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Chapter 101. The problem asks us to assume vcom and are constant. For consistency of units, wewrite5280 ft mivcom 85 mi h 7480 ft min .60 min hg FGHbIJKThus, with x 60 ft , the time of flight ist x vcom (60 ft) /(7480 ft/min) 0.00802min .During that time, the angular displacement of a point on the ball’s surface isbgbg t 1800 rev min 0.00802 min 14 rev .2. (a) The second hand of the smoothly running watch turns through 2 radians during60 s . Thus,2 0.105 rad/s.60(b) The minute hand of the smoothly running watch turns through 2 radians during3600 s . Thus,2 175. 10 3 rad / s.3600(c) The hour hand of the smoothly running 12-hour watch turns through 2 radiansduring 43200 s. Thus,2 145. 10 4 rad / s.432003. The falling is the type of constant-acceleration motion you had in Chapter 2. The timeit takes for the buttered toast to hit the floor is t 2h2(0.76 m) 0.394 s.g9.8 m/s 2(a) The smallest angle turned for the toast to land butter-side down is min 0.25 rev / 2 rad. This corresponds to an angular speed of477
478CHAPTER 10 min min / 2 rad 4.0 rad/s. t0.394 s(b) The largest angle (less than 1 revolution) turned for the toast to land butter-side downis max 0.75 rev 3 / 2 rad. This corresponds to an angular speed of max max 3 / 2 rad 12.0 rad/s. t0.394 s4. If we make the units explicit, the function is 2.0 rad 4.0 rad/s2 t 2 2.0 rad/s3 t 3but in some places we will proceed as indicated in the problem—by letting these units beunderstood.(a) We evaluate the function at t 0 to obtain 0 2.0 rad.(b) The angular velocity as a function of time is given by Eq. 10-6: d 8.0 rad/s 2 t 6.0 rad/s3 t 2dtwhich we evaluate at t 0 to obtain 0 0.(c) For t 4.0 s, the function found in the previous part is 4 (8.0)(4.0) (6.0)(4.0)2 128 rad/s.If we round this to two figures, we obtain 4 1.3 102 rad/s.(d) The angular acceleration as a function of time is given by Eq. 10-8: d 8.0 rad/s 2 12 rad/s3 tdtwhich yields 2 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, dependson time; it is not constant.5. Applying Eq. 2-15 to the vertical axis (with y downward) we obtain the free-fall time:
47912(10 m) y v0 y t gt 2 t 1.4 s.29.8 m/s 2Thus, by Eq. 10-5, the magnitude of the average angular velocity is avg (2.5 rev) (2 rad/rev) 11 rad/s.1.4 s6. If we make the units explicit, the function isbg ch ch 4.0 rad / s t 3.0 rad / s2 t 2 10. rad / s3 t 3but generally we will proceed as shown in the problem—letting these units be understood.Also, in our manipulations we will generally not display the coefficients with their propernumber of significant figures.(a) Equation 10-6 leads to chd4t 3t 2 t 3 4 6t 3t 2 .dtEvaluating this at t 2 s yields 2 4.0 rad/s.(b) Evaluating the expression in part (a) at t 4 s gives 4 28 rad/s.(c) Consequently, Eq. 10-7 gives avg 4 24 2 12 rad / s2 .(d) And Eq. 10-8 gives d d 4 6t 3t 2 6 6t .dt dtchEvaluating this at t 2 s produces 2 6.0 rad/s2.(e) Evaluating the expression in part (d) at t 4 s yields 4 18 rad/s2. We note that ouranswer for avg does turn out to be the arithmetic average of 2 and 4 but point out thatthis will not always be the case.7. (a) To avoid touching the spokes, the arrow must go through the wheel in not morethan1 / 8 rev t 0.050 s.2.5 rev / s
480CHAPTER 10The minimum speed of the arrow is then vmin 20 cm 400 cm / s 4.0 m / s.0.050 s(b) No—there is no dependence on radial position in the above computation.8. (a) We integrate (with respect to time) the 6.0t4 – 4.0t2 expression, taking intoaccount that the initial angular velocity is 2.0 rad/s. The result is 1.2 t5 – 1.33 t3 2.0.(b) Integrating again (and keeping in mind that o 1) we get 0.20t6 – 0.33 t4 2.0 t 1.0 .9. (a) With 0 and – 4.2 rad/s2, Eq. 10-12 yields t – o/ 3.00 s.(b) Eq. 10-4 gives o o2 / 2 18.9 rad.10. We assume the sense of rotation is positive, which (since it starts from rest) means allquantities (angular displacements, accelerations, etc.) are positive-valued.(a) The angular acceleration satisfies Eq. 10-13:125 rad (5.0 s)2 2.0 rad/s 2 .2(b) The average angular velocity is given by Eq. 10-5: avg 25 rad 5.0 rad / s. t5.0 s(c) Using Eq. 10-12, the instantaneous angular velocity at t 5.0 s is 2.0 rad/s2 (5.0 s) 10 rad/s .(d) According to Eq. 10-13, the angular displacement at t 10 s is1212 0 t 2 0 (2.0 rad/s2 ) (10 s)2 100 rad.Thus, the displacement between t 5 s and t 10 s is 100 rad – 25 rad 75 rad.
48111. We assume the sense of initial rotation is positive. Then, with 0 120 rad/s and 0 (since it stops at time t), our angular acceleration (‘‘deceleration’’) will be negativevalued: – 4.0 rad/s2.(a) We apply Eq. 10-12 to obtain t. 0 t t 0 120 rad/s 30 s. 4.0 rad/s2(b) And Eq. 10-15 gives1212 ( 0 ) t (120 rad/s 0) (30 s) 1.8 103 rad.Alternatively, Eq. 10-14 could be used if it is desired to only use the given information(as opposed to using the result from part (a)) in obtaining . If using the result of part (a)is acceptable, then any angular equation in Table 10-1 (except Eq. 10-12) can be used tofind .12. (a) We assume the sense of rotation is positive. Applying Eq. 10-12, we obtain 0 t (3000 1200) rev/min 9.0 103 rev/min 2 .(12 / 60) min(b) And Eq. 10-15 gives12 12 min 4.2 102 rev. 60 12 ( 0 ) t (1200 rev/min 3000 rev/min) 13. The wheel has angular velocity 0 1.5 rad/s 0.239 rev/s at t 0, and hasconstant value of angular acceleration 0, which indicates our choice for positivesense of rotation. At t1 its angular displacement (relative to its orientation at t 0) is 1 20 rev, and at t2 its angular displacement is 2 40 rev and its angular velocity is 2 0 .(a) We obtain t2 using Eq. 10-15: 2 12(40 rev) 335 s 0 2 t2 t2 20.239 rev/swhich we round off to t2 3.4 102 s .(b) Any equation in Table 10-1 involving can be used to find the angular acceleration;we select Eq. 10-16.
482CHAPTER 1012 2 2t2 t22 2(40 rev) 7.12 10 4 rev/s22(335 s)which we convert to – 4.5 10–3 rad/s2.(c) Using 1 0t1 21 t12 (Eq. 10-13) and the quadratic formula, we havet1 0 02 2 1 (0.239 rev/s) (0.239 rev/s)2 2(20 rev)( 7.12 10 4 rev/s2 ) 7.12 10 4 rev/s 2which 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.14. The wheel starts turning from rest ( 0 0) at t 0, and accelerates uniformly at 0,which makes our choice for positive sense of rotation. At t1 its angular velocity is 1 10 rev/s, and at t2 its angular velocity is 2 15 rev/s. Between t1 and t2 it turnsthrough 60 rev, where t2 – t1 t.(a) We find using Eq. 10-14: 22 12 2 (15 rev/s)2 (10 rev/s)2 1.04 rev/s22(60 rev)which we round off to 1.0 rev/s2.(b) We find t using Eq. 10-15: 12(60 rev) 4.8 s. 1 2 t t 210 rev/s 15 rev/s(c) We obtain t1 using Eq. 10-12: 1 0 t1 t1 10 rev/s 9.6 s.1.04 rev/s2(d) Any equation in Table 10-1 involving can be used to find 1 (the angulardisplacement during 0 t t1); we select Eq. 10-14.(10 rev/s)2 2 1 1 48 rev.2(1.04 rev/s2 )212015. THINK We have a wheel rotating with constant angular acceleration. We can applythe equations given in Table 10-1 to analyze the motion.EXPRESS Since the wheel starts from rest, its angular displacement as a function oftime is given by 12 t 2 . We take t1 to be the start time of the interval so thatt2 t1 4.0 s . The corresponding angular displacements at these times are
4831212 1 t12 , 2 t22Given 2 1 , we can solve for t1 , which tells us how long the wheel has been inmotion up to the beginning of the 4.0 s-interval.ANALYZE The above expressions can be combined to give11 2 1 t22 t12 (t2 t1 )(t2 t1 )22With 120 rad , 3.0 rad/s 2 , and t2 t1 4.0 s , we obtaint2 t1 2( )2(120 rad) 20 s , (t2 t1 ) (3.0 rad/s 2 )(4.0 s)which can be further solved to give t2 12.0 s and t1 8.0 s . So, the wheel started fromrest 8.0 s before the start of the described 4.0 s interval.LEARN We can readily verify the results by calculating 1 and 2 explicitly:11221 2 1 2 t2 (3.0 rad/s 2 )(12.0 s) 2 216 rad.22 1 t12 (3.0 rad/s 2 )(8.0 s) 2 96 radIndeed the difference is 2 1 120 rad .16. (a) Eq. 10-13 gives o o t 1 2 t21 0 2 (1.5 rad/s²) t12where o (2 rev)(2 rad/rev). Therefore, t1 4.09 s.(b) We can find the time to go through a full 4 rev (using the same equation to solve for anew time t2) and then subtract the result of part (a) for t1 in order to find this answer.(4 rev)(2 rad/rev) 0 2 (1.5 rad/s²) t221 t2 5.789 s.Thus, the answer is 5.789 s – 4.093 s 1.70 s.17. The problem has (implicitly) specified the positive sense of rotation. The angularacceleration of magnitude 0.25 rad/s2 in the negative direction is assumed to be constantover a large time interval, including negative values (for t).
484CHAPTER 10(a) We specify max with the condition 0 (this is when the wheel reverses frompositive rotation to rotation in the negative direction). We obtain max using Eq. 10-14: max o2(4.7 rad/s)2 44 rad.2 2( 0.25 rad/s 2 )(b) We find values for t1 when the angular displacement (relative to its orientation at t 0)is 1 22 rad (or 22.09 rad if we wish to keep track of accurate values in all intermediatesteps and only round off on the final answers). Using Eq. 10-13 and the quadratic formula,we have12 1 ot1 t12 t1 o o2 2 1 which yields the two roots 5.5 s and 32 s. Thus, the first time the reference line will be at 1 22 rad is t 5.5 s.(c) The second time the reference line will be at 1 22 rad is t 32 s.(d) We find values for t2 when the angular displacement (relative to its orientation at t 0)is 2 –10.5 rad. Using Eq. 10-13 and the quadratic formula, we have o o2 2 2 1 2 2 o t2 t2 t2 2 which yields the two roots –2.1 s and 40 s. Thus, at t –2.1 s the reference line will be at 2 –10.5 rad.(e) At t 40 s the reference line will be at 2 –10.5 rad.(f) With radians and seconds understood, the graph of versus t is shown below (with thepoints found in the previous parts indicated as small dots).
48518. (a) A complete revolution is an angular displacement of 2 rad, so the angularvelocity in rad/s is given by /T 2 /T. The angular acceleration is given by d 2 dT 2.dtT dtFor the pulsar described in the problem, we havedT 126. 10 5 s / y 4.00 10 13 .7dt316. 10 s / yTherefore, FG 2 IJ (4.00 10H (0.033 s) K2 13) 2.3 10 9 rad / s2 .The negative sign indicates that the angular acceleration is opposite the angular velocityand the pulsar is slowing down.(b) We solve 0 t for the time t when 0:t 02 2 8.3 1010 s 2.6 103 years 9 T( 2.3 10 rad/s 2 )(0.033 s)(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 0 t 2 2 t ( 2.3 10 9 rad/s2 )( 2.96 1010 s) 258 rad/s.T0.033 sIts period wasT 2 2 2.4 10 2 s.258 rad / s19. (a) Converting from hours to seconds, we find the angular velocity (assuming it ispositive) from Eq. 10-18:4v 2.90 10 km/h 1.000 h / 3600 s 2.50 10 3 rad/s.3r3.22 10 km(b) The radial (or centripetal) acceleration is computed according to Eq. 10-23:ch c3.22 10 mh 20.2 m / s .ar 2 r 2.50 10 3 rad / s262
486CHAPTER 10(c) Assuming the angular velocity is constant, then the angular acceleration and thetangential acceleration vanish, since d 0 and at r 0.dt20. The function e t where 0.40 rad and 2 s–1 is describing the angularcoordinate of a line (which is marked in such a way that all points on it have the samevalue of angle at a given time) on the object. Taking derivatives with respect to time2leads to ddt e t and ddt 2 2 e t .(a) Using Eq. 10-22, we have at r d 2 r 6.4 cm / s2 .2dt(b) Using Eq. 10-23, we get ar 2 r FG d IJ r 2.6 cm / s .H dt K2221. We assume the given rate of 1.2 10–3 m/y is the linear speed of the top; it is alsopossible to interpret it as just the horizontal component of the linear speed but thedifference between these interpretations is arguably negligible. Thus, Eq. 10-18 leads to12. 10 3 m / y 2.18 10 5 rad / y55 mwhich we convert (since there are about 3.16 107 s in a year) to 6.9 10–13 rad/s.22. (a) Using Eq. 10-6, the angular velocity at t 5.0s is d dt t 5.0cd0.30t 2dth 2(0.30)(5.0) 3.0 rad / s.t 5.0(b) Equation 10-18 gives the linear speed at t 5.0s: v r (3.0 rad/s)(10 m) 30 m/s.(c) The angular acceleration is, from Eq. 10-8, d d (0.60t ) 0.60 rad / s2 .dt dtThen, the tangential acceleration at t 5.0s is, using Eq. 10-22,chat r (10 m) 0.60 rad / s2 6.0 m / s2 .
487(d) The radial (centripetal) acceleration is given by Eq. 10-23:bgb gar 2 r 3.0 rad / s 10 m 90 m / s2 .223. THINK A positive angular acceleration is required in order to increase the angularspeed of the flywheel.EXPRESS The linear speed of the flywheel is related to its angular speed by v r ,where r is the radius of the wheel. As the wheel is accelerated, its angular speed at a latertime is 0 t .ANALYZE (a) The angular speed of the wheel, expressed in rad/s, is 0 (200 rev/min)(2 rad/rev) 20.9 rad/s.60 s / min(b) With r (1.20 m)/2 0.60 m, using Eq. 10-18, we find the linear speed to bev r 0 (0.60 m)(20.9 rad/s) 12.5 m/s.(c) With t 1 min, 1000 rev/min and 0 200 rev/min, Eq. 10-12 gives the requiredacceleration: 0 800 rev / min 2 .t(d) With the same values used in part (c), Eq. 10-15 becomes 11 0 t (200 rev/min 1000 rev/min)(1.0 min) 600 rev.22LEARN An alternative way to solve for (d) is to use Eq. 10-13:1212 0 0t t 2 0 (200 rev/min)(1.0 min) (800 rev/min 2 )(1.0 min)2 600 rev.24. Converting 33 3 rev/min to radians-per-second, we get 3.49 rad/s. Combining1v r (Eq. 10-18) with t d/v where t is the time between bumps (a distance d apart),we arrive at the rate of striking bumps:1 r 199 / s . t d25. THINK The linear speed of a point on Earth’s surface depends on its distance fromthe Earth’s axis of rotation.
488CHAPTER 10EXPRESS To solve for the linear speed, we use v r, where r is the radius of its orbit.A point on Earth at a latitude of 40 moves along a circular path of radius r R cos40 ,where R is the radius of Earth (6.4 106 m). On the other hand, r R at the equator.ANALYZE (a) Earth makes one rotation per day and 1 d is (24 h) (3600 s/h) 8.64 104 s, so the angular speed of Earth is2 rad 7.3 10 5 rad/s.8.64 104 s(b) At latitude of 40 , the linear speed isv ( R cos 40 ) (7.3 10 5 rad/s)(6.4 106 m)cos40 3.5 102 m/s.(c) At the equator (and all other points on Earth) the value of is the same (7.3 10–5rad/s).(d) The latitude at the equator is 0 and the speed isv R (7.3 10 5 rad/s)(6.4 106 m) 4.6 102 m/s.LEARN The linear speed at the poles is zero since r R cos90 0 .26. (a) The angular acceleration is 0 150 rev / min 114. rev / min2 . t (2.2 h)(60 min / 1h)(b) Using Eq. 10-13 with t (2.2) (60) 132 min, the number of revolutions is12 0t t 2 (150 rev/min)(132 min) 12 1.14 rev/min 2 132 min 9.9 103 rev.2(c) With r 500 mm, the tangential acceleration isF 2 rad IJ FG 1 min IJa r c 114. rev / min h GH 1 rev K H 60 s K2t2(500 mm)which yields at –0.99 mm/s2.(d) The angular speed of the flywheel is (75 rev/min)(2 rad/rev)(1 min/ 60 s) 7.85 rad/s.
489With r 0.50 m, the radial (or centripetal) acceleration is given by Eq. 10-23:ar 2 r (7.85 rad/s)2 (0.50 m) 31 m/s2which is much bigger than at. Consequently, the magnitude of the acceleration is a ar2 at2 ar 31 m / s2 .27. (a) The angular speed in rad/s isFGHIJ FG 2 rad / rev IJ 3.49 rad / s.K H 60 s / min K13 33 rev / minConsequently, the radial (centripetal) acceleration is (using Eq. 10-23)bga 2 r 3.49 rad / s (6.0 10 2 m) 0.73 m / s2 .2(b) Using Ch. 6 methods, we have ma fs fs,max s mg, which is used to obtain the(minimum allowable) coefficient of friction: s,min a 0.73 0.075.g 9.8(c) The radial acceleration of the object is ar 2r, while the tangential acceleration is at r. Thus, a ar2 at2 ( 2 r ) 2 ( r ) 2 r 4 2 .If the object is not to slip at any time, we requiref s,max smg mamax mr 4max 2 .Thus, since t (from Eq. 10-12), we find s ,min 4r max 2g 4r max ( max / t )2g (0.060) 3.494 (3.4 / 0.25) 2 0.11.9.828. Since the belt does not slip, a point on the rim of wheel C has the same tangentialacceleration as a point on the rim of wheel A. This means that ArA CrC, where A isthe angular acceleration of wheel A and C is the angular acceleration of wheel C. Thus,
490CHAPTER 10 C FG r IJ FG 10 cmIJ (16. rad / s ) 0.64 rad / s .H r K H 25 cmK2A2CCWith the angular speed of wheel C given by C C t , the time for it to reach an angularspeed of 100 rev/min 10.5 rad/s starting from rest ist C 10.5 rad / s 16 s. C 0.64 rad / s229. (a) In the time light takes to go from the wheel to the mirror and back again, thewheel turns through an angle of 2 /500 1.26 10–2 rad. That time ist 2 2(500 m) 3.34 10 6 s8c 2.998 10 m / sso the angular velocity of the wheel is 126. 10 2 rad 38. 103 rad / s. 6t3.34 10 s(b) If r is the radius of the wheel, the linear speed of a point on its rim is v r 3.8 103 rad/s 0.050 m 1.9 102 m/s.30. (a) The tangential acceleration, using Eq. 10-22, ischat r 14.2 rad / s2 (2.83 cm) 40.2 cm / s2 .(b) In rad/s, the angular velocity is (2760)(2 /60) 289 rad/s, soar 2r (289 rad / s) 2 (0.0283 m) 2.36 103 m / s2 .(c) The angular displacement is, using Eq. 10-14, 2(289 rad/s)2 2.94 103 rad.22 2(14.2 rad/s )Then, using Eq. 10-1, the distance traveled iss r (0.0283 m) (2.94 103 rad) 83.2 m.
49131. (a) The upper limit for centripetal acceleration (same as the radial acceleration – seeEq. 10-23) places an upper limit of the rate of spin (the angular velocity ) byconsidering a point at the rim (r 0.25 m). Thus, max a/r 40 rad/s. Now we applyEq. 10-15 to first half of the motion (where o 0): o 1( o2 )t 400 rad 1(02 40 rad/s)twhich 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 to40 rad/s20 s o t 2.0 rad/s2 .32. (a) The linear speed at t 15.0 s isdv at t 0.500 m s2i b15.0 sg 7.50 m s .The radial (centripetal) acceleration at that moment isbg7.50 m sv2ar r30.0 mThus, the net acceleration has magnitude:a at2 ar2 2 1.875m s2 .ms hc0.500 m s h c18752 22 2 194. m s2 . (b) We note that at v . Therefore, the angle between v and a istan 1FG a IJ tan FG 1875. IJ 751. H0.5 KHa Kr 1tso that the vector is pointing more toward the center of the track than in the direction ofmotion.33. THINK We want to calculate the rotational inertia of a wheel, given its rotationalenergy and rotational speed.EXPRESS The kinetic energy (in J) is given by K 21 I 2 , where I is the rotationalinertia (in kg m2 ) and is the angular velocity (in rad/s).
492CHAPTER 10ANALYZE Expressing the angular speed as (602 rev/min)(2 rad/rev) 63.0 rad/s,60 s/minwe find the rotational inertia to be I 2K 2 2(24400 J) 12.3 kg m2 .(63.0 rad / s) 2LEARN Note the analogy between rotational kinetic energykinetic energy associated with linear motion.12I 2 and12mv 2 , the34. (a) Equation 10-12 implies that the angular acceleration should be the slope of the vs t graph. Thus, 9/6 1.5 rad/s2.(b) By Eq. 10-34, K is proportional to 2. Since the angular velocity at t 0 is –2 rad/s(and this value squared is 4) and the angular velocity at t 4 s is 4 rad/s (and this valuesquared is 16), then the ratio of the corresponding kinetic energies must beKo4 K416 Ko K4/4 0.40 J .35. THINK The rotational inertia of a rigid body depends on how its mass is distributed.EXPRESS Since the rotational inertia of a cylinder is I 21 MR 2 (Table 10-2(c)), itsrotational kinetic energy is11K I 2 MR 2 2 .24ANALYZE (a) For the smaller cylinder, we have1K1 (1.25 kg)(0.25 m) 2 (235 rad/s) 2 1.08 103 J.4(b) For the larger cylinder, we obtain1K 2 (1.25 kg)(0.75 m) 2 (235 rad/s) 2 9.71 103 J.4LEARN The ratio of the rotational kinetic energies of the two cylinders having the samemass and angular speed is2K 2 R2 0.75 m 2 (3) 9.K1 R1 0.25 m 2
49336. The parallel axis theorem (Eq. 10-36) shows that I increases with h. The phrase “outto the edge of the disk” (in the problem statement) implies that the maximum h in thegraph is, in fact, the radius R of the disk. Thus, R 0.20 m. Now we can examine, say,the h 0 datum and use the formula for Icom (see Table 10-2(c)) for a solid disk, or(which might be a little better, since this is independent of whether it is really a solid disk)we can the difference between the h 0 datum and the h hmax R datum and relate thatdifference to the parallel axis theorem (thus the difference is M(hmax)2 0.10 kg m2 ). Ineither case, we arrive at M 2.5 kg.37. THINK We want to calculate the rotational inertia of a meter stick about an axisperpendicular to the stick but not through its center.EXPRESS We use the parallel-axis theorem: I Icom Mh2, where Icom is the rotationalinertia about the center of mass (see Table 10-2(d)), M is the mass, and h is the distancebetween the center of mass and the chosen rotation axis. The center of mass is at thecenter of the meter stick, which implies h 0.50 m – 0.20 m 0.30 m.ANALYZE With M 0.56 kg and L 1.0 m, we haveI com bgbg112ML2 0.56 kg 10. m 4.67 10 2 kg m2 .1212Consequently, the parallel-axis theorem yieldsbgbgI 4.67 10 2 kg m2 0.56 kg 0.30 m 9.7 10 2 kg m2 .2LEARN A greater moment of inertia I I com means that it is more difficult to rotate themeter stick about this axis than the case where the axis passes through the center.38. (a) Equation 10-33 gives22Itotal md m(2d)2 m(3d)2 14 md .2If the innermost one is removed then we would only obtain m(2d)2 m(3d)2 13 md .The percentage difference between these is (13 – 14)/14 0.0714 7.1%.22(b) If, instead, the outermost particle is removed, we would have md m(2d)2 5 md .The percentage difference in this case is 0.643 64%.39. (a) Using Table 10-2(c) and Eq. 10-34, the rotational kinetic energy isK 1 2 1 11 I MR 2 2 (500 kg)(200 rad/s) 2 (1.0 m) 2 4.9 10 7 J.22 24 (b) We solve P K/t (where P is the average power) for the operating time t.
494CHAPTER 10t K4.9 107 J 6.2 103 s3P 8.0 10 Wwhich we rewrite as t 1.0 102 min.40. (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 item21(c) in Table 10-2) I 2 mR . The next one (using the parallel-axis theorem) has21I 2 mR mh1222where h 2R. The third one has I 2 mR m(4R) . If we had considered five of thedisks OO OO with the one at O in the middle, then the total rotational inertia is1222I 5(2 mR ) 2(m(2R) m(4R) ).The pattern is now clear and we can write down the total I for the collection of fifteendisks:22222212255I 15(2 mR ) 2(m(2R) m(4R) m(6R) m(14R) ) 2 mR .The generalization to N disks (where N is assumed to be an odd number) is122I 6(2N 1)NmR .In terms of the total mass (m M/15) and the total length (R L/30), we obtainI 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.41. The particles are treated “point-like” in the sense that Eq. 10-33 yields their rotationalinertia, and the rotational inertia for the rods is figured using Table 10-2(e) and theparallel-axis theorem (Eq. 10-36).(a) With subscript 1 standing for the rod nearest the axis and 4 for the particle farthestfrom it, we have
49522 1 1 1 3 I I1 I 2 I 3 I 4 Md 2 M d md 2 Md 2 M d m(2d ) 2 12 12 2 2 88 Md 2 5md 2 (1.2 kg)(0.056 m)2 5(0.85 kg)(0.056 m)2332 0.023 kg m .(b) Using Eq. 10-34, we have1 2 45 5 4 I M m d 2 2 (1.2 kg) (0.85 kg) (0.056 m)2 (0.30 rad/s)222 2 3 3 3 1.1 10 J.K 42. (a) We apply Eq. 10-33:42222I x mi yi2 50 2.0 25 4.0 25 3.0 30 4.0 g cm2 1.3 103 g cm2 . i 1(b) For rotation about the y axis we obtainb g b gb g4b gb gI y mi xi2 50 2.0 25 0 25 3.0 30 2.0 55. 102 g cm2 .2i 1222(c) And about the z axis, we find (using the fact that the distance from the z axis isx2 y2 )4chI z mi xi2 yi2 I x I y 1.3 103 5.5 102 1.9 102 g cm2 .i 1(d) Clearly, the answer to part (c) is A B.43. THINK Since the rotation axis does not pass through the center of the block, we usethe parallel-axis theorem to calculate the rotational inertia.EXPRESS According to Table 10-2(i), the rotational inertia of a uniform slab about anaxis through the center and perpendicular to the large faces is given byM 222I com a b 2 . A parallel axis through the corner is a distance h a / 2 b / 212from the center. Therefore,MMMI I com Mh2 a 2 b2 a 2 b 2 a 2 b 2 .1243chb g b gANALYZE With M 0.172 kg, a 3.5 cm and b 8.4 cm, we have
496CHAPTER 10I M 20.172 kga b2 [(0.035 m)2 (0.084 m)2 ] 4.7 10 4 kg m2 . 33LEARN A greater moment of inertia I I com means that it is more difficult to rotate theblock about the axis through the corner than the case where the axis passes through thecenter.44. (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 obtainI mi ri 2 4 (0.50 kg) (1.0 m)2 2.0 kg m2 .(b) In this case, the two masses nearest the axis are r 1.0 m away from it, but the twofurthest from the axis are r (1.0 m)2 (2.0 m)2 from it. Here, then, Eq. 10-33 leads tobgch bgchI mi ri 2 2 0.50 kg 10. m2 2 0.50 kg 5.0 m2 6.0 kg m2 .(c) Now, two masses are on the axis (with r 0) and the other two are a distancer (1.0 m)2 (1.0 m)2 away. Now we obtain I 2.0 kg m2 .45. THINK Torque is the product of the force applied and the moment arm. When twotorques act on a body, the net torque is their vector sum.EXPRESS We take a torque that tends to cause a counterclockwise rotation from rest tobe positive and a torque tending to cause a clockwise rotation to be negative. Thus, apositive torque of magnitude r1 F1 sin 1 is associated with F1 and a negative torque of magnitude r2F2 sin 2 is associated with F2 . The net torque is consequently r1 F1 sin 1 r2 F2 sin 2 .ANALYZE Substituting the given values, we obtain r1F1 sin 1 r2 F2 sin 2 (1.30 m)(4.20 N)sin 75 (2.15 m)(4.90 N)sin 60 3.85 N m.
497LEARN Since 0 , the body will rotate clockwise about the pivot point.46. The net torque is A B C FA rA sin A FB rB sin B FC rC sin C (10)(8.0)sin135 (16)(4.0)sin 90 (19)(3.0)sin160 12 N m.47. THINK In this problem we have a pendulum made up of a ball attached to a masslessrod. There are two forces acting on the ball, the force of the rod and the force of gravity.EXPRESS No torque about the pivot pointis associated with the force of the rod sincethat force is along the line from the pivotpoint to the ball. As can be seen from thediagram, the component of the force ofgravity that is perpendicular to the rod ismg sin . If is the length of the rod, thenthe torque associated with this force hasmagnitude mg sin .ANALYZE With m 0.75 kg , 1.25 m and 30 , we find the torque to be mg sin (0.75)(9.8)(1.25)sin 30 4.6 N m .LEARN The moment arm of the gravitational force mg is sin . Alternatively, we maysay that is the moment arm of mg sin , the tangential component of the gravitationalforce. Both interpretations lead to the same result: (mg )( sin ) (mg sin )( ) .48. We compute the torques using rF sin .(a) For 30 a (0.152 m)(111 N)sin 30 8.4 N m.(b) For 90 , b (0.152 m)(111 N)sin 90 17 N m .(c) For 180 , c (0.152 m)(111N)sin180 0 .49. THINK Since the angular velocity of the diver changes with time, there must be anon-vanishing angular acceleration.
498CHAPTER 10EXPRESS To calculate the angular acceleration , we use the kinematic equation 0 t , where 0 is the initial angular velocity, is the final angular velocity and tis the time. If I is the rotational inertia of the diver, then the magnitude of the torqueacting on her is I .ANALYZE (a) Using the values given, the angular acceleration is 06.20 rad / s 28.2 rad / s2 . 3t220 10 s(b) Similarly, we find the magnitude of the torque on the diver to be c hch I 12.0 kg m2 28.2 rad / s2 3.38 102 N m.LEARN A net toque results in an angular acceleration that changes angular velocity. Theequation I implies that t
2.0 rad 4.0 rad/s 2.0 rad/s 2 2 3 3 tt but in some places we will proceed as indicated in the problem—by letting these units be understood. (a) We evaluate the function at t 0 to obtain 0 2.0 rad. (b) The
