ch10
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ch10

Course Number: PHYSIC 2, Spring 2008

College/University: Lehigh

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1. (a) The second hand of the smoothly running watch turns through 2 radians during 60 s . Thus, = 2 = 0.105 rad/s. 60 (b) The minute hand of the smoothly running watch turns through 2 radians during 3600 s . Thus, = 2 = 1.75 10 -3 rad / s. 3600 (c) The hour hand of the smoothly running 12-hour watch turns through 2 radians during 43200 s. Thus, = 2 = 145 10 -4 rad / s. . 43200 2. The problem asks us to...

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(a) 1. The second hand of the smoothly running watch turns through 2 radians during 60 s . Thus, = 2 = 0.105 rad/s. 60 (b) The minute hand of the smoothly running watch turns through 2 radians during 3600 s . Thus, = 2 = 1.75 10 -3 rad / s. 3600 (c) The hour hand of the smoothly running 12-hour watch turns through 2 radians during 43200 s. Thus, = 2 = 145 10 -4 rad / s. . 43200 2. The problem asks us to assume vcom and are constant. For consistency of units, we write vcom = 85 mi h b g FGH 5280 ft mi IJK = 7480 ft min . 60 min h Thus, with x = 60 ft , the time of flight is t = x vcom = 60 7480 = 0.00802 min . During that time, the angular displacement of a point on the ball's surface is = t = 1800 rev min 0.00802 min 14 rev . b gb g 3. We have = 10 rad/s. Since = 0, Eq. 10-13 gives = t = (10 rad/s)(n t), for n = 1, 2, 3, 4, 5, .... For t = 0.20 s, we always get an integer multiple of 2 (and 2 radians corresponds to 1 revolution). (a) At f1 = 2 rad the dot appears at the "12:00" (straight up) position. (b) At f2, = 4 rad and the dot appears at the "12:00" position. t = 0.050 s, and we explicitly include the 1/2 conversion (to revolutions) in this calculation: = t = (10 rad/s)n(0.050 s) 1 2 = , , , 1, ... (revs) (c) At f1(n=1), = 1/4 rev and the dot appears at the "3:00" position. (d) At f2(n=2), = 1/2 rev and the dot appears at the "6:00" position. (e) At f3(n=3), = 3/4 rev and the dot appears at the "9:00" position. (f) At f4(n=4), = 1 rev and the dot appears at the "12:00" position. Now t = 0.040 s, and we have = t = (10 rad/s)n(0.040 s) 1 2 = 0.2 , 0.4 , 0.6 , 0.8, 1, ... (revs) Note that 20% of 12 hours is 2.4 h = 2 h and 24 min. (g) At f1(n=1), = 0.2 rev and the dot appears at the "2:24" position. (h) At f2(n=2), = 0.4 rev and the dot appears at the "4:48" position. (i) At f3(n=3), = 0.6 rev and the dot appears at the "7:12" position. (j) At f4(n=4), = 0.8 rev and the dot appears at the "9:36" position. (k) At f5(n=5), = 1.0 rev and the dot appears at the "12:00" position. 4. If we make the units explicit, the function is = 4.0 rad / s t - 3.0 rad / s2 t 2 + 10 rad / s3 t 3 . but generally we will proceed as shown in the problem--letting these units be understood. Also, in our manipulations we will generally not display the coefficients with their proper number of significant figures. (a) Eq. 10-6 leads to b g c h c h = d 4t - 3t 2 + t 3 = 4 - 6t + 3t 2 . dt c h Evaluating this at t = 2 s yields 2 = 4.0 rad/s. (b) Evaluating the expression in part (a) at t = 4 s gives 4 = 28 rad/s. (c) Consequently, Eq. 10-7 gives avg = (d) And Eq. 10-8 gives 4 -2 4-2 = 12 rad / s2 . = d d = 4 - 6t + 3t 2 = -6 + 6t . dt dt c h Evaluating this at t = 2 s produces 2 = 6.0 rad/s2. (e) Evaluating the expression in part (d) at t = 4 s yields 4 = 18 rad/s2. We note that our answer for avg does turn out to be the arithmetic average of 2 and 4 but point out that this will not always be the case. 5. Applying Eq. 2-15 to the vertical axis (with +y downward) we obtain the free-fall time: y = v0 y t + 1 2 gt 2 t= 2(10) = 14 s. . 9.8 Thus, by Eq. 10-5, the magnitude of the average angular velocity is avg = ( 2.5) ( 2 ) = 11 rad / s. 14 . 6. If we make the units explicit, the function is = 2.0 rad + ( 4.0 rad/s 2 ) t 2 + ( 2.0 rad/s3 ) t 3 but in some places we will proceed as indicated in the problem--by letting these units be understood. (a) We evaluate the function at t = 0 to obtain 0 = 2.0 rad. (b) The angular velocity as a function of time is given by Eq. 10-6: = d = ( 8.0 rad/s 2 ) t + ( 6.0 rad/s3 ) t 2 dt which we evaluate at t = 0 to obtain 0 = 0. (c) For t = 4.0 s, the function found in the previous part is 4 = (8.0)(4.0) + (6.0)(4.0)2 = 128 rad/s. If we round this to two figures, we obtain 4 1.3 102 rad/s. (d) The angular acceleration as a function of time is given by Eq. 10-8: = d = 8.0 rad/s 2 + (12 rad/s3 ) t dt which yields 2 = 8.0 + (12)(2.0) = 32 rad/s2 at t = 2.0 s. (e) The angular acceleration, given by the function obtained in the previous part, depends on time; it is not constant. 7. (a) To avoid touching the spokes, the arrow must go through the wheel in not more than t = 1 / 8 rev = 0.050 s. 2.5 rev / s 20 cm = 400 cm / s = 4.0 m / s. 0.050 s The minimum speed of the arrow is then vmin = (b) No--there is no dependence on radial position in the above computation. 8. (a) With = 0 and = 4.2 rad/s2, Eq. 10-12 yields t = o/ = 3.00 s. (b) Eq. 10-4 gives - o = - o2 / 2 = 18.9 rad. 9. (a) We assume the sense of rotation is positive. Applying Eq. 10-12, we obtain = 0 + t (b) And Eq. 10-15 gives = 3000 - 1200 = 9.0 103 rev/min 2 . 12 / 60 = ( 0 + ) t = (1200 + 3000) 1 2 1 2 FG 12 IJ = 4.2 10 H 60K 2 rev. 10. We assume the sense of initial rotation is positive. Then, with 0 = +120 rad/s and = 0 (since it stops at time t), our angular acceleration (``deceleration'') will be negativevalued: = 4.0 rad/s2. (a) We apply Eq. 10-12 to obtain t. = 0 + t (b) And Eq. 10-15 gives t= 0 - 120 = 30 s. -4.0 1 1 = ( 0 + ) t = (120 + 0) (30) = 1.8 103 rad. 2 2 Alternatively, Eq. 10-14 could be used if it is desired to only use the given information (as opposed to using the result from part (a)) in obtaining . If using the result of part (a) is acceptable, then any angular equation in Table 10-1 (except Eq. 10-12) can be used to find . 11. We assume the sense of rotation is positive, which (since it starts from rest) means all quantities (angular displacements, accelerations, etc.) are positive-valued. (a) The angular acceleration satisfies Eq. 10-13: 1 25 rad = (5.0 s) 2 2 = 2.0 rad/s 2 . (b) The average angular velocity is given by Eq. 10-5: avg = 25 rad = = 5.0 rad / s. t 5.0 s (c) Using Eq. 10-12, the instantaneous angular velocity at t = 5.0 s is = 2.0 rad / s2 (5.0 s) = 10 rad / s . (d) According to Eq. 10-13, the angular displacement at t = 10 s is 1 1 = 0 + t 2 = 0 + (2.0) (10) 2 = 100 rad. 2 2 c h Thus, the displacement between t = 5 s and t = 10 s is = 100 25 = 75 rad. 12. (a) Eq. 10-13 gives - o = o t + 1 2 t 2 = 0 + 2 (1.5 rad/s) t12 1 where - o = (2 rev)(2 rad/rev). Therefore, t1 = 4.09 s. (b) We can find the time to go through a full 4 rev (using the same equation to solve for a new time t2) and then subtract the result of part (a) for t1 in order to find this answer. (4 rev)(2 rad/rev) = 0 + 2 (1.5 rad/s) t22 Thus, the answer is 5.789 4.093 1.70 s. 1 t2 = 5.789 s. 13. We take t = 0 at the start of the interval and take the sense of rotation as positive. 1 Then at the end of the t = 4.0 s interval, the angular displacement is = 0 t + 2 t 2 . We solve for the angular velocity at the start of the interval: 0 = - 1 t2 2 t = 120 rad - 1 3.0 rad/s 2 2 4.0 s ( ) ( 4.0 s ) 2 = 24 rad/s. We now use = 0 + t (Eq. 10-12) to find the time when the wheel is at rest: t=- 0 24 rad / s =- = -8.0 s. 3.0 rad / s2 That is, the wheel started from rest 8.0 s before the start of the described 4.0 s interval. 14. (a) The upper limit for centripetal acceleration (same as the radial acceleration see Eq. 10-23) places an upper limit of the rate of spin (the angular velocity ) by considering a point at the rim (r = 0.25 m). Thus, max = a/r = 40 rad/s. Now we apply Eq. 10-15 to first half of the motion (where o = 0): - o = 1 (o 2 + )t 400 rad = 1 (0 2 + 40 rad/s)t which leads to t = 20 s. The second half of the motion takes the same amount of time (the process is essentially the reverse of the first); the total time is therefore 40 s. (b) Considering the first half of the motion again, Eq. 10-11 leads to = o + t = 40 rad/s = 2.0 rad/s2 . 20 s 15. The wheel has angular velocity 0 = +1.5 rad/s = +0.239 rev/s2 at t = 0, and has constant value of angular acceleration < 0, which indicates our choice for positive sense of rotation. At t1 its angular displacement (relative to its orientation at t = 0) is 1 = +20 rev, and at t2 its angular displacement is 2 = +40 rev and its angular velocity is 2 = 0 . (a) We obtain t2 using Eq. 10-15: 2 = 1 0 + 2 t2 2 b g t2 = 2(40) 0.239 which yields t2 = 335 s which we round off to t2 3.4 10 2 s . (b) Any equation in Table 10-1 involving can be used to find the angular acceleration; we select Eq. 10-16. 1 2 2 = 2t2 - t2 2 =- 2(40) 3352 which yields = 7.12 104 rev/s2 which we convert to = 4.5 103 rad/s2. 1 (c) Using 1 = 0t1 + 2 t12 (Eq. 10-13) and the quadratic formula, we have t1 = - 0 2 + 2 1 0 = -0.239 0.239 2 + 2 20 -7.12 10 -4 -7.12 10 -4 b gc h which yields two positive roots: 98 s and 572 s. Since the question makes sense only if t1 < t2 we conclude the correct result is t1 = 98 s. 16. The wheel starts turning from rest (0 = 0) at t = 0, and accelerates uniformly at > 0, which makes our choice for positive sense of rotation. At t1 its angular velocity is 1 = +10 rev/s, and at t2 its angular velocity is 2 = +15 rev/s. Between t1 and t2 it turns through = 60 rev, where t2 t1 = t. (a) We find using Eq. 10-14: 2 2 = 12 + 2 = 152 - 102 2(60) which yields = 1.04 rev/s2 which we round off to 1.0 rev/s2. (b) We find t using Eq. 10-15: = 1 1 + 2 t 2 b g t = 2(60) = 4.8 s. 10 + 15 10 = 9.6 s. 1.04 (c) We obtain t1 using Eq. 10-12: 1 = 0 + t1 t1 = (d) Any equation in Table 10-1 involving can be used to find 1 (the angular displacement during 0 t t1); we select Eq. 10-14. 12 = 02 + 21 1 = 102 = 48 rev. 2(1.04) 17. The problem has (implicitly) specified the positive sense of rotation. The angular acceleration of magnitude 0.25 rad/s2 in the negative direction is assumed to be constant over a large time interval, including negative values (for t). (a) We specify max with the condition = 0 (this is when the wheel reverses from positive rotation to rotation in the negative direction). We obtain max using Eq. 10-14: max = - 4.7 2 2 o =- = 44 rad. 2 2( -0.25) (b) We find values for t1 when the angular displacement (relative to its orientation at t = 0) is 1 = 22 rad (or 22.09 rad if we wish to keep track of accurate values in all intermediate steps and only round off on the final answers). Using Eq. 10-13 and the quadratic formula, we have 1 1 = ot1 + t12 2 t1 = 2 - o o + 21 which yields the two roots 5.5 s and 32 s. Thus, the first time the reference line will be at 1 = 22 rad is t = 5.5 s. (c) The second time the reference line will be at 1 = 22 rad is t = 32 s. (d) We find values for t2 when the angular displacement (relative to its orientation at t = 0) is 2 = 10.5 rad. Using Eq. 10-13 and the quadratic formula, we have 2 2 = o t 2 + t2 1 2 t2 = 2 - o o + 2 2 which yields the two roots 2.1 s and 40 s. Thus, at t = 2.1 s the reference line will be at 2 = 10.5 rad. (e) At t = 40 s the reference line will be at 2 = 10.5 rad. (f) With radians and seconds understood, the graph of versus t is shown below (with the points found in the previous parts indicated as small circles). 18. The wheel starts turning from rest (0 = 0) at t = 0, and accelerates uniformly at = 2.00 rad/s 2 .Between t1 and t2 the wheel turns through = 90.0 rad, where t2 t1 = t = 3.00 s. (a) We use Eq. 10-13 (with a slight change in notation) to describe the motion for t1 t t2: 1 = 1 t + ( t ) 2 2 1 = t - t 2 which we plug into Eq. 10-12, set up to describe the motion during 0 t t1: 1 = 0 + t1 yielding t1 = 13.5 s. t - = t1 t 2 90.0 (2.00) (3.00) - = (2.00) t1 3.00 2 (b) Plugging into our expression for 1 (in previous part) we obtain 1 = t 90.0 (2.00)(3.00) - = - = 27.0 rad / s. t 2 3.00 2 19. We assume the given rate of 1.2 103 m/y is the linear speed of the top; it is also possible to interpret it as just the horizontal component of the linear speed but the difference between these interpretations is arguably negligible. Thus, Eq. 10-18 leads to 12 10-3 m / y . = = 2.18 10-5 rad / y 55 m which we convert (since there are about 3.16 107 s in a year) to = 6.9 1013 rad/s. 20. Converting 333 rev/min to radians-per-second, we get = 3.49 rad/s. Combining 1 v = r (Eq. 10-18) with t = d/v where t is the time between bumps (a distance d apart), we arrive at the rate of striking bumps: 1 r = d t 199 /s. 21. (a) We obtain = (200 rev / min)(2 rad / rev) = 20.9 rad / s. 60 s / min (b) With r = 1.20/2 = 0.60 m, Eq. 10-18 leads to v = r = (0.60)(20.9) = 12.5 m/s. (c) With t = 1 min, = 1000 rev/min and 0 = 200 rev/min, Eq. 10-12 gives = -o = 800 rev / min 2 . t (d) With the same values used in part (c), Eq. 10-15 becomes = 1 1 o + t = (200 + 100)(1) = 600 rev. 2 2 b g 22. (a) Using Eq. 10-6, the angular velocity at t = 5.0s is = d dt = t =5.0 d 0.30t 2 dt c h t =5.0 = 2(0.30)(5.0) = 3.0 rad / s. (b) Eq. 10-18 gives the linear speed at t = 5.0s: v = r = (3.0 rad/s)(10 m) = 30 m/s. (c) The angular acceleration is, from Eq. 10-8, = d d = (0.60t ) = 0.60 rad / s2 . dt dt Then, the tangential acceleration at t = 5.0s is, using Eq. 10-22, at = r = (10 m) 0.60 rad / s2 = 6.0 m / s2 . c h (d) The radial (centripetal) acceleration is given by Eq. 10-23: ar = 2 r = 3.0 rad / s 10 m = 90 m / s2 . 2 b gb g 23. (a) Converting from hours to seconds, we find the angular velocity (assuming it is positive) from Eq. 10-18: 1.00 2.90 104 km / h 3600hs v = 2.50 10 -3 rad / s. = = 3.22 103 km r c hd i (b) The radial (or centripetal) acceleration is computed according to Eq. 10-23: ar = 2 r = 2.50 10-3 rad / s c h c3.22 10 mh = 20.2 m / s . 6 2 2 (c) Assuming the angular velocity is constant, then the angular acceleration and the tangential acceleration vanish, since = d = 0 and at = r = 0. dt 24. First, we convert the angular velocity: = (2000) (2 /60) = 209 rad/s. Also, we convert the plane's speed to SI units: (480)(1000/3600) = 133 m/s. We use Eq. 10-18 in part (a) and (implicitly) Eq. 4-39 in part (b). (a) The speed of the tip as seen by the pilot is vt = r = 209 rad s 15 m = 314 m s , . which (since the radius is given to only two significant figures) we write as vt = 3.1 102 m s . (b) The plane's velocity v p and the velocity of the tip vt (found in the plane's frame of reference), in any of the tip's positions, must be perpendicular to each other. Thus, the speed as seen by an observer on the ground is v = v 2 + vt2 = p b gb g b133 m sg + b314 m sg 2 2 = 3.4 102 m s . 25. The function = e t where = 0.40 rad and = 2 s1 is describing the angular coordinate of a line (which is marked in such a way that all points on it have the same value of angle at a given time) on the object. Taking derivatives with respect to time 2 leads to d = e t and d = 2 e t . dt dt 2 (a) Using Eq. 10-22, we have d 2 at = r = 2 r = 6.4 cm / s2 . dt (b) Using Eq. 10-23, we have ar = 2 r = FG d IJ r = 2.6 cm / s . H dt K 2 2 26. (a) The tangential acceleration, using Eq. 10-22, is at = r = 14.2 rad / s2 ( 2.83 cm) = 40.2 cm / s2 . c h (b) In rad/s, the angular velocity is = (2760)(2/60) = 289, so ar = 2 r = (289 rad / s) 2 (0.0283 m) = 2.36 103 m / s2 . (c) The angular displacement is, using Eq. 10-14, = 2 289 2 = = 2.94 103 rad. 2 2(14.2) Then, using Eq. 10-1, the distance traveled is s = r = (0.0283 m) (2.94 103 rad) = 83.2 m. 28. Since the belt does not slip, a point on the rim of wheel C has the same tangential acceleration as a point on the rim of wheel A. This means that ArA = CrC, where A is the angular acceleration of wheel A and C is the angular acceleration of wheel C. Thus, C = FG r IJ = FG 10 cm IJ (16 rad / s ) = 0.64 rad / s . H r K H 25 cmK . A 2 2 C C Since the angular speed of wheel C is given by C = Ct, the time for it to reach an angular speed of = 100 rev/min = 10.5 rad/s starting from rest is t= C 10.5 rad / s = = 16 s. C 0.64 rad / s2 29. (a) In the time light takes to go from the wheel to the mirror and back again, the wheel turns through an angle of = 2/500 = 1.26 102 rad. That time is t= 2 2(500 m) = = 3.34 10-6 s 8 c 2.998 10 m / s so the angular velocity of the wheel is = t = 126 10-2 rad . = 38 103 rad / s. . 3.34 10-6 s (b) If r is the radius of the wheel, the linear speed of a point on its rim is v = r = ( 3.8 103 rad/s ) ( 0.050 m ) = 1.9 102 m/s. 30. (a) The angular acceleration is = 0 - 150 rev / min = = -114 rev / min 2 . . t (2.2 h)(60 min / 1h) (b) Using Eq. 10-13 with t = (2.2) (60) = 132 min, the number of revolutions is = 0t + t 2 = (150 rev/min)(132 min) + 1 2 1 2 -1.14 rev/min 2 (132 min ) = 9.9 103 rev. 2 ( ) 2 (c) With r = 500 mm, the tangential acceleration is F 2 rad IJ FG 1 min IJ a = r = c-114 rev / min h G . H 1 rev K H 60 s K 2 t (500 mm) which yields at = 0.99 mm/s2. (d) With r = 0.50 m, the radial (or centripetal) acceleration is given by Eq. 10-23: ar F F 2 rad / rev IJ I = r = G (75 rev / min) G H 1 min/ 60 s K JK H 2 2 (0.50 m) which yields ar = 31 in SI units--and is seen to be much bigger than at. Consequently, the magnitude of the acceleration is | a | = ar2 + at2 ar = 31 m / s2 . 31. (a) The angular speed in rad/s is = 33 rev / min FG H 1 3 IJ FG 2 rad / rev IJ = 3.49 rad / s. K H 60 s / min K g 2 Consequently, the radial (centripetal) acceleration is (using Eq. 10-23) a = 2 r = 3.49 rad / s (6.0 10-2 m) = 0.73 m / s2 . (b) Using Ch. 6 methods, we have ma = fs fs,max = s mg, which is used to obtain the (minimum allowable) coefficient of friction: b s,min = a 0.73 = = 0.075. g 9.8 (c) The radial acceleration of the object is ar = 2r, while the tangential acceleration is at = r. Thus | a | = ar2 + at2 = ( 2 r ) 2 + (r ) 2 = r 4 + 2 . If the object is not to slip at any time, we require f s,max = smg = mamax = mr 4 + 2 . max Thus, since = /t (from Eq. 10-12), we find s ,min = 4 r max + 2 g = 4 r max + (max / t )2 g = (0.060) 3.494 + (3.4 / 0.25) 2 = 0.11. 9.8 32. (a) A complete revolution is an angular displacement of = 2 rad, so the angular velocity in rad/s is given by = /T = 2/T. The angular acceleration is given by = d 2 dT =- 2 . dt T dt For the pulsar described in the problem, we have dT 126 10-5 s / y . = = 4.00 10-13 . 7 316 10 s / y . dt Therefore, =- FG 2 IJ (4.00 10 H (0.033 s) K 2 -13 ) = -2.3 10-9 rad / s2 . The negative sign indicates that the angular acceleration is opposite the angular velocity and the pulsar is slowing down. (b) We solve = 0 + t for the time t when = 0: t=- 0 2 2 =- =- = 8.3 1010 s 2.6 103 years 2 -9 T (-2.3 10 rad/s )(0.033 s) (c) The pulsar was born 19921054 = 938 years ago. This is equivalent to (938 y)(3.16 107 s/y) = 2.96 1010 s. Its angular velocity at that time was = 0 + t + Its period was 2 2 + t = + (-2.3 10-9 rad/s 2 )(-2.96 1010 s) = 258 rad/s. T 0.033 s T= 2 = 2 = 2.4 10 -2 s. 258 rad / s 1 33. The kinetic energy (in J) is given by K = 2 I 2 , where I is the rotational inertia (in kg m2 ) and is the angular velocity (in rad/s). We have = (602 rev / min)(2 rad / rev) = 63.0 rad / s. 60 s / min Consequently, the rotational inertia is I= 2K = 2(24400 J) = 12.3 kg m2 . 2 (63.0 rad / s) 2 34. (a) Eq. 10-12 implies that the angular acceleration should be the slope of the vs t graph. Thus, = 9/6 = 1.5 rad/s2. (b) By Eq. 10-34, K is proportional to 2. Since the angular velocity at t = 0 is 2 rad/s (and this value squared is 4) and the angular velocity at t = 4 s is 4 rad/s (and this value squared is 16), then the ratio of the corresponding kinetic energies must be Ko 4 K4 = 16 Ko = K4 = 0.40 J . 1 35. Since the rotational inertia of a cylinder is I = 2 MR 2 (Table 10-2(c)), its rotational kinetic energy is K= 1 2 1 I = MR 2 2 . 2 4 1 (a) For the smaller cylinder, we have K = 4 (125)(0.25) 2 ( 235) 2 = 11 103 J. . . 1 (b) For the larger cylinder, we obtain K = 4 (125)(0.75) 2 ( 235) 2 = 9.7 103 J. . 36. (a) Eq. 10-33 gives Itotal = md + m(2d)2 + m(3d)2 = 14 md . If the innermost one is removed then we would only obtain m(2d)2 + m(3d)2 = 13 md . The percentage difference between these is (13 14)/14 = 0.0714 7.1%. (b) If, instead, the outermost particle is removed, we would have md + m(2d)2 = 5 md . The percentage difference in this case is 0.643 64%. 2 2 2 2 2 37. We use the parallel axis theorem: I = Icom + Mh2, where Icom is the rotational inertia about the center of mass (see Table 10-2(d)), M is the mass, and h is the distance between the center of mass and the chosen rotation axis. The center of mass is at the center of the meter stick, which implies h = 0.50 m 0.20 m = 0.30 m. We find I com = 1 1 2 ML2 = 0.56 kg 1.0 m = 4.67 10 -2 kg m2 . 12 12 b gb g Consequently, the parallel axis theorem yields I = 4.67 10-2 kg m2 + 0.56 kg 0.30 m = 9.7 10-2 kg m2 . b gb g 2 38. The parallel axis theorem (Eq. 10-36) shows that I increases with h. The phrase "out to the edge of the disk" (in the problem statement) implies that the maximum h in the graph is, in fact, the radius R of the disk. Thus, R = 0.20 m. Now we can examine, say, the h = 0 datum and use the formula for Icom (see Table 10-2(c)) for a solid disk, or (which might be a little better, since this is independent of whether it is really a solid disk) we can the difference between the h = 0 datum and the h = hmax =R datum and relate that difference to the parallel axis theorem (thus the difference is M(hmax)2 = 0.10 kgm2). In either case, we arrive at M = 2.5 kg. 39. The particles are treated "point-like" in the sense that Eq. 10-33 yields their rotational inertia, and the rotational inertia for the rods is figured using Table 10-2(e) and the parallel-axis theorem (Eq. 10-36). (a) With subscript 1 standing for the rod nearest the axis and 4 for the particle farthest from it, we have I = I1 + I 2 + I 3 + I 4 = 1 1 Md 2 + M d 12 2 2 + md 2 + 1 3 Md 2 + M d 12 2 2 + m(2d ) 2 8 8 = Md 2 + 5md 2 = (1.2 kg)(0.056 m) 2 +5(0.85 kg)(0.056 m) 2 3 3 2 =0.023 kg m . (b) Using Eq. 10-34, we have K= 1 2 4 5 4 5 I = M + m d 2 2 = (1.2 kg) + (0.85 kg) (0.056 m) 2 (0.30 rad/s) 2 2 3 2 3 2 -3 = 1.1 10 J. 40. (a) Consider three of the disks (starting with the one at point O): OO . The first one (the one at point O shown here with the plus sign inside) has rotational inertial (see 2 1 item (c) in Table 10-2) I = 2 mR . The next one (using the parallel-axis theorem) has I = 2 mR + mh 1 2 1 2 2 where h = 2R. The third one has I = 2 mR + m(4R) . If we had considered five of the disks OOOO with the one at O in the middle, then the total rotational inertia is 2 I = 5(2 mR ) + 2(m(2R) + m(4R) ). The pattern is now clear and we can write down the total I for the collection of fifteen disks: 1 2 2 2 I = 15(2 mR ) + 2(m(2R) + m(4R) + m(6R) + ... + m(14R) ) = 1 2 2 2 2 2 2 2255 mR . 2 The generalization to N disks (where N is assumed to be an odd number) is I = 6(2N + 1)NmR . In terms of the total mass (m = M/15) and the total length (R = L/30), we obtain 1 2 2 I = 0.083519ML2 (0.08352)(0.1000 kg)(1.0000 m)2 = 8.352 10-3 kg m2. (b) Comparing to the formula (e) in Table 10-2 (which gives roughly I =0.08333 ML2), we find our answer to part (a) is 0.22% lower. 41. We use the parallel-axis theorem. According to Table 10-2(i), the rotational inertia of a uniform slab about an axis through the center and perpendicular to the large faces is given by I com = M 2 a + b2 . 12 c h A parallel axis through the corner is a distance h = Therefore, I = I com + Mh 2 = = M 12 ba / 2g + bb / 2g 2 2 from the center. (a 2 + b2 ) + M 4 (a 2 + b2 ) = M 3 (a 2 + b2 ) 0.172 kg [(0.035 m) 2 +(0.084 m) 2 ] = 4.7 10-4 kg m 2 . 3 42. (a) We show the figure with its axis of rotation (the thin horizontal line). We note that each mass is r = 1.0 m from the axis. Therefore, using Eq. 10-26, we obtain I= mi ri 2 = 4 (0.50 kg) (1.0 m) 2 = 2.0 kg m 2 . (b) In this case, the two masses nearest the axis are r = 1.0 m away from it, but the two furthest from the axis are r = 1.02 + 2.02 m from it. Here, then, Eq. 10-33 leads to I= mi ri 2 = 2 0.50 kg 10 m2 + 2 0.50 kg 5.0 m2 = 6.0 kg m2 . . b gc h b gc h (c) Now, two masses are on the axis (with r = 0) and the other two are a distance r = 102 + 102 m away. Now we obtain I = 2.0 kg m 2 . . . 43. (a) We apply Eq. 10-33: Ix = 4 mi yi2 = 50 2.0 + 25 4.0 + 25 -3.0 + 30 4.0 = 13 103 g cm2 . . b g b gb g 2 2 b g 2 2 b g 2 2 i =1 (b) For rotation about the y axis we obtain Iy = 4 mi xi2 = 50 2.0 + 25 0 + 25 3.0 + 30 2.0 = 5.5 102 g cm2 . b g b gb g 2 2 b g b g i =1 (c) And about the z axis, we find (using the fact that the distance from the z axis is x2 + y2 ) Iz = 4 mi xi2 + yi2 = I x + I y =1.3 103 + 5.5 102 = 1.9 102 g cm2 . c h i =1 (d) Clearly, the answer to part (c) is A + B. 44. (a) Using Table 10-2(c) and Eq. 10-34, the rotational kinetic energy is K= 1 2 1 1 1 I = MR 2 2 = (500 kg)(200 rad/s) 2 (1.0 m) 2 = 4.9 107 J. 2 2 2 4 (b) We solve P = K/t (where P is the average power) for the operating time t. t= 4.9 107 J K = = 6.2 103 s 3 P 8.0 10 W which we rewrite as t 1.0 102 min. 45. Two forces act on the ball, the force of the rod and the force of gravity. No torque about the pivot point is associated with the force of the rod since that force is along the line from the pivot point to the ball. As can be seen from the diagram, the component of the force of gravity that is perpendicular to the rod is mg sin . If is the length of the rod, then the torque associated with this force has magnitude = mg sin = (0.75)(9.8)(1.25) sin 30 = 4.6 N m . For the position shown, the torque is counter-clockwise. 46. We compute the torques using = rF sin . (a) For = 30 , a = (0.152 m)(111 N) sin 30 = 8.4 N m . (b) For = 90 , b = (0.152 m)(111 N) sin 90 = 17 N m . (c) For = 180 , c = (0.152 m)(111N) sin180 = 0 . 47. We take a torque that tends to cause a counterclockwise rotation from rest to be positive and a torque tending to cause a clockwise rotation to be negative. Thus, a positive torque of magnitude r1 F1 sin 1 is associated with F1 and a negative torque of magnitude r2F2 sin 2 is associated with F2 . The net torque is consequently = r1 F1 sin 1 - r2 F2 sin 2 . Substituting the given values, we obtain = (1.30 m)(4.20 N) sin 75 - (2.15 m)(4.90 N) sin 60 = -3.85 N m. 48. The net torque is = A + B + C = FA rA sin A - FB rB sin B + FC rC sin C = (10)(8.0) sin135 - (16)(4.0) sin 90 + (19)(3.0) sin160 = 12 N m. 49. (a) We use the kinematic equation = 0 + t , where 0 is the initial angular velocity, is the final angular velocity, is the angular acceleration, and t is the time. This gives = -0 t = 6.20 rad / s = 28.2 rad / s2 . 220 10 -3 s (b) If I is the rotational inertia of the diver, then the magnitude of the torque acting on her is = I = 12.0 kg m2 28.2 rad / s2 = 3.38 102 N m. c hc h 50. The rotational inertia is found from Eq. 10-45. I= 32.0 . = = 128 kg m2 25.0 51. Combining Eq. 10-45 (net = I ) with Eq. 10-38 gives RF2 RF1 = I , where = / t by Eq. 10-12 (with = 0). Using item (c) in Table 10-2 and solving for F2 we find F2 = MR (0.02)(0.02)(250) + F1 = + 0.1 = 0.140 N. 2t 2(1.25) 52. (a) In this case, the force is mg = (70)(9.8), and the "lever arm" (the perpendicular distance from point O to the line of action of the force) is 0.28 meter. Thus, the torque (in absolute value) is (70)(9.8)(0.28). Since the moment-of-inertia is I = 65 kgm2, then Eq. 10-45 gives || = 2.955 3.0 rad/s2. (b) Now we have another contribution (1.4 m x 300 N) to the net torque, so |net| = (70)(9.8)(0.28) + (1.4)(300 N) = (65) || which leads to || = 9.4 rad/s2. 53. According to the sign conventions used in the book, the magnitude of the net torque exerted on the cylinder of mass m and radius R is net = F1 R - F2 R - F3 r = (6.0 N)(0.12 m) - (4.0 N)(0.12 m) - (2.0 N)(0.050 m) = 71N m. 1 (a) The resulting angular acceleration of the cylinder (with I = 2 MR 2 according to Table 10-2(c)) is = net I = 1 2 71N m = 9.7 rad/s 2 . (2.0 kg)(0.12 m) 2 (b) The direction is counterclockwise (which is the positive sense of rotation). 54. With counterclockwise positive, the angular acceleration for both masses satisfies 2 = mgL1 - mgL2 = I = ( mL1 + mL2 ) , by combining Eq. 10-45 with Eq. 10-39 and 2 Eq. 10-33. Therefore, using SI units, g ( L1 - L2 ) ( 9.8 )( 0.20 - 0.80 ) = = - 8.65 rad/s 2 2 2 2 2 L1 + L2 (0.20) + (0.80) = where the negative sign indicates the system starts turning in the clockwise sense. The magnitude of the acceleration vector involves no radial component (yet) since it is evaluated at t = 0 when the instantaneous velocity is zero. Thus, for the two masses, we apply Eq. 10-22: (a) a1 = | |L1 = ( 8.65 rad/s 2 ) ( 0.20 m ) = 1.7 m/s. (b) a2 = | |L2 = ( 8.65 rad/s 2 ) ( 0.80 m ) = 6.9 m/s 2 . 55. (a) We use constant acceleration kinematics. If down is taken to be positive and a is 1 the acceleration of the heavier block, then its coordinate is given by y = 2 at 2 , so a= 2 y 2(0.750 m) = = 6.00 10 -2 m / s2 . 2 2 (5.00 s) t The lighter block has an acceleration of 6.00 102 m/s2 upward. (b) Newton's second law for the heavier block is mh g - Th = mh a , where mh is its mass and Th is the tension force on the block. Thus, Th = mh ( g - a ) = ( 0.500 kg) 9.8 m / s2 - 6.00 10-2 m / s2 = 4.87 N. c h (c) Newton's second law for the lighter block is ml g - Tl = - ml a , where Tl is the tension force on the block. Thus, Tl = ml ( g + a ) = ( 0.460 kg) 9.8 m / s2 + 6.00 10 -2 m / s2 = 4.54 N. c h (d) Since the cord does not slip on the pulley, the tangential acceleration of a point on the rim of the pulley must be the same as the acceleration of the blocks, so = a 6.00 10-2 m / s2 . = = 120 rad / s2 . -2 R 5.00 10 m (e) The net torque acting on the pulley is = (Th Tl)R. Equating this to I we solve for the rotational inertia: (T - T ) R = ( 4.87 N - 4.54 N ) ( 5.00 10 I= h l 1.20 rad/s 2 -2 m) = 1.38 10-2 kg m 2 . 56. Combining Eq. 10-34 and Eq. 10-45, we have RF = I, where is given by /t (according to Eq. 10-12, since o = 0 in this case). We also use the fact that I = Iplate + Idisk where Idisk = 2 MR (item (c) in Table 10-2). Therefore, Iplate = RFt 1 2 2 1 MR 2 = 2.51 10-4 kg.m2. 57. Since the force acts tangentially at r = 0.10 m, the angular acceleration (presumed positive) is 0.5t + 0.3t 2 010 . Fr = = = = 50t + 30t 2 -3 I I 10 10 . c hb g in SI units (rad/s2). (a) At t = 3 s, the above expression becomes = 4.2 102 rad/s2. (b) We integrate the above expression, noting that o = 0, to obtain the angular speed at t = 3 s: = 3 0 dt = ( 25t 2 + 10t 3 ) 3 0 = 5.0 102 rad/s. 58. With = (1800)(2/60) = 188.5 rad/s, we apply Eq. 10-55: P = = 74600 W = 396 N m . 188.5 rad/s 59. (a) The speed of v of the mass m after it has descended d = 50 cm is given by v2 = 2ad (Eq. 2-16). Thus, using g = 980 cm/s2, we have v = 2ad = 2(2mg )d = M + 2m 4(50)(980)(50) = 1.4 102 cm / s. 400 + 2(50) (b) The answer is still 1.4 102 cm/s = 1.4 m/s, since it is independent of R. 60. The initial angular speed is = (280)(2/60) = 29.3 rad/s. (a) Since the rotational inertia is (Table 10-2(a)) I = (32) (1.2) 2 = 46.1 kg m2 , the work done is W = K = 0 - 1 2 1 I = - (461) (29.3) 2 . 2 2 which yields |W| = 19.8 103 J. (b) The average power (in absolute value) is therefore | P| = |W | 19.8 103 = = 1.32 103 W. t 15 61. (a) We apply Eq. 10-34: K= 1 2 1 1 2 1 1 I = mL 2 = mL2 2 = (0.42 kg)(0.75 m)2 (4.0 rad/s)2 = 0.63 J. 2 2 3 6 6 (b) Simple conservation of mechanical energy leads to K = mgh. Consequently, the center of mass rises by h= K mL2 2 L2 2 (0.75 m) 2 (4.0 rad/s) 2 = = = = 0.153 m 0.15 m. mg 6mg 6g 6(9.8 m/s 2 ) 62. (a) Eq. 10-33 gives Itotal = md + m(2d)2 + m(3d)2 = 14 md , where d = 0.020 m and m = 0.010 kg. The work done is W = K = 2 If 1 2 2 2 1 Ii2, 2 where f = 20 rad/s and i = 0. This gives W = 11.2 mJ. (b) Now, f = 40 rad/s and i = 20 rad/s, and we get W = 33.6 mJ. (c) In this case, f = 60 rad/s and i = 40 rad/s. This gives W = 56.0 mJ. (d) Eq. 10-34 indicates that the slope should be 2 I. Therefore, it should be 7md = 2.80 10-5 J.s2. 2 1 63. We use to denote the length of the stick. Since its center of mass is / 2 from 1 either end, its initial potential energy is 2 mg , where m is its mass. Its initial kinetic 1 energy is zero. Its final potential energy is zero, and its final kinetic energy is 2 I 2 , where I is its rotational inertia about an axis passing through one end of the stick and is the angular velocity just before it hits the floor. Conservation of energy yields 1 1 mg = I 2 2 2 The free end of the stick is a distance floor is (from Eq. 10-18) = mg . I from the rotation axis, so its speed as it hits the v = = mg 3 . I Using Table 10-2 and the parallel-axis theorem, the rotational inertial is I = 1 m 2 , so 3 v = 3g = 3 9.8 m / s2 1.00 m = 5.42 m / s. c hb g 64. (a) We use the parallel-axis theorem to find the rotational inertia: I = I com + Mh 2 = 1 1 2 2 MR 2 + Mh 2 = ( 20 kg )( 0.10 m ) + ( 20 kg )( 0.50 m ) = 0.15 kg m 2 . 2 2 1 (b) Conservation of energy requires Mgh that = 2 I 2 , where is the angular speed of the cylinder as it passes through the lowest position. Therefore, = 2 Mgh 2(20) (9.8) (0.050) = = 11 rad / s. I 015 . 65. Using the parallel axis theorem and items (e) and (h) in Table 10-2, the rotational inertia is I = 2 1 mL 12 + m(L/2) + 2 2 1 mR 2 + m(R + L) = 10.83mR , 2 2 where L = 2R has been used. If we take the base of the rod to be at the coordinate origin (x = 0, y = 0) then the center of mass is at y= mL/2 + m(L + R) = 2R . m+m Comparing the position shown in the textbook figure to its upside down (inverted) position shows that the change in center of mass position (in absolute value) is |y| = 4R. The corresponding loss in gravitational potential energy is converted into kinetic energy. Thus K = (2m)g(4R) where Eq. 10-34 has been used. = 9.82 rad/s . 66. (a) We use conservation of mechanical energy to find an expression for 2 as a function of the angle that the chimney makes with the vertical. The potential energy of the chimney is given by U = Mgh, where M is its mass and h is the altitude of its center of mass above the ground. When the chimney makes the angle with the vertical, h = (H/2) cos . Initially the potential energy is Ui = Mg(H/2) and the kinetic energy is zero. 1 The kinetic energy is 2 I 2 when the chimney makes the angle with the vertical, where I is its rotational inertia about its bottom edge. Conservation of energy then leads to MgH / 2 = Mg ( H / 2)cos + 1 2 I 2 2 = ( MgH / I ) (1 - cos ). The rotational inertia of the chimney about its base is I = MH2/3 (found using Table 10-2(e) with the parallel axis theorem). Thus = 3g 3(9.80 m/s 2 ) (1 - cos ) = (1 - cos 35.0) = 0.311 rad/s. H 55.0 m (b) The radial component of the acceleration of the chimney top is given by ar = H2, so ar = 3g (1 cos ) = 3 (9.80 m/s2)(1 cos 35.0 ) = 5.32 m/s2 . (c) The tangential component of the acceleration of the chimney top is given by at = H, where is the angular acceleration. We are unable to use Table 10-1 since the acceleration is not uniform. Hence, we differentiate 2 = (3g/H)(1 cos ) with respect to time, replacing d / dt with , and d / dt with , and obtain d 2 = 2 = (3g / H ) sin dt Consequently, 3(9.80 m/s 2 ) 3g a = H = sin = sin 35.0 = 8.43 m/s 2 . t 2 2 = (3g / 2 H )sin . (d) The angle at which at = g is the solution to obtain = 41.8. 3g 2 sin = g. Thus, sin = 2/3 and we 67. From Table 10-2, the rotational inertia of the spherical shell is 2MR2/3, so the kinetic energy (after the object has descended distance h) is K= 1 2 1 1 MR 2 2 + I 2 + mv 2 . sphere pulley 2 3 2 2 FG H IJ K Since it started from rest, then this energy must be equal (in the absence of friction) to the potential energy mgh with which the system started. We substitute v/r for the pulley's angular speed and v/R for that of the sphere and solve for v. v= = mgh 2 gh = 2 M 1 I 1 + ( I / mr ) + (2 M / 3m) m + 2 r2 + 3 -3 1 2 2(9.8)(0.82) = 1.4 m/s 1 + 3.0 10 /((0.60)(0.050) 2 ) + 2(4.5) / 3(0.60) 68. (a) We integrate (with respect to time) the = 6.0t4 4.0t2 expression, taking into account that the initial angular velocity is 2.0 rad/s. The result is = 1.2 t5 1.33 t3 + 2.0. (b) Integrating again (and keeping in mind that o = 1) we get = 0.20t6 0.33 t4 + 2.0 t + 1.0 . 69. We choose positive coordinate directions (different choices for each item) so that each is accelerating positively, which will allow us to set a2 = a1 = R (for simplicity, we denote this as a). Thus, we choose rightward positive for m2 = M (the block on the table), downward positive for m1 = M (the block at the end of the string) and (somewhat unconventionally) clockwise for positive sense of disk rotation. This means that we interpret given in the problem as a positive-valued quantity. Applying Newton's second law to m1, m2 and (in the form of Eq. 10-45) to M, respectively, we arrive at the following three equations (where we allow for the possibility of friction f2 acting on m2). m1 g - T1 = m1a1 T2 - f 2 = m2 a2 T1 R - T2 R = I (a) From Eq. 10-13 (with 0 = 0) we find = 0 t + t 2 1 2 = 2 2(1.30 rad) = = 314 rad/s 2 . t 2 (0.0910 s) 2 (b) From the fact that a = R (noted above), we obtain a= 2 R 2(0.024 m)(1.30 rad) = = 7.54 m/s 2 . t2 (0.0910 s) 2 (c) From the first of the above equations, we find T1 = m1 ( g - a1 ) = M g - 2 R 2(0.024 m)(1.30 rad) = (6.20 kg) 9.80 m/s 2 - = 14.0 N. 2 t (0.0910 s) 2 (d) From the last of the above equations, we obtain the second tension: T2 = T1 - I 2 R 2 I (7.40 10 -4 kg m 2 )(314 rad/s 2 ) = M g - 2 - 2 = 14.0 N - = 4.36 N. R t Rt 0.024 m 70. The rotational inertia of the passengers is (to a good approximation) given by Eq. 1053: I = mR 2 = NmR 2 where N is the number of people and m is the (estimated) mass per person. We apply Eq. 10-52: W= 1 2 1 I = NmR 2 2 2 2 where R = 38 m and N = 36 60 = 2160 persons. The rotation rate is constant so that = /t which leads to = 2/120 = 0.052 rad/s. The mass (in kg) of the average person is probably in the range 50 m 100, so the work should be in the range 1 2160 50 38 2 b gb gb g b0.052g 2 1 2160 100 38 2 2 105 J W 4 105 J. 2 W b gb gb g b0.052g 2 2 71. The volume of each disk is r2h where we are using h to denote the thickness (which equals 0.00500 m). If we use R (which equals 0.0400 m) for the radius of the larger disk and r (which equals 0.0200 m) for the radius of the smaller one, then the mass of each is 2 m = r2h and M = R h where = 1400 kg/m3 is the given density. We now use the parallel axis theorem as well as item (c) in Table 10-2 to obtain the rotation inertia of the two-disk assembly: I = 2 MR + 1 2 1 mr2 2 + m(r + R) = h[ 2 R + 2 1 4 1 4 r 2 + r2(r + R) ] = 6.16 10-5 kg.m2. 2 72. In the calculation below, M1 and M2 are the ring masses, R1i and R2i are their inner radii, and R1o and R2o are their outer radii. Referring to item (b) in Table 10-2, we compute I = 2 M1 (R1i + R1o ) + 2 M2 (R2i + R2o ) = 0.00346 kgm2 . Thus, with Eq. 10-38 ( = rF where r = R2o) and = I (Eq. 10-45), we find 1 2 2 1 2 2 = (0.140)(12.0) = 485 rad/s2 . 0.00346 Then Eq. 10-12 gives = t = 146 rad/s. 73. (a) The longitudinal separation between Helsinki and the explosion site is = 102 - 25 = 77 . The spin of the earth is constant at = 1 rev 360 = 1 day 24 h so that an angular displacement of corresponds to a time interval of t = 77 24 b g FGH 360h IJK = 5.1 h. (b) Now = 102- -20 = 122 so the required time shift would be t = 122 b g b g FGH 24 hIJK = 8.1 h. 360 74. In the figure below, we show a pull tab of a beverage can. Since the tab is pivoted, when pulling on one end upward with a force F1 , a force F2 will be exerted on the other end. The torque produced by F1 must be balanced by the torque produced by F2 so that the tab does not rotate. The two forces are related by r1 F1 = r2 F2 where r1 1.8 cm and r2 0.73 cm . Thus, if F1 = 10 N, F2 = r1 1.8 F1 (10 N) 25 N. r2 0.73 75. (a) We apply Eq. 10-18, using the subscript J for the Jeep. = v J 114 km h = rJ 0.100 km which yields 1140 rad/h or (dividing by 3600) 0.32 rad/s for the value of the angular speed . (b) Since the cheetah has the same angular speed, we again apply Eq. 10-18, using the subscript c for the cheetah. vc = rc = ( 92 m ) (1140 rad h ) = 1.048 105 m h 1.0 102 km/h for the cheetah's speed. 76. The angular displacements of disks A and B can be written as: A = At , B = B t 2 . (a) The time when A = B is given by 1 2 At = B t 2 1 2 t= 2 A B = 2(9.5 rad/s) = 8.6 s. (2.2 rad/s 2 ) (b) The difference in the angular displacement is 1 = A - B = At - B t 2 = 9.5t - 1.1t 2 . 2 For their reference lines to align momentarily, we only require = 2 N , where N is an integer. The quadratic equation can be readily solve to yield tN = 9.5 (9.5) 2 - 4(1.1)(2 N ) 9.5 90.25 - 27.6 N = . 2(1.1) 2.2 The solution t0 = 8.63 s (taking the positive root) coincides with the result obtained in (a), while t0 = 0 (taking the negative root) is the moment when both disks begin to rotate. In fact, two solutions exist for N = 0, 1, 2, and 3. 77. (a) The rotational inertia relative to the specified axis is I= mi ri 2 = 2 M L2 + 2 M L2 + M 2 L b g b g b g 2 which is found to be I = 4.6 kg m2. Then, with = 1.2 rad/s, we obtain the kinetic energy from Eq. 10-34: K= 1 2 I = 3.3 J. 2 (b) In this case the axis of rotation would appear as a standard y axis with origin at P. Each of the 2M balls are a distance of r = L cos 30 from that axis. Thus, the rotational inertia in this case is I= mi ri 2 = 2 M r 2 + 2 M r 2 + M 2 L b g b g b g 2 which is found to be I = 4.0 kg m2. Again, from Eq. 10-34 we obtain the kinetic energy K= 1 2 I = 2.9 J. 2 78. We choose positive coordinate directions (different choices for each item) so that each is accelerating positively, which will allow us to set a1 = a2 = R (for simplicity, we denote this as a). Thus, we choose upward positive for m1, downward positive for m2 and (somewhat unconventionally) clockwise for positive sense of disk rotation. Applying Newton's second law to m1m2 and (in the form of Eq. 10-45) to M, respectively, we arrive at the following three equations. T1 - m1 g = m1a1 m2 g - T2 = m2 a2 T2 R - T1 R = I 1 (a) The rotational inertia of the disk is I = 2 MR 2 (Table 10-2(c)), so we divide the third equation (above) by R, add them all, and use the earlier equality among accelerations -- to obtain: m2 g - m1 g = m1 + m2 + FG H 1 M a 2 IJ K which yields a = 4 g = 1.57 m/s 2 . 25 (b) Plugging back in to the first equation, we find T1 = 29 m1 g = 4.55 N 25 where it is important in this step to have the mass in SI units: m1 = 0.40 kg. (c) Similarly, with m2 = 0.60 kg, we find T2 = 5 m2 g = 4.94 N. 6 79. (a) Constant angular acceleration kinematics can be used to compute the angular acceleration . If 0 is the initial angular velocity and t is the time to come to rest, then 0 = 0 + t =- 0 t which yields 39/32 = 1.2 rev/s or (multiplying by 2) 7.66 rad/s2 for the value of . (b) We use = I, where is the torque and I is the rotational inertia. The contribution of the rod to I is M 2 / 12 (Table 10-2(e)), where M is its mass and is its length. The 2 contribution of each ball is m / 2 , where m is the mass of a ball. The total rotational inertia is b g b 6.40 kg 120 m . 106 kg 120 m . . M 2 m 2 I= +2 = + 12 4 12 2 which yields I = 1.53 kg m2. The torque, therefore, is gb g b 2 gb g 2 = 153 kg m 2 -7.66 rad / s2 = -117 N m. . . (c) Since the system comes to rest the mechanical energy that is converted to thermal energy is simply the initial kinetic energy Ki = 1 2 1 I 0 = 153 kg m2 . 2 2 c hc h c hcb2gb39g rad / sh hb 2 = 4.59 104 J. (d) We apply Eq. 10-13: = 0t + t 2 = 2 39 rad / s 32.0 s + 1 2 cb gb g 1 g 2 c-7.66 rad / s hb32.0 sg 2 2 which yields 3920 rad or (dividing by 2) 624 rev for the value of angular displacement . (e) Only the mechanical energy that is converted to thermal energy can still be computed without additional information. It is 4.59 104 J no matter how varies with time, as long as the system comes to rest. 80. The Hint given in the problem would make the computation in part (a) very straightforward (without doing the integration as we show here), but we present this further level of detail in case that hint is not obvious or -- simply -- in case one wishes to see how the calculus supports our intuition. (a) The (centripetal) force exerted on an infinitesimal portion of the blade with mass dm located a distance r from the rotational axis is (Newton's second law) dF = (dm)2r, where dm can be written as (M/L)dr and the angular speed is = ( 320 )( 2 60 ) = 33.5 rad s . Thus for the entire blade of mass M and length L the total force is given by M F = dF = rdm = L 5 = 4.8110 N. 2 M 2 L (110kg ) ( 33.5 rad s ) rdr = = 0 2 2 L 2 2 ( 7.80m ) (b) About its center of mass, the blade has I = ML2 / 12 according to Table 10-2(e), and using the parallel-axis theorem to "move" the axis of rotation to its end-point, we find the rotational inertia becomes I = ML2 / 3. Using Eq. 10-45, the torque (assumed constant) is = I = 1 ML2 3 1 2 33.5rad/s = (110 kg )( 7.8 m ) = 1.12 10 4 N m. t 3 6.7 s (c) Using Eq. 10-52, the work done is W = K = 1 2 1 1 1 2 2 I - 0 = ML2 2 = (110kg )( 7.80m ) ( 33.5rad/s ) = 1.25 106 J. 2 2 3 6 81. (a) The linear speed of a point on belt 1 is v1 = rA A = (15 cm)(10 rad/s) = 1.5 10 2 cm/s . (b) The angular speed of pulley B is rBB = rA A B = rA A 15 cm (10 rad/s) = 15 rad/s . = 10 cm rB (c) Since the two pulleys are rigidly attached to each other, the angular speed of pulley B is the same as that of pulley B, i.e., B = 15 rad/s . (d) The linear speed of a point on belt 2 is v2 = rBB = (5 cm)(15 rad/s) = 75 cm/s . (e) The angular speed of pulley C is rCC = rBB C = rBB 5 cm (15 rad/s) = 3 rad/s = rC 25 cm 82. To get the time to reach the maximum height, we use Eq. 4-23, setting the left-hand side to zero. Thus, we find t= (60 m/s)sin(20o) = 2.094 s. 9.8 m/s2 Then (assuming = 0) Eq. 10-13 gives - o = o t = (90 rad/s)(2.094 s) = 188 rad, which is equivalent to roughly 30 rev. 83. With rightward positive for the block and clockwise negative for the wheel (as is conventional), then we note that the tangential acceleration of the wheel is of opposite sign from the block's acceleration (which we simply denote as a); that is, at = a. Applying Newton's second law to the block leads to P - T = ma , where m = 2.0 kg. Applying Newton's second law (for rotation) to the wheel leads to -TR = I , where I = 0.050 kg m 2 . Noting that R = at = a, we multiply this equation by R and obtain -TR 2 = - Ia T =a I . R2 Adding this to the above equation (for the block) leads to P = ( m + I / R 2 ) a. Thus, a = 0.92 m/s2 and therefore = 4.6 rad/s2 (or || = 4.6 rad/s2 ), where the negative sign in should not be mistaken for a deceleration (it simply indicates the clockwise sense to the motion). 84. We use conservation of mechanical energy. The center of mass is at the midpoint of the cross bar of the H and it drops by L/2, where L is the length of any one of the rods. The gravitational potential energy decreases by MgL/2, where M is the mass of the body. 1 The initial kinetic energy is zero and the final kinetic energy may be written 2 I 2 , where I is the rotational inertia of the body and is its angular velocity when it is vertical. Thus, 0 = - MgL / 2 + 1 2 I 2 = MgL / I . Since the rods are thin the one along the axis of rotation does not contribute to the rotational inertia. All points on the other leg are the same distance from the axis of rotation, so that leg contributes (M/3)L2, where M/3 is its mass. The cross bar is a rod that rotates around one end, so its contribution is (M/3)L2/3 = ML2/9. The total rotational inertia is I = (ML2/3) + (ML2/9) = 4ML2/9. Consequently, the angular velocity is = MgL MgL 9g 9(9.800 m/s 2 ) = = = = 6.06 rad/s. I 4ML2 / 9 4L 4(0.600 m) 85. (a) According to Table 10-2, the rotational inertia formulas for the cylinder (radius R) and the hoop (radius r) are given by IC = 1 MR 2 and I H = Mr 2 . 2 Since the two bodies have the same mass, then they will have the same rotational inertia if 2 R 2 / 2 = RH RH = R / 2 . (b) We require the rotational inertia to be written as I = Mk 2 , where M is the mass of the given body and k is the radius of the "equivalent hoop." It follows directly that k= I/M . 86. (a) The axis of rotation is at the bottom right edge of the rod along the ground, a horizontal distance of d 3 + d 2 + d1 / 2 from the middle of the table assembly (mass m = 90 kg ). The linebacker's center of mass at that critical moment was a horizontal distance of d 4 + d5 from the axis of rotation. For the clockwise torque caused by the linebacker (mass M) to overcome the counterclockwise torque of the table assembly, we require (using Eq. 10-41) Mg d 4 + d5 > mg d 3 + d 2 + b g FG H d1 . 2 IJ K With the values given in the problem, we do indeed find the inequality is satisfied. (b) Replacing our inequality with an equality and solving for M, we obtain M =m d3 + d 2 + 1 d1 2 = 114 kg 1.1102 kg. d 4 + d5 87. We choose directions such that the initial angular velocity is 0 = 317 rad/s and the values for , and F are positive. (a) Combining Eq. 10-12 with Eq. 10-45 and Table 10-2(f) (and using the fact that = 0) we arrive at the expression = FG 2 MR IJ FG - IJ = - 2 MR H5 K H t K 5 t 2 2 0 0 . With t = 15.5 s, R = 0.226 m and M = 1.65 kg, we obtain = 0.689 N m. (b) From Eq. 10-40, we find F = /R = 3.05 N. (c) Using again the expression found in part (a), but this time with R = 0.854 m, we get = 9.84 N m . (d) Now, F = / R = 11.5 N. 88. We choose positive coordinate directions so that each is accelerating positively, which will allow us to set abox = R (for simplicity, we denote this as a). Thus, we choose downhill positive for the m = 2.0 kg box and (as is conventional) counterclockwise for positive sense of wheel rotation. Applying Newton's second law to the box and (in the form of Eq. 10-45) to the wheel, respectively, we arrive at the following two equations (using as the incline angle 20, not as the angular displacement of the wheel). mg sin - T = ma TR = I Since the problem gives a = 2.0 m/s2, the first equation gives the tension T = m (g sin a) = 2.7 N. Plugging this and R = 0.20 m into the second equation (along with the fact that = a/R) we find the rotational inertia I = TR2/a = 0.054 kg m2. L 89. The center of mass is initially at height h = 2 sin 40 when the system is released (where L = 2.0 m). The corresponding potential energy Mgh (where M = 1.5 kg) becomes 1 rotational kinetic energy 2 I 2 as it passes the horizontal position (where I is the rotational inertia about the pin). Using Table 10-2 (e) and the parallel axis theorem, we find 1 I = 12 ML2 + M ( L / 2) 2 = 1 ML2 . 3 Therefore, Mg L 1 1 2 2 ML sin 40 = 2 2 3 = 3g sin 40 = 3.1 rad/s. L 90. (a) The particle at A has r = 0 with respect to the axis of rotation. The particle at B is r = L = 0.50 m from the axis; similarly for the particle directly above A in the figure. The particle diagonally opposite A is a distance r = 2 L = 0.71 m from the axis. Therefore, I= mi ri 2 = 2mL2 + m d 2 Li 2 = 0.20 kg m2 . (b) One imagines rotating the figure (about point A) clockwise by 90 and noting that the center of mass has fallen a distance equal to L as a result. If we let our reference position for gravitational potential be the height of the center of mass at the instant AB swings through vertical orientation, then K0 + U 0 = K + U 0 + ( 4m ) gh0 = K + 0. Since h0 = L = 0.50 m, we find K = 3.9 J. Then, using Eq. 10-34, we obtain K= 1 I A 2 2 = 6.3 rad/s. 91. (a) Eq. 10-12 leads to = - o / t = -25.0 / 20.0 = -125 rad / s2 . . 1 1 (b) Eq. 10-15 leads to = 2 o t = 2 (25.0)(20.0) = 250 rad. (c) Dividing the previous result by 2 we obtain = 39.8 rev. 92. The centripetal acceleration at a point P which is r away from the axis of rotation is given by Eq. 10-23: a = v 2 / r = 2 r , where v = r , with = 2000 rev/min 209.4 rad/s. (a) If points A and P are at a radial distance rA=1.50 m and r = 0.150 m from the axis, the difference in their acceleration is a = a A - a = 2 ( rA - r ) = (209.4 rad/s) 2 (1.50 m - 0.150 m) 5.92 104 m/s 2 (b) The slope is given by a / r = 2 = 4.39 104 / s 2 . 93. (a) With r = 0.780 m, the rotational inertia is I = Mr 2 = 130 kg 0.780 m = 0.791 kg m2 . . (b) The torque that must be applied to counteract the effect of the drag is b gb g 2 = rf = 0.780 m 2.30 10 -2 N = 179 10-2 N m. . b gc h 94. Let T be the tension on the rope. From Newton's second law, we have T - mg = ma T = m( g + a ) . Since the box has an upward acceleration a = 0.80 m/s2, the tension is given by T = (30 kg)(9.8 m/s 2 + 0.8 m/s 2 ) = 318 N. The rotation of the device is described by Fapp R - Tr = I = Ia / r . The moment of inertia can then be obtained as I= r ( Fapp R - Tr ) a = (0.20 m)[(140 N)(0.50 m) - (318 N)(0.20 m)] = 1.6 kg m 2 2 0.80 m/s 95. The motion consists of two stages. The first, the interval 0 t 20 s, consists of constant angular acceleration given by = 5.0 rad s 2 = 2.5 rad s . 2.0 s The second stage, 20 < t 40 s, consists of constant angular velocity = / t . Analyzing the first stage, we find 1 = t 2 t = 20 1 2 = 500 rad, = t t =20 = 50 rad s. Analyzing the second stage, we obtain 2 = 1 + t = 500 + ( 50 )( 20 ) = 1.5 103 rad. 96. Using Eq. 10-12, we have = 0 + t Using this value in Eq. 10-14 leads to = 2.6 - 8.0 = - 1.8 rad/s 2 . 3.0 = + 2 2 2 0 02 - 8.02 = 18 rad. = . 2 -18 b g 97. (a) Using Eq. 10-15 with = 0, we have = 0 + 2 t = 2.8 rad. (b) One ingredient in this calculation is = (0 3.5 rad/s)/(1.6 s) = 2.2 rad/s2, so that the tangential acceleration is r = 0.33 m/s2. Another ingredient is = 0 + t = 1.3 rad/s for t = 1.0 s, so that the radial (centripetal) acceleration is 2 r = 0.26 m/s2. Thus, the magnitude of the acceleration is a = 0.332 + 0.262 = 0.42 m s . 2 98. We make use of Table 10-2(e) as well as the parallel-axis theorem, Eq. 10-34, where needed. We use (as a subscript) to refer to the long rod and s to refer to the short rod. (a) The rotational inertia is I = Is + I = 1 1 ms L2 + m L2 = 0.019 kg m2 . s 12 3 (b) We note that the center of the short rod is a distance of h = 0.25 m from the axis. The rotational inertia is I = Is + I = 1 1 ms L2 + ms h 2 + m L2 s 12 12 which again yields I = 0.019 kg m2. 99. (a) One particle is on the axis, so r = 0 for it. For each of the others, the distance from the axis is r = (0.60 m) sin 60 = 0.52 m. Therefore, the rotational inertia is I = mi ri 2 = 0.27 kg m2 . (b) The two particles that are nearest the axis are each a distance of r = 0.30 m from it. The particle "opposite" from that side is a distance r = (0.60 m) sin 60 = 0.52 m from the axis. Thus, the rotational inertia is I= mi ri 2 = 0.22 kg m2 . 1 (c) The distance from the axis for each of the particles is r = 2 (0.60 m) sin 60 . Now, I = 3(0.50 kg)(0.26 m) 2 = 010 kg m2 . . 100. We use the rotational inertia formula for particles (or "point-masses"): I = mr2 (Eq. 10-33), being careful in each case to use the distances which are perpendicular to the axis of rotation. (a) Here we use the y values (for r) and get I = 3.4 105 g.cm2. (b) Now we use the x values (for r) and get I =2.9 105 g.cm2. (c) In this case, we use the Pythagorean theorem (r2 = x2 + y2) and get I = 6.3 105 g.cm2. ^ ^ (d) Eq. 9-8 yields (1.2 cm)i + (5.9 cm)j for the center of mass position. 101. We employ energy methods in this solution; thus, considerations of positive versus negative sense (regarding the rotation of the wheel) are not relevant. (a) The speed of the box is related to the angular speed of the wheel by v = R, so that Kbox = 1 mbox v 2 2 v= 2 Kbox = 141 m / s . mbox implies that the angular speed is = 1.41/0.20 = 0.71 rad/s. Thus, the kinetic energy of 1 rotation is 2 I 2 = 10.0 J. (b) Since it was released from rest at what we will consider to be the reference position for gravitational potential, then (with SI units understood) energy conservation requires K0 + U 0 = K + U Therefore, h = 16.0/58.8 = 0.27 m. 0 + 0 = ( 6.0 + 10.0 ) + mbox g ( -h ) . 102. We make use of Table 10-2(e) and the parallel-axis theorem in Eq. 10-36. (a) The moment of inertia is I= 1 1 mL2 + mh 2 = (2.0 kg)(3.0 m) 2 + (2.0 kg)(0.50 m) 2 = 2.0 kg m 2 . 12 12 The maximum angular speed is attained when the rod is in a vertical position with all its potential energy transformed into kinetic energy. Moving from horizontal to vertical position, the center of mass is lowered by h = 0.50 m. Thus, the change (decrease) in potential energy is U = mgh . The maximum angular speed can be obtained as 1 U = K rot = I 2 2 which yields 2U 2mgh 2gh 2(9.8 m/s 2 )(0.50 m) = = 2 2 = = 3.1 rad/s = I mh 2 + mL2 /12 h + L /12 (0.50 m)2 + (3.0 m) 2 /12 (b) The answer remains the same since is independent of the mass m. 103. Except for using the relation v = r (Eq. 10-18), this problem has already been analyzed in sample problem 6-9. Plugging v = r into Eq. 6-24, then, leads to 0 = s g (0.40)(9.8 m/s 2 ) = = 10.6 rad/s 11 rad/s. 0.035 m R 104. The distances from P to the particles are as follows: g r = b - a for m = M b topg r = a for m = 2 M blower right g r1 = a for m1 = 2 M lower left 2 2 2 3 2 1 b The rotational inertia of the system about P is I= 3 i =1 mi ri 2 = 3a 2 + b 2 M c h which yields I = 0.208 kg m2 for M = 0.40 kg, a = 0.30 m and b = 0.50 m. Applying Eq. 10-52, we find W= 1 2 1 2 I = 0.208 5.0 = 2.6 J. 2 2 b gb g 105. (a) Using Eq. 10-15, we have 60.0 rad = 2 (1 + 2)(6.00 s) . With 2 = 15.0 rad/s, then 1 = 5.00 rad/s. (b) Eq. 10-12 gives = (15.0 5.0)/6.00 = 1.67 rad/s2. (c) Interpreting now as 1 and as 1 = 10.0 rad (and o = 0) Eq. 10-14 leads to 1 o = 12 + 1 = 2.50 rad . 2 106. (a) The time for one revolution is the circumference of the orbit divided by the speed v of the Sun: T = 2R/v, where R is the radius of the orbit. We convert the radius: R = 2.3 104 ly 9.46 1012 km / ly = 2.18 1017 km c hc h where the ly km conversion can be found in Appendix D or figured "from basics" (knowing the speed of light). Therefore, we obtain T= 2 2.18 1017 km 250 km / s c h = 55 10 . 15 s. (b) The number of revolutions N is the total time t divided by the time T for one revolution; that is, N = t/T. We convert the total time from years to seconds and obtain c4.5 10 yhc3.16 10 N= 9 7 s/ y . 55 10 s 15 h = 26. 107. We assume the sense of initial rotation is positive. Then, with 0 > 0 and = 0 (since it stops at time t), our angular acceleration is negative-valued. (a) The angular acceleration is constant, so we can apply Eq. 10-12 ( = 0 + t). To obtain the requested units, we have t = 30/60 = 0.50 min. Thus, =- 33.33 rev/min = - 66.7 rev/min 2 - 67 rev/min 2 . 0.50 min (b) We use Eq. 10-13: = 0t + t 2 = (33.33) (0.50) + ( -66.7) (0.50) 2 = 8.3 rev. 1 2 1 2 108. (a) We use = I, where is the net torque acting on the shell, I is the rotational inertia of the shell, and is its angular acceleration. Therefore, I= 960 N m = = 155 kg m2 . 6.20 rad / s2 (b) The rotational inertia of the shell is given by I = (2/3) MR2 (see Table 10-2 of the text). This implies M= 3 155 kg m2 3I = = 64.4 kg. 2 2 R2 2 190 m . c h b g 109. (a) We integrate the angular acceleration (as a function of with respect to to find the angular velocity as a function of t > 0. = 0 + zc t 0 4a 3 - 3b 2 d = 0 + at 4 - bt 3 . h (b) We integrate the angular velocity (as a function of ) with respect to to find the angular position as a function of t > 0. = 0 + zc t 0 4a 3 - 3b 2 d = 0 + 0 t + h a 5 b 4 t - t . 5 4 110. (a) Eq. 10-6 leads to = (b) And Eq. 10-8 gives d at + bt 3 - ct 4 = a + 3bt 2 - 4ct 3 . dt c h = d a + 3bt 2 - 4ct 3 = 6bt - 12ct 2 . dt c h 111. Analyzing the forces tending to drag the M = 5124 kg stone down the oak beam, we find F = Mg sin + s cos b g where s = 0.22 (static friction is assumed to be at its maximum value) and the incline angle for the oak beam is sin -1 3.9 10 = 23 (but the incline angle for the spruce log is the complement of that). We note that the component of the weight of the workers (N of them) which is perpendicular to the spruce log is Nmg cos(90 ) = Nmg sin , where m = 85 kg. The corresponding torque is therefore Nmg sin where = 4.5 - 0.7 = 38 m . . This must (at least) equal the magnitude of torque due to F, so with r = 0.7 m, we have b g Mgr sin + s cos = Ngm sin . b g This expression yields N 17 for the number of workers. 112. In SI unit, the moment of inertia can be written as I = 14, 000 u pm 2 = (14, 000)(1.6 10-27 kg)(10-12 m) 2 = 2.24 10-47 kg m 2 . Thus, the rotational kinetic energy is given by 1 1 K rot = I 2 = (2.2 10 -47 kg m 2 )(4.3 1012 rad/s) 2 = 2.110-22 J. 2 2 113. Eq. 10-40 leads to = mgr = (70) (9.8) (0.20) = 1.4 102 N m. 114. (a) Eq. 10-15 gives 90 rev = 1 0 + 10 rev s 15 s 2 b gb g which leads to 0 = 2.0 rev/s. (b) From Eq. 10-12, the angular acceleration is = 10 rev s - 2.0 rev s 2 = 0.53 rev s . 15 s Using the equation again (with the same value for ) we seek a negative value of t (meaning an earlier time than that when 0 = 2.0 rev/s) such that = 0. Thus, t=- 0 2.0 rev s =- = -3.8 s 2 0.53 rev s which means that the wheel was at rest 3.8 s before the 15 s interval began. 115. Using Eq. 10-7 and Eq. 10-18, the average angular acceleration is avg = v 25 - 12 = = = 5.6 rad / s2 . t rt 0.75 2 6.2 b gb g 116. We make use of Table 10-2(e) and the parallel-axis theorem in Eq. 10-36. (a) The moment of inertia is I= 1 1 ML2 + Mh 2 = (3.0 kg)(4.0 m) 2 + (3.0 kg)(1.0 m) 2 = 7.0 kg m 2 . 12 12 (b) The rotational kinetic energy is 1 K rot = I 2 2 = 2K rot 2(20 J) = = 2.4 rad/s I 7 kg m 2 The linear speed of the end B is given by vB = rAB = (2.4 rad/s)(3.00 m) = 7.2 m/s , where rAB is the distance between A and B. (c) The maximum angle is attained when all the rotational kinetic energy is transformed into potential energy. Moving from the vertical position ( = 0) to the maximum angle , the center of mass is elevated by y = d AC (1 - cos ) , where dAC = 1.00 m is the distance between A and the center of mass of the rod. Thus, the change in potential energy is U = mg y = mgd AC (1 - cos ) 20 J = (3.0 kg)(9.8 m/s 2 )(1.0 m)(1 - cos ) which yields cos = 0.32 , or 71 . 117. (a) The linear speed at t = 15.0 s is v = at t = 0.500 m s2 15.0 s = 7.50 m s . The radial (centripetal) acceleration at that moment is 7.50 m s v2 ar = = 30.0 m r Thus, the net acceleration has magnitude: a = at2 + ar2 = . c0.500 m s h + c1875 m s h 2 2 2 2 d ib g g b 2 = 1.875m s2 . = 1.94 m s2 . (b) We note that at || v . Therefore, the angle between v and a is tan -1 FG a IJ = tan FG 1875IJ = 751 . H 0.5 K . Ha K r t -1 so that the vector is pointing more toward the center of the track than in the direction of motion. 118. (a) Using Eq. 10-1, the angular displacement is = 5.6 m = 14 102 rad . . 8.0 10 -2 m 1 (b) We use = 2 t 2 (Eq. 10-13) to obtain t: t= 2 = . 2 14 102 rad . 15 rad s 2 c h = 14 s . 119. We apply Eq. 10-12 twice, assuming the sense of rotation is positive. We have > 0 and < 0. Since the angular velocity at t = 1 min is 1 = (0.90)(250) = 225 rev/min, we have 1 = 0 + t a= 225 - 250 = -25 rev / min 2 . 1 Next, between t = 1 min and t = 2 min we have the interval t = 1 min. Consequently, the angular velocity at t = 2 min is 2 = 1 + t = 225 + ( -25) (1) = 200 rev / min . 120. (a) Using Table 10-2(c), the rotational inertia is 1 1 1.21 m I = mR 2 = (1210 kg) 2 2 2 (b) The rotational kinetic energy is, by Eq. 10-34, K= 1 2 1 I = (2.21 102 kg m 2 )[(1.52 rev/s)(2 rad/rev)]2 = 1.10 104 J. 2 2 FG H IJ K 2 = 221 kg m2 . 121. (a) We obtain = (33.33 rev / min) (2 rad/rev) = 3.5 rad/s. 60 s/min (b) Using Eq. 10-18, we have v = r = (15)(3.49) = 52 cm/s. (c) Similarly, when r = 7.4 cm we find v = r = 26 cm/s. The goal of this exercise is to observe what is and is not the same at different locations on a body in rotational motion ( is the same, v is not), as well as to emphasize the importance of radians when working with equations such as Eq. 10-18. 122. With v = 50(1000/3600) = 13.9 m/s, Eq. 10-18 leads to = v 13.9 = = 013 rad / s. . r 110 123. The translational kinetic energy of the molecule is Kt = 1 2 1 mv = (5.30 10 -26 ) (500) 2 = 6.63 10-21 J. 2 2 With I = 194 10-46 kg m2 , we employ Eq. 10-34: . Kr = 2 Kt 3 1 2 2 I = (6.63 10 -21 ) 2 3 which leads to =6.75 1012 rad/s. 124. (a) The angular speed associated with Earth's spin is = 2/T, where T = 86400s (one day). Thus = 2 = 7.3 10- 5 rad/s 86400 s and the angular acceleration required to accelerate the Earth from rest to in one day is = /T. The torque needed is then = I = where we used I (9.7 1027 ) (7.3 10- 5 ) = = 8.2 1028 N m T 86400 I= 2 2 M R 2 = 5.98 1024 6.37 106 5 5 c hc h 2 for Earth's rotational inertia. (b) Using the values from part (a), the kinetic energy of the Earth associated with its rotation about its own axis is K = 1 I 2 = 2.6 1029 J . This is how much energy would 2 need to be supplied to bring it (starting from rest) to the current angular speed. (c) The associated power is K 2.57 1029 J = 3.0 1024 W. P= = T 86400 s 125. The mass of the Earth is M = 5.98 1024 kg and the radius is R = 6.37 106 m. (a) Assuming the Earth to be a sphere of uniform density, its moment of inertia is 2 2 I = MR 2 = (5.98 10 24 kg)(6.37 106 m) 2 = 9.711037 kg m 2 . 5 5 (b) The angular speed of the Earth is = 2 2 2 = = = 7.27 10-5 rad/s T 24 hr 8.64 10 4 s Thus, its rotational kinetic energy is 1 1 K rot = I 2 = (9.711037 kg m 2 )(7.27 10-5 rad/s) 2 = 2.57 1029 J 2 2 (c) The amount of time the rotational energy could be supplied to at a rate of P = 1.0 kW = 1.0 103 J/s to a population of approximately N = 5.0 109 people is t = K rot 2.57 1029 J = = 5.14 1016 s 1.6 109 y 9 3 NP (5.0 10 )(1.0 10 J/s)

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Lehigh - PHYSIC - 2
1. (a) All the energy in the circuit resides in the capacitor when it has its maximum charge. The current is then zero. If Q is the maximum charge on the capacitor, then the total energy is2.90 10-6 C Q2 U= = = 117 10-6 J. . -6 2C 2 3.60 10 Fc
Lehigh - PHYSIC - 2
1. The vector area A and the electric field E are shown on the diagram below. The angle between them is 180 35 = 145, so the electric flux through the area is = E A = EA cos = (1800 N C ) 3.2 10-3 m cos145 = -1.5 10-2 N m 2 C.2()2. We u
Lehigh - PHYSIC - 2
1. (a) The amplitude is half the range of the displacement, or xm = 1.0 mm. (b) The maximum speed vm is related to the amplitude xm by vm = xm, where is the angular frequency. Since = 2f, where f is the frequency,vm = 2 fxm = 2 (120 Hz ) (1.0 10 -
Lehigh - PHYSIC - 2
1. The time it takes for a soldier in the rear end of the column to switch from the left to the right foot to stride forward is t = 1 min/120 = 1/120 min = 0.50 s. This is also the time for the sound of the music to reach from the musicians (who are
Lehigh - PHYSIC - 2
1. We take p3 to be 80 kPa for both thermometers. According to Fig. 18-6, the nitrogen thermometer gives 373.35 K for the boiling point of water. Use Eq. 18-5 to compute the pressure:pN =T 373.35 K p3 = 273.16 K 273.16 K(80 kPa) = 109.343kPa.
Lehigh - PHYSIC - 2
1. According to Eq. 39-4 En L 2. As a consequence, the new energy level E'n satisfiesEn L = En LFG IJ = FG L IJ H K H L K-22=1 , 2which gives L = 2 L. Thus, the ratio is L / L = 2 = 1.41.2. (a) The ground-state energy is( 6.63 10
Lehigh - PHYSIC - 2
1. Our calculation is similar to that shown in Sample Problem 42-1. We set K = 5.30 MeV=U = (1/ 4 0 )( q qCu / rmin ) and solve for the closest separation, rmin:rmin-19 9 q qCu kq qCu ( 2e )( 29 ) (1.60 10 C )( 8.99 10 V m/C ) = = = 4 0 K 4 0 K
Lehigh - PHYSIC - 2
1. The x and the y components of a vector a lying on the xy plane are given byax = a cos , a y = a sin where a =| a | is the magnitude and is the angle between a and the positive x axis. (a) The x component of a is given by ax = 7.3 cos 250 = 2
A.T. Still University - SDF - asdf
Economics 100b Professor Wood1/31/08 Lecture 4ASUC Lecture Notes Online is the only authorized note-taking service at UC Berkeley. Please do not share, copy or illegally distribute these notes. Our non-profit, student-run program depends on your i
Lehigh - PHYSIC - 2
1. Conservation of momentum requires that the gamma ray particles move in opposite directions with momenta of the same magnitude. Since the magnitude p of the momentum of a gamma ray particle is related to its energy by p = E/c, the particles have th
Lehigh - PHYSIC - 2
1. Each atom has a mass of m = M/NA, where M is the molar mass and NA is the Avogadro constant. The molar mass of arsenic is 74.9 g/mol or 74.9 103 kg/mol. 7.50 1024 arsenic atoms have a total mass of (7.50 1024) (74.9 103 kg/mol)/(6.02 1023 mol
Lehigh - PHYSIC - 2
1. From the time dilation equation t = t0 (where t0 is the proper time interval, = 1 / 1 - 2 , and = v/c), we obtain = 1-FG t IJ . H t K2 0The proper time interval is measured by a clock at rest relative to the muon. Specifically, t0 = 2.2
Lehigh - PHYSIC - 2
1. (a) The flux through the top is +(0.30 T)r2 where r = 0.020 m. The flux through the bottom is +0.70 mWb as given in the problem statement. Since the net flux must be zero then the flux through the sides must be negative and exactly cancel the tota
Lehigh - PHYSIC - 2
1. The air inside pushes outward with a force given by piA, where pi is the pressure inside the room and A is the area of the window. Similarly, the air on the outside pushes inward with a force given by poA, where po is the pressure outside. The mag
Lehigh - PHYSIC - 2
1 1. The potential energy stored by the spring is given by U = 2 kx 2 , where k is the spring constant and x is the displacement of the end of the spring from its position when the spring is in equilibrium. Thusk=2 25 J 2U = 2 x 0.075 mb g b g
Lehigh - PHYSIC - 2
1. If R is the fission rate, then the power output is P = RQ, where Q is the energy released in each fission event. Hence, R = P/Q = (1.0 W)/(200 106 eV)(1.60 10 19 J/eV) = 3.1 1010 fissions/s.2. We note that the sum of superscripts (mass number
Lehigh - PHYSIC - 2
1. (a) An Ampere is a Coulomb per second, so84 A h = 84FG HCh sIJ FG 3600 s IJ = 3.0 10 K H hK5C.(b) The change in potential energy is U = qV = (3.0 105 C)(12 V) = 3.6 106 J.2. The magnitude is U = eV = 1.2 109 eV = 1.2 GeV.3. T
Lehigh - PHYSIC - 2
1. In air, light travels at roughly c = 3.0 108 m/s. Therefore, for t = 1.0 ns, we have a distance of d = ct = (3.0 108 m / s) (1.0 10-9 s) = 0.30 m.2. (a) From Fig. 33-2 we find the smaller wavelength in question to be about 515 nm, (b) and the
Lehigh - PHYSIC - 2
1. We are only concerned with horizontal forces in this problem (gravity plays no direct role). We take East as the +x direction and North as +y. This calculation is efficiently implemented on a vector-capable calculator, using magnitude-angle notati
Lehigh - PHYSIC - 2
1. (a) Let E = 1240 eVnm/min = 0.6 eV to get = 2.1 103 nm = 2.1 m. (b) It is in the infrared region.2. The energy of a photon is given by E = hf, where h is the Planck constant and f is the frequency. The wavelength is related to the frequency b
Lehigh - PHYSIC - 2
1. Comparing the light speeds in sapphire and diamond, we obtainv = vs - vd = cFG 1 - 1 IJ Hn n K F 1 - 1 IJ = 4.55 10 m s. = c2.998 10 m sh G H 177 2.42 K .s d 8 72. (a) The frequency of yellow sodium light is c 2.998 108 m s f = = = 5.09
Lehigh - PHYSIC - 2
1. Charge flows until the potential difference across the capacitor is the same as the potential difference across the battery. The charge on the capacitor is then q = CV, and this is the same as the total charge that has passed through the battery.
Lehigh - PHYSIC - 2
1. (a) For a given value of the principal quantum number n, the orbital quantum number ranges from 0 to n 1. For n = 3, there are three possible values: 0, 1, and 2. (b) For a given value of , the magnetic quantum number m ranges from - to + . For =
Lehigh - PHYSIC - 2
1. With speed v = 11200 m/s, we findK= 1 2 1 mv = (2.9 105 ) (11200) 2 = 18 1013 J. . 2 22. (a) The change in kinetic energy for the meteorite would be1 1 K = K f - K i = - K i = - mi vi2 = - 4 106 kg 15 103 m/s 2 2()()2= -5 1014 J
Lehigh - PHYSIC - 2
Instructor's Manual forFUNDAMENTALS OF PHYSICSSeventh Edition by David Halliday, Robert Resnick, and Jearl WalkerPrepared byJ. Richard ChristmanProfessor Emeritus United States Coast Guard Academywith the assistance of Stanley A. Williams I
Lehigh - PHYSIC - 2
1. (a) The motion from maximum displacement to zero is one-fourth of a cycle so 0.170 s is one-fourth of a period. The period is T = 4(0.170 s) = 0.680 s. (b) The frequency is the reciprocal of the period:f = 1 1 = = 1.47 Hz. T 0.680 s(c) A sinuso
Lehigh - PHY - 2
1. (a) The magnitude of the magnetic field due to the current in the wire, at a point a distance r from the wire, is given byB= 0i2r.With r = 20 ft = 6.10 m, we havec4 10 B=hb 2 b6.10 mg-7T m A 100 Ag = 3.3 10-6T = 3.3 T.(b
Lehigh - PHYSIC - 2
Chapter 1:MEASUREMENT1. The SI standard of time is based on: A. the daily rotation of the earth B. the frequency of light emitted by Kr86 C. the yearly revolution of the earth about the sun D. a precision pendulum clock E. none of these Ans: E 2.
Lehigh - PHYSIC - 2
1. The image is 10 cm behind the mirror and you are 30 cm in front of the mirror. You must focus your eyes for a distance of 10 cm + 30 cm = 40 cm.2. The bird is a distance d2 in front of the mirror; the plane of its image is that same distance d2
Lehigh - PHYSIC - 2
1. The magnitude of the force of one particle on the other is given by F = Gm1m2/r2, where m1 and m2 are the masses, r is their separation, and G is the universal gravitational constant. We solve for r:Gm1m2 = r= F( 6.67 10-11N m 2 / kg 2 ) (
Lehigh - PHYSIC - 2
1. (a) The charge that passes through any cross section is the product of the current and time. Since 4.0 min = (4.0 min)(60 s/min) = 240 s, q = it = (5.0 A)(240 s) = 1.2 103 C. (b) The number of electrons N is given by q = Ne, where e is the magnitu
Cornell - HADM - 2225
HADM225 Practice FinalName:_1. A stock has an expected rate of return of 8.3 percent and a standard deviation of 6.4 percent. Which one of the following best describes the probability that this stock will lose 11 percent or more in any one given y
Cal Poly Pomona - HIST - 101
Seung Yeo Per:2 Chapter 3 VocabsExercise 1 1. C 2. C 3. A 4. D 5. B 6. D 7. B 8. C 9. B 10. D Exercise 2 1. Delightful 2. Arousing pity 3. Plight 4. Blandishment 5. Careworn 6. Self-denying 7. Perfect happiness 8. Weakened 9. Slavish Follower 10. Ho
USC - MUHL - 232
Ch. 59 Music and Ballet in 19th Cent. Russia No. 161 Modest Mussorgsky, &quot;with in four walls&quot; from Sunless 1874 Genre: Russian song cycle - 1st song of song cycle - chords hover over D major triad; D is constant - m2-4 contrast in textures between clo
USC - MUHL - 232
Terms Ballets Russes Russian Ballet; established by Sergei Diaghilev in 1909; toured in Europe and Americas. &quot;Beethoven's 10th&quot; Brahms 3rd symphony illustrates his modern extension of Beethoven symphonic style. Ex of this are non tradition tonal pl
USC - GEOG - 100gm
HOLY LAND (Pg. 61, # 116) I cannot tell you what I care for, I can only tell you what I fear to lose. This quote embodies what Lakewood is all about, and how the people there lived their lives, and why they would even want to go there in the first pl
Cal Poly Pomona - ENG - 1B
Seung Yeo Per:2 English IB Mr.Costa CeremonyThe struggles engulfed in Ceremony had plenty to do with before, during, and after the war. It not only captured the people who were physically involved in the war but also, among others around them. Even
USC - MUHL - 232
Modest Mussorsgsky Russia. The mighty handful: folksong were more imp than anything that they learned in school. was totally Russian Peter Ilyich Tchaikovsky Born in Russia. Western influenced. Did not exemplify Russian nationalism. Bedrich Smetana C
USC - MUHL - 232
Ch.55 German Opera of the 19th Century: Weber &amp; Wagner Singspiel play w/ singing; far more modest Carl Maria Von Weber 1786-1826 Friderich kind was librettist for Der Freischutz Romantic opera term used by Weber for certain of his operas and by W
Cal Poly Pomona - COM - 413
Man Kiu Chu COM 413 Mariusz Ozminkowski, Ph.D. 2:06 p.m. 3:40pm 10/10/07 &quot;Though Letter&quot;At each naturalization ceremony, every newly &quot;naturalized&quot; U.S. citizens have to declare if they are Democratic or Republican, etc. So it is very hard to be un
USC - POSC - 130g
Question 1: Is Galanter wrong? Galanter is right 1. same argument that Galanter made, RP's have relationships with judges, know the system, have more resources 2. you can't reform A-L, due to capture, the distrust of agencies, bargaining in the shado
Cal Poly Pomona - COM - 106
COM 106 Writing for Communication PractitionersEssay 1: Topic 50 pointsWe sometimes say or think: &quot;There needs to be a change.&quot; About the 2008 presidential elections, we think: there needs to be a change in the way people vote. There needs to be a
USC - POSC - 130g
Nation of strangers o ID: idea that America is diverse and not close-knit o SIG: need for more A-L in America since we sue others more (not neighborneighbor); comparison of American courts to other nations with smaller, more homogeneous populations w
USC - BUAD - 304
CHAPTER 14 POWER AND POLITICS Formal power based on indiv position in org 1. coercive power dependent on fear; one reacts out of fear of the negative results that might occur if one failed to comply 2. reward power opposite of coercive; people co
USC - BUAD - 304
Leader as Sense Maker Biographical Features Age: older turnover less than younger, older are absent less Gender: women don't turnover/perform/etc. less than men; women absent more than men Tenure: the greater the tenure the more productive, happier,
USC - POSC - 130g
Essay Questions 1. When are courts most likely to be constrained/dynamic? CCV: Definition: courts are limited in ability to make a difference and thus &quot;constrained&quot; When: Basically when all forms of constraints are strong: Strong doctrinal constraint
USC - POSC - 130g
Part I: Identify and give significance of 3 of 5 terms and concepts, which will be selected from the following list: The logic of the triad o ID: Triadic relationship between the Courts and Litigants 1 and 2, propped up by consent (agreement by sides
Cal Poly Pomona - COM - 413
Man Kiu Chu COM 413 Maurisz Ozminkowski, Ph.D. 2:06 p.m. 3:40 p.m. 10/07/07 Propaganda Example: MUFON (Mutual UFO Network) The propaganda example I am using is the Mutual UFO Network of United States. From reporting of the latest daily sightings to
Cal Poly Pomona - COM - 413
PUBLIC OPINION Definitional preliminaries Measurements Formation Ideological differences in the U.S.What constitutes a &quot;public&quot;? a group of people who are confronted by an issue who are divided in the ideas as how to meet the issue who
Cal Poly Pomona - COM - 260
Com260:Access to Places and InformationPower Point Presentations for Cal Poly Pomona COM 260 Based on J.D. Zelezny's &quot;Communications Law&quot; Copyright 2006 by Mariusz Ozminkowski For permission to use email mozminkowski@csupomona.eduThe First Amen
Binghamton - ENG - 117D
1The novel One flew Over The Cuckoo's Nest focuses on ideas of conformity and power distribution. The best and easiest way to achieve these two things in society is through Jeremy Bentham's idea and strategies involved with the Panopticon. Characte
Cal Poly Pomona - HIST - 370
In terms of their work force, American farmers and ranchers a. continued to use Indian laborers to work their lands b. hired Chinese immigrants d. relied on black workersThe Big Four included all of the following except a. Leland Stanford b. Collis
Binghamton - ENG - 117G
The Big Sleep's SignificanceThe Big Sleep by Raymond Chandler contains many symbols and themes that are socially, culturally, and historically significant in real life. Much of the novel is focused around the demand for money and the countless acti
Cal Poly Pomona - HIST - 370
What event greatly contributed to California's increasing importance in world trade? the opening of the Suez Canal the opening of the Panama Canal the international exposition held in Los Angeles in 1915 all of the above Which of the following statem
USC - ANTH - 200Lg
ANTH Exam 1 ReviewEvolution9/26/2007 8:56:00 PMGradual change over time Change in gene allele frequencies over generations Framework used to explain the origin of adaptations Mechanisms: o Mutation o Gene flow o Genetic drift o Natural selection
Cal Poly Pomona - HIST - 370
What has been the result of extensive water development in California? a. It has augmented the power of both state and local governments. b. It has tendedto favor large-scale commercial farming over small family farms. c. It has intensified political
Cal Poly Pomona - HIST - 370
The foreigners who arrived in California in the late 1840s could best be described as a. people who came mostly in families or as part of communities and, as aresult, were unwilling to assimilate into Californio society b. mainly single men who often
Binghamton - ENG - 117G
Real Life Significance of &quot;The Birds&quot; In the short story ,The Birds&quot; by Daphne du Maurier, there are many symbols and events which closely resemble society in Britain during this time period. Many aspects of the story are socially, politically, and t
Binghamton - ENG - 117D
Foucault begins his story by describing the seventeenth century plaque. He tells how houses were closed off, there was separation, inspection, and how things were quarantined. From this, concepts of power and discipline were created. Since that time,
USC - HIST - 200gm
American Midterm ReviewPart I Essay Question 1: Please characterize the changes in American politics and the American economy from the beginning of the 19th century through the 1828 election of Andrew Jackson. What were these changes, and how did t
USC - HIST - 200gm
History midterm PART I Essay #1 Major changes within political party system with both the rise of new parties and division amongst them. 1800 Thomas Jefferson elected president Jefferson and Burr tied in the electoral college so the vote went to the