωw + Itableωtable = 0, and ωtable = - Iw Itable ωwoman = - m w r2 vwoman x , or Itable r (60 kg)(2 m)2 1.5 m/s X = - 0.36 rad/s. (0.36 rad/s CW) 2m 500 kg m2 1 1 work done = ΔKE = KEf - 0 = mwomanv2woman + Iω2table 2 2 1 1 = (60 kg) (1.5 m/s) 2 + (500 kg m2)(0.36 rad/s) 2 = 99.9 J. 2 2 ωtable = - (b) 8.63 (a) From conservation of angular momentum: (I1 + I2)ω = I1ω0, or I1 ω= ω I1 + I2 0. 1 1 (b) Kf = (I1 + I2) ω2 , and K i = I1ω 02 , 2 2 Kf I1 so = (after some algebra) = which is less than 1. Ki I1 + I2 8.64 Consider the total weight of each hand to act at the center of gravity (mid-point) of that hand. Then the total torque (taking CCW as positive) of these hands about the center of the clock is given by 118 CHAPTER EIGHT SOLUTIONS Lh Lm -g τ = -mhg sinθh - mmg sinθm = (mhLhsinθh + mmLmsinθm) . 2 2 2 If we take t = 0 at 12 o'clock, then the angular positions of the hands at time t are θh = ωht, where π ωh = rad/h and θm = ωmt, where ωm = 2π rad/h. 6 9.8 m/s2 π Therefore, τ = [(60 kg)(2.7 m)sin( t) + (100 kg)(4.5 m)sin 2πt] 2 6 π or τ = -2.205 X 103 N m[0.36sin( t) + sin2πt], where time, t is expressed in hours. 6 (a) At 3:00, t = 3 h, so that π τ = -2.205 X 103 N m[[ 0.36sin ]+ sin6π] = -793.8 N m. 2 15 (b) At 5:15, t = 5 h + h= 5.25 h, and substitution as above gives: 60 τ = - 2508.8 N m. (c) At 6:00 τ = 0 N m. (d) At 8:20 τ = -1163.7 N m. (e) At 9:45 τ = -2938.4 N m. 8.65 We will use conservation of energy in the form 1 1 1 1 mvi2 + Iωi2 + mgyi = mvf2 + Iωf2 + mgyf 2 2 2 2 1 1 2 0 + 0 + mg(6 m)sin37° = m(0.2 m)2ω2 + ⎛ ⎞ m(0.2)2ω2 + 0 2 ⎝2⎠ 5 2 where we have used vf = rω and I = mR2. 5 The mass cancels from the equation, and we find the angular velocity to be ω = 36 rad/s. 8.66 The free body diagram is shown at the right. Ry Ty = Tsin12° Στ = 0 , yields Spine R 2L/3 x L 2L - (350 N) + (Tsin 12°)( ) - (200 N)L = 0. 2 3 Tx = Tcos12° L/2 From which, T = 2705 N. 200 N Compression force along spine = Rx, and we find this 350 N Pivot L from ΣFx = 0, which gives Rx = Tx = T cos12° = 2646 N. 8.67 Use the lower left hand corner as the pivot point, and apply Στ = 0. -(10 N)(.15 m) - (T1cos50°)(.15 m) + (T1sin50°)(0.3 m) = 0 Which gives, T1 = 11.2 N. Now use ΣFx = 0: -F + (11.2)cos50° = 0, and F = 7.23 N. Finally, ΣFy = 0 yields T2 - 10 N + (11.2 N)sin50° = 0 so, T2 = 1.39 N. 8.68 The free body diagram is shown. For a maximum value of x, the rod is on the verge of slipping, so f = µkN = 0.5 N. Now, use ΣFx = 0 N = Tx = Tcos37°, so N = 0.799 T. 2m N f 119 T 2 0.3 m 0.15 m 0.15 m Ty = Tsin37° = 0.602 T Tx = Tcos37° = 0.799 T W W T1x 10 N 2m x T1y F CHAPTER EIGHT SOLUTIONS Therefore, f = 0.5 N. = 0.399 T . 0.399 T + 0.602 T = 2W which reduces to Taking torques about the left end, we have -Wx = W(2 m) + Ty (4 m) = 0, which gives 8.69 From ΣFy = 0, f + Ty - 2w = 0, or T = 2W. x = 2.8 m. The free body diagram is shown at the right: ΣFy = 0 gives Point A Ay - 29,400 N - 98,000 N = 0, so Ax Ay = 127,400 N. Summing torques about point A, gives 1m Bx(1 m) - 29,400 N(2 m) - 98,000 N(6 m) = 0, and Bx = 646,800 N. B x Finally, ΣFx = 0 gives Ax = Bx = 646,800 N. The resultant force for A can be found from the Pythagorean theorem, and an appropriate trig function, to be FA = 6.59 X 105 N at an angle of 78.9° to the left of vertical. Ay 98,000 N W = 29,400 N 2m 6m 8.70 The initial moment of inertia of the system is: I0 = Σmiri2 = (4M)(1 m)2. The moment of inertia of the system after the spokes are shortened is If = Σmfrf2 = (4M)(0.5 m)2 . Ii We conserve angular momentum as: ωf = ωi = (4)(2 rev/s) = 8.0 rev/s. If 8.71 From Newton's second law, we have mgsin37° - T = ma, which gives 2 (12 kg)(9.8 m/s )sin37° - T = (12 kg)a, or T = 70.56 N - (12 kg)a. a 2 m/s2 Also α= = = 20 rad/s2 (2) r 0.1 m (a) Thus, from (1), T = 70.56 N - (12 kg)(2 m/s2) = 46.6 N. τ Tr (46.6 N)(0.1 m) (b) From τ = Iα, we have: I = = = = 0.233 kg m2. 20 rad/s α α 8.72 (1) ω = 0 + (20 rad/s2)(2 s) = 40 rad/s. (c) ω = ω0 + αt becomes: (a) The center of mass of the wheel and the mass m move with the same velocity, v. Using conservation of energy, we have 1 1 v 1 mgH = Iω2 + Mv2 + mv2 , where ω= , so 2 2 R 2 I M m 2 2 kg m2 8 kg 4 kg 2 + + ]v = [ + + ]v = (10 kg)v2, or 2 2 2 2 2 2R 2(0.5 m)2 mgH (4 kg)(9.8 m/s2)(2 m) = = 7.84 m2/s2 and v = 2.8 m/s. 10 kg 10 kg mgH = [ (b) Consider the mass and write Δy = vyave(t) = t= (c) v2 = v+0 t, so 2 2Δy 2(2 m) = = 1.43 s. v 2.8 m/s Since the center of mass moves a distance equal to H, the angular displacement is 2m 1 rev = 4 rad = 0.637 rev. 0.5 m 2π rad 120 θ= H = r CHAPTER EIGHT SOLUTIONS 8.73 Since the ladder is about to slip, f = (fs)max = µsN at each contact point. Because the ladder is still (barely) in equilibrium: ΣFx = 0, which gives ff - Nw = 0, or Nw µsN f = Nw giving Nf= . µs N f, N w = Nf - mg + fw = 0, or the above to eliminate L/ 2 µsmg . (1) 1 + µ 2s Στlower end = 0 gives L - mg cosθ + NwLsinθ + fwLcosθ = 0, 2 mg which can be written as + Nwtanθ + µsNw = 0, or 2 mg = 2Nw(tanθ + µs). (2) Substituting equation (1) into equation (2) gives 2 µsmg(tanθ + µs) mg = , which reduces to 1 + µ 2s 1 + µ2s = 2µstanθ + 2µ2s, or With θ = 60°, this becomes which has one positive solution: 8.74 NW L/ 2 ΣFy = 0 ⇒ fW = µsNW θ NF µ2s + (2tanθ)µ2s - 1 = 0. µ2s + 3.646µs - 1 = 0, µs = 0.268. (a) and (b) First, we write down the second law for the 2 kg mass. T1 - 19.6 N = (2 kg)a. (1) Now, we write the second law for the 5 kg mass. 49 N - T2 = (5 kg)a. (2) Finally, τ = Iα for the pulley becomes (T2 - T1)(0.5 m) = (5 kg m2)(a/0.5 m) or T2 - T1 = (20 kg)a. (3) Equations (1), (2), and (3) can be solved simultaneously to give a = 1.1 m/s2, T1 = 22 N, and T2 = 44 N. 121 W = mg fF = µs NF CHAPTER EIGHT SOLUTIONS 8.75 (a) and (b) First, we write down Newton's second law for the 3 kg mass. T2 = (3 kg)a. (1) The second law for the 4 kg mass is 39.2 N - T1 = (4 kg)a. (2) Now apply τ = Iα to the pulley (axis of rotation at its center) (T1 - T2)r = (0.5 kg m2)(a/r), or T1 - T2 = (5.56 kg)a. (3) Equations (1), (2) and (3) can be solved simultaneously to give: a = 3.12 m/s2, T1 = 26.7 N, and T2 = 9.37 N. 8.76 (a) and (b) 1 a MR2 , 2 R where a is the acceleration of the falling masses. For each of the falling masses: ΣF = mg - T = ma. Combining Equations (1) and (2), we find: Mmg 4mg T= , and a = . M + 4m (M + 4m) For the cylinder; Στ = (2T)R = 122 (1) (2)