Chapter12 - t"roDlem I' ! __ ._ ... 1. Determine the...

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Unformatted text preview: t"roDlem I' ! __ ._ ... 1. Determine the angular speed, in rad/s, of (a) Earth about its axis; (b) the minute hand of a clock; (c) the hour hand of a clock; (d) an egg beater ','turning at 300 rpm. Splution The angular speed is w A8/At ..(a) WE 1 rev+ 1d 211"/86,400 7.27x10-5 S-I. (b) Wmin s 1 rev/1 h = 21t/3600 s 1.75x10-3 S-I. (c) Wh'r = 1 rev/12 h = l2Wmin = 1.45xlO-4 S-I. (d) W = 390 rev/min = 300x211"/60 = 31.4\s-l. (Note: s Radians are a dimensionless angular measure, i.e., pure numbersj therefor~ angular speed can be expressed in units of inverse secondS.) = = = = = = 35. A neutron star is an extremely dense, rapidly spinning object that results from the collapse of a star at the end of its life. A neutron star of 1.8 times the Sun's mass has an approximately uniform density of 1x1018 kg/m3. (a) What is its rotational inertia? (b) The neutron star's spin rate slowly decreases as a result of torque associated with magnetiC forces. If the spin-down rate is 5x10-5 radii, what is the magnetic torque? Solution :~ Pro~lem 7. A:icoinpact disc (CD) player varies the rotation rate ofi ~he ~sc in order to keep the part of the disc from w~Ich mformation is being read moving at a const.ant linear speed of 1.30 m/s. Compare the ~otatlOn.rates of a 12.O-cm-diameterCD when mformatlOn is being read from (a) its outer edge and (b) ~.point 3.75 cm from the center. Give your answers In rad/s and rpm. (a) For a uniform sphei~ about an axis through its center, I = ~MR2. Since the density is p = M+ (l7l'R3), the radius can be eliminated to yield I = M(3M/47l'p)2/3 =0.4M5/3(O.75/7l'p)2/3 = . 0.4[(1.8)(1.99xl03o kg)]5/3[0.75/7l'(1018 kg/m3)]2/3 = 1.29x 1038 kg.m2. (b) The torque responsible for the spin-down rate (i.e., the angular deceleration) is T = Ia (1.29x 1038 kg.m2)(-5x10-5 s-2) = -6.45x1033 N.m, the negative value indicating a direction opposite to the angular velocity. i = Problem 39. A space station is constructed in the shape of a wheel 22 m in di~etdr, with essentially all of its 5.0x105-kg mass at the rim (Fig. 12-45). Once the station is completed, it is set rotating at a rate that requires an object at the rim to have radial acceleration g, thereby simulating Earth's surface gravity. This is accomplished using two small rockets, each with 100 N thrust, that are mounted on the rim of the station as shown. (8.) How long will it take to reach the desired spin rate? (b) How many revolutions will the station make in this time? So'lution Eq~ationll2-4 gives the relation between linear speed angu ar speed, w = v /T, where r is the distance frOJ;Il center of rotation. With v a Constant the 13q cm/s, (a) w = (130 cm/s)/(6 em) = 21.7 S-1 = 20~1pm for a point on the CD's outer edge, and r (b):w = (130 cm!s)/(3.7? cm) = 34.7 s-1 = 331 rpm at th~ second pomt specIfied. (See the note to the solutIon of Problem 1 and solution of Problem 3 the conversion of angular units.) or 8Du Solution (a) Suppose that any difference in the distance to the axis from the center toithe rockets, or any part of the ring, is negligible comp~red to the radius of the ring, R = 11 m. The angular acceleration (about the axis perpendicular to the ring and through its center) of the ring is a = T / I = 2FR/MR2 Starting from rest (wo = 0), it would take time t = w/a for the ring to reach'a final angular speed w; Since 9 = w2 R specifies the desired rate of rotation, we find t Problem 19. A 55-g mouse runs out to the end of the 17-cm-Iongminute hand of a grandfather clock when the clock reads 10 minutes past the hour. What torque does the mouse's weight exert about ~he rotation axis of the clock hand? Solution The.!angle between the minute hand (for the clock in upright position) and the weight of the mo~se (vertically downward) is 1200 at ten past the hour so the magnitude of the torque exerted is T = rmg s~ 8 = (17 cm)(55 g)(9.8 m/s2) sin 120 = 7.94x 10-2 N'm. . The direction of the torque is clockwise~ 0 !I = = = ~=~(~:) -----2 (5x105 kg)V(9.8 m/s )(11 m) ,! 2(100 N) 2.60x 10 s = 7.21 h. 4 (b) From eith~r Equation 12-8 or 12-10, 8 - 80 = ~at2 = ~wt = ~Jg/Rt = h/(9.8 m/s2)/(11 m)x (2.60~104 s)(l rev/27l') ~ 1950 rev. Problem 19 Solution. F , , Problem 44. At the MIT Magnet Laboratory, energy is stored 4 in huge solid flywheels of mass 7.7x 10 kg and radius 2.4 m. The flywheels ride on shafts 41 cm in diameter. If a frictional force of 34 kN acts tangentially on the shaft, how iong will it take the flywheel to coast to a stop from its normal . rotation rate of 360 rpm? Problem 56. A hollow ball is rolling along a horizontal surface at 3.7 mfs when it encounters an upward incline. If it rolls without slipping up the incline, what maximum height will it reach? Solution The conservation of mechanical energy, of course, also applies to an object rolling without slipping up an incline, .so 1f h is the maximum vertical height (where Vern at the bottotn is given by the expression derived in the previous p~oblem, and Vern 0 at the top), then h (1 + a)v~mf2'g. In this problem, a for a hollow ball, so h (1 + :~)(3.7 mfs)2f2(9.8 mfs ) 1.16 m. Solution The frictional torque decelerates the flywheel, so Equations 12-9 and 14 yield WI = 0 = Wo + at, and 1" == -heRs [a, or t = -wofa = wolf fkRs, where R is the radius of the shaft. Since the radius of the Si flyWheel, Rw, is much larger than Rs, the rotational inertia of the wheel is approximately that of a solid disk, ~M.n;lherefore t = (360 rpm)(lI' S-1 f30 rpm)x t H7.7x104 kg)(2.4 m)2f(34 kN)(O.205 m) = 1.20x 103 s 20.0 min. (Note: The result in Problem 30 gives the exact I for the flywheel as !M(R~ + R~), which is just 0.7% different from ~M~ for the given = = = = =~ = Problem 69. A. so~id marble starts from rest and rolls without s~ppmg on the loop-the-loop track shown in F1g. 12-53. Find the minimum starting height of . the marble from which it will remain on the track through the loop. Assume the marble radius is small compared with R. = radii.) Problem 47. A 25-cm-diameter circular saw blade has a mass of 0.85 kg, distributed uniformly as in a disk. (a) What is its rotational kinetic energy at 3500 rpm? (b) What average power must be applied to bring the blade from rest to 3500 rpm in 3.2 s? II Solution The eM of the marble travels in a circle of ra.dius R - r inside the loop-the-Ioop, so at the top, mg + N = mv2 f(R - r). To remain in contact with the track, N ~ 0, or v2 ~ g(R - r). If we assume that energy is conserved between points A (the start) and B (the top of the loop), and use KA = 0 and KB = (1 + ~Hmv~, we find UA +KA mg(h + r) = UB +KB = mg(2R - r) + {omvt, or h 2(R - r) + {o(vtjg) ~ 2(R - r) + {oCR- r) = 2.7(R - r). i S61ution II (~) K (0,.125 m~(35001l'f30 S)2 = 446 J. (b) "fIt = tlKft 446 Jj3.2 s = 139 W. = !Iw2 = !(~MR2)W2 = ~(0.85 kg) x = Pay = = = Problem 52. IA flywheel for energy storage in a hybrid gasoline-electric car is a ring-shaped rotor of mass 48 kg, designed to rotate at 30,000 rpm; at that speed it stores 10 MJ of energy (equivalent to about a cup of gasoline). (a) What is the ring's diameter? (b) If the flyWheel can supply power at the rate of 25 kW, how long will it take for its stored energy to be exhausted? 2R-r ~ t Solution (a) The rotational kinetic energy of a ring (about a perpendicular axis through its center, which is 2 2 implicitly assumed here) is Krot VW2 HMR )w , so the diameter 2(30,000'ITj30 S)-1 J2xl07 Jf48 kg = 41.1 em. (b) If the entire stored energy were depleted at the average rate given, it would take t = WfP 10 MJj25 kW = = is 2R = (2fw)J2KTM = = = 400 s = 6.67 min. ...
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This note was uploaded on 07/27/2009 for the course PHY 557 taught by Professor Rijssenbeek during the Spring '08 term at Adelphi.

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