10.1: Equation (10.2) or Eq. (10.3) is used for all parts. a) m,N00.4090sin N)m)(10.000.4(  out of the page. b) m,N6.34120sin N)m)(10.000.4(  out of the page. c) m,N0.2030sin N)m)(10.000.4(  out of the page. d) m,N3.1760sin N)m)(10.0000.2(  into the page. e) The force is applied at the origin, so 0.τ f) .0180sin N)m)(10.000.4(  10.2: m,N40.0m)N)(5.0000.8( 1 τ m,N0.1230sin m)N)(2.00(12.0 2 τ where positive torques are taken counterclockwise, so the net torque is m,N0.28  with the minus sign indicating a clockwise torque, or a torque into the page. 10.3: Taking positive torques to be counterclockwise (out of the page), m,N2.34N)m)(26.0(0.09m,N1.62N)(180.0m)090.0( 21  ττ   m,N78.1N)(14.0m)090.0(2 3   so the net torque is m,N50.2  with the direction counterclockwise (out of the page). Note that for 3  the applied force is perpendicular to the lever arm. 10.4: RFFRFRFττ )( 122121  m.N726.0m)N)(0.3307.50N30.5(  10.5: a) b) Into the plane of the page. c) ] ˆ N)00.4( ˆ n)00.5[(] ˆ m)150.0( ˆ m)450.0[( iijiFr    ˆ m)N1.05( ˆ N)00.5m)(150.0(N)00.4(m)450.0( k k   10.6: (a) CCWm,N8.7m)2.0)(60N)(sin 50( A τ CWm,N10m)2.0(N)50( CWm,N5m)2.0)(30N)(sin 50( 0 D C B    τ τ τ (b) mN10mN5mN8.7 τ CWm,N3.6  10.7: kg00.2kg,40.8 where,2 22 3 2  mMmRMRI 2 mkg600.0 I mN0524.0 , ;srad0.08726gives ?s,0.30;srad5.236rpm0.50;srad7.854rpm75.0 2 0 0    IατIατ ααtωω α tωω f 10.8:       m.N1.13 s00.8 minrev400 mkg50.2 a) minrev srad 60 2 2        t II α b) .J102.19 minrev srad 60 2 minrev400)mkg50.2( 2 1 2 1 3 2 22           π I 10.9:     in found that assame the,sm2.1m0.2sm36.022 2  asv Example 9-8. 10.10:      .srad00.2 mkg0.5 m250.0N0.40 2 2    I FR I τ α 10.11: a)                  Mm mM g Mm m MgTMgn 21 3 21 b) This is less than the total weight; the suspended mass is accelerating down, so the tension is less than mg. c) As long as the cable remains taut, the velocity of the mass does not affect the acceleration, and the tension and normal force are unchanged. 10.12: a) The cylinder does not move, so the net force must be zero. The cable exerts a horizontal force to the right, and gravity exerts a downward force, so the normal force must exert a force up and to the left, as shown in Fig. (10.9). b)      N,490sm80.9kg)50(N0.9 2 2 2 n at an angle of arctan    1.1 490 0.9 from the vertical (the weight is much larger than the applied force F ). 10.13:   n tMR Rn I n R n f 2 0 k               .482.0 N160s50.72 minrev850m260.0kg0.50 minrev srad 30   10.14: (a) Falling stone: 2 2 1 atg      )2( ::Pulley )1(::Stone sm80.2 s00.3m6.12 2 1 R 2 2 1 2 2 1 2 2 2 1 MaT MR αMRTRIατ maTmgmaF a a a      Solve (1) and (2): kg00.2 m/s80.2m/s80.9 m/s80.2 2 kg0.10 2 22 2                            M ag aM M (b)From (2):     N0.14 m/s80.2kg0.10 2 1 2 1 2   T MaT 10.15:    2 2 2 1 2 2 1 mkg02320.0m0750.0kg8.25  mRI   N47.7so rad/s046.8gives2 ?rad,33.0rev25.5;0rad/s;23.04rpm220 k k kk 2 0 2 0 2 00      Rμ Iα nIαnRμ nRμRfττ Iατ αθθαωω αθθωω f 10.16: This is the same situtation as in Example 10.3. a) N.0.42)21(  MmmgT b) s.m11.82(12  mMghv c) There are many ways to find the time of fall. Rather than make the intermediate calculation of the acceleration, the time is the distance divided by the average speed, or   s.69.12 vh d) The normal force in Fig. (10.10(b)) is the sum of the tension found in part (a) and the weight of the windlass, a total 159.6 N (keeping extra figures in part ( a)). 10.17: See Example 10.4. In this case, the moment of inertia I is unknown, so      . 2 2121 RImmgma  a)     ,m/s75.3s80.0m20.12 2 2 1 a   N.2.18andN50.7so 122111  agmTamT b) The torque on the pulley is   m,N0.803 12  RTT and the angular acceleration is .mkg0.016so,rad/s50 22 1  IRaα 10.18: . 3 2 3 1 Ml F Ml Fl I    10.19: The acceleration of the mass is related to the tension by , cm TMgMa  and the angular acceleration is related to the torque by 2 cmcm and/ where,/or , MRIRStaαMTaTRτIα  have been used. a) Solving these for N.882.02/gives  MgTT b) Substituting the expression for T into either of the above relations gives which from,2/ cm ga  s.553.042 cm  ghaht rad/s.9.33c) cmcm  RtaRvω 10.20: See Example 10.6 and Exercise 10.21. In this case, .srad33.9,and cmcm 2 cm2  RvωghvMvK 10.21: From Eq. (10.11), the fraction of the total kinetic energy that is rotational is         , 1 1 /1 1 2121 21 cm 2 22 cmcm 2 cm 2 cm 2 cm I MR vIMIMv I        where  Rv cm for an object that is rolling without slipping has been used. a) ,5)MR2(Ib).31isratioabove theso,)21( 22 cm  MRI so the above ratio is ,32c) .72 2 MRI  so the ratio is ,MR85d) .52 2 I so the ratio is .135 10.22: a) The acceleration down the slope is ,sin M f θga  the torque about the center of the shell is , 3 2 3 2 2 MRa R a MR R a II αRfτ  .so 3 2 a M f  Solving these relations a for f and simultaneously gives or,sin 3 5 θga  .N83.4)smkg)(3.6200.2( 3 2 3 2 ,sm62.30.38sin)sm80.9( 5 3 sin 5 3 2 22   Maf θga The normal force is Mg cos θ , and since , s nμf  .313.0tan 5 2 cos sin 3 2 cos3 2 cos 5 3 3 2 s  θ θ g θg θg a θMg Ma n f μ b) 2 sm62.3a since it does not depend on the mass. The frictional force, however, is twice as large, 9.65 N, since it does depend on the mass. The minimum value of s μ also does not change. 10.23: )1eq.()cos(sin cossin cos s s aθμθg ma θmgμθmg mgn    n and mg act at the center of the ball and provide no torque. αmRθmgμIατ mRIθRmgμττ f 2 5 2 s 2 5 2 s cosgives ;cos   No slipping means (eq.2) cos so , 5 2 s agμRaα  We have two equations in the two unknowns a and . s  Solving gives 613.00.65 tan tanandsin 7 2 7 2 7 5  θμθga s b) Repeat the calculation of part (a), but now . 2 3 2 mRI  858.00.65 tan tanandsin 5 2 5 2 s 5 3  θμθga The value of s  calculated in part (a) is not large enough to prevent slipping for the hollow ball. c) There is no slipping at the point of contact. 10.24: slippingnofor cm  Rv  a) Get v at bottom: ghv R v mRmvmgh I ωmvmgh 7 10 5 2 2 1 2 1 2 1 2 1 2 22 22                Now use energy conservation. Rotational KE does not change h g gh g v h KEhmgKEmv 7 5 22 2 1 7 10 2 RotRot 2      hhhmgmgh     (b) With friction on both halves, all the PE gets converted back to PE. With one smooth side, some of the PE remains as rotational KE. 10.25: cm 2 2 1 0 2 cm1 )2/1( mvwIKWwh f  Solving for h with Rwv  cm   m.7.11 N392 J3500 ]s)rad0.25(m)600.0()rad0.25(m)600.0)(800.0([ 2222 sm80.9 2 1 2    w s h w 10.26: a) The angular speed of the ball must decrease, and so the torque is provided by a friction force that acts up the hill. b) The friction force results in an angular acceleration, related by .fRI   The equation of motion is cm, sin mafβmg  and the acceleration and angular acceleration are related by Rαa  cm (note that positive acceleration is taken to be down the incline, and relation between cm a and  is correct for a friction force directed uphill). Combining,   ,571sin 2 ma mR I ma βmg           .sin75 which from cm βga  c) From either of the above relations between if f and , cm a ,cossin 7 2 5 2 sscm βmgμnμβmgmaf  from which   . tan72 s  μ 10.27: a)           rad/s,3086.0s0.15mkg2100m40.2N0.18 2  tIFRtαω or 0.309 rad/s to three figures. b)     2121 2 2  IωKW     J.100srad3086.0mkg00.2 2 2  c) W.67.6,or either From ave  PtWPτωP 10.28: a)      m.N519 rev/min rad/s 30 rev/min2400 hp/ W746hp175         π ω P τ b)    J.32612mN519  πθτW 10.29: a) t ω IIατ            m.N377.0 s5.2 minrev srad 30 minrev1200m100.0kg50.121 2           π b)    rad.157rev0.25 s/min60 s5.2rev/min600 ave t  c) J.2.59θτ d) 2 2 1 IωK    J, 2.59 rev/min rad/s 30 rev/min)1200(m)kg)(0.1005.1)(2/1( 2 1 2 2                 π the same as in part (c). 10.30: From Eq. (10.26), the power output is W,2161 rev/min rad/s 60 2 rev/min 4800m)N30.4(          τωP which is 2.9 hp. 10.31: a) With no load, the only torque to be overcome is friction in the bearings (neglecting air friction), and the bearing radius is small compared to the blade radius, so any frictional torque could be neglected. b) N.6.65 m)086.0( rev/min rad/s 30 rev/min)(2400 W/hp)hp)(7469.1(/         π R ωP R τ F 10.32: 22 2 1 2 2 1 mkg2.42m)kg)(2.08117(  mLI a) .rad/s2.46 mkg42.2 mN1950 2 2     I τ α b) rad/s.9.53rev)2rev0.5)(rad/s2.46(22 2   c) From either (10.24),Eq.or 2 1 2   KW J. 106.13rad/rev)2revN.m)(5.001950( 4  πτW d), e) The time may be found from the angular acceleration and the total angle, but the instantaneous power is also found from hp).kW(141105 τωP The average power is half of this, or kW.6.52 10.33: a) m.N358 rev/min rad/s 30 rev/min)(400 W)10150(/ 3                 π ω Pτ b) If the tension in the rope is N. 1079.1/soand , 3  RτwwFF c) Assuming ideal efficiency, the rate at which the weight gains potential energy is the power output of the motor, or m/s.8.83so,  wPvPwv Equivalently, .Rv   10.34: As a point, the woman’s moment of inertia with respect to the disk axis is 2 mR , and so the total angular momentum is s./mkg1028.5 rad/rev)2rev/s500.0(m)00.4(kg50.0kg110 2 1 2 1 )( 23 2 2 womandiskwomandisk                 π ω RmMωIILLL 10.35: a) ,s/mkg115sin 2 φmvr with a direction from the right hand rule of into the page. b)        ,smkg125mN12536.990sinm8kgN8.9kg2 2 2  τdtdL out of the page. [...].. .10. 36: For both parts, L  Iω Also,   v r , so L  I (v r ) a) L  (mr 2 )(v r )  mvr L  (5.97  102 4 kg)(2.98  104 m s) (1.50  101 1 m)  2.67  104 0 kg  m 2 s b) L  (2 5mr 2 )(ω) L  (2 5)(5.97  102 4 kg)(6.38  106 m) 2 (2 rad (24.0 hr  3600 s hr))  7.07  103 3 kg  m 2 s 10. 37: The period of a second hand is one minute, so the angular momentum is M 2 2 l 3 T 3  6.0  10 kg ... rev/min   360   1 .10  105 N  m 10. 50: Using Eq (10. 36) for all parts, a) halved b) doubled (assuming that the added weight is distributed in such a way that r and I are not changed) c) halved (assuming that w and r are not changed) d) doubled e) unchanged 10. 51:  2 rad a) Solving Eq (10. 36) for τ , τ  Iω   (2 / 5) MR 2ω  Using ω  86,400 s and 2 ( 26 , 000 y)(3.175 10 7 s/y) and the mass... (15.0  10 2 m) 2   4.71  10 6 kg  m 2 s   3 60 s   L  Iω  10. 38: The moment of inertia is proportional to the square of the radius, and so the angular velocity will be proportional to the inverse of the square of the radius, and the final angular velocity is 2 2 R    7.0  10 5 km  2 rad 3  2  1  1     (30 d)(86,400 s d  16 km   4.6  10 rad s   R     2  10. 39:... conserved 10. 47: 10. 48: a) Since the gyroscope is precessing in a horizontal plane, there can be no net vertical force on the gyroscope, so the force that the pivot exerts must be equal in magnitude to the weight of the gyroscope, F  ω  mg  0.165 kg  9.80 m s 2  1.617 N, 1.62 N to three figures b) Solving Eq (10. 36) for ω,     ωR 1.617 N  4.00  10 2 m  ω  188.7 rad s , I 1.20  10 4 kg... 2)m((ω2 r2 ) 2  (ω1r1 ) 2 )  1.03  10 2 J d) No other force does work, so 1.03  10 2 J of work were done in pulling the cord 10. 40: The skater’s initial moment of inertia is 1 I1  (0.400 kg  m 2 )  (8.00 kg )(1.80 m) 2  2.56 kg  m 2 , 2 and her final moment of inertia is I 2  (0.400 kg  m 2 )  (8.00 kg)(25  10 2 m)  0.9 kg  m 2 Then from Eq (10. 33), ω2  ω1 2.56 kg  m 2 I1  (0.40... then states mgy  (7 10) mv 2 , and combining, canceling the factors of m and g leads to (7 10) ( R  r )  h  r  2 R, and solving for h gives h  (27 10) R  (17 10) r b) In the absence of friction, mgy  (1 2)mv 2 , and substitution of the expressions for y and v 2 in terms of the other parameters gives (1 2)( R  r )  h  r  2 R, which is solved for h  (5 2) R  (3 2)r  10. 70: In the first case,... sections Rough : mgh1  1 mv 2  1 I 2 2 2 1 2 12  v  mv   mR 2   2 25  R  10 v 2  gh1 7 Smooth: Rotational KE does not change 2 mgh1  1 mgh2  mv 2  KE Rot  2 1  10  gh2   gh1   2 7  vB  1 2 mv Bottom  KE Rot 2 1 2 vB 2 10 gh1  2 gh2 7 10 (9.80 m/s 2 )(25 m)  2(9.80 m/s 2 )(25 m) 7  29.0 m/s  10. 75: a) Use conservation of energy to find the speed v 2 of the ball just before... (b) d) The greater the angular acceleration of the upper end of the cue, the faster you would have to react to overcome deviations from the vertical 10. 60: In Fig (10. 22) and Eq (10. 22), with the angle  measured from the vertical, sin θ  cos θ in Eq (10. 2) The torque is then τ  FR cos  a)  2 W  0 FR cos θ d θ  FR b) In Eq (6.14), dl is the horizontal distance the point moves, and so W  F ... 2 2π  rad 2.20 s   which is 1.80  10 rev min Note that in this and similar situations, since  appears in the denominator of the expression for  , the conversion from rev s and back to rev min must be made 3 c) 10. 49: a) K (1 / 2)((1 / 2) MR 2 ) 2  P P 2  rad/s (1 / 2)(1 / 2)(60,000 kg)(2.00 m)2 (500 rev/min) 30 rev/min   7.46  104 W  2.21  103 s, or 36.8 min b) τ  I  π rad/s... slipping means ω  v r I  5 mr 2 , so 1 I2  1 mv 2 2 5 7 mgh  10 mv 2 h 7v 2 7(17.82 m/s)   23 m 10 g 10( 9.80 m/s 2 ) b) 1 2 I 2  1 mv 2 , Independent of r 5 c) All is the same, except there is no rotational kinetic energy term in K : K  1 mv 2 2 mgh  1 mv 2 2 h v2  16 m, 0.7 times smaller than the answer in part ( a) 2g 10. 74: Break into 2 parts, the rough and smooth sections Rough : mgh1 . 10. 1: Equation (10. 2) or Eq. (10. 3) is used for all parts. a) m,N00.4090sin N)m) (10. 000.4(  out of the page. b) m,N6.34120sin N)m) (10. 000.4(. rad/s.9.33c) cmcm  RtaRvω 10. 20: See Example 10. 6 and Exercise 10. 21. In this case, .srad33.9,and cmcm 2 cm2  RvωghvMvK 10. 21: From Eq. (10. 11), the fraction