 # Physics exercises_solution: Chapter 09

152
lượt xem
14 ## Physics exercises_solution: Chapter 09

Mô tả tài liệu

Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 09

Chủ đề:

## Nội dung Text: Physics exercises_solution: Chapter 09

4. 9.15: From Eq. (9.11), 2 θ  θ0 α z t 60.0 rad (2.25 rad s )(4.00 s) ω0 z      10.5 rad s. t 2 4.00 s 2 2 9.16: From Eq. (9.7), with ω0 z  0, α z  ωz t  140.00 s s  23.33 rad s  The angle is most 6 rad 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.
5. 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: (a ) ( b) (c ) t rev's θ rev's θ rev's θ 0.05 0.50 180 0.03 11.3 0.44 158 0.10 1.00 360 0.13 45 0.75 270 0.15 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. vr 5.00 m s 9.33: The angular velocity of the rear wheel is ωr    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 9.34: The distances of the masses from the axis are L 4 , L and 34L , and so from Eq. 9.16, 4 the moment of inertia is 2 2 2 L L  3L  11 I  m   m   m   mL2 . 4 4  4  16 L2 2 9.35: The moment of inertia of the cylinder is M 12 and that of each cap is m L4 , so the moment of inertia of the combination is 12  m L . M 2 2 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 , 1 1 ML2  0.042 kg 1.50 m   7.88  10  3 kg  m 2 . 2 I 12 12 b) From Table 9.2b , 1 1 I  ML2  0.042 kg 1.50 m   3.15  10 2 kg  m 2 . 2 3 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 , 1 1 I MR 2  (0.042 kg) (1.5  10 3 m)2  4.73  10 8 kg  m 2 . 2 2
10. 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 2 , and the moment of inertia is 4(0.200 kg) (0.800  102 m 2 )  6.40  102 kg  m 2 . 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 2 . 2 a) The two masses through which the axis passes do not contribute to the moment of inertia.   2 I  2(0.2 kg) 0.2 2 m  0.032 kg  m 2 . 2 1  L 9.38: (a) I  I bar  I balls  M bar L2  2mballs   12 2 1  4.00 kg 2.00 m 2  20.500 kg 1.00 m 2  2.33 kg  m 2 12 1 (b) I  mbar L2  mball L2 3 1  4.00 kg 2.00 m 2  0.500 kg 2.00 m 2  7.33 kg  m 2 3 c) I  0 because all masses are on the axis (d) I  mbar d 2  2mballd 2  M Total d 2  (5.00 kg)(0.500 m) 2  1.25 kg  m 2 9.39: I  I d  I r (d  disk, r  ring) 3 disk : md  (3.00 g cm )πrd2  23.56 kg 1 Id  md rd2  2.945 kg  m 2 2 ring : mr  (2.00 g cm3 ) π (r22  r12 )  15.08 kg (r1  50.0 cm, r2  70.0 cm ) 1 Ir  mr (r12  r22 )  5.580 kg  m 2 2 I  I d  I r  8.52 kg  m 2
11. 9.40: a) In the expression of Eq. (9.16), each term will have the mass multiplied by f 3 and the distance multiplied by f , and so the moment of inertia is multiplied by f 3 ( f ) 2  f 5 . b) (2.5)(48)5  6.37  108. 9.41: Each of the eight spokes may be treated as a slender rod about an axis through an end, so the moment of inertia of the combination is m  I  mrim R 2  8  spoke  R 2  3     8   (1.40 kg )  (0.20 kg) (0.300 m)2  3   0.193 kg  m 2 9.42: a) From Eq. (9.17), with I from Table (9.2(a)), 1 1 1 rev 2π rad rev 2 K mL2ω2  (117 kg )(2.08 m) 2 (2400  )  1.3  10 J. 2 12 24 min 60 s min b) From mgy  K , y K   1.3  10 6 J   1.16  10 3 m  1.16 km. 2 mg (117 kg)(9.80 m s ) 9.43: a) The units of moment of inertia are [kg] [m 2 ] and the units of ω are equivalent to [s 1 ] and so the product 1 Iω 2 has units equivalent to [kg  m  s 2 ]  [kg  (m s) 2 ] , 2 which are the units of Joules. A radian is a ratio of distances and is therefore unitless. b) K  π 2 Iω2 1800 , when is in rev min. 9.44: Solving Eq. (9.17) for I, 2K 2(0.025J) I 2  2 π rad s 2  2.25  103 kg  m 2 . ω (45 rev min  60 rev min )
12. 9.45: From Eq. (9.17), K 2  K  1 I (ω2  ω12 ), and solving for I, 2 2 I 2 K 2  K1  (ω2  ω12 ) 2 (500 J) 2 ((520 rev min) 2  (650 rev min ) 2 )  rad s 30 rev min 2  0.600 kg  m 2 . 9.46: The work done on the cylinder is PL, where L is the length of the rope. Combining Equations (9.17), (9.13) and the expression for I from Table (9.2(g)), 1ω 2 1 ω v2 (40.0 N)(6.00 m s) 2 PL  v , or P    14.7 N. 2g 2 g L 2(9.80 m s 2 )(5.00 m) a rad 1 9.47: Expressing ω in terms of αrad , ω2  R . Combining with I  MR 2 , Eq. (9.17) 2 11 (70.0 kg)(1.20 m)(3500 m s) 2 becomes K  MRarad   7.35  104 J. 22 4 9.48: a) With I  MR 2 , with expression for v is 2 gh v . 1 M m b) This expression is smaller than that for the solid cylinder; more of the cylinder’s mass is concentrated at its edge, so for a given speed, the kinetic energy of the cylinder is larger. A larger fraction of the potential energy is converted to the kinetic energy of the cylinder, and so less is available for the falling mass.
13. 9.49: a) ω  2 T , so Eq. (9.17) becomes K  22 I T 2 . b) Differentiating the expression found in part (a) with respect to T, dK  (4 2 I T 3 ) dT . dt dt c) 2 2 (8.0 kg  m 2 ) (1.5 s) 2  70.2 J, or 70 to two figures. d) (4 2 (8.0 kg  m 2 ) (1.5 s) 3 )(0.0060)  0.56 W. 9.50: The center of mass has fallen half of the length of the rope, so the change in gravitational potential energy is 1 1  mgL   (3.00 kg )(9.80 m s 2 )(10.0 m)  147 J. 2 2 9.51: (120 kg)(9.80 m s 2 )(0.700 m)  823 J. 9.52: In Eq; (9.19), I cm  MR 2 and d  R 2 , so I P  2 MR 2 . 2 2 4 9.53: MR 2  MR 2  Md 2 , so d 2  R 2 , and the axis comes nearest to the center of 3 5 15 the sphere at a distance d  (2 15 ) R  (0.516) R. 9.54: Using the parallel-axis theorem to find the moment of inertia of a thin rod about an axis through its end and perpendicular to the rod, 2 M 2  L M 2 I p  I cm  Md 2 L  M   L. 12 2 3 9.55: Ιp  Ι cm  md 2 ,so Ι  12 Μ a 2  b 2   Μ ( a ) 2  ( b ) 2  , which gives 1 2 2 1   1   Ι  Μ a 2  b 2   a 2  b 2 , or Ι  Μ a 2  b 2 . 12 4 1 3 
14. 9.56: a) Ι  12 Μa 2 1 b) Ι  12 Μb 2 1 9.57: In Eq. 9.19 , Ι cm  M 12 L2 and d  L 2  h , so 1 2 L   2 Ιp  Μ  L    h   12  2   1 1   Μ  L2  L2  Lh  h 2  12 4  1   Μ  L2  Lh  h 2  , 3  which is the same as found in Example 9.12. 9.58: The analysis is identical to that of Example 9.13, with the lower limit in the integral being zero and the upper limit being R, and the mass Μ  πLρR 2 . The result is Ι  1 ΜR 2 , as given in Table 9.2(f) . 2 9.59: With dm  M L dx L L M M x3 M 2 Ι  x 2 dx   L. 0 L L 3 0 3
15. 9.60: For this case, dm   dx. L L x2 yL2 a) Μ   dm   γx dx  γ  0 2 0 2 L 4 L x Ι   x 2 (γx)dx  γ 4 b)  γL  Μ L2 . 4 2 0 4 0 This is larger than the moment of inertia of a uniform rod of the same mass and length, since the mass density is greater further away from the axis than nearer the axis. L c) Ι   ( L  x) 2 γxdx 0 L    ( L2 x  2 Lx 2  x 3 )dx 0 L  x2 x3 x 4    L2  2 L    2  3 4 0  L4  12 M 2  L. 6 This is a third of the result of part (b), reflecting the fact that more of the mass is concentrated at the right end.  9.61: a) For a clockwise rotation, ω will be out of the page. b ) The upward direction   crossed into the radial direction is, by the right-hand rule, counterclockwise. ω and r are    perpendicular, so the magnitude of ω  r is ωr  v. c) Geometrically, ω is    perpendicular to v , and so ω  v has magnitude ωv  arad, and from the right-hand rule, the upward direction crossed into the counterclockwise direction is inward, the  direction of arad . Algebraically,       a rad  ω  v  ω  ω  r         ω ω  r   r ω  ω      2r ,   where the fact that ω and r are perpendicular has been used to eliminate their dot product.
16. 9.62: For planetary alignment, earth must go through 60 more than Mars: θE  θM  60 wE t  ωM t  60 60 t ωE  ωM 360 360 wE  and wM  1yr 1.9 yr 60  365d  t  0.352 yr   1yr   128 d  360 1yr  1.9 yr 360   9.63: a) v  60 mph  26.82 m s r  12 in.  0.3048 m v ω   88.0 rad s  14.0 rev s  840 rpm r b) same ω as in part (a) since speedometer reads same r  15 in.  0.381 m v  rω  (0.381m)(88.0 rad s)  33.5 m s  75 mph c) v  50 mph  22.35 m s r  10 in.  0.254 m v ω   88.0 rad s; . this is the same as for 60 mph with correct tires, so r speedometer read 60 mph. 9.64: a) For constant angular acceleration θ  ωα , and so arad  ω2 r  2αθr. 2 2 b) Denoting the angle that the acceleration vector makes with the radial direction as β , and using Equations (9.14) and (9.15), a αr αr 1 tan β  tan  2   , arad ω r 2 αθ r 2θ so   2 tan β  2 tan136.9  0.666 rad. 1 