13. The final angular speed is w = (1 rpm)(2p rad/rev)/(60 s/min) = 0.105 rad/s. (a) We find the angular acceleration from a = Dw/Dt = (0.105 rad/s - 0)/(10.0 min)(60 s/min) = 1.75 x10^{-4}rad/s^{2}. (b) We find the angular speed after 5.0 min: w = w_{0}+ at = 0 + (1.75 x10^{-4}rad/s^{2})(5.0 min)(60 s/min) = 5.25 x10^{-2}rad/s. At this time the radial acceleration of a point on the skin is a_{R}= w^{2}r = (5.25 x10^{-2}rad/s)^{2}(4.25 m) = 1.17 x10^{-2}m/s^{2}. The tangential acceleration is a_{tan}= ar = (1.75 x10^{-4}rad/s^{2})(4.25 m) = 7.44 x10^{-4}m/s^{2}.

23. We assume clockwise motion, so the frictional torque is counterclockwise. If we take the clockwise direction as positive, we have t_{net}= rF_{1}- RF_{2}+ RF_{3}- t_{fr}= (0.10 m)(35 N) - (0.20 m)(30 N) + (0.20 m)(20 N) - 0.30 m · N = 1.2 m · N (clockwise).

30. (a) For the moment of inertia about the y-axis, we have I_{a}= Sm_{i}R_{i}^{2}= md_{1}^{2}+ Md_{1}^{2}+ m(d_{2}- d_{1})^{2}+ M(d_{2}- d_{1})^{2}= (1.8 kg)(0.50 m)^{2}+ (3.1 kg)(0.50 m)^{2}+ (1.8 kg)(1.00 m)^{2}+ (3.1 kg)(1.00 m)^{2}= 6.1 kg · m^{2}. (b) For the moment of inertia about the x-axis, all the masses are the same distance from the axis, so we have I_{b}= Sm_{i}R_{i}2 = (2m + 2M)(1/2h)^{2}= [2(1.8 kg) + 2(3.1 kg)](0.25 m)^{2}= 0.61 kg · m^{2}. It will be harder to accelerate the array around the y-axis, because the moment of inertia is greater.

45. (a) We use the parallel-axis theorem: I = 2(I_{CM}+ Mh^{2}) = 2[(2MR_{0}^{2}/5) + M(3R_{0}/2)^{2}] = 5.30MR_{0}^{2}. (b) If we treat the spheres as point masses, we get I' = 2[M(3R_{0}/2)^{2}] = 4.50MR_{0}^{2}. The error is error = (I' - I)/I = (4.50 - 5.30)/(5.30) = - 0.15 = - 15%.

65. For the system of the two blocks and pulley, no work will be done by nonconservative forces. The rope ensures that each block has the same speed v and the angular speed of the pulley is w = v/R_{0}. We choose the reference level for gravitational potential energy at the floor. The rotational inertia of the pulley is I = 1/2MR_{0}^{2}. For the work-energy principle we have W_{net}= DK + DU; 0 = [(1/2m_{1}v^{2}+ 1/2m_{2}v^{2}+ 1/2Iw^{2}) - 0] + m_{1}g(h - 0) + m_{2}g(0 - h); 1/2m_{1}v^{2}+ 1/2m_{2}v^{2}+ 1/2(1/2MR_{0}^{2})(v/R_{0})^{2}= (m_{2}- m_{1})gh; 1/2[m_{1}+ m_{2}+ 1/2M]v^{2}= (m_{2}- m_{1})gh; 1/2[35.0 kg + 38.0 kg + 1/2(4.8 kg)]v^{2}= (38.0 kg - 35.0 kg)(9.80 m/s^{2})(2.5 m), which gives v = 1.4 m/s.

75. If the ball rolls without slipping, the speed of the center of mass is v = r_{0}w. Because energy is conserved, for the motion from A to B we have 0 = DK + DU; 0 = [(1/2mv^{2}+ 1/2Iw^{2}) - 0] + (mgr_{0}- mgR_{0}), or 1/2mv^{2}+ 1/2(2/5 mr_{0}^{2})(v/r_{0})^{2}= mg(R_{0}- r_{0}), which gives v = [10g(R_{0}- r_{0})/7]^{1/2}.

18. We find the initial and final angular velocities of the wheel from the rolling condition: w_{0}= v_{0}/r = [(90.0 km/h)/(3.6 ks/h)]/(0.45 m) = 55.6 rad/s; w = v/r = [(60.0 km/h)/(3.6 ks/h)]/(0.45 m) = 37.0 rad/s. (a) We find the angular acceleration from w^{2}= w_{0}^{2}+ 2aq; (37.0 rad/s)^{2}= (55.6 rad/s)^{2}+ 2a (85 rev)(2p rad/rev), which gives a= - 1.6 rad/s^{2}. (b) We find the additional time from w_{final}= w + at; 0 = 37.0 rad/s + (- 1.6 rad/s^{2})t, which gives t = 23 s.

24. (a) We let L be the length of the beam and take clockwise as the positive direction. For the net torque about point C, we have t_{C}= 1/2L(F_{1}sin q_{1}) - 1/2L(F_{3}sin q_{3}) = 1/2(2.0 m)(50 N) sin 30° - 1/2(2.0 m)(50 N) sin 60° = - 18 m · N (CCW=counter-clock-wise). (b) For the net torque about point P, we have t_{P}= L(F_{1}sin q_{1}) - 1/2L(F_{2}sin q_{2}) = (2.0 m)(50 N) sin 30° - 1/2(2.0 m)(60 N) sin 45° = 7.6 m · N (CW=clockwise).

(See FIGURE 10-56)

31. For the moment of inertia of the rotor blades we have I = 3(1/3m_{blade}L^{2}) = m_{blade}L^{2}= (160 kg)(3.75 m)^{2}= 2.25 x10^{3}kg · m^{2}. We find the required torque from t = Ia = I(w - w_{0})/t = (2.25 x10^{-3}kg · m^{2})[(5.0 rev/s)(2p rad/rev) - 0]/(8.0 s) = 8.8 x10^{3}m · N.

46. (a) If we treat the sphere as a point mass, we get I_{a}= MR_{0}^{2}. (b) We use the parallel-axis theorem: I_{b}= I_{CM}+ Mh^{2}= (2MR_{1}^{2}/5) + MR_{0}^{2}. (c) The error is error = (I_{a}- I_{b})/I_{b}= [MR_{0}^{2}- (2MR_{1}^{2}/5) - MR_{0}^{2}]/[(2MR_{1}^{2}/5) + MR_{0}^{2}] = [- (2R_{1}^{2}/5)]/[(2R_{1}^{2}/5) + R_{0}^{2}] = - (0.40)(0.10 m)^{2}/[(0.40)(0.10 m)^{2}+ (1.0 m)^{2}] = - 0.0040 = - 0.40%.

92. (a) The angular acceleration of the ball-arm system is a = a_{tan}/d_{1}= (7.0 m/s^{2})/(0.30 m) = 23.3 rad/s^{2}. Because we ignore the mass of the arm, for the moment of inertia we have I = m_{ball}d_{1}^{2}= (1.00 kg)(0.30 m)^{2}= 0.090 kg · m^{2}. Thus we find the required torque from t = Ia = (0.090 kg · m^{2})(23.3 rad/s^{2}) = 2.1 m · N. (b) Because the force from the triceps muscle is perpendicular to the line from the axis, we find the force from F = t/d_{2}= (2.1 m · N)/(0.025 m) = 84 N.

105. (a) The speed of the block is the tangential speed of the cylinder, so v = Rw. For the system of block and cylinder, if there is no friction, energy is conserved. After the block has moved a distance D, we have 0 = DK + DU; 0 = [(1/2mv^{2}+ 1/2I_{1}w2) - 0] + (0 - mgD sin q); 1/2mv^{2}+ 1/2(1/2MR^{2})(v/R)^{2}= mgD sin q; 1/2mv^{2}+ 1/4Mv^{2}= mgD sin q; [1/2(3.0 kg) + 1/4(30 kg)]v^{2}= (3.0 kg)(9.80 m/s^{2})(1.80 m) sin 30°, which gives v = 1.7 m/s. (b) From the force diagram for the block we see that F_{fr2}= mF_{N2}= mmg cos q. For the cylinder, we assume a single vertical normal force. Because there is no linear acceleration of the center of mass, we have F_{N1}= Mg + F_{T}sin q. Because the block accelerates down the plane, F_{T}< mg sin D, and m = 0.1M. Thus we ignore the contribution of F_{T}sin q (and we are justified in ignoring the horizontal component of F_{N1}) to get F_{N1}~ Mg, and F_{fr1}= mF_{N1}= mMg. When the block moves a distance D, the surface of the rotating cylinder will move a distance D through the depression. For the work-energy theorem we have W_{fr}= DK + DU; - F_{fr1}D - F_{fr2}D = [(1/2mv^{2}+ 1/2I_{1}w^{2}) - 0] + (0 - mgD sin q); - m(M + m cos q)gD = 1/2mv^{2}+ 1/4Mv^{2}- mgD sin q; - (0.035)[30 kg + (3.0 kg) cos 30°](9.80 m/s^{2})(1.80 m) = [1/2(3.0 kg) + 1/4(30 kg)]v^{2}- (3.0 kg)(9.80 m/s^{2})(1.80 m) sin 30°, which gives v = 0.84 m/s.

(See FIGURE 10-44)

79. (a) From Example 10-24, we see that while the ball is slipping, the acceleration of the center of mass is - m_{k}g and is constant. The ball slips for a time T = 2v_{0}/7m_{k}g, so we find the distance from x_{slip}= v_{0}t + 1/2at^{2}= v_{0}(2v_{0}/7m_{k}g) + 1/2(- m_{k}g)(2v_{0}/7m_{k}g)^{2}= 12v_{0}2/49m_{k}g. (b) Once the ball starts rolling at time T, the linear speed is constant v = v_{0}+ at = v_{0}+ (- m_{k}g)(2v_{0}/7m_{k}g) = 5v_{0}/7. We can find the angular speed from w = w_{0}+ at, or from w = v/R = (5v_{0}/7)/R = 5v_{0}/7R