Physics exercises_solution: Chapter 10

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Physics exercises_solution: Chapter 10

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 10

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Nội dung Text: Physics exercises_solution: Chapter 10

  1. 10.1: Equation (10.2) or Eq. (10.3) is used for all parts. a) (4.00 m)(10.0 N) sin 90  40.00 N  m, out of the page. b) (4.00 m)(10.0 N) sin 120  34.6 N  m, out of the page. c) (4.00 m)(10.0 N) sin 30  20.0 N  m, out of the page. d) (2.00 m)(10.00 N) sin 60  17.3 N  m, into the page. e) The force is applied at the origin, so τ  0. f) (4.00 m)(10.0 N) sin 180  0. 10.2: τ1  (8.00 N)(5.00 m)  40.0 N  m, τ 2  (12.0 N)(2.00 m) sin 30  12.0 N  m, where positive torques are taken counterclockwise, so the net torque is  28.0 N  m, 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), τ1  (0.090 m)  (180.0 N)  1.62 N  m, τ 2  (0.09 m)(26.0 N)  2.34 N  m,    3  2 (0.090 m) (14.0 N)  1.78 N  m, so the net torque is 2.50 N  m, with the direction counterclockwise (out of the page). Note that for  3 the applied force is perpendicular to the lever arm. 10.4: τ1  τ 2   F1R  F2 R  ( F2  F1 ) R  (5.30 N  7.50 N)(0.330 m)  0.726 N  m. 10.5: a) b) Into the plane of the page.   c) r  F  [(0.450 m) i  (0.150 m) ˆ]  [(5.00 n)i  (4.00 N)i ] ˆ j ˆ ˆ  (0.450 m) (4.00 N)  (0.150 m)(5.00 N)k ˆ ˆ  (1.05 N  m)k
  2. 10.6: (a) τ A  (50 N)(sin 60)(0.2 m)  8.7 N  m, CCW τB  0 τ C  (50 N)(sin 30)(0.2 m)  5 N  m, CW τ D  (50 N)(0.2 m)  10 N  m, CW (b)  τ  8.7 N  m  5 N  m  10 N  m  6.3 N  m, CW 10.7: I  2 MR 2  2mR 2 , where M  8.40 kg, m  2.00 kg 3 I  0.600 kg  m 2 ω0  75.0 rpm  7.854 rad s ; ω  50.0 rpm  5.236 rad s ; t  30.0 s, α  ? 2 ω  ω0  αt gives α  0.08726 rad s ; τ  Iα, τ f  Iα  0.0524 N  m    13.1 N  m. rad s  400 rev min  2  10.8: a)   Iα  I t   2.50 kg  m 2 8.00 s  60 rev min 2 1 1  2π rad s  b) I2  (2.50 kg  m 2 )  400 rev min     2.19  103 J. 2 2  60 rev min     10.9: v  2as  2 0.36 m s 2 2.0 m   1.2 m s , the same as that found in Example 9-8. τ FR 40.0 N 0.250 m  10.10: α    2.00 rad s 2 . I I  5.0 kg  m 2   m   M  3m  10.11: a) n  Mg  T  g  M  g   1  2m M  1  2m M  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.
  3. 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  9.0 N   (50 kg) 9.80 m s 2 490 N, at an angle of arctan 2 2  490   1.1 from the 9.0 vertical (the weight is much larger than the applied force F ). f  R I MR 0 t  10.13: k     n n Rn 2n 50.0 kg 0.260 m 850 rev min  30 rev min   rad s   0.482. 27.50 s 160 N  10.14: (a) Falling stone: g  1 at 2 2 12.6 m  1 a3.00 s  2 2 a  2.80 m s 2 Stone :  F  ma : mg  T  ma(1) Pulley :  τ  Iα : TR  1 MR 2α  1 MR 2  R  2 2 a T  1 Ma(2) 2 Solve (1) and (2): M a   10.0 kg   2.80 m/s 2  M    g a      2   9.80 m/s  2.80 m/s 2 2 2  M  2.00 kg (b)From (2): T 1 2 1  Ma  10.0 kg  2.80 m/s 2 2  T  14.0 N
  4. I  1 mR 2  1 8.25 kg 0.0750 m   0.02320 kg  m 2 2 10.15: 2 2 ω0  220 rpm  23.04 rad/s; ω  0; θ  θ0  5.25 rev  33.0 rad, α  ? ω2  ω0  2α θ  θ0  gives α  8.046 rad/s 2 2  τ  Iα  τ  τ f   f k R   μk nR Iα  μk nR  Iα so n   7.47 N μk R 10.16: This is the same situtation as in Example 10.3. a) T  mg (1  2m M )  42.0 N. b) v  2 gh (1  M 2m  11.8 m s. 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 h v 2  1.69 s. 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 a1  m2 g  m1  m2  I R 2 . a) a1  2 1.20 m  0.80 s   3.75 m/s 2 , 2 so T1 m1a1  7.50 N and T2  m2  g  a1   18.2 N. b) The torque on the pulley is T2  T1 R  0.803 N  m, and the angular acceleration is α  a1 R  50 rad/s 2 , so I     0.016 kg  m 2 .  Fl 3F 10.18:   1 2  . I 3 Ml Ml 10.19: The acceleration of the mass is related to the tension by Macm  Mg  T , and the angular acceleration is related to the torque by Iα  τ  TR, or acm  T / M , where α  acm / RSt and I  MR 2 have been used. a) Solving these for T gives T  Mg / 2  0.882 N. b) Substituting the expression for T into either of the above relations gives acm  g / 2, from which t  2h acm  4h g  0.553 s. c) ω  vcm R  acmt R  33.9 rad/s. 10.20: See Example 10.6 and Exercise 10.21. In this case, K 2  Mvcm and vcm  gh , ω  vcm R  33.9 rad s . 2
  5. 10.21: From Eq. (10.11), the fraction of the total kinetic energy that is rotational is 1 2I cm 2  1  1 2 , 1 2Mvcm  1 2I cm 2 2 1  M I cm  v /  2 cm 2 1  MR I cm where vcm  R for an object that is rolling without slipping has been used. a) I cm  (1 2) MR 2 , so the above ratio is 1 3. b) I  (2 5)MR 2 , so the above ratio is 2 7. c) I  2 3 MR 2 , so the ratio is 2 5. d) I  5 8 MR 2 , so the ratio is 5 13. 10.22: a) The acceleration down the slope is a  g sin θ  M , the torque about the f center of the shell is a 2 a 2 τ  Rf  Iα  I  MR 2  MRa, R 3 R 3 so M  3 a. Solving these relations a for f and simultaneously gives 5 a  g sin θ , or f 2 3 3 3 a  g sin θ  (9.80 m s 2 ) sin 38.0  3.62 m s 2 , 5 5 2 2 f  Ma  (2.00 kg)(3.62 m s 2 )  4.83 N. 3 3 The normal force is Mg cos θ , and since f  μs n, f 2 Ma 2 a 2 3 g sin θ 2 μs   3   5  tan θ  0.313. n Mg cos θ 3 g cos θ 3 g cos θ 5 b) a  3.62 m s 2 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 μs also does not change.
  6. 10.23: n  mg cos  mg sin θ  μs mg cos θ  ma g (sin θ  μs cos θ )  a (eq.1) n and mg act at the center of the ball and provide no torque.  τ  τ f  μs mg cos θR; I  5 mR 2 2  τ  Iα gives μs mg cos θ  5 mR 2α 2 No slipping means α  a R , so μs g cos  5 a (eq.2) 2 We have two equations in the two unknowns a and  s . Solving gives a  5 g sin θ and μs  7 tan θ  7 tan 65.0  0.613 7 2 2 b) Repeat the calculation of part (a), but now I  2 mR 2 . 3 a  5 g sin θ and μs  5 tan θ  5 tan 65.0  0.858 3 2 2 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.
  7. 10.24: vcm  R for no slipping a) Get v at bottom: 1 2 1 mgh  mv  Iω 2 2 2 2 1 12  v  mgh  mv 2   mR 2    2 2 5  R 10 v gh 7 Now use energy conservation. Rotational KE does not change 1 2 mv  KERot  mgh  KERot 2 v 2 10 gh 5 h   7  h 2g 2g 7 (b) mgh  mgh  h  h 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: wh  W f  K 1  (1 / 2) I cm w 2 0  1 mv 2 cm 2 Solving for h with vcm  Rw h 1  w 2 9.80 m s 2 [(0.800)(0.600 m) (25.0 rad s) 2 2  (0.600 m) 2 (25.0 rad s) 2 ] w 3500 J   11.7 m. 392 N
  8. 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 I  fR. The equation of motion is mg sin β  f  macm, and the acceleration and angular acceleration are related by acm  Rα (note that positive acceleration is taken to be down the incline, and relation between acm and  is correct for a friction force directed uphill). Combining,  I  mg sinβ  ma1  2   ma7 5,  mR  from which acm  5 7 g sinβ. c) From either of the above relations between if f and acm , 2 2 f  macm  mg sin β  μs n  μs mg cos β , 5 7 from which μs  2 7  tan  .    10.27: a) ω  αt  FR I t  18.0 N 2.40 m  2100 kg  m 2 15.0 s   0.3086 rad/s, or 0.309 rad/s to three figures. b) W  K 2  1 2Iω2  1 2   2.00 kg  m 2 0.3086 rad s   100 J. 2 c) From either P  τωave or P  W t , P  6.67 W. 10.28: a) τ P  175 hp 746 W / hp   519 N  m. ω 2400 rev/min π rad/s     30 rev/min  b) W  τθ  519 N  m 2π   3261 J.
  9. ω 10.29: a) τ  Iα  I t 1 21.50 kg 0.100 m 1200 rev min  30 rev min  2   π rad s      2.5 s  0.377 N  m. b)  avet  600 rev/min2.5 s  25.0 rev  157 rad. 60 s/min c) τθ  59.2 J. 1 d) K  Iω2 2 2   π rad/s   1 2   (1 / 2)(1.5 kg)(0.100 m)2   (1200 rev/min)     30 rev/min    59.2 J, the same as in part (c). 10.30: From Eq. (10.26), the power output is  2 rad/s  P  τω  (4.30 N  m) 4800 rev/min    2161 W,  60 rev/min  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. τ P/ω (1.9 hp)(746 W/hp) b) F    65.6 N. R R  π rad/s  (2400 rev/min) (0.086 m)  30 rev/min 
  10. 10.32: I  1 mL2  1 (117 kg)(2.08 m)2  42.2 kg  m 2 2 2 τ 1950 N  m a) α   46.2 rad/s2 . I 42.2 kg  m 2 b)   2  2(46.2 rad/s 2 )(5.0 rev  2 rev)  53.9 rad/s. 1 c) From either W  K   2 or Eq. (10.24), 2 W  τ  (1950 N.m)(5.00 rev  2π rad/rev)  6.13  104 J. d), e) The time may be found from the angular acceleration and the total angle, but the instantaneous power is also found from P  τω  105 kW(141 hp). The average power is half of this, or 52.6 kW.   π rad/s   10.33: a) τ  P / ω  (150  103 W)  (400 rev/min)     358 N  m.    30 rev/min   b) If the tension in the rope is F , F  w and so w  τ/R  1.79  103 N. c) Assuming ideal efficiency, the rate at which the weight gains potential energy is the power output of the motor, or wv  P, so v  P w  83.8 m/s. Equivalently, v  R. 10.34: As a point, the woman’s moment of inertia with respect to the disk axis is mR 2 , and so the total angular momentum is 1  L  Ldisk  Lwoman  ( I disk  I woman )ω   M  m  R 2ω 2  1    110 kg  50.0 kg (4.00 m)2 (0.500 rev/s  2π rad/rev) 2   5.28  103 kg  m 2 / s. 10.35: a) mvr sinφ  115 kg  m 2 / s, with a direction from the right hand rule of into the page. b) dL dt  τ  2 kg 9.8 N kg   8 m   sin 90  36.9  125 N  m  125 kg  m 2 s , 2 out of the page.
  11. 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  1024 kg)(2.98  104 m s) (1.50  1011 m)  2.67  1040 kg  m 2 s b) L  (2 5mr 2 )(ω) L  (2 5)(5.97  1024 kg)(6.38  106 m) 2 (2 rad (24.0 hr  3600 s hr))  7.07  1033 kg  m 2 s 10.37: The period of a second hand is one minute, so the angular momentum is M 2 2 L  Iω  l 3 T  6.0  10 kg  3 2   (15.0  10 2 m) 2   4.71  10 6 kg  m 2 s.  3  60 s 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   2 rad  7.0  10 5 km   2  1  1    R   (30 d)(86,400 s d  16 km   4.6  10 rad s.   3  2    10.39: a) The net force is due to the tension in the rope, which always acts in the radial direction, so the angular momentum with respect to the hole is constant. b) L1  m 1 r 21 , L2  m 2 r2 , and with L1  L2 , ω2  ω1 (r1 r2 ) 2  7.00 rad s. 2 c) K  (1 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.
  12. 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), I1 2.56 kg  m 2 ω2  ω1  (0.40 rev s)  1.14 rev s. I2 0.9 kg  m 2 Note that conversion from rev/s to rad s is not necessary. 10.41: If she had tucked, she would have made (2) (3.6 kg  m 2 ) 18 kg  m 2 )  0.40 rev in the last 1.0 s, so she would have made (0.40 rev)(1.5 1.0)  0.60 rev in the total 1.5 s. 10.42: Let I1  I 0  1200kg  m 2 , I 2  I 0  mR 2  1200kg  m 2  (40.0kg )(2.00 m) 2  1360kg  m 2 . Then, from Eq. (10.33), I1  2π rad  1200 kg.m 2 2  1     0.924 rad s. I 2  6.00 s  1360 kg.m 2
  13. 10.43: a) From conservation of angular momentum, ω2  ω1 I1  ω1 1 2MR 2  ω 1 I 0  mR 2 1 2MR  mR 2 2 1 1  2m M 3.0 rad s   1.385 rad s 1  270 120 or 1.39 rad s to three figures b) K1  1 21 2120 kg 2.00 m  3.00 rad s   1.80 kJ, and 2 2   K 2  1 2  I 0  70 kg 2.00 m  ω2  499 J. In changing the parachutist’s horizontal 2 2 component of velocity and slowing down the turntable, friction does negative work. 10.44: Let the width of the door be l; L mvl 2 ω  I 1 3Ml 2  ml 22  0.500 kg 12.0 m s0.500 m   0.223 rad s. 1 340.0 kg 1.00 m 2  0.500 kg 0.500 m 2 Ignoring the mass of the mud in the denominator of the above expression gives ω  0.225 rad s , so the mass of the mud in the moment of inertia does affect the third significant figure.  10.45: Apply conservation of angular momentum L, with the axis at the nail. Let object A be the bug and object B be the bar. Initially, all objects are at rest and L1  0. Just after the bug jumps, it has angular momentum in one direction of rotation and the bar is rotating with angular velocity ωB in the opposite direction. L2  mAv Ar  I B ωB where r  1.00 m and I B  1 mB r 2 3 L1  L2 gives mAv Ar  1 mB r 2B 3 3mAv A ωB   0.120 rad s mB r
  14. 10.46: (a) Conservation of angular momentum: m1v0 d   m1vd  1 m2 L2ω 3 1  90.0 N  (3.00 kg)(10.0 m s) (1.50 m)  (3.00 kg)(6.00 m s)(1.50 m)    (2.00 m) 2 ω 3  9.80 m s 2    ω  5.88 rad s (b) There are no unbalanced torques about the pivot, so angular momentum is conserved. But the pivot exerts an unbalanced horizontal external force on the system, so the linear momentum is not conserved. 10.47:
  15. 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, ω ωR   1.617 N  4.00  102 m   188.7 rad s ,   I 1.20  10 4 kg  m 2 2π  rad 2.20 s which is 1.80  10 rev min . Note that in this and similar situations, since  appears in 3 the denominator of the expression for  , the conversion from rev s and back to rev min must be made. c) K (1 / 2)((1 / 2) MR 2 ) 2 10.49: a)  P P (1 / 2)(1 / 2)(60,000 kg)(2.00 m)2 (500 rev/min) 30 rev/min   rad/s 2  7.46  104 W  2.21  103 s, or 36.8 min. b) τ  I  π rad/s   2π rad   (1 / 2)(60,000 kg)(2.00 m)2 (500 rev/min) (1.00/s)   30 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.
  16. 10.51: a) Solving Eq. (10.36) for τ , τ  Iω   (2 / 5) MR 2ω . Using ω  86,400 s and 2 rad  2 ( 26 , 000 y)(3.17510 7 s/y) and the mass and radius of the earth from Appendix F, τ ~ 5.4  1022 N  m. 10.52: a) The net torque must be  2 rad/s  120 rev/min   ω 2  60 rev/min  τ  Iα  I  (1.86 kg  m )  2.60 N  m. t (9.00 s) This torque must be the sum of the applied force FR and the opposing frictional torques τ f at the axle and fr  μk nr due to the knife. Combining, 1 F ( τ  τ f  μk nr ) R 1  (2.60 N  m)  (6.50 N  m)  (0.60)(160 N)(0.260 m) 0.500 m  68.1 N. b) To maintain a constant angular velocity, the net torque τ is zero, and the force F  is F   0.500 m (6.50 N  m  24.96 N  m)  62.9 N. c) The time t needed to come to a stop is 1 found by taking the magnitudes in Eq. (10.27), with τ  τ f constant; t   60  L ωI 120 rev/min  2  rev/min  1.86 kg  m 2 rad/s   3.6 s. τf τf 6.50 N  m Note that this time can also be found as t  9.00 s  2.60 Nm . 6.50N m 10.53: a) I  τ τt   5.0 N  m 2.0 s   0.955 kg  m 2 . α ω 100 rev/min   π rad/s     30 rev/min  b) Rather than use the result of part (a), the magnitude of the torque is proportional to  and hence inversely proportional to | t | ; equivalently, the magnitude of the change in angular momentum is the same and so the magnitude of the torque is again proportional 2s to 1 / | t | . Either way, τ f  5.0 N  m   0.080 N  m. 125 s c) ωave t  50.0 rev/min 125 s 1 min/60 s   104.2 rev.
  17. 10.54: a) The moment of inertia is not given, so the angular acceleration must be found from kinematics; 2θ 2s 25.00 m  α  2   8.33 rad / s 2 . t 2 rt 0.30 m 2.00 s 2   b) αt  8.33 rad/s 2 2.00 s   16.67 rad/s. c) The work done by the rope on the flywheel will be the final kinetic energy; K  W  Fs  40.0 N 5.0 m   200 J. 2K 2200 J  d) I   1.44 kg  m 2 .  2 16.67 rad/s  2 τ t 10.55: a) P  τω  ταt  τ  t  τ 2  . I I b) From the result of part (a), the power is 500 W   60..0   4.50 kW. 2 20 0 c) P  τω  τ 2αθ  τ 2τ / I  θ  τ 3 / 2 2θ / I . d) From the result of part (c), the power is 500 W  20.00   2.6 kW. e) No; the 6 3/ 2 .00 power is proportional to the time t or proportional to the square root of the angle. 10.56: a) From the right-hand rule, the direction of the torque is i  ˆ  k , the  z ˆ j ˆ direction. b), c) d) The magnitude of the torque is F0 ( x  x 2 l ), which has it maximum at l 2. The torque at x  l 2 is F0 l 4.
  18. 2θ 2θ 2θ I 10.57: t2    .  ( I )  The angle in radiants is  2, the moment of inertia is (1 3) ((750 N) (9.80 m s 2 )(1.25 m))3  39.9 kg  m 2 and the torque is (220 N)(1.25 m)  275 N  m. Using these in the above expression gives t 2  0.455 s 2 , so t  0.675 s. 10.58: a) From geometric consideration, the lever arm and the sine of the angle  between F and r are both maximum if the string is attached at the end of the rod. b) In terms of the distance x where the string is attached, the magnitude of the torque is Fxh x 2  h 2 . This function attains its maximum at the boundary, where x  h, so the string should be attached at the right end of the rod. c) As a function of x, l and h, the torque has magnitude xh  F . ( x  l 2) 2  h 2 This form shows that there are two aspects to increasing the torque; maximizing the lever arm l and maximizing sin  . Differentiating  with respect to x and setting equal to zero gives x max  (l 2)(1  (2 h l ) 2 ). This will be the point at which to attach the string unless 2h > l, in which case the string should be attached at the furthest point to the right, x  l. 10.59: a) A distance L 4 from the end with the clay. b) In this case I  (4 3) ML2 and the gravitational torque is (3L 4)(2Mg ) sin  (3Mg L 2) sin , so   (9 g 8 L) sin  . c) In this case I  (1 3) ML2 and the gravitational torque is ( L 4)(2Mg ) sin   ( Mg L 2) sin  , so   (3 g 2 L) sin  . This is greater than in part (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.
  19. 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 .  2 a) W  FR cos θ d θ  FR 0 b) In Eq. (6.14), dl is the horizontal distance the point moves, and so W  F  dl  FR, the same as part (a). c) From K 2  W  (MR 2 4)ω2 , ω  4 F MR . d) The torque, and hence the angular acceleration, is greatest when   0, at which point α  ( τ I )  2 F MR , and so the maximum tangential acceleration is 2 F M . e) Using the value for  found in part (c), arad  ω2 R  4 F M . 10.61: The tension in the rope must be m( g  a)  530 N. The angular acceleration of the cylinder is a R  3.2 rad/s 2 , and so the net torque on the cylinder must be 9.28 N  m. Thus, the torque supplied by the crank is (530 N)(0.25 m)  (9.28 N  m)  141.8 N  m, and the force applied to the crank handle is 0.12 m  1.2 kN to two figures. 141.8 Nm 10.62: At the point of contact, the wall exerts a friction force f directed downward and a normal force n directed to the right. This is a situation where the net force on the roll is zero, but the net torque is not zero, so balancing torques would not be correct. Balancing vertical forces, Frod cos  f  w  F , and balacing horizontal forces Frod sin θ  n. With f  μk n, these equations become Frod cos θ  μk n  F  w, Frod sin θ  n. (a) Eliminating n and solving for Frod gives ω F (16.0 kg) (9.80 m/s2 )  (40.0 N) Frod    266 N. cos θ  μk sin θ cos 30  (0.25) sin 30 b) With respect to the center of the roll, the rod and the normal force exert zero torque. The magnitude of the net torque is ( F  f ) R, and f  μk n may be found insertion of the value found for Frod into either of the above relations; i.e., f  μk Frod sin θ  33.2 N. Then, τ (40.0 N  31.54 N)(18.0  102 m) α   4.71 rad/s 2 . I (0.260 kg  m 2 )
  20. 10.63: The net torque on the pulley is TR, where T is the tension in the string, and α  TR I . The net force on the block down the ramp is mg (sin β  μk cos β )  T  ma. The acceleration of the block and the angular acceleration of the pulley are related by α  αR. a) Multiplying the first of these relations by I R and eliminating  in terms of a, and then adding to the second to eliminate T gives a  mg sin β  μk cos β   g sin β  μk cos β  , m  I / R2 1  I / mR  2 and substitution of numerical values given 1.12 m/s 2 . b) Substitution of this result into either of the above expressions involving the tension gives T = 14.0 N. 10.64: For a tension T in the string, mg  T  ma and TR  Iα  I R . Eliminating T and a solving for a gives m g ag  , mI/R 2 1  I / mR 2 where m is the mass of the hanging weight, I is the moment of inertia of the disk   combination I  2.25  10 3 kg  m 2 from Problem 9.89 and R is the radius of the disk to which the string is attached. a) With m = 1.50 kg, R  2.50  102 m, a  2.88 m/s 2 . b) With m = 1.50 kg, R  5.00  10 2 m, a  6.13 m/s 2 . The acceleration is larger in case (b); with the string attached to the larger disk, the tension in the string is capable of applying a larger torque. 10.65: Taking the torque about the center of the roller, the net torque is fR  αI , I  MR 2 for a hollow cylinder, and with   a / R, f  Ma (note that this is a relation   between magnitudes; the vectors f and a are in opposite directions). The net force is F  f  Ma, from which F  2Ma and so a  F 2M and f  F 2 .
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