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9. A 12-N force is applied at the point x = 3 m, y= 1 rn. Find the torque about the origin if the force points in (a) the x ,direction, (b) the y iidirection,and (c) the z ditection. Problem
27. 'Fwo identical 1800-kg cars are traveling in opposite directions at 90 km/h. Each car's center of mass is 3.0 m from the center of the highway (Fig. 13.31). What are the magnitude and direction of the angular momentum of the system consisting ~f the two cars, about a point on the center-line of the highway? Solution
Equa.tion 13-2 and the definition of the cross pr~duct give:(a) r = r x F = (311:j) x 121N.~= ::-12kN'm, (b)J3i+j) x 12j N.m=36kN:F' .(c) (31+J)X . 12k 'iN.m (12i - 36j) N.m, (magnitude 37.9 N.m ill = the ~y plane, 71.6 clockwisefrom the x axis). Solution
The position of a particle with respect to some point ..l'.~"t,~ can always be expresse\." ':~":::':f components 0 Prdbleni
21.;A gymnast of rotational inertia 62 kg.m2 is tumbling head over heels. If her angular momentum is 470 kg.m2/s, what is her angular speed? " Solution If ~~ regard the gymnast as a rigid body rotating . about a fixed (instantaneous) axis, Equation 13-4 gIVes w = L/ 1= (470 kg.m2/s)/(62 kg.m2) = 7.58 8.-1 = 1.21 rev/s as her angular speed about that axlS. perpendicular and parallel to its direction of motion (Le., momentum pl. Thus, Equation 13-3 can be written as L=r x p= (r.L+ rll) x p = r.L x p, since rll x p=O. For straight line motion, r.L is a constant (this is an alternate solution to Problem 26). For the two cars (regarded as particles located at their respective centers of mass) PI = -P2 and r1.L = -r2.L (for any point on the center line), so the total angular momentum of their centers of mass is L=LI +L2 =ru x PI + (-ru) x (-PI) = 2r1.L x Pl. This has magnitude 2r1.LmVl = 2(3 m)x (1800 kg)(90 m/3.6 s) = 2.70 X 105 J.s and direction out of the plane of Fig. 13-30. Problem
32. Two ice skaters, both of mass 60 kg, approach on parallel paths 1.4 m apart. Both are moving at 3.2 m/s with their arms outstretched (Fig. 13-32). They join hands as they pass, still maintaining their l.4-m separation, and begin rotating about one.another. What is their angular speed? 33. In Fig. 13-33 the lower disk, of mass 440 g and radius 3.5 em, is rotating at 180 rpm on a frictionless shaft of negligible radius. The upper disk, of mass 270 g and radius 2.3 em, is initially not rotating. It drops freely down the shaft onto the lower disk, arid frictional forces act to bring the two disks to a common rotational speed. (a) What is that speed? (b) What fraction of the initial kinetic energy is ~ostto friction? Solution
(a) There are no external torques acting about the frictionless shaft, so the total angular momentum of the two-disk system in this direction is conserved, or L1 + L2 = const.(The frictional torques between the disks are internal to the system.) Initially, L1 =11Wli and L2 = 0, and finally Ll + L2 = (II + 12)wf, therefore wf = WliIl/(I1 + 12) = (180 rpm) x [440x 3.52/(440 x 3.52 + 270x2.32)) = 142 rpm. (We canceled common factors of ~ g.cm2 from the rotational inertias.) (b) The fraction of initial kinetic energy lost is
- Kf = Ki 1_ Kf Ki = 1_ ~ h +12 = 1_ wf
Wli = 1- O.7~1 = 20.9%. (We used K = !Iw2 = L2/2I for the kinetic energy of a rigid body rotating about a fixed axis, and the result of part (a).) FIGURE 13-32Problem 32. Solution
Since P em = ml VI + m2v2 = 0, both skaters rotate about a vertical axis through their center of mass 'i when joined. If the skater.sexert no horizontal forces on the ice surface, there is no vertical external torque and their angular momentum about this axis is conserved. Before they join hands, L = 2mvb (see . Problems 26 and 27), where b is the perpendicular distance from the eM to either of the parallel paths. After they join, L = Iemwcm = 2mb2wem. Therefore 2 Wcm = 2mvb/2mb = v/b,i= (3.2 m/s)/(1.4 m/2) = i 4.57 S-I. '
:1 (a) (b) FIGURE 13-33 Problems 33 and 34. ...
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