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Chapter 8 (1, 3, 6, 7, 13, 19, 22, 39, 40, 44, 45, 52, 54, 56, 57, 63, 65 & 66) 8.1 Since the friction force is tangential to a point on the rim of the wheel, it is perpendicular to the radius line connecting this point with the center of the wheel. The torque of this force about the axis through the center of the wheel is then = rf sin 90.0º = rf, and the friction force is f 8.3 r 76.0 N m 217 N 0.350 m First resolve all of the forces shown in Figure P8.3 into components parallel to and perpendicular to the beam as shown in the sketch below. (a) O 25 N cos 30 2.0 m 10 N sin 20 4.0 m 30 N m or (b) C 30 N sin 45 2.0 m 10 N sin 20 2.0 m 36 N m or 8.6 0 = 30 N m counterclockwise c = 30 N m counterclockwise The object is in both translational and rotational equilibrium. Thus, we may write: Fx 0 Fy 0 and Fx Rx 0 Fy Ry Fg 0 O 0 Fy cos Fx sin Fg cos 0 2 8.7 Requiring that = 0, using the shoulder joint at point O as a pivot, gives Ft sin 12.0 0.080 m 41.5 N 0.290 m 0 or Ft 724 N Then Fy 0 Fsy 724 N sin 12.0 41.5 N = 0 , yielding Fsy 109 N Fy 0 gives Fsx 724 N cos12.0 = 0 , or Fsx 708 N Therefore, Fs 8.13 Fsx2 Fsy2 708 N 2 109 N 2 716 N Requiring that xcg mi xi mi 0 gives 5.0 kg 0 3.0 kg 0 4.0 kg 3.0 m 8.0 kg x 5.0 3.0 4.0 8.0 kg 0 or 8.0 x + 12 m = 0 which yields x = – 1.5 m Also, requiring that ycg mi yi mi 0 gives 5.0 kg 0 3.0 kg 4.0 m 4.0 kg 0 8.0 kg y 5.0 3.0 4.0 8.0 kg or 8.0y + 12 m = 0 yielding y = – 1.5 m 0 Thus, the 8.0-kg object should be placed at coordinates (–1.5 m, –1.5m) . 8.19 Consider the torques about an axis perpendicular to the page through the left end of the rod. 0 T 100 N 3.00 m 500 N 4.00 m 6.00 m cos 30.0 T 443 N Fx 0 Rx T sin 30.0 443 N sin 30.0 Rx = 221 N toward the right Fy 0 Ry T cos 30.0 100 N 500 N 0 Ry = 600 N (443 N) cos 30.0 = 217 N upward 8.22 (a) See the diagram below (b) If x = 1.00 m, then left end 0 700 N 1.00 m 200 N 3.00 m 80.0 N 6.00 m T sin 60.0° 6.00 m 0 giving T = 434 N. Then, Fx 0 H T cos 60.0° 0 , or H 343 N cos 60.0° 172 N and Fy 0 V 980 N + 343 N sin 60.0° 0 , or V = 683 N. (c) When the wire is on the verge of breaking, T = 900 N and left end 700 N xmax 200 N 3.00 m 80.0 N 6.00 m 900 N sin 60.0° 6.00 m 0 which gives xmax = 5.14 m 8.39 I 1 1 2 MR2 150 kg 1.50 m 169 kg m 2 2 2 and f i t 0.500 rev s 0 2 rad rad s2 1 rev 2 .00 s 2 Thus, F r I gives F 8.40 (a) I r 169 kg m2 2 rad s2 1.50 m 177 N It is necessary that the tensions T1 and T2 be different in order to provide a net torque about the axis of the pulley and produce and angular acceleration of the pulley. Since intuition tells us that the system will accelerate in the directions shown in the diagrams at the right when m2 > m1, it is necessary that T2 > T1. (b) We adopt a sign convention for each object with the positive direction being the indicated direction of the acceleration of that object in the diagrams at the right. Then, apply Newton’s second law to each object: For m1 : Fy m1a T1 m1g m1a or T1 m1 g a [1] For m2 : Fy m2 a For M : I m2 g T2 m2 a m2 rT2 rT1 I or T2 m2 g a [2] or T2 T1 I r [3] Substitute Equations [1] and [2], along with the relations I Mr 2 2 and a r , into Equation [3] to obtain m2 g a m1 g a Mr 2 a Ma 2r r 2 or M m1 m2 2 a m2 m1 g and a (c) m2 m1 g m1 m2 M 2 20.0 kg 10.0 kg 9.80 m s2 20.0 kg 10.0 kg + 8.00 kg 2 2.88 m s2 From Equation [1]: T1 10.0 kg 9.80 m s2 2.88 m s2 127 N . From Equation [2]: T2 20.0 kg 9.80 m s2 2.88 m s2 138 N . 8.44 (a) (b) Hoop: I MR2 4.80 kg 0.230 m 0.254 kg m2 Solid Cylinder: I 1 1 2 MR2 4.80 kg 0.230 m 0.127 kg m 2 2 2 Solid Sphere: I 2 2 2 MR2 4.80 kg 0.230 m 0.102 kg m 2 5 5 Thin, Spherical, Shell: I 2 2 2 MR2 4.80 kg 0.230 m 0.169 kg m 2 3 3 2 When different objects of mass M and radius R roll without slipping a R down a ramp, the one with the largest translational acceleration a will have the highest translational speed at the bottom. To determine the translational acceleration for the various objects, consider the free-body diagram at the right: Fx Ma I Mg sin f Ma f R I a R [1] or f Ia R2 [2] Substitute Equation [2] into [1] to obtain Mg sin Ia R 2 Ma or a Mg sin M I R2 Since M, R, g are the same for all of the objects, we see that the translational acceleration (and hence the translational speed) increases as the moment of inertia decreases. Thus, the proper rankings from highest to lowest by translational speed will be: Solid sphere; solid cylinder; thin, spherical, shell; and hoop (c) When an object rolls down the ramp without slipping, the friction force does no work and mechanical energy is conserved. Then, the total kinetic energy gained equals the gravitational potential energy given up: KEr KEt PEg Mgh and KEr Mgh 1 2 M v2 , where h is the vertical drop of the ramp and is the translational speed at the bottom. Since M, g, and h are the same for all of the objects, the rotational kinetic energy decreases as the translational speed increases. Using this fact, along with the result of Part (b), we rank the object’s final rotational kinetic energies, from highest to lowest, as: hoop; thin, spherical, shell; solid cylinder; and solid sphere 8.45 (a) Treating the particles on the ends of the rod as point masses, the total moment of inertia of the rotating system is I I rod I 3 I 4 mrod L2 12 m3 (L 2)2 m4 (L 2)2 . If the mass of the rod can be ignored, this reduces to L I 0 m3 m4 2 2 3.00 kg 4.00 kg 0.500 m 1.75 kg m 2 and the rotational kinetic energy is 2 KEr (b) 1 2 1 I 1.75 kg m 2 2 2 2.50 rad s 5.47 J 2 If the rod has mass mrod 2.00 kg I 1 2.00 kg 1.00 m 2 1.75 kg m2 1.92 kg m 2 12 and KEr 8.52 1 2 1 I 1.92 kg m 2 2 2 2.50 rad s 6.00 J 2 As the bucket drops, it loses gravitational potential energy. The spool gains rotational kinetic energy and the bucket gains translational kinetic energy. Since the string does not slip on the spool, r where r is the radius of the spool. The moment of inertia of the spool is I 1 2 Mr 2 , where M is the mass of the spool. Conservation of energy gives KE t KEr PEg f KEt KEr PEg i 1 1 m 2 I 2 mgy f 0 0 mgyi 2 2 or 1 11 2 m r Mr 2 2 mg yi y f 2 22 This gives 2mg yi y f m 21 M r 2 2 3.00 kg 9.80 m s2 4.00 m 3.00 kg+ 21 5.00 kg 0.600 m 2 10.9 rad s 8.54 8.56 (a) L I MR2 2.40 kg 0.180 m (b) 1 1 2 L I MR 2 2.40 kg 0.180 m 35.0 rad s 1.36 kg m 2 s 2 2 (c) 2 2 2 L I MR 2 2.40 kg 0.180 m 35.0 rad s 1.09 kg m 2 s 5 5 (d) 2 2 2 L I MR 2 2.40 kg 0.180 m 35.0 rad s 1.81 kg m 2 s 3 3 (a) Yes, the bullet has angular momentum about an axis through the 2 35.0 rad s 2.72 kg m 2 s hinges of the door before the collision. Consider the sketch at the right, showing the bullet the instant before it hits the door. The physical situation is identical to that of a point mass mg moving in a circular path of radius r with tangential speed t = i. For that situation the angular momentum is Li I i i mB r 2 i mB r i r and this is also the angular momentum of the bullet about the axis through the hinge at the instant just before impact. (b) No, mechanical energy is not conserved in the collision. The bullet embeds itself in the door with the two moving as a unit after impact. This is a perfectly inelastic collision in which a significant amount of mechanic cal energy is converted to other forms, notably thermal energy. (c) Apply conservation of angular momentum with Li mBri as discussed in part (a). After impact, L f I f f I door Ibullet f 1 2 Mdoor L2 mBr 2 f where L = 1.00 m = the width of the door and r = L – 10.0 cm = 0.900 m. Then, L f Li f mBri 1 Mdoor L2 mBr 2 3 0.005 kg 0.900 m 1.00 103 ms 1 18.0 kg 1.00 m 2 0.005 kg 0.900 m 2 3 yielding f 0.749 rad s . (d) The kinetic energy of the door-bullet system immediately after impact is KE f or 1 1 1 2 2 2 I f 2f 18.0 kg 1.00 m 0.005 kg 0.900 m 0.749 rad s 2 2 3 KE f 1.68 J . The kinetic energy (of the bullet) just before impact was KEi 8.57 1 1 mBi2 0.005 kg 1.00 103 m s 2 2 2 2.50 10 3 J Each mass moves in a circular path of radius r = 0.500 m/s about the center of the connecting rod. Their angular speed is r 5.00 m s 10.0 m s 0.500 m Neglecting the moment of inertia of the light connecting rod, the angular momentum of this rotating system is L I m1r 2 m2r 2 4.00 kg 3.00 kg 0.500 m 10.0 rad s 17.5 J s 2 8.63 The initial angular velocity of the puck is i t i ri 0.800 m s rad 2 .00 0.400 m s Since the tension in the string does not exert a torque about the axis of revolution, the angular momentum of the puck is conserved, or If f = Ii i. Thus, Ii If f 2 mri2 0.400 m i 2 i 0.250 m 2 .00 rad s 5.12 rad s mrf The net work done on the puck is Wnet KE f KEi 1 1 1 m 2 2 rf f ri2i2 I f 2f I i i2 mrf2 2f mri2 i2 2 2 2 2 or 0.120 kg Wnet 2 2 2 2 2 0.250 m 5.12 rad s 0.400 m 2 .00 rad s This yields Wnet 5.99 102 J . 8.65 (a) From conservation of angular momentum, If f = Ii i, so Ii I1 o i I1 I 2 If f 2 (b) KE f I I 1 1 1 I1 I f 2f I1 I 2 1 o2 1 I1 o2 KEi 2 2 I1 I 2 I1 I 2 2 I1 I 2 or KE f KEi I1 I1 I 2 Since this is less than 1.0, kinetic energy was lost. 8.66 The initial angular velocity of the system is rev 2 rad 0.40 rad s s 1 rev i 0.20 The total moment of inertia is given by I I man I cylinder mr 2 1 1 2 M R2 80 kg r 2 25 kg 2 .0 m 2 2 Initially, the man is at r 2 .0 m from the axis, and this gives I i 3.7 102 kg m 2 . At the end, when r = 1.0, the moment of inertia is I f 1.3 102 kg m 2 . (a) From conservation of angular momentum, If f = Ii i, or Ii 3.7 102 kg m 2 0.40 rad s 1.14 rad s 3.6 rad s i 1.3 102 kg m 2 If f (b) The change in kinetic energy is KE KE 1 I 2 f 2f 21 I f i2 , or 2 1 rad 1 rad 1.3 102 kg m 2 1.14 3.7 102 kg m 2 0.40 2 s 2 s or KE 5.4 102 J . The difference is the work done by the man as he walks inward. 2