1.50 m  0.60 rad  34.4 2.50 m 9.1: a) b) (14.0 cm)  6.27 cm (128)(π rad 180) c) (1.50 m)(0.70 rad)  1.05 m rev   2π rad     1900     199 rad s   rev   60 s   (35   rad 180) (199 rad s)  3.07  10 3 s 9.2: a) b) dωz  (12.0 rad s3 ) t , so at t  3.5 s, α  42 rad s The angular acceleration dt is proportional to the time, so the average angular acceleration between any two times is the arithmetic average of the angular accelerations b) ωz  (6.0 rad s )t , so at t  3.5 s, ωz  73.5 rad s The angular velocity is not linear function of time, so the average angular velocity is not the arithmetic average or the angular velocity at the midpoint of the interval 9.3: a) αz  9.4: a) b) αz ( t)  dω z dt  2 βt  (1.60 rad s )t α z (3.0 s)  (1.60 rad s )(3.0 s)  4.80 rad s ω(3.0 s)  ω(0)  2.20 rad s  5.00 rad s   2.40 rad s , 3.0 s 3.0 s which is half as large (in magnitude) as the acceleration at t  3.0 s αav  z  9.5: a) ωz  γ  βt  (0.400 rad s)  (0.036 rad s )t b) At t  0, ωz  γ  50 rad 0.400 rad s c) At t  5.00 s, ωz  1.3 rad s, θ  3.50 rad, so ωav  z  35.00 s  0.70 rad s The acceleration is not constant, but increasing, so the angular velocity is larger than the average angular velocity 9.6: ωz  (250 rad s)  (40.0 rad s )t  (4.50 rad s3 )t , αz  (40.0 rad s )  (9.00 rad s3 )t a) Setting ωz  results in a quadratic in t ; the only positive time at which ωz  is t  4.23 s b) At t  4.23 s, α z  78.1 rad s c) At t  4.23 s, θ  586 rad  93.3 rev d) At t  0, ωz  250 rad s e) ωav  z  586 rad  138 rad s 4.23 s 9.7: a) ωz  dθ dt  2bt  3ct and α z  dw z dt  2b  6ct b) Setting α z  0, t  b 3c 9.8: (a) The angular acceleration is positive, since the angular velocity increases steadily from a negative value to a positive value (b) The angular acceleration is α ω  ω0 8.00 rad s  (6.00 rad s)   2.00 rad s t 7.00 s Thus it takes 3.00 seconds for the wheel to stop (ω z  0) During this time its speed is decreasing For the next 4.00 s its speed is increasing from rad s to  8.00 rad s (c) We have θ  0  0t  t 2   (6.00 rad s) (7.00 s)  (2.00 rad s ) (7.00 s) 2  42.0 rad  49.0 rad   7.00 rad Alternatively, the average angular velocity is  6.00 rad s  8.00 rad s  1.00 rad s Which leads to displacement of 7.00 rad after 7.00 s 9.9: a) ω  θ0  200 rev, ω0  500 rev  8.333 rev s, t  30.0 s, ω  ? ω  ω θ  θ0    t gives ω  5.00 rev s  300 rpm   b) Use the information in part (a) to find α : ω  ω0  t gives α  0.1111 rev s Then ω  0,   0.1111 rev s , ω0  8.333 rev s, t  ? ω  ω0  αt gives t  75.0 and  ω  ω θ  θ0    t gives   0  312 rev   9.10: a) ωz  ω0 z  α z t  1.50 rad s  (0.300 rad s )(2.50 s)  2.25 rad s b) θ  ω0 z t  α z t  (1.50 rad s)(2.50 s)  (0.300 rad s )(2.50 s)  4.69 rad (200 rev  500 rev min)  160 s  rev  1.25 (4.00 s) s The number of revolutions is the average angular velocity, 350 rev min, times the time interval of 0.067 min, or 23.33 rev b) The angular velocity will decrease by another 200 rev in a time 200 revmin  1.25 rev s  2.67 s 60 s 9.11: a) 9.12: a) Solving Eq (9.7) for t gives t  ω z  ω0 z αz Rewriting Eq (9.11) as θ  θ0  t (ω0 z  α z t ) and substituting for t gives  ω  ω0 z   ω0 z  (ωz  ω0 z )  θ  θ0   z   α     z  ω  ω0 z   (ωz  ω0 z ) z      (ωz2  ω20 z ), 2 which when rearranged gives Eq (9.12)   b) α z  1 1 θ ωz2  ω0 z   1 1 (7.00 rad)  16.0 rad s   12.0 rad s   2 rad s 9.13: a) From Eq 9 , with ω0 z  0, t  ωz z b ) From Eq 9.12, with ω0 z  0, θ  θ0  rad  136.0rad ss2  24.0 s .50 ( 36.0 rad s) 2 (1.50 rad) s  432 rad  68.8 rev 9.14: a) The average angular velocity is 162 rad  40.5 rad s, and so the initial angular 4.00 s velocity is 2ωav  z  ω2 z  ω0 z , ω0z   27 rad s ωz 108 rad s  (27 rad s) αz  b)   33.8 rad s t 4.00 s 9.15: From Eq (9.11), θ  θ0 α z t 60.0 rad (2.25 rad s )(4.00 s) ω0 z      10.5 rad s 4.00 s t 9.16: From Eq (9.7), with ω0 z  0, α z  ωz t rad  140.00 s s  23.33 rad s  The angle is most easily found from θ  ωav  z t  (70 rad s)(6.00 s)  420 rad 9.17: From Eq (9.12), with ω z  0, the number of revolutions is proportional to the square of the initial angular velocity, so tripling the initial angular velocity increases the number of revolutions by 9, to 9.0 rev 9.18: The following table gives the revolutions and the angle θ through which the wheel has rotated for each instant in time and each of the three situations: t (a ) rev's θ ( b) rev's 0.05 0.50 180 0.10 0.15 θ (c ) rev's θ 0.03 11.3 0.44 158 1.00 360 0.13 45 0.75 270 1.50 540 0.28 101 0.94 338 0.20 2.00 720 0.50 180 1.00 360 –––––––––––––––––––––––––––––––––––––– The θ and ωz graphs are as follows: a) b) c) 9.19: a) Before the circuit breaker trips, the angle through which the wheel turned was (24.0 rad s) (2.00 s)  (30.0 rad s ) (2.00 s) 2  108 rad, so the total angle is 108 rad  432 rad  540 rad b) The angular velocity when the circuit breaker trips is 24.0 rad s   30.0 rad s 2.00 s   84 rad s, so the average angular velocity while the 432 rad wheel is slowing is 42.0 rad s, and the time to slow to a stop is 42.0 rad s  10.3 s, so the   time when the wheel stops is 12.3 s c) Of the many ways to find the angular acceleration, the most direct is to use the intermediate calculation of part (b) to find that while slowing down ωz   84 rad s so  z  84 rad s   8.17 rad s 10 s α 9.20: a) Equation (9.7) is solved for ω0 z  ωz  αz t , which gives ωz  ave  ωz  2z t , or θ  θ0  ωz t  α z t b) 2  ωz t   2θ   0.125 rad s c) ωz  α z t  5.5 rad s t 9.21: The horizontal component of velocity is rω , so the magnitude of the velocity is a) 47.1 m/s b)   π rad/s    (5.0 m)(90 rev / )     (4.0 m/s)  47.3 m/s    30 rev/min    9.22: a) 1.25 m s 25.0 10 3 m  50.0 rad s , 1.25 m 58.0 10 3  21.55 rad s , or 21.6 rad s to three figures b) (1.25 m s ) (74.0 min) (60 s min) = 5.55 km c)  z  50.0 rad s  21.55 rad s ( 74.0 min) (60 s min)  6.41  10 3 rad s 9.23: a) ω2 r  (6.00 rad s) (0.500 m)  18 m s b) v  ωr  (6.00 rad s) (0.500 m)  3.00 m s , and v2 r  ( 3.00 m s ) ( 0.500 m )  18 m s 9.24: From arad  ω2 r , 400,000  9.80 m s a   1.25  10 rad s, r 2.50  10 2 m rev π rad which is (1.25  10 rad s) 60 s  1.20  10 rev ω   9.25: a) arad  0, atan  αr  0.600 rad s  0.300 m   0.180 m s and so a  0.180 m s b)θ  π rad, so arad  ω2 r  20.600 rad s  π rad 0.300 m   0.377 m s The tangential acceleration is still 0.180 m s , and so on a 0.180 m s   0.377 m s  2 2  0.418 m s c) For an angle of 120, a rad  0.754 m s , and a  0.775 m s , since a tan is still 0.180 m/s 9.26: a) ωz  ω0z  α z t  0.250 rev s  0.900 rev s  0.200 s   0.430 rev s (note that since ω0 z and α z are given in terms of revolutions, it’s not necessary to convert to radians) b) ωav  z t  (0.340 rev s) (0.2s)  0.068 rev c) Here, the conversion to radians must be made to use Eq (9.13), and  0.750 m  v  rω    0.430 rev/s  2π rad rev  1.01 m s   d) Combining equations (9.14) and (9.15), a  a rad  a tan  (ω2 r )  (αr )  [((0.430 rev s  2π rad rev)2 (0.375 m))  ((0.900 rev s  2π rad rev)(0.375 m))2  3.46 m s 9.27: r (3000)(9.80 m s ) arad   rad s ω2 (5000 rev min) 30 rev     10.7 cm, so the diameter is more than 12.7 cm, contrary to the claim  9.28: a) Combining Equations (9.13) and (9.15), v arad  ω2 r  ω2    ωv ω a rad 0.500 m s b) From the result of part (a), ω  v  2.00 m s  0.250 rad s 9.29: a) ωr  (1250 rev min) b) v2 r  ( 0.831 m s ) a tan r 50.0 m s 0.200 rad s π 30 rev  12.710 3 m   0.831 m s  109 m s (12.7 10 3 m ) 9.30: a) α   m  10.200 ms   50.0 rad s b) At t  3.00 s, v  50.0 m s and ω v   250 rad s and at t  0, v  50.0 m s  (10.0 m s ) r (0  3.00 s)  80.0 m s , so ω  400 rad s c) ωavet  (325 rad s)(3.00 s)  975 rad  155 rev d) v  arad r  (9.80 m s )(0.200 m)  1.40 m s This speed will be reached at time 50.0 m s 1.40 m s 10.0 m s  4.86 s after t  3.00 s, or at t  7.86 s (There are many equivalent ways to this calculation.) 9.31: (a) For a given radius and mass, the force is proportional to the square of the angular velocity;    2.29 (note that conversion to rad s is not necessary for this 640 rev 423 rev part) b) For a given radius, the tangential speed is proportional to the angular velocity; 640  1.51 (again conversion of the units of angular speed is not necessary) 423 c) (640 rev min) arad  v2 r 9.32: (a)  (15.75 m s) ( 0.470 m )  π rad s 30 rev  0.470 m   15.75 m s, or 15.7 m s to three figures, and  1.06  103 m s  108 g vT  Rω  7.5 rev     2π rad  2.00 cm s  R        60 s   rev  R  2.55 cm b) D  R  5.09 cm aT  Rα α aT 0.400 m s   15.7 rad s R 0.0255 m vr 5.00 m s   15.15 rad s r 0.330 m The angular velocity of the front wheel is ωf  0.600 rev s  3.77 rad s Points on the chain all move at the same speed, so rr ωr  rf ωf rr  rr ωf ωr   2.99 cm 9.33: The angular velocity of the rear wheel is ωr  9.34: The distances of the masses from the axis are the moment of inertia is 2 L , L and 34L , and so from Eq 9.16, L L  3L  11 I  m   m   m   mL2 4 4   16 9.35: The moment of inertia of the cylinder is M moment of inertia of the combination is L2 12 M 12  m L 2 and that of each cap is m L4 , so the 9.36: Since the rod is 500 times as long as it is wide, it can be considered slender a) From Table 9.2a , I 1 ML2  0.042 kg 1.50 m   7.88  10  kg  m 12 12 b) From Table 9.2b , 1 I  ML2  0.042 kg 1.50 m   3.15  10 kg  m 3 c) For this slender rod, the moment of inertia about the axis is obtained by considering it as a solid cylinder, and from Table 9.2f , I 1 MR  (0.042 kg) (1.5  10 m)2  4.73  10 8 kg  m 2 9.37: a) For each mass, the square of the distance from the axis is 2(0.200 m)2  8.00  102 m , and the moment of inertia is 4(0.200 kg) (0.800  102 m )  6.40  102 kg  m b) Each sphere is 0.200 m from the axis, so the moment of inertia is 40.200 kg 0.200 m   3.20  102 kg  m a) The two masses through which the axis passes not contribute to the moment of   I  2(0.2 kg) 0.2 m  0.032 kg  m inertia 9.38: (a) I  I bar  I balls    L M bar L2  2mballs   12 2 4.00 kg 2.00 m 2  20.500 kg 1.00 m 2  2.33 kg  m 12 (b) I  mbar L2  mball L2  4.00 kg 2.00 m 2  0.500 kg 2.00 m 2  7.33 kg  m c) I  because all masses are on the axis (d) I  mbar d  2mballd  M Total d  (5.00 kg)(0.500 m)  1.25 kg  m 9.39: I  I d  I r (d  disk, r  ring) disk : md  (3.00 g cm )πrd2  23.56 kg md rd2  2.945 kg  m 2 ring : mr  (2.00 g cm3 ) π (r22  r12 )  15.08 kg Id  mr (r12  r22 )  5.580 kg  m 2 I  I d  I r  8.52 kg  m Ir  (r1  50.0 cm, r2  70.0 cm ) 9.68: a) α  a tan r  3.00 m s 60.0 m  0.050 rad s b) αt  (0.05 rad s )(6.00s)  0.300 rad s c) arad  2 r  (0.300 rad s) (60.0 m)  5.40 m s d) e) a  a rad  a tan  (5.40 m s )  (3.00 m s )  6.18 m s , and the magnitude of the force is F  ma  (1240 kg)(6.18 m s )  7.66 kN f) arctan    arctan   60.9 a rad a tan 5.40 3.00 9.69: a) Expressing angular frequencies in units of revolutions per minute may be accomodated by changing the units of the dynamic quantities; specifically, 2W ω2  ω12  I  2(4000 J )   (300 rev min)    16.0 kg  m      211 rev   rad s    30 rev     b) At the initial speed, the 4000 J will be recovered; if this is to be done is 5.00 s, the power must be 4000sJ  800 W 5.00 9.70: a) The angular acceleration will be zero when the speed is a maximum, which is at the bottom of the circle The speed, from energy considerations, is v  gh  gR (1  cos ), where  is the angle from the vertical at release, and 2g 2(9.80 m s ) v (1  cos )   (1  cos 36.9)  1.25 rad s (2.50 m) R R b)  will again be when the meatball again passes through the lowest point c) arad is directed toward the center, and arad  ω2 R, arad  (1.25 rad s ) (2.50 m)  3.93 m  d) arad  2 R  (2 g R )(1  cos β ) R  (2 g )(1  cos β ), independent of R 9.71: a) (60.0 rev s)(2π rad rev)(0.45  102 m)  1.696 m s b) ω  v  r 1.696 m s 2.00  10  m  84.8 rad s 9.72: The second pulley, with half the diameter of the first, must have twice the angular velocity, and this is the angular velocity of the saw blade a)   rad s   0.208 m  (2(3450 rev min))    75.1 m s  30 rev          rad s    0.208 m  b) arad   r   2(3450 rev min)    5.43  10 m s ,  30 rev           so the force holding sawdust on the blade would have to be about 5500 times as strong as gravity 2 arad  2 r  0 r  (2  0 )r 9.73: a)    0   0  r    0   (  0 )t  r  t    [2 (  0 )r b) From the above, αr  arad (85.0 m s  25.0 m s )   2.00 m s 2θ 2(15.0 rad) c) Similar to the derivation of part (a), 1 K  ω2 I  ω0 I  [α ][2θ ]I  I 2 d) Using the result of part (c), (45.0 J  20.0 J) K   0.208 kg  m  ((2.00 m s ) /(0.250 m))(15.0 rad) I 9.74: I  I wood  I lead 2 mw R  mL R mw  ρwVw  ρw πR 3  mL  σ L AL  σ L 4π R 2  I   ρ w π R  R  ( L π R ) R 5 3  ρ R   π R4  w  σL     (800 kg m3 )(0.20 m)  8  20 kg m  (0.20 m)4     0.70 kgm  9.75: I approximate my body as a vertical cylinder with mass 80 kg, length 1.7 m, and diameter 0.30 m (radius 0.15 m) I 1 mR  (80 kg) (0.15 m)2  0.9 kg  m 2 9.76: Treat the V like two thin 0.160 kg bars, each 25 cm long 1  1   2 mL2   2  (0.160 kg)(0.250 m)2 3   3 3  6.67  10 kg  m 9.77: a)   90.0 rpm  9.425 rad s K 2 K 2( 10.0  106 J) Ιω so Ι    2.252  105 kg  m 2 ( 9.425 rad s) ω m  ρV  ρπR 2t ( ρ  7800 kg m is the density of iron and t=0.100 m is the thickness of the flywheel) 1 Ι  mR  ρπtR 2 R  (2 I ρπt )1  3.68 m; diameter  7.36 m b) ac  Rω2  327 m s 9.78: Quantitatively, from Table (9.2), I A  mR , Ι B  mR and Ι C  mR  a) Object A has the smallest moment of inertia because, of the three objects, its mass is the most concentrated near its axis b) Conversely, object B’s mass is concentrated and farthest from its axis c) Because Ι sphere  mR , the sphere would replace the disk as having the smallest moment of inertia 9.79: a) See Exercise 9.50 2π Ι 2π (0.3308)(5.97  1024 kg)(6.38  106 m)2 K   2.14  1029 J (86, 164 s) T b)  2πR  2π (5.97  1024 kg)(1.50  1011 m)2 M  2.66  1033 J    T  (3.156  107 s) c) Since the Earth’s moment on inertia is less than that of a uniform sphere, more of the Earth’s mass must be concentrated near its center 9.80: Using energy considerations, the system gains as kinetic energy the lost potential energy, mgR The kinetic energy is 1 1 K  Ιω2  mv  Ιω2  m(ωR)  ( Ι  mR )   2 2 2 Using Ι  mR and solving for ω, ω2  4g 4g , and   3R 3R 9.81: a) Consider a small strip of width dy and a distance y below the top of the triangle The length of the strip is x   y h b The strip has area x dy and the area of the sign is bh, so the mass of the strip is  xdy   yb  dy   2M  dm  M    M       y dy  bh   h  bh   h  2  dI  dm x  2Mb 3h y dy 2Mb h 2Mb y dy  3h 0 3h b) I  Mb  2.304 kg  m ω  2.00 rev s  4.00π rad s h I   dI  K  Iω2  182 J 1 h  y |0   Mb   9.82: (a) The kinetic energy of the falling mass after 2.00 m is KE  mv  8.00 kg 5.00 m/s   100 J The change in its potential energy while 2   falling is mgh  8.00 kg  9.8 m/s2 2.00 m   156.8 J The wheel must have the “missing” 56.8 J in the form of rotational KE Since its outer rim is moving at the same speed as the falling mass, 5.00 m s : v  r  KE  v 5.00m/s   13.51 rad/s r 0.370m I ; therefore 2 KE 256.8 J  I   0.6224 kg  m or 0.622 kg  m 2  13.51 rad s  (b) The wheel’s mass is 280 N 9.8 m s = 28.6 kg The wheel with the largest possible moment of inertia would have all this mass concentrated in its rim Its moment of inertia would be I  MR  28.6kg 0.370m   3.92 kg  m The boss’s wheel is physically impossible   9.83: a) 0.160 kg  0.500 m  9.80 m s  0.784 J b) The kinetic energy of the stick is 0.784 J, and so the angular velocity is  2k 2k   I ML2 2(0.784 J)  5.42 rad s 0.160 kg 1.00 m 2 This result may also be found by using the algebraic form for the kinetic energy, K  MgL 2, from which   g L , giving the same result Note that ω is independent of the mass c) v  L  5.42 rad s 1.00 m   5.42 m s d) 2gL  4.43 m s ; This is of the result of part (c) 9.84: Taking the zero of gravitational potential energy to be at the axle, the initial potential energy is zero (the rope is wrapped in a circle with center on the axle).When the rope has unwound, its center of mass is a distance R below the axle, since the length of the rope is 2R and half this distance is the position of the center of the mass Initially, every part of the rope is moving with speed  R, and when the rope has unwound, and the cylinder has angular speed  , the speed of the rope is R (the upper end of the rope has the same tangential speed at the edge of the cylinder) From conservation of energy, using I  (1 2) MR for a uniform cylinder,  M m 2  M m 2    R 0     R   mgπ R  2  2 Solving for  gives 4πmg R  , ω  ω0  M  m  and the speed of any part of the rope is v  R 9.85: In descending a distance d, gravity has done work mB gd and friction has done work   K m A gd , and so the total kinetic energy of the system is gd mB  μK mA  In terms of the speed v of the blocks, the kinetic energy is 1 K  m A  m B v  I  m A  m B  I R v , 2 where ω  v R, and condition that the rope not slip, have been used Setting the kinetic energy equal to the work done and solving for the speed v, gd mB   k m A  v mA  mB  I R  9.86: The gravitational potential energy which has become kinetic energy is K  4.00 kg  2.00 kg  9.80 m s 5.00 m   98.0 J In terms of the common speed v of the blocks, the kinetic energy of the system is   1 v K  (m1  m2 )v  I   2  R  v2 (0.480 kg  m )  1  4.00 kg  2.00 kg    v (12.4 kg) 2 (0.160 m)    Solving for v gives v  98.0 J 12.4 kg  2.81 m s 9.87: The moment of inertia of the hoop about the nail is 2MR (see Exercise 9.52), and the initial potential energy with respect to the center of the loop when its center is directly below the nail is gR (1  cos β ) From the work-energy theorem, K Iω  Mω2 R  MgR (1  cos ), from which ω  ( g R )(1  cos ) 9.88: a) K  Iω 11 2π rad s      (1000 kg)(0.90 m)2  3000 rev   22 60 rev     2.00  107 J 2.00  107 J K   1075 s, Pave 1.86  104 W b) which is about 18 9.89: a) 1 M 1R12  M R2  ((0.80 kg)(2.50  10  m)2  (1.60 kg)(5.00  10 m) ) 2  2.25  10 3 kg  m b) See Example 9.9 In this case,   v R1 , and so the expression for v becomes v  gh  ( I mR ) 2(9.80 m s )(2.00 m)  3.40 m s (1  ((2.25  103 kg  m ) (1.50 kg)(0.025 m)2 )) c) The same calculation, with R2 instead of R1 gives v  4.95 m s This does make sense, because for a given total energy, the disk combination will have a larger fraction of the kinetic energy with the string of the larger radius, and with this larger fraction, the disk combination must be moving faster 9.90: a) In the case that no energy is lost, the rebound height h is related to the speed υ2 v by h  g , and with the form for h given in Example 9.9, h  1 Mh m b) Considering the system as a whole, some of the initial potential energy of the mass went into the kinetic energy of the cylinder Considering the mass alone, the tension in the string did work on the mass, so its total energy is not conserved 9.91: We can use Κ (cylinder)  250 J to find  for the cylinder and v for the mass Ι  MR  (10.0 kg)(0.150 m)  0.1125 kg  m 2 K  I2 so   K I  66.67 rad s v  R  10.0 m s Use conservation of energy K1  U  K  U Take y  at lowest point of the mass, so y  and y1  h, the distance the mass descends K1  U  so U1  K mgh  mv  I , where m  12.0 kg 2 For the cylinder, I  MR and   v R , so mgh  mv  Mv 2 h v2  M  1    7.23 m g  2m  I  Mv 9.92: Energy conservation: Loss of PE of box equals gain in KE of system 1 2 mbox gh  mbox vbox  I pulley ωpulley 2  I cylinder ωcylinder v v ωpulley  Box and ωcylinder  Box rcylinder rp 1 1 v  mB gh  mBvB   mP rp2   B  22   rp    1 v    mC rC2   B  22   rC    1 2 mBvB  mP vB  mCvB 4 mB gh vB  mB  m p  mC 4 mB gh   (3.00 kg)(9.80 m s )(1.50 m)  3.68 m s 1.50 kg  (7.00 kg) 9.93: a) The initial moment of inertia is I  MR The piece punched has a mass of M 16 and a moment of inertia with respect to the axis of the original disk of 2 M 1  R   R   MR        16       512   The moment of inertia of the remaining piece is then I 247 MR  MR  MR 512 512 b) I  MR  M ( R / 2)  ( M / 16)( R / 4)  2 383 512 MR 9.94: a) From the parallel-axis theorem, the moment of inertia is I P  (2 5) MR  ML2 , and    R   IP  1      ML2    L     If R = (0.05)L, the difference is (2 5)(0.05)  0.001 b) ( I rod ML2 )  (mrod 3M ), which is 0.33% when mrod  (0.01) M 2 9.95: a) With respect to O, each element ri in Eq (9.17) is xi  y i , and so I O   mi ri   mi ( xi  y i )   mi xi   mi y i  I x  I y i 2 i 2 i i b) Two perpendicular axes, both perpendicular to the washer’s axis, will have the same moment of inertia about those axes, and the perpendicular-axis theorem predicts 2 that they will sum to the moment of inertia about the washer axis, which is M ( R1  R2 ), and so I x  I y  M 2 ( R1  R2 ) c) From Table (9.2), I  12 m( L  L2 )  mL2 Since I  I x  I y , and I x  I y , both I x and I y must be 12 mL2 9.96: Each side has length a and mass M , and the moment of inertia of each side about an axis perpendicular to the side and through its center is 12 M a  Ma The moment of 48 inertia of each side about the axis through the center of the square is, from the 2 perpendicular axis theorem, Ma  M  a   Ma The total moment of inertia is the sum of 48 12 the contributions from the four sides, or  Ma  12 Ma 9.97: Introduce the auxiliary variable L, the length of the cylinder, and consider thin cylindrical shells of thickness dr and radius r; the cross-sectional area of such a shell is 2 r dr , and the mass of shell is dm  2π rLρ dr  2π α Lr dr The total mass of the cylinder is then R R3 M   dm  2π L  r dr  2π L and the moment of inertia is R R5 I   r dm  2π L  r dr  2π L  MR o 5 b) This is less than the moment of inertia if all the mass were concentrated at the edge, as with a thin shell with I  MR , and is greater than that for a uniform cylinder with I  MR , as expected 9.98: a) From Exercise 9.49, the rate of energy loss is π Ι dΤ Τ dt ; solving for the moment of inertia I in terms of the power P, (5  1031 W )(0.0331 s)3 1s ΡΤ  Ι  1.09  1038 kg  m 13 4π dΤ dt 4π 4.22  10 s b) R  c) d) 5Ι  2Μ 5(1.08  1038 kg  m )  9.9  103 m, about 10 km 2(1.4)(1.99  1030 kg) 2πR 2(9.9  103 m)   1.9  106 m s  6.3  103 c (0.0331 s) Τ Μ Μ   6.9  1017 kg m3 , V (4 3) R which is much higher than the density of ordinary rock by 14 orders of magnitude, and is comparable to nuclear mass densities 9.99: a) Following the hint, the moment of inertia of a uniform sphere in terms of the mass density is I  ΜR  8 ρR , and so the difference in the moments of inertia of 15 two spheres with the same density  but different radii 5 R2 and R1 is Ι  ρ(8π 15)( R2  R1 ) b) A rather tedious calculation, summing the product of the densities times the difference in the cubes of the radii that bound the regions and multiplying by 4 , gives M  5.97  10 24 kg c) A similar calculation, summing the product of the densities times the difference in the fifth powers of the radii that bound the regions and multiplying by 8 15, gives I  8.02  1022 kg  m  0.334MR 9.100: Following the procedure used in Example 9.14 (and using z as the coordinate along the vertical axis) r(z)  z R , dm  πρ R2 z dz and dΙ  πρ R4 z dz Then, h h h Ι   dΙ  πρ R h  h z dz    πρ R h z  πρR h 10 h 10 The volume of a right circular cone is V  R h, the mass is R hand so 3 Ι  πR h    R  ΜR 10   10   9.101: a) ds  r d  r0 d   d , so s( )  r0    b) Setting s  vt  r0    2 gives a quadratic in θ The positive solution is θ(t )    r0  βvt  r0       (The negative solution would be going backwards, to values of r smaller than r0 ) c) Differentiating, ωz (t )  z  dθ v ,  dt r02  βvt d v  dt r02  vt   32 The angular acceleration  z is not constant d) r0  25.0 mm; It is crucial that θ is measured in radians, so β  1.55 μm rev1 rev 2π rad   0.247 μm rad The total angle turned in 74.0 = 4440 s is     2.47  10 m/rad 1.25 m/s 4440 s     2.47  10 m/rad   25.0  10 m  25.0  10 m     1.337  10 rad θ which is 2.13  10 rev e)  ... b) v2 r  ( 0.831 m s ) a tan r 50.0 m s 0.200 rad s π 30 rev  12.710 3 m   0.831 m s  109 m s (12.7 10 3 m ) 9.30: a) α   m  10.200 ms   50.0 rad s b) At t  3.00 s, v  50.0... 7.5 rev     2π rad  2.00 cm s  R        60 s   rev  R  2.55 cm b) D  R  5 .09 cm aT  Rα α aT 0.400 m s   15.7 rad s R 0.0255 m vr 5.00 m s   15.15 rad s r 0.330 m The... solving for the moment of inertia I in terms of the power P, (5  1031 W )(0.0331 s)3 1s ΡΤ  Ι  1 .09  1038 kg  m 13 4π dΤ dt 4π 4.22  10 s b) R  c) d) 5Ι  2Μ 5(1.08  1038 kg  m )  9.9