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RESOLUÇÃO HALLIDAY, 9ªEDIÇÃO, VOLUME 1, CAPITULO 10

Tipologia: Exercícios

2019

1 / 116

Baixe RESOLUÇÃO HALLIDAY, 9ªEDIÇÃO, VOLUME 1, CAPITULO 10 e outras Exercícios em PDF para Física, somente na Docsity! 1. The problem asks us to assume vcom and ω are constant. For consistency of units, we write vcom mi h ft mi 60min h ft min= F HG I KJ =85 5280 7480b g . Thus, with Δx = 60 ft , the time of flight is com (60 ft) /(7480 ft/min) 0.00802 mint x v= Δ = = . During that time, the angular displacement of a point on the ball’s surface is θ ω= = ≈t 1800 0 00802 14rev min rev .b gb g. min 2. (a) The second hand of the smoothly running watch turns through 2π radians during 60 s . Thus, 2 0.105 rad/s. 60 πω = = (b) The minute hand of the smoothly running watch turns through 2π radians during 3600 s . Thus, ω = = × −2 3600 175 10 3π . rad / s. (c) The hour hand of the smoothly running 12-hour watch turns through 2π radians during 43200 s. Thus, ω = = × −2 43200 145 10 4π . rad / s. (b) The largest angle (less than 1 revolution) turned for the toast to land butter-side down is max 0.75 rev 3 / 2 rad.θ πΔ = = This corresponds to an angular speed of max max 3 / 2 rad 12.0 rad/s. 0.394 st θ πω Δ= = = Δ 5. The falling is the type of constant-acceleration motion you had in Chapter 2. The time it takes for the buttered toast to hit the floor is 2 2 2(0.76 m) 0.394 s. 9.8 m/s ht g Δ = = = (a) The smallest angle turned for the toast to land butter-side down is min 0.25 rev / 2 rad.θ πΔ = = This corresponds to an angular speed of min min / 2 rad 4.0 rad/s. 0.394 st θ πω Δ= = = Δ 6. If we make the units explicit, the function is ( ) ( )2 2 3 32.0 rad 4.0 rad/s 2.0 rad/st tθ = + + but in some places we will proceed as indicated in the problem—by letting these units be understood. (a) We evaluate the function θ at t = 0 to obtain θ0 = 2.0 rad. (b) The angular velocity as a function of time is given by Eq. 10-6: ( ) ( )2 3 28.0 rad/s 6.0 rad/sd t tdt θω = = + which we evaluate at t = 0 to obtain ω0 = 0. (c) For t = 4.0 s, the function found in the previous part is ω4 = (8.0)(4.0) + (6.0)(4.0)2 = 128 rad/s. If we round this to two figures, we obtain ω4 ≈ 1.3×102 rad/s. (d) The angular acceleration as a function of time is given by Eq. 10-8: ( )2 38.0 rad/s 12 rad/sd tdt ωα = = + which yields α2 = 8.0 + (12)(2.0) = 32 rad/s2 at t = 2.0 s. (e) The angular acceleration, given by the function obtained in the previous part, depends on time; it is not constant. The minimum speed of the arrow is then vmin .= = = 20 400 4 0 cm 0.050 s cm / s m / s. (b) No—there is no dependence on radial position in the above computation. 7. (a) To avoid touching the spokes, the arrow must go through the wheel in not more than Δt = =1 8 0 050/ . rev 2.5 rev / s s. 10. (a) We assume the sense of rotation is positive. Applying Eq. 10-12, we obtain 3 2 0 (3000 1200) rev/min 9.0 10 rev/min . (12 / 60) min tω ω α α −= + = = × (b) And Eq. 10-15 gives 0 1 1 12( ) (1200 rev/min 3000 rev/min) min 2 2 60 tθ ω ω= + = + = 24.2 10× rev. 11. (a) With ω = 0 and α = – 4.2 rad/s2, Eq. 10-12 yields t = –ωo/α = 3.00 s. (b) Eq. 10-4 gives θ − θo = − ωo2 / 2α = 18.9 rad. 12. We assume the sense of rotation is positive, which (since it starts from rest) means all quantities (angular displacements, accelerations, etc.) are positive-valued. (a) The angular acceleration satisfies Eq. 10-13: 2 2125 rad (5.0 s) 2.0 rad/s . 2 α α= = (b) The average angular velocity is given by Eq. 10-5: ω θavg rad 5.0 s rad / s.= = =Δ Δt 25 5 0. (c) Using Eq. 10-12, the instantaneous angular velocity at t = 5.0 s is ( )22.0 rad/s (5.0 s) 10 rad/s .ω = = (d) According to Eq. 10-13, the angular displacement at t = 10 s is 2 2 2 0 1 10 (2.0 rad/s ) (10 s) 100 rad. 2 2 tθ ω α= + = + = Thus, the displacement between t = 5 s and t = 10 s is Δθ = 100 rad – 25 rad = 75 rad. (f) With radians and seconds understood, the graph of θ versus t is shown below (with the points found in the previous parts indicated as small circles). 15. The problem has (implicitly) specified the positive sense of rotation. The angular acceleration of magnitude 0.25 rad/s2 in the negative direction is assumed to be constant over a large time interval, including negative values (for t). (a) We specify θmax with the condition ω = 0 (this is when the wheel reverses from positive rotation to rotation in the negative direction). We obtain θmax using Eq. 10-14: 2 2 o max 2 (4.7 rad/s) 44 rad. 2 2( 0.25 rad/s ) ωθ α = − = − = − (b) We find values for t1 when the angular displacement (relative to its orientation at t = 0) is θ1 = 22 rad (or 22.09 rad if we wish to keep track of accurate values in all intermediate steps and only round off on the final answers). Using Eq. 10-13 and the quadratic formula, we have 2 o o 12 1 o 1 1 1 21 2 t t t ω ω θ α θ ω α α − ± + = + = which yields the two roots 5.5 s and 32 s. Thus, the first time the reference line will be at θ1 = 22 rad is t = 5.5 s. (c) The second time the reference line will be at θ1 = 22 rad is t = 32 s. (d) We find values for t2 when the angular displacement (relative to its orientation at t = 0) is θ2 = –10.5 rad. Using Eq. 10-13 and the quadratic formula, we have 2 o o 22 2 o 2 2 2 21 2 t t t ω ω θ α θ ω α α − ± + = + = which yields the two roots –2.1 s and 40 s. Thus, at t = –2.1 s the reference line will be at θ2 = –10.5 rad. (e) At t = 40 s the reference line will be at θ2 = –10.5 rad. 16. The wheel starts turning from rest (ω0 = 0) at t = 0, and accelerates uniformly at α > 0, which makes our choice for positive sense of rotation. At t1 its angular velocity is ω1 = +10 rev/s, and at t2 its angular velocity is ω2 = +15 rev/s. Between t1 and t2 it turns through Δθ = 60 rev, where t2 – t1 = Δt. (a) We find α using Eq. 10-14: 2 2 2 2 2 2 1 (15 rev/s) (10 rev/s)2 1.04 rev/s 2(60 rev) ω ω α θ α −= + Δ = = which we round off to 1.0 rev/s2. (b) We find Δt using Eq. 10-15: ( )1 2 1 2(60 rev) 4.8 s. 2 10 rev/s 15 rev/s t tθ ω ωΔ = + Δ Δ = = + (c) We obtain t1 using Eq. 10-12: 1 0 1 1 2 10 rev/s 9.6 s. 1.04 rev/s t tω ω α= + = = (d) Any equation in Table 10-1 involving θ can be used to find θ1 (the angular displacement during 0 ≤ t ≤ t1); we select Eq. 10-14. 2 2 2 1 0 1 1 2 (10 rev/s)2 48 rev. 2(1.04 rev/s ) ω ω αθ θ= + = = +20 rev, and at t2 its angular displacement is θ2 = +40 rev and its angular velocity is 2 0ω = . (a) We obtain t2 using Eq. 10-15: ( )2 0 2 2 2 1 2(40 rev) 335 s 2 0.239 rev/s t tθ ω ω= + = = which we round off to 22 3.4 10 st ≈ × . (b) Any equation in Table 10-1 involving α can be used to find the angular acceleration; we select Eq. 10-16. 2 4 2 2 2 2 2 2 1 2(40 rev) 7.12 10 rev/s 2 (335 s) t tθ ω α α −= − = − = − × which we convert to α = – 4.5 × 10–3 rad/s2. (c) Using θ ω α1 0 1 12 1 2= +t t (Eq. 10-13) and the quadratic formula, we have 2 2 4 2 0 0 1 1 4 2 2 (0.239 rev/s) (0.239 rev/s) 2(20 rev)( 7.12 10 rev/s ) 7.12 10 rev/s t ω ω θ α α − − − ± + − ± + − × = = − × which yields two positive roots: 98 s and 572 s. Since the question makes sense only if t1 < t2 we conclude the correct result is t1 = 98 s. 17. The wheel has angular velocity ω0 = +1.5 rad/s = +0.239 rev/s at t = 0, and has constant value of angular acceleration α < 0, which indicates our choice for positive sense of rotation. At t1 its angular displacement (relative to its orientation at t = 0) is θ1 = 20. (a) Using Eq. 10-6, the angular velocity at t = 5.0s is ω θ= = = = = = d dt d dt t t t5 0 2 5 0 0 30 2 0 30 50 30 . . . ( . )( . ) .c h rad / s. (b) Eq. 10-18 gives the linear speed at t = 5.0s: (3.0 rad/s)(10 m) 30 m/s.v rω= = = (c) The angular acceleration is, from Eq. 10-8, α ω= = =d dt d dt t( . ) . .0 60 0 60 rad / s2 Then, the tangential acceleration at t = 5.0s is, using Eq. 10-22, a rt = = =α ( . .10 6 0m) 0.60 rad / s m / s 2 2c h (d) The radial (centripetal) acceleration is given by Eq. 10-23: a rr = = =ω 2 230 10 90. .rad / s m m / s2b g b g 21. (a) We obtain ω = =( / min .200 60 20 9rev / min)(2 rad / rev) s rad / s.π (b) With r = 1.20/2 = 0.60 m, Eq. 10-18 gives (0.60 m)(20.9 rad/s) 12.5 m/s.v rω= = = (c) With t = 1 min, ω = 1000 rev/min and ω0 = 200 rev/min, Eq. 10-12 gives α ω ω= − =o rev t 800 2/ min . (d) With the same values used in part (c), Eq. 10-15 becomes ( )o 1 1 (200 rev/min 1000 rev/min)(1.0 min) 600 rev. 2 2 tθ ω ω= + = + = (b) The plane’s velocity vp and the velocity of the tip vt (found in the plane’s frame of reference), in any of the tip’s positions, must be perpendicular to each other. Thus, the speed as seen by an observer on the ground is v v vp t= + = + = × 2 2 2 2 2133 314 3 4 10m s m s m sb g b g . . 22. First, we convert the angular velocity: ω = (2000 rev/min)(2π /60) = 209 rad/s. Also, we convert the plane’s speed to SI units: (480)(1000/3600) = 133 m/s. We use Eq. 10-18 in part (a) and (implicitly) Eq. 4-39 in part (b). (a) The speed of the tip as seen by the pilot is v rt = = =ω 209 15 314rad s m m sb gb g. , which (since the radius is given to only two significant figures) we write as 23.1 10 m stv = × . which leads to t = 20 s. The second half of the motion takes the same amount of time (the process is essentially the reverse of the first); the total time is therefore 40 s. (b) Considering the first half of the motion again, Eq. 10-11 leads to ω = ωo + α t α = 40 rad/s 20 s = 2.0 rad/s 2 . 25. (a) The upper limit for centripetal acceleration (same as the radial acceleration – see Eq. 10-23) places an upper limit of the rate of spin (the angular velocity ω) by considering a point at the rim (r = 0.25 m). Thus, ωmax = a/r = 40 rad/s. Now we apply Eq. 10-15 to first half of the motion (where ωo = 0): θ − θo = 12 (ωo + ω)t 400 rad = 1 2 (0 + 40 rad/s)t 26. (a) The tangential acceleration, using Eq. 10-22, is a rt = = =α 14 2 2 83. ( . .rad / s cm) 40.2 cm / s 2 2c h (b) In rad/s, the angular velocity is ω = (2760)(2π/60) = 289 rad/s, so a rr = = = ×ω 2 289 0 0283( ( . .rad / s) m) 2.36 10 m / s2 3 2 (c) The angular displacement is, using Eq. 10-14, 2 2 3 2 (289 rad/s) 2.94 10 rad. 2 2(14.2 rad/s ) ωθ α = = = × Then, using Eq. 10-1, the distance traveled is s r= = × =θ ( .0 0283 m) (2.94 10 rad) 83.2 m.3 27. (a) In the time light takes to go from the wheel to the mirror and back again, the wheel turns through an angle of θ = 2π/500 = 1.26 × 10–2 rad. That time is t c = = × = × −2 2 500 2 998 10 3 34 108 6( . .m) m / s s so the angular velocity of the wheel is ω θ= = × × = × − −t 126 10 334 10 38 10 2 6 3. . .rad s rad / s. (b) If r is the radius of the wheel, the linear speed of a point on its rim is ( ) ( )3 23.8 10 rad/s 0.050 m 1.9 10 m/s.v rω= = × = × 30. Since the belt does not slip, a point on the rim of wheel C has the same tangential acceleration as a point on the rim of wheel A. This means that αArA = αCrC, where αA is the angular acceleration of wheel A and αC is the angular acceleration of wheel C. Thus, α αC A C C r r = F HG I KJ = F HG I KJ = 10 25 16 0 64cm cm rad / s rad / s2 2( . ) . . Since the angular speed of wheel C is given by ωC = αCt, the time for it to reach an angular speed of ω = 100 rev/min = 10.5 rad/s starting from rest is t C C = = =ω α 10 5 0 64 16. . rad / s rad / s s.2 31. (a) The angular speed in rad/s is ω = FHG I KJ F HG I KJ =33 1 3 2 60 349rev / min rad / rev s / min rad / s.π . Consequently, the radial (centripetal) acceleration is (using Eq. 10-23) a r= = × =−ω 2 2 2349 6 0 10. ( . .rad / s m) 0.73 m / s2b g (b) Using Ch. 6 methods, we have ma = fs ≤ fs,max = μs mg, which is used to obtain the (minimum allowable) coefficient of friction: μ s a g,min . . . .= = =0 73 9 8 0 075 (c) The radial acceleration of the object is ar = ω2r, while the tangential acceleration is at = αr. Thus, | | ( ) ( ) .a a a r r rr t= + = + = + 2 2 2 2 2 4 2ω α ω α If the object is not to slip at any time, we require f mg ma mrs s,max max max .= = = +μ ω α 4 2 Thus, since α = ω/t (from Eq. 10-12), we find 4 2 4 2 4 2 max max max ,min ( / ) (0.060) 3.49 (3.4 / 0.25) 0.11. 9.8s r r t g g ω α ω ω μ + + + = = = = 32. (a) A complete revolution is an angular displacement of Δθ = 2π rad, so the angular velocity in rad/s is given by ω = Δθ/T = 2π/T. The angular acceleration is given by α ω= = −d dt T dT dt 2 2 π . For the pulsar described in the problem, we have dT dt = × × = × − −126 10 316 10 4 00 10 5 7 13. . . .s / y s / y Therefore, α = − F HG I KJ × = − × − −2 0 033 4 00 10 2 3 1013 9π ( . ( . ) . . s) rad / s2 2 The negative sign indicates that the angular acceleration is opposite the angular velocity and the pulsar is slowing down. (b) We solve ω = ω0 + αt for the time t when ω = 0: 10 30 9 2 2 2 8.3 10 s 2.6 10 years ( 2.3 10 rad/s )(0.033 s) t T ω π π α α − = − = − = − = × ≈ × − × (c) The pulsar was born 1992–1054 = 938 years ago. This is equivalent to (938 y)(3.16 × 107 s/y) = 2.96 × 1010 s. Its angular velocity at that time was 9 2 10 0 2 2 ( 2.3 10 rad/s )( 2.96 10 s) 258 rad/s. 0.033 s t t T ω ω α α −π π= + + + = + − × − × = Its period was T = = = × −2 2 258 2 4 10 2π π ω rad / s s.. 35. We use the parallel axis theorem: I = Icom + Mh2, where Icom is the rotational inertia about the center of mass (see Table 10-2(d)), M is the mass, and h is the distance between the center of mass and the chosen rotation axis. The center of mass is at the center of the meter stick, which implies h = 0.50 m – 0.20 m = 0.30 m. We find I MLcom 2kg m kg m= = = × ⋅−1 12 1 12 0 56 10 4 67 102 2 2. . . .b gb g Consequently, the parallel axis theorem yields I = × ⋅ + = × ⋅− −4 67 10 056 0 30 9 7 102 2 2. . . . .kg m kg m kg m2 2b gb g 36. (a) Eq. 10-33 gives Itotal = md2 + m(2d)2 + m(3d)2 = 14 md2. If the innermost one is removed then we would only obtain m(2d)2 + m(3d)2 = 13 md2. The percentage difference between these is (13 – 14)/14 = 0.0714 ≈ 7.1%. (b) If, instead, the outermost particle is removed, we would have md2 + m(2d)2 = 5 md2. The percentage difference in this case is 0.643 ≈ 64%. 37. Since the rotational inertia of a cylinder is I MR= 12 2 (Table 10-2(c)), its rotational kinetic energy is 2 2 21 1 . 2 4 K I MRω ω= = (a) For the smaller cylinder, we have K = = ×14 2 2 3125 0 25 235 11 10( . )( . ) ( ) . J. (b) For the larger cylinder, we obtain K = = ×14 2 2 3125 0 75 235 9 7 10( . )( . ) ( ) . J. I m ri i= = + = ⋅ 2 2 22 050 10 2 0 50 5 0 6 0. . . . . . kg m kg m kg m2b g c h b g c h (c) Now, two masses are on the axis (with r = 0) and the other two are a distance 2 2(1.0 m) (1.0 m)r = + away. Now we obtain 22.0 kg m .I = ⋅ 40. (a) We show the figure with its axis of rotation (the thin horizontal line). We note that each mass is r = 1.0 m from the axis. Therefore, using Eq. 10-26, we obtain 2 2 24 (0.50 kg) (1.0 m) 2.0 kg m .i iI m r= = = ⋅ (b) In this case, the two masses nearest the axis are r = 1.0 m away from it, but the two furthest from the axis are 2 2(1.0 m) (2.0 m)r = + from it. Here, then, Eq. 10-33 leads to 41. We use the parallel-axis theorem. According to Table 10-2(i), the rotational inertia of a uniform slab about an axis through the center and perpendicular to the large faces is given by I M a bcom = +12 2 2c h. A parallel axis through the corner is a distance h a b= +/ /2 22 2b g b g from the center. Therefore, ( ) ( ) ( )2 2 2 2 2 2 2com 2 2 4 2 12 4 3 0.172 kg [(0.035 m) (0.084 m) ] 3 4.7 10 kg m . M M MI I Mh a b a b a b − = + = + + + = + = + = × ⋅ I = 0.083519ML2 ≈ (0.08352)(0.1000 kg)(1.0000 m)2 = 8.352 10−3 kg m2. (b) Comparing to the formula (e) in Table 10-2 (which gives roughly I =0.08333 ML2), we find our answer to part (a) is 0.22% lower. 42. (a) Consider three of the disks (starting with the one at point O): ⊕OO . The first one (the one at point O – shown here with the plus sign inside) has rotational inertial (see item (c) in Table 10-2) I = 12 mR 2. The next one (using the parallel-axis theorem) has I = 12 mR 2 + mh2 where h = 2R. The third one has I = 12 mR 2 + m(4R)2. If we had considered five of the disks OO⊕OO with the one at O in the middle, then the total rotational inertia is I = 5(12 mR 2) + 2(m(2R)2 + m(4R)2). The pattern is now clear and we can write down the total I for the collection of fifteen disks: I = 15(12 mR 2) + 2(m(2R)2 + m(4R)2 + m(6R)2+ … + m(14R)2) = 22552 mR 2. The generalization to N disks (where N is assumed to be an odd number) is I = 16(2N 2 + 1)NmR2. In terms of the total mass (m = M/15) and the total length (R = L/30), we obtain 45. Two forces act on the ball, the force of the rod and the force of gravity. No torque about the pivot point is associated with the force of the rod since that force is along the line from the pivot point to the ball. As can be seen from the diagram, the component of the force of gravity that is perpendicular to the rod is mg sin θ. If is the length of the rod, then the torque associated with this force has magnitude sinmgτ θ= = (0.75)(9.8)(1.25)sin 30° = 4.6 N m⋅ . For the position shown, the torque is counter-clockwise. 46. We compute the torques using τ = rF sin φ. (a) For 30φ = ° , (0.152 m)(111 N)sin 30 8.4 N maτ = ° = ⋅ . (b) For 90φ = ° , (0.152 m)(111 N)sin 90 17 N mbτ = ° = ⋅ . (c) For 180φ = ° , (0.152 m)(111N)sin180 0cτ = ° = . 47. We take a torque that tends to cause a counterclockwise rotation from rest to be positive and a torque tending to cause a clockwise rotation to be negative. Thus, a positive torque of magnitude r1 F1 sin θ1 is associated with F1 and a negative torque of magnitude r2F2 sin θ2 is associated with F2 . The net torque is consequently τ θ θ= −r F r F1 1 1 2 2 2sin sin . Substituting the given values, we obtain (1.30 m)(4.20 N)sin 75 (2.15 m)(4.90 N)sin 60 3.85 N m.τ = ° − ° = − ⋅ 50. The rotational inertia is found from Eq. 10-45. I = = = ⋅τ α 32 0 25 0 128. . . kg m2 51. Combining Eq. 10-45 (τnet = I α) with Eq. 10-38 gives RF2 – RF1 = Iα , where / tα ω= by Eq. 10-12 (with ωο = 0). Using item (c) in Table 10-2 and solving for F2 we find F2 = MRω 2t + F1 = (0.02)(0.02)(250) 2(1.25) + 0.1 = 0.140 N. 52. With counterclockwise positive, the angular acceleration α for both masses satisfies ( )2 21 2 1 2 ,mgL mgL I mL mLτ α α= − = = + by combining Eq. 10-45 with Eq. 10-39 and Eq. 10-33. Therefore, using SI units, ( ) ( )( )21 2 2 2 2 2 2 1 2 9.8 m/s 0.20 m 0.80 m 8.65 rad/s (0.20 m) (0.80 m) g L L L L α −− = = = − + + where the negative sign indicates the system starts turning in the clockwise sense. The magnitude of the acceleration vector involves no radial component (yet) since it is evaluated at t = 0 when the instantaneous velocity is zero. Thus, for the two masses, we apply Eq. 10-22: (a) ( )( )21 1| | 8.65 rad/s 0.20 m 1.7 m/s.a Lα= = = (b) ( )( )2 22 2| | 8.65 rad/s 0.80 m 6.9 m/s .a Lα= = = 55. (a) We use constant acceleration kinematics. If down is taken to be positive and a is the acceleration of the heavier block m2, then its coordinate is given by y at= 12 2 , so a y t = = = × −2 2 0 750 5 00 6 00 102 2 2 2( . ) ( . ) . .m s m / s Block 1 has an acceleration of 6.00 × 10–2 m/s2 upward. (b) Newton’s second law for block 2 is 2 2 2m g T m a− = , where m2 is its mass and T2 is the tension force on the block. Thus, ( )2 2 22 2 ( ) (0.500 kg) 9.8 m/s 6.00 10 m/s 4.87 N.T m g a −= − = − × = (c) Newton’s second law for block 1 is 1 1 1 ,m g T m a− = − where T1 is the tension force on the block. Thus, ( )2 2 21 1( ) (0.460 kg) 9.8 m/s 6.00 10 m/s 4.54 N.T m g a −= + = + × = (d) Since the cord does not slip on the pulley, the tangential acceleration of a point on the rim of the pulley must be the same as the acceleration of the blocks, so α = = × × = − − a R 6 00 10 500 10 120 2 2 2 2. . . .m / s m rad / s (e) The net torque acting on the pulley is 2 1( )T T Rτ = − . Equating this to Iα we solve for the rotational inertia: ( ) ( )( )22 1 2 2 2 4.87 N 4.54 N 5.00 10 m 1.38 10 kg m . 1.20 rad/s T T R I α − − − ×− = = = × ⋅ 56. (a) In this case, the force is mg = (70 kg)(9.8 m/s2), and the “lever arm” (the perpendicular distance from point O to the line of action of the force) is 0.28 m. Thus, the torque (in absolute value) is (70 kg)(9.8 m/s2)(0.28 m). Since the moment-of-inertia is I = 65 kg·m2, then Eq. 10-45 gives |α| = 2.955 ≈ 3.0 rad/s2. (b) Now we have another contribution (1.4 m × 300 N) to the net torque, so |τnet| = (70 kg)(9.8 m/s2)(0.28 m) + (1.4 m)(300 N) = (65 kg·m2) |α| which leads to |α| = 9.4 rad/s2. 57. Since the force acts tangentially at r = 0.10 m, the angular acceleration (presumed positive) is α τ= = = + × = +−I Fr I t t t t 05 0 3 010 10 10 50 30 2 3 2 . . . . c hb g in SI units (rad/s2). (a) At t = 3 s, the above expression becomes α = 4.2 102 rad/s2. (b) We integrate the above expression, noting that ωo = 0, to obtain the angular speed at t = 3 s: ( )3 2 3 3 200 25 10 5.0 10 rad/s.dt t tω α= = + = × 60. (a) The speed of v of the mass m after it has descended d = 50 cm is given by v2 = 2ad (Eq. 2-16). Thus, using g = 980 cm/s2, we have v ad mg d M m = = + = + = ×2 2(2 2 4(50)(980)(50) 2(50) 1.4 10 cm / s.2) 400 (b) The answer is still 1.4 × 102 cm/s = 1.4 m/s, since it is independent of R. 61. With ω = (1800)(2π/60) = 188.5 rad/s, we apply Eq. 10-55: 74600 W 396 N m 188.5 rad/s P τω τ= = = ⋅ . 62. (a) We use the parallel-axis theorem to find the rotational inertia: ( )( ) ( )( )2 22 2 2 2com 1 1 20 kg 0.10 m 20 kg 0.50 m 0.15 kg m . 2 2 I I Mh MR Mh= + = + = + = ⋅ (b) Conservation of energy requires that Mgh I= 12 ω 2 , where ω is the angular speed of the cylinder as it passes through the lowest position. Therefore, 2 2 2 2(20 kg) (9.8 m/s ) (0.050 m) 11 rad/s. 0.15 kg m Mgh I ω = = = ⋅ 65. Using the parallel axis theorem and items (e) and (h) in Table 10-2, the rotational inertia is I = 112 mL 2 + m(L/2)2 + 12 mR 2 + m(R + L)2 = 10.83mR2 , where L = 2R has been used. If we take the base of the rod to be at the coordinate origin (x = 0, y = 0) then the center of mass is at y = mL/2 + m(L + R) m + m = 2R . Comparing the position shown in the textbook figure to its upside down (inverted) position shows that the change in center of mass position (in absolute value) is |Δy| = 4R. The corresponding loss in gravitational potential energy is converted into kinetic energy. Thus, K = (2m)g(4R) ω = 9.82 rad/s . where Eq. 10-34 has been used. 2 21 1 2 2 3 3 2 2 1 ( / ) (2 / 3 ) 2(9.8)(0.82) 1.4 m/s 1 3.0 10 /((0.60)(0.050) ) 2(4.5) / 3(0.60) I M r mgh ghv m I mr M m − = = + + + + = = + × + 66. From Table 10-2, the rotational inertia of the spherical shell is 2MR2/3, so the kinetic energy (after the object has descended distance h) is K MR I mv= FHG I KJ + + 1 2 2 3 1 2 1 2sphere pulley 2 2 2 2ω ω . Since it started from rest, then this energy must be equal (in the absence of friction) to the potential energy mgh with which the system started. We substitute v/r for the pulley’s angular speed and v/R for that of the sphere and solve for v. 67. (a) We use conservation of mechanical energy to find an expression for ω2 as a function of the angle θ that the chimney makes with the vertical. The potential energy of the chimney is given by U = Mgh, where M is its mass and h is the altitude of its center of mass above the ground. When the chimney makes the angle θ with the vertical, h = (H/2) cos θ. Initially the potential energy is Ui = Mg(H/2) and the kinetic energy is zero. The kinetic energy is 12 2Iω when the chimney makes the angle θ with the vertical, where I is its rotational inertia about its bottom edge. Conservation of energy then leads to MgH Mg H I MgH I/ ( / ) / ) (2 2 12= + = −cos 1 2 ( cos ).2θ ω ω θ The rotational inertia of the chimney about its base is I = MH2/3 (found using Table 10-2(e) with the parallel axis theorem). Thus 23 3(9.80 m/s )(1 cos ) (1 cos35.0 ) 0.311 rad/s. 55.0 m g H ω θ= − = − ° = (b) The radial component of the acceleration of the chimney top is given by ar = Hω2, so ar = 3g (1 – cos θ) = 3 (9.80 m/s2)(1– cos 35.0 ° ) = 5.32 m/s2 . (c) The tangential component of the acceleration of the chimney top is given by at = Hα, where α is the angular acceleration. We are unable to use Table 10-1 since the acceleration is not uniform. Hence, we differentiate ω2 = (3g/H)(1 – cos θ) with respect to time, replacing dω / dt with α, and dθ / dt with ω, and obtain d dt g H g Hω ωα ω θ α θ 2 2= = =2 (3 sin (3 sin ./ ) / ) Consequently, 2 23(9.80 m/s )3 sin sin 35.0 8.43 m/s .2 2 ga Ht α θ= = = ° = (d) The angle θ at which at = g is the solution to 32 g g sin θ = . Thus, sin θ = 2/3 and we obtain θ = 41.8°. 70. In the calculation below, M1 and M2 are the ring masses, R1i and R2i are their inner radii, and R1o and R2o are their outer radii. Referring to item (b) in Table 10-2, we compute I = 12 M1 (R1i 2 + R1o2) + 12 M2 (R2i 2 + R2o2) = 0.00346 kg·m2 . Thus, with Eq. 10-38 (τ = rF where r = R2o) and τ = Iα (Eq. 10-45), we find α = (0.140)(12.0)0.00346 = 485 rad/s 2 . Then Eq. 10-12 gives ω = αt = 146 rad/s. 71. The volume of each disk is πr2h where we are using h to denote the thickness (which equals 0.00500 m). If we use R (which equals 0.0400 m) for the radius of the larger disk and r (which equals 0.0200 m) for the radius of the smaller one, then the mass of each is m = ρπr2h and M = ρπR2h where ρ = 1400 kg/m3 is the given density. We now use the parallel axis theorem as well as item (c) in Table 10-2 to obtain the rotation inertia of the two-disk assembly: I = 12 MR 2 + 12 mr 2 + m(r + R)2 = ρπh[ 12 R 4 + 12 r 4 + r2(r + R)2 ] = 6.16 × 10−5 kg.m2. 72. (a) The longitudinal separation between Helsinki and the explosion site is 102 25 77 .θΔ = ° − ° = ° The spin of the earth is constant at ω = = °1 rev 1 day 360 24 h so that an angular displacement of Δθ corresponds to a time interval of Δt = ° ° F HG I KJ =77 24 h 360 5.1 h.b g (b) Now Δθ = °− − ° = °102 20 122b g so the required time shift would be Δt = ° ° F HG I KJ =122 24 360 81b g h h. . ( ) ( ) ( )222 2 0 5 110kg 33.5 rad s 7.80m 2 2 4.81 10 N. LM M LF dF rdm rdr L ωω ω= = = = = = × (b) About its center of mass, the blade has I ML= 2 12/ according to Table 10-2(e), and using the parallel-axis theorem to “move” the axis of rotation to its end-point, we find the rotational inertia becomes I ML= 2 / 3. Using Eq. 10-45, the torque (assumed constant) is ( )( )22 41 1 33.5rad/s110kg 7.8 m 1.12 10 N m. 3 3 6.7s I ML t ωτ α Δ= = = = × ⋅ Δ (c) Using Eq. 10-52, the work done is ( )( ) ( )2 22 2 2 61 1 1 10 110kg 7.80m 33.5rad/s 1.25 10 J. 2 2 3 6 W K I MLω ω= Δ = − = = = × 75. The Hint given in the problem would make the computation in part (a) very straightforward (without doing the integration as we show here), but we present this further level of detail in case that hint is not obvious or — simply — in case one wishes to see how the calculus supports our intuition. (a) The (centripetal) force exerted on an infinitesimal portion of the blade with mass dm located a distance r from the rotational axis is (Newton’s second law) dF = (dm)ω2r, where dm can be written as (M/L)dr and the angular speed is ( )( )320 2 60ω = π = 33.5 rad s . Thus for the entire blade of mass M and length L the total force is given by 76. The wheel starts turning from rest (ω0 = 0) at t = 0, and accelerates uniformly at 22.00 rad/sα = .Between t1 and t2 the wheel turns through Δθ = 90.0 rad, where t2 – t1 = Δt = 3.00 s. We solve (b) first. (b) We use Eq. 10-13 (with a slight change in notation) to describe the motion for t1 ≤ t ≤ t2: Δ Δ Δ Δ Δ Δθ ω α ω θ α= + = −1 2 1 1 2 2 t t t t( ) which we plug into Eq. 10-12, set up to describe the motion during 0 ≤ t ≤ t1: 1 0 1 1 1 90.0 (2.00) (3.00) (2.00) 2 3.00 2 tt t t t θ αω ω α αΔ Δ= + − = − = Δ yielding t1 = 13.5 s. (a) Plugging into our expression for ω1 (in previous part) we obtain ω θ α1 2 90 0 3 00 2 00 3 00 2 27 0= − = − =Δ Δ Δ t t . . ( . )( . ) . rad / s. Then (assuming α = 0) Eq. 10-13 gives θ − θo = ωo t = (90 rad/s)(2.094 s) = 188 rad, which is equivalent to roughly 30 rev. 77. To get the time to reach the maximum height, we use Eq. 4-23, setting the left-hand side to zero. Thus, we find t = (60 m/s)sin(20o) 9.8 m/s2 = 2.094 s. 80. (a) Eq. 10-12 leads to 2o / (25.0 rad/s) /(20.0 s) 1.25 rad/s .tα ω= − = − = − (b) Eq. 10-15 leads to o 1 1 (25.0 rad/s)(20.0 s) 250 rad. 2 2 tθ ω= = = (c) Dividing the previous result by 2π we obtain θ = 39.8 rev. I Mr= = = ⋅2 2130 0 780 0 791. . . .kg m kg m2b gb g (b) The torque that must be applied to counteract the effect of the drag is τ = = × = × ⋅− −rf 0 780 2 30 10 179 102 2. . .m N N m.b gc h 81. (a) With r = 0.780 m, the rotational inertia is 82. The motion consists of two stages. The first, the interval 0 ≤ t ≤ 20 s, consists of constant angular acceleration given by α = =5 0 2 0 2 5 2. . . .rad s s rad s The second stage, 20 < t ≤ 40 s, consists of constant angular velocity ω θ= Δ Δ/ .t Analyzing the first stage, we find 2 1 20 20 1 500 rad, 50 rad s. 2 tt t tθ α ω α = = = = = = Analyzing the second stage, we obtain ( )( ) 32 1 500 rad 50 rad/s 20 s 1.5 10 rad.tθ θ ω= + Δ = + = × 85. Eq. 10-40 leads to τ = mgr = (70 kg) (9.8 m/s2) (0.20 m) = 1.4 × 102 N·m. 86. (a) Using Eq. 10-15, we have 60.0 rad = 12 (ω1 + ω2)(6.00 s) . With ω2 = 15.0 rad/s, then ω1 = 5.00 rad/s. (b) Eq. 10-12 gives α = (15.0 rad/s – 5.0 rad/s)/(6.00 s) = 1.67 rad/s2. (c) Interpreting ω now as ω1 and θ as θ1 = 10.0 rad (and ωo = 0) Eq. 10-14 leads to θo = – 2 1 2 ω α + θ1 = 2.50 rad . sign from the block’s acceleration (which we simply denote as a); that is, at = – a. Applying Newton’s second law to the block leads to P T ma− = , where 2.0 kg.m = Applying Newton’s second law (for rotation) to the wheel leads to TR Iα− = , where 20.050 kg m .I = ⋅ Noting that Rα = at = – a, we multiply this equation by R and obtain 2 2 . ITR Ia T a R − = − = Adding this to the above equation (for the block) leads to 2( / ) .P m I R a= + Thus, a = 0.92 m/s2 and therefore α = – 4.6 rad/s2 (or |α| = 4.6 rad/s2 ), where the negative sign in α should not be mistaken for a deceleration (it simply indicates the clockwise sense to the motion). 87. With rightward positive for the block and clockwise negative for the wheel (as is conventional), then we note that the tangential acceleration of the wheel is of opposite 90. We use conservation of mechanical energy. The center of mass is at the midpoint of the cross bar of the H and it drops by L/2, where L is the length of any one of the rods. The gravitational potential energy decreases by MgL/2, where M is the mass of the body. The initial kinetic energy is zero and the final kinetic energy may be written 12 2Iω , where I is the rotational inertia of the body and ω is its angular velocity when it is vertical. Thus, 0 2 2= − + =MgL I MgL I/ / .1 2 ω ω Since the rods are thin the one along the axis of rotation does not contribute to the rotational inertia. All points on the other leg are the same distance from the axis of rotation, so that leg contributes (M/3)L2, where M/3 is its mass. The cross bar is a rod that rotates around one end, so its contribution is (M/3)L2/3 = ML2/9. The total rotational inertia is I = (ML2/3) + (ML2/9) = 4ML2/9. Consequently, the angular velocity is 2 2 9 9(9.800 m/s ) 6.06 rad/s. 4 / 9 4 4(0.600 m) MgL MgL g I ML L ω = = = = = 91. (a) According to Table 10-2, the rotational inertia formulas for the cylinder (radius R) and the hoop (radius r) are given by I MR I MrC H= = 1 2 2 2and . Since the two bodies have the same mass, then they will have the same rotational inertia if R RH 2 22/ = R RH = / 2 . (b) We require the rotational inertia to be written as I Mk= 2 , where M is the mass of the given body and k is the radius of the “equivalent hoop.” It follows directly that k I M= / . I = = ⋅ = ⋅τ α 960 6 20 155N m rad / s kg m2 2 . . (b) The rotational inertia of the shell is given by I = (2/3) MR2 (see Table 10-2 of the text). This implies M I R = = ⋅ =3 2 3 155 2 190 64 42 2 2 kg m m kg c h b g. . . 92. (a) We use τ = Iα, where τ is the net torque acting on the shell, I is the rotational inertia of the shell, and α is its angular acceleration. Therefore, (a) If points A and P are at a radial distance rA=1.50 m and r = 0.150 m from the axis, the difference in their acceleration is 2 2 4 2( ) (209.4 rad/s) (1.50 m 0.150 m) 5.92 10 m/sA Aa a a r rωΔ = − = − = − ≈ × (b) The slope is given by 2 4 2/ 4.39 10 / sa r ω= = × . 95. The centripetal acceleration at a point P which is r away from the axis of rotation is given by Eq. 10-23: 2 2/a v r rω= = , where v rω= , with 2000 rev/min 209.4 rad/s.ω = ≈ 96. Let T be the tension on the rope. From Newton’s second law, we have ( )T mg ma T m g a− = = + . Since the box has an upward acceleration a = 0.80 m/s2, the tension is given by 2 2(30 kg)(9.8 m/s 0.8 m/s ) 318 N.T = + = The rotation of the device is described by app /F R Tr I Ia rα− = = . The moment of inertia can then be obtained as app 2 2 ( ) (0.20 m)[(140 N)(0.50 m) (318 N)(0.20 m)] 1.6 kg m 0.80 m/s r F R Tr I a − −= = = ⋅ 97. The distances from P to the particles are as follows: r a m M r b a m M r a m M 1 2 2 2 3 2 2 = = = − = = = for lower left for top for lower right 1 2 1 b g b g b g The rotational inertia of the system about P is I m r a b Mi i i = = + = 2 1 3 2 23c h which yields 20.208 kg mI = ⋅ for M = 0.40 kg, a = 0.30 m and b = 0.50 m. Applying Eq. 10-52, we find ( )( )22 21 1 0.208 kg m 5.0 rad/s 2.6 J. 2 2 W Iω= = ⋅ =