(1) [(p/180)×(a),(b),(c),(d), and (e)]® 0.524, 0.995,1.571, 6.283, and 7.33. respectively. As fractions of p, these are 0.167, 0.317, 0.5, 2.0, and 2.33 respectively.
(3) The subtended angles of the Moon (q_M) and the Sun (q_S are, respectively (using text data):
q_M  =  2×1.74×10³/384×10³  =  9.1 mrad
q_S  =  2×6.96×10⁵/1.5×10⁸  =  9.3 mrad
(5) D  =  R q  =  380,000 km×1.8×10^-5  =  6.8 km
(6) 1800 rev/min ×1 min/60 s×6.2832 rad/rev  =  189 rad/s
(7) The speed of a point on the rim is v  =  wR  =  189 s^-1×0.35 m/2  =  33 m/s. The acceleration is centripetal, given by a_c  =  v²/R  =  (33²/0.175) m/s²  =  6200 m/s².
(11) Distance per revolution is 2p R. Now 15×2p R  =  4.5 m, and solving for R gives R  =  4.78 cm, so the diameter is D  =  2×4.78 cm  =  9.5 cm.
(15) The angular speed of the Earth is given by w  =  6.2832/(24×3600) rad/s  =  7.27×10^-5 s^-1.
(a) At the equator v_eq  =  w R_E  =  7.27×10^-5×6.38×10⁶ m  =  460 m/s.
(b) At latitude l other than zero, R =  R_E cos l. Thus for the arctic circle, V_a.c.  =  7.27×10^-5×6.38×10⁶×cos(66.5^°)  =  180 m/s.
(c) As in (b), get (using cos(45^°),  330 m/s
(17) 160 rpm  =  16.8 s^-1, 280 rpm  =  29.3 s^-1.
(a) The angular acceleration is a  =  Dw/Dt  =  (29.3 - 16.8)s^-1/4 s  =  3.1 s^-2.
(b) After 2 s, the angular speed in rad/s is (16.8 + 29.3)/2  =  23. At this time, the centripetal acceleration is given by a_c  =  a_radial  =  w² R  =  23²×35×10^-2  =  190 m/s². The tangential acceleration is given by a_t  =  (v_f - v_i)/Dt  =  R (w_f - w_i)/Dt  =  (10.3 - 5.9)/4 s  =  1.1 m/s²
(21) 15,000 rpm  =  1.57×10³ rad/s,  a  =  1.57×10³/220 s  =  7.14 rad/s². The total angle traveled in this 220 s is given by q  =  w²/(2a) from the kinematics for constant angular acceleration. Thus q  =  1.73×10⁵ rad×1 rev/(2p rad)  =  2.75×10⁴ revolutions.
(29) F  =  55×9.8 N  =  540 N. The torque from one pedal is t  =  F×d  =  540×0.17 m  =  92 N m. Since she can put her weight on only one pedal at a time (not a racing bike where one can also pull up with the `caged' foot), this is the answer.
(31) The 35 N force and the 20 N force yield a torque in the same direction, call it +. The moment arm of the 35 N force is 0.1 m and that of both the others is 0.2 m. Thus the torque from the three indicated forces is given by t  =  [35×0.1 + (30 - 20)×0.2] N m  =  1.5 N m. The friction (not shown in the figure) provides a torque of -0.4 N m, the negative sign being necessary since friction opposes the rotation. Thus the net torque is (1.5 - 0.4)N m  =  1.1 N m.
Note: The angles indicated in the figure are irrelevant-a ``confusion'' factor to see if you understand the concept of torque.
(34) The moment of inertia of a sphere, for an axis through the center, is given by 2 M R²/5, as seen from Figure 8-20, pg 223 of the text. Thus I  =  2×12.2 kg×0.623² m²/5  =  1.89 kg m².
(35) The mass of the hub can be ignored because its radius is small, and the moment of inertia is proportional to radius squared.
I  =  M R²  =  1.25 kg×0.3335² m²  =  0.139 kg m².
(51) Energy at any time is given by I w²/2. 8000 rpm  =  838 s^-1. Thus get 3.15×10^-2×838²/2 J  =  1.11×10⁴ J.
(57) Angular momentum is given by M R² w  =  0.210 kg×1.10² m²×10.4 s^-1  =  2.64 kg m²/s
(59) Conservation of angular momentum gives I_i w_i  =  I_f w_f or w_f  =  I_i w_i/I_f. Since the w's in rad/s are proportional to rev's in s, the final number of rev/s is just (1.33 rev/s)/3.5  =  0.38 rev/s

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On 25 Oct 1999, 11:29.