7. (a) For the vectors A = 7.0i - 3.5j and B = - 8.5i + 7.0j + 2.0k, we have

(b) The magnitudes of the vectors are A = [(7.0)^{2}+ (- 3.5)^{2}]^{1/2}= 7.83; B = [(- 8.5)^{2}+ (7.0)^{2}+ (2.0)^{2}]^{1/2}= 11.2;|A x B|= [(- 7.0)^{2}+ (- 14.0)^{2}+ (19.3)^{2}]^{1/2}= 24.8. We find the angle between A and B from|A x B|= |A| |B| sinq ; 24.8 = (7.83)(11.2) sinq , which gives sinq = 0.283, which gives q = 16^{o}, 164^{o}. From the diagram, we see that q = 164^{o}.

18. For the angular momentum we haveL=r xmv= m(r x v) = (0.060 kg)[(7.0 m)i+ (- 6.0 m)j] ? [(2.0 m/s)i- (8.0 m/s)k] = (0.060 kg){[(- 6.0 m)(- 8.0 m/s) - 0]i+ [0 - (7.0 m)(- 8.0 m/s)]j+ [0 - (- 6.0 m)(2.0 m/s)]k} = (2.9i+ 3.4j+ 0.72k) kg * m^{2}/s.

25. (a) With the positive direction CCW, for the angular momentum about the axis of the pulley we have L = R_{0}M_{1}v + R_{0}M_{2}v + Iw = R_{0}M_{1}v + R_{0}M_{2}v + I(v/R_{0}) = [R_{0}M_{1}+ R_{0}M_{2}+ (I/R_{0})]v. (b) Because M_{1}g is balanced by the normal force on the horizontal surface, the net torque is from M_{2}g only: t = dL/dt; M_{2}gR_{0}= [R_{0}M_{1}+ R_{0}M_{2}+ (I/R_{0})] dv/dt, which gives a = M_{2}g/[M_{1}+ M_{2}+ (I/R_{0}^{2})].

35. In the collision, during which we ignore any motion of the rod, angular momentum about the pivot point will be conserved: L_{i}= L_{f}; mv(r/2) + 0 = I_{total}w = [1/3Mr^{2}+ m(1/2r)^{2}]w, which gives w = 6mv/(4M + 3m)r. For the rotation about the pivot after the collision, energy will be conserved. I_{f}the center of the rod reaches a height h, the bottom of the rod will swing to a height H = 2h, so we have K_{i}+ U_{i}= K_{f}+ U_{f}; 1/2I_{total}w^{2}+ 0 = 0 + (m + M)gh; 1/2[1/3Mr^{2}+ m(1/2r)^{2}][6mv/(4M + 3m)r]^{2}= (m + M)gH/2, which gives H = 3m^{2}v^{2}/g(3m + 4M)(m + M).

57. (a) The angular momentum delivered to the waterwheel in a time Dt is that lost by the water: DL/Dt = Dm R(v_{i}- v_{f})/Dt = (Dm/Dt)R(v_{i}- v_{f}) = (150 kg/s)(3.0 m)(7.0 m/s - 3.0 m/s) = 1.8 x 10^{3 }kg * m^{2}/s^{2}. (b) The torque applied to the waterwheel increases the angular momentum: t = DL/Dt = 1.8 x 10^{3 }m * N. (c) The power delivered is P = tw = (1.8 x 10^{3 }m * N)(2p/5.5 s) = 2.1 x 10^{3 }W.

59. Because there is no friction, the center of mass must fall straight down. The vertical velocity of the right end of the stick must always be zero. I_{f}w is the angular velocity of the stick just before it hits the table, the velocity of the right end with respect to the center of mass will be w(r/2) up. Thus we have w(r/2) - v_{CM}= 0, or w = 2v_{CM}/r. The kinetic energy will be the translational energy of the center of mass and the rotational energy about the center of mass. With the reference level for potential energy at the ground, we use energy conservation to find the speed of the center of mass just before the stick hits the ground: K_{i}+ U_{i}= K_{f}+ U_{f}; 0 + Mg1/2r = 1/2Mv_{CM}^{2}+ 1/2(Mr^{2}/12)w^{2}+ 0; Mg1/2r = 1/2Mv_{CM}^{2}+ 1/2(Mr^{2}/12)(2v_{CM}/r)^{2}= 1/2(4Mv_{CM}^{2}/3), which gives v_{CM}= (3gr/4)^{1/2}.

10. For the torque we have t =r x F= [(4.0i+ 8.0j+ 6.0k) m] x [(16.0j- 4.0k) N] = {[(8.0)(- 4.0) - (6.0)(16.0)]i+ [0 - (4.0)(- 4.0)]j+ [(4.0)(16.0) - 0]k} m * N = (- 128i+ 16j+ 64k) m * N.

19. For the angular momentum we have

L=

26. (a) With the positive direction CCW, we have the same formula for the angular momentum: L = [R_{0}M_{1}+ R_{0}M_{2}+ (I/R_{0})]v. (b) Because M_{1}g is balanced by the normal force on the horizontal surface, the friction force is F_{fr}= m_{k}M_{1}g. If we consider only the block M_{1}, which does not rotate, the net torque must be zero, which means that the normal force and the weight are separated by a distance d. The torque of this couple must be balanced by the torque from the tension and friction. If we take the point A as the axis, we have m_{k}M_{1}gR_{0}- M_{1}gd = 0, or - M_{1}gd = - m_{k}M_{1}gR_{0}. For the entire system the tension is an internal force. Because the line of the friction force passes through the center of the pulley, the net torque is produced by M_{2}g and the couple from F_{N}and M_{1}g: t = dL/dt; M_{2}gR_{0}- m_{k}M_{1}gR_{0}= [R_{0}M_{1}+ R_{0}M_{2}+ (I/R_{0})] dv/dt, which gives a = (M_{2}- m_{k}M_{1})g/[M_{1}+ M_{2}+ (I/R_{0}^{2})].

36. For the system of stick and bullet during the collision, angular momentum about the center of mass is conserved: L_{i}= L_{f}; mv_{i}(1/4r) + 0 = mv_{i}(1/4r) + I_{rod}w ; (3.0 g)(250 m/s)1/4(1.0 m) = (3.0 g)(160 m/s)1/4(1.0 m) + [(300 g)(1.0 m)^{2}/12)]t, which gives w = 2.7 rad/s.

60. (a) Because the hoop is rolling down the string, the acceleration of the center of mass is related to the angular acceleration: a = aR. For the linear motion of the center of mass we have SF = Ma; Mg - F_{T}= Ma = MRa. For the angular motion about the center of mass we have St = dL/dt = Ia; F_{T}R = MR^{2}a = MRa. When we combine these equations, we get a = a/R = 1/2g/R. Thus we have dL/dt = (MR^{2})(1/2g/R) = 1/2MRg, so L = 1/2MRgt. (b) The tension in the string is F_{T}= Ma = M(1/2g) = 1/2Mg.

63. (a) If the thrown-off mass carries off no angular momentum, from conservation of angular momentum for the star we have I_{i}w_{i}= I_{f}w_{f}; 2/5Mr_{s}^{2}w_{1}= 2/5(1/4M)r_{n}^{2}w_{2}, which gives w_{2}= 4(r_{s}/r_{n})^{2}w_{1}= 4(7.0 x 10^{5 }km/10 km)^{2}(1 rev/10 days)/(86,400 s/day) = 2.3 x 10^{4 }rev/s. (b) If the thrown-off mass carries off & of the initial angular momentum, from conservation of angular momentum for the star we have 1/4I_{i}w_{i}= I_{f}w_{f}; 1/4(2/5Mr_{s}^{2})w_{1}= 2/5(1/4M)r_{n}^{2}w_{2}, which gives w_{2}= (r_{s}/r_{n})^{2}w_{1}= (7.0 x 10^{5 }km/10 km)^{2}(1 rev/10 days)/(86,400 s/day) = 5.7 x 10^{3 }rev/s.

39. (a) We find the speed of the center of mass from the conservation of linear momentum: Mv + 0 = (M + m)v_{CM}; (200 kg)(18 m/s) = (200 kg + 50 kg)v_{CM}, which gives v_{CM}= 14 m/s. (b) During the collision, angular momentum about the center of mass will be conserved. We find the location of the center of mass relative to the center of the beam: d = m(1/2r)/(M + m) = (50 kg)^{1/2}(2.0 m)/(200 kg + 50 kg) = 0.20 m. When we use the parallel-axis theorem for the moment of inertia of the beam, angular momentum conservation gives us L_{i}= L_{f}; Mvd + 0 = I_{total}w = [(Mr^{2}/12) + Md^{2}+ m(1/2r - d)^{2}]w; (200 kg)(18 m/s)(0.20 m) = ((200 kg){[(2.0 m)^{2}/12] + (0.20 m)^{2}} + (50 kg)(1.0 m - 0.20 m)^{2})w, which gives w = 6.8 rad/s.