f5ch11 - 296 CHAPTER 11 ROTATIONAL MOTION CHAPTER 11 Answer...

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Unformatted text preview: 296 CHAPTER 11 ROTATIONAL MOTION CHAPTER 11 Answer to Checkpoint Questions 1. 2. 3. 4. 5. 6. 7. (b) and (c) (a) and (d) (a) yes; (b) no; (c) yes; (d) yes all tie 1, 2, 4, 3 (a) 1 and 3 tie, 4, then 2 and 5 tie (zero) (a) downward in the gure; (b) less Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. (a) positive; (b) zero; (c) negative; (d) negative (a) clockwise; (b) counterclockwise; (c) yes; (d) positive; (e) constant (a) 2 and 3; (b) 1 and 3; (c) 4 all tie (a) and (c) (a) c, a, then b and d tie; (b) b, then a and c tie, then d (a) all tie; (b) 2, 3, and then 1 and 4 tie less b, c, a (a) hoop (mass at greatest distance from rotational axis); (b) prism (mass at least distance) less larger 90 , then 70 and 110 tie all tie CHAPTER 11 ROTATIONAL MOTION 297 15. nite angular displacements are not commutative Solutions to Exercises & Problems (a) The angle is = 1:8 m=1:2 m = 1:50 rad: (b) = (1:5 rad)(180 = rad) = 85:9 : (c) The arc length is l = r = (0:62 rad)(2:40 m) = 1:49 m. (a) The angular velocity is 2E 1E d ! = d = dt (at + bt3 ct4 ) = a + 3bt2 4ct3 : dt (b) The angular acceleration is d = d! = dt (a + 3bt2 4ct3 ) = 6bt 12ct2 : dt Use ! = 2=T . (a) For the second hand T = 60 s so ! = 2=60 s = 0:105 rad/s: (b) For the minute hand T = 60 min = 3600 s so ! = 2=3600 s = 1:75 10 3 rad/s: (c) For the hour hand T = 24 h so ! = 2=[(24 h)(3600 s= h)] = 1:45 10 4 rad/s: (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. Since R = 2:3 104 ly = (2:3 104 ly)(9:460 1012 km/ly) = 2:18 1017 km, 18 T = 2(2:250 10 km) = 5:5 1015 s : km/s 17 4E 3E (b) The number of revolutions is the total time t divided by the time T for one revolution: N = t=T . Convert the total time from years to seconds. The result for the number of revolutions is 9 y)(3:16 7 N = (4:5 105:5 1015 10 s/y) = 26 : s 298 CHAPTER 11 ROTATIONAL MOTION (a) The angular velocity is given by 5E At t = 2:0 s and at t = 4:0 s d ! = d = dt (4:0t 3:0t2 + t3 ) = 4:0 6:0t + 3t2 : dt !(2:0 s) = 4:0 6:0(2:0) + 3(2:0)2 = 4:0 rad/s ; !(4:0 s) = 4:0 6:0(4:0) + 3(4:0)2 = 28 rad/s : = ! = 28 rad/ss 4::0 rad/s = 12 rad/s2 : t 4:0 2 0 s (b) The average angular acceleration is (c) The angular acceleration is At t = 2:0 s and at t = 4:0 s d = d! = dt (4:0 6:0t + 3t2 ) = 6:0 + 6t : dt (2:0 s) = 6:0 + 6(2:0) = 6:0 rad/s2 ; (4:0 s) = 6:0 + 6(4:0) = 18 rad/s2 : (a) Evaluate (t) = 2 + 4t2 + 2t3 for t = 0 to obtain (0) = 2 rad. (b) The angular velocity (in rad/s) is given by !(t) = d=dt = 8t + 6t2 . Evaluate this expression for t = 0 to obtain !(0) = 0. (c) For t = 4:0 s, ! = 8(4:0) + 6(4:0)2 = 130 rad/s. (d) The angular acceleration (in rad/s2 ) is given by = d!=dt = 8 + 12t. For t = 2:0 s, = 8 + 12(2:0) = 32 rad/s2 . (e) The angular acceleration, given by = 8 + 12t, depends on the time and so is not constant. (a) The angular speed is given by 7P 6E !(t) = !0 + dt = !0 + 0 = !0 + at4 bt3 : Z t Z t 0 (4at3 3bt2 )dt CHAPTER 11 ROTATIONAL MOTION 299 (b) The angular displacement is (t) = 0 + Z t 1 = 0 + !0 t + 1 at5 4 bt4 ; 5 where 0 is the initial angular displacement at t=0. The number of spins is (60 ft)(1800 rev/min)(1 n = (85 mi/h)(5279 ft/mi)(1min=60 s) = 14 rev : h=3600 s) 1 The time t during which the diver makes 2.5 revolutions is given by 2 g(t)2 = h, where h = 10 m. Thus the average angular speed of the diver is ! = p :5 rev)(2 rad/rev) = 11 rad/s : (2 2(10 m)=(9:80 m/s2 ) 10P 9P 8P 0 !(t)dt = 0 + Z t 0 (!0 + at4 bt3 )dt (a) To avoid touching the spokes, the arrow must go through the wheel in not more than = t = (15 8) rev = 0:050 s : 2: rev/s The minimum speed of the arrow is then vmin = 20 cm=0:050 s = 400 cm/s = 4:0 m/s: (b) No (as is apparent from the calculation above). (a) 11E (b) The number of revolutions in t = 12 s = 0:20 min is n = !0 t + 1 t2 = (1200 rev/min)(0:20 min) + 1 (1200 rev/min2 )(0:20 min)2 2 2 2 rev : = 4:2 10 1200 = ! = 3000 rev/min min=60rev/min = 9000 rev/min2 : t (12 s)(1 s) 300 CHAPTER 11 ROTATIONAL MOTION (a) For constant angular acceleration ! = !0 + t, so = (! !0 )=t. Take ! = 0 and to obtain the units requested use t = (30 s)=(60 s/min) = 0:50 min. Then 33 rev/min = 33:0:50 min = 67 rev/min2 : 12E 1 1 (b) Use = !0 t + 2 t2 = (33:33 rev/min)(0:50 min) + 2 ( 66:7 rev/min2 )(0:50 min)2 = 8:3 rev. (a) The time t is given by t = != = (120 rad/s)=(4:0 ras/s2 ) = 30 s: 1 (b) The angle is = 1 t2 = 2 (4:0 rad/s2 )(30 s)2 = 1:8 103 rad: 2 (a) = 5:6 m=(8:0 10 2 m) = 1:4 102 rad: (b) Use = 1 t2 to obtain t: 2 14E 13E t= r 2 = s 2(1:4 102 rad) = 14 s : 1:5 rad/s2 The angular acceleration of the wheel is = ! = (0:90 1:00)(250 rev/min)=1:0 min = 25 rev/min2 ; t so the angular speed at the end of the second minute is 15E ! = (0:90)(250 rev/min) (25 rev/min2 )(1:0 min) = 200 rev/min : 16E (a) = !=t = 25:0 rad/s2 =20:0 s = 1:25 rad/s2 : (b) = !0 t + 1 t2 = (25:0 rad/s2 )(20:0 s) + ( 1:25 rad/s2 )(20:0 s)2 =2 = 250 rad: 2 (c) The number of revolutions is n = =2 rad = 250 rad=2 rad = 39:8 rev: (a) The angular acceleration satis es 25 rad = 1 (5:0 s)2 ; which gives = 2:0 rad/s2 : 2 17E CHAPTER 11 ROTATIONAL MOTION 301 (b) The average angular velocity is ! = = 25:0 s = 5:0 rad/s : t 5 rad (c) The instantaneous angular velocity is ! = (2:0 rad/s2 )(5:0 s) = 10 rad/s : (d) The additional angle turned is = (10:0 s) (5:0 s) = 1 (2:0 rad/s2 )[(10:0 s)2 (5:0 s)2 ] = 75 rad : 2 The angular acceleration during the rst 20 s is = (5:0 rad/s)=2:0 s = 2:5 rad/s2 : The angular displacement at t = 20 s is then 1 = 1 t2 = 1 (2:5 rad/s2 )(20 s)2 = 500 rad: From 2 2 t = 20 s to t = 40 s, the angular speed is ! = (2:5 rad/s2 )(20 s) = 50 rad/s: The angular displacement from t = 20 s to t = 40 s is then 2 = (50 rad/s)(40 s 20 s) = 1000 rad. The total angle turned is thus = 1 + 2 = 500 rad + 1000 rad = 1500 rad. (a) Suppose that the wheel has already been turning for a time t. Then its initial angular speed at the beginning of the t = 3:00 s interval is !i = t, where = 2:00 rad/s2 . For the interval = 90:0 rad = !i t + 1 (t)2 = tt + 1 (t)2 ; 2 2 so 1 90:0 t = 1 t = (2:00 rad/srad :00 s) 3:00 s = 13:5 s : 2 )(3 t 2 2 (b) !i = t = (2:00 rad/s2 )(13:5 s) = 27:0 rad/s: 20P 19P 18P Now use ! = !0 + t to nd the time when the wheel was at rest (! = 0): Take t = 0 at the start of the interval. Then at the end of the interval t = 4:0 s, and the angle of rotation is = !0 t + 1 t2 . Solve for !0 : 2 1 (3:0 rad/s2 )(4:0 s)2 1 t2 120 rad 2 = 24 rad/s : !0 = t2 = 4:0 s t = !0 = 24 rad/s2 = 8:0 s : 3:0 rad/s 302 CHAPTER 11 ROTATIONAL MOTION That is, the wheel started from rest 8:0 s before the start of the 4:0 s interval. (a) Use = !t = 1 !0 t. The time required is 2 2 t = ! = 2(40 rev)(2 rad/rev) = 3:4 102 s : 1:5 rad/s 0 21P (b) = !0 =t = ( 1:5 rad/s)=(3:4 102 s) = 4:5 10 3 rad/s2 : 2 (c) Let the angular speed of the wheel be ! as it nishes the rst 20 turns. Then !2 !0 = p 2 2(=2) = , or ! = !0 + . Thus the time required is t0 = ! 2 !0 = !0 + !0 p (1:5 rad/s)2 + ( 4:5 10 3 rad/s2 )(40 rev)(2 rad/rev) 1:5 rad/s = 4:5 10 3 rad/s2 = 98 s : p (a) The angle through which the line will turn as a function of time t is (t) = 0 + !0 t + 1 t2 : To maximize (t), set d=dt=0: 2 22P d = d ( + ! t + 1 t2 ) = ! + t = 0 ; 0 dt t=t0 dt 0 0 2 t=t0 0 which gives t0 = !0 =. Thus !0 max = (t0 ) = 0 2 : rad/s) = 2((40725 rad/s2 ) = 44 rad : : 2 !0 + 1 2 2 2 !0 = 2 (b) Let b = max =2 = !0 tb + 1 t2 and solve for tb . Use max = 44 rad, !0 = 4:7 rad/s, and 2 b = 0:25 rad/s2 . You should get tb = 5:5 s or 32 s. (c) Let 0 = 10:5 rad = !0 tc + 1 t2 and solve for tc . You should get tc = 2:1 s or 40 s. 2 c CHAPTER 11 ROTATIONAL MOTION 303 (d) 44rad 22rad -2.1s 5.5s -10.5rad 32s t 40s 2 (a) Use !f !i2 = 2 to calculate : 2 (10 2 !2 !2 = f 2 i = (15 rev/s) rev) rev/s) = 1:0 rev/s2 : 2(60 23P (b) Use = !t = [(!i + !f )=2]t. The time required is 2(60 t = ! 2 ! = 15 rev/s +rev)rev/s = 4:8 s : 10 i+ f (c) The time required is (c) The number of revolutions is t0 = ! = 1:10 rev/s 2 = 9:6 s : 04 rev/s 0 1 n = 1 t02 = 2 (1:0 rev/s2 )(9:6 s)2 = 48 rev : 2 24P (a) Let the initial angular speeds be !i at the beginning of the 15-s period. Then 90 rev = 1 (!i + 10 rev/s)(15 s); which gives !i = 2:0 rev/s: 2 (b) The angular acceleration is = (10 rev/s 2:0 rev/s)=15 s = 0:53 rev/s2 : The time elapsed is then t = !i = = (2:0 rev/s)(0:53 rev/s2 ) = 3:8 s: 304 CHAPTER 11 ROTATIONAL MOTION The magnitude of the acceleration is given by a = !2 r, where r is the distance from the center of rotation and ! is the angular velocity. You must convert the given angular velocity to rad/s: ! = (33:33 rev/min)(2 rad/rev)=(60 s/min) = 3:49 rad/s. Thus 25E a = r!2 = (0:15 m)(3:49 rad/s2 )2 = 1:8 m/s2 : The acceleration vector is toward the center of the record. (a) ! = [(33 + 1=3) rev/min](2 rad/rev)(1:0 min=60 s) = 3:5 rad/s: (b) v1 = r1 = (3:5 rad/s)(5:9 in.) = 21 in./s: (c) v2 = r2 = (3:5 rad/s)(2:9 in.) = 10 in./s: 27E 26E Use v = r!. First convert 50 km/h to m/s: (50 km/h)(1000 m= km)=(3600 s=h) = 13:9 m/s. Then ! = v=r = (13:9 m/s)=(110 m) = 0:13 rad/s. 28E (a) ! = (200 rev/min)(2 rad/rev)(1:00 min=60 s) = 20:9 rad/s: (b) v = !r = (20:9 rad/s)(1:20 m=2) = 12:5 m/s: (c) Use = !=t to obtain : 200 rev/min = ! = 1000 rev/min min=60 s) = 800 rev/min2 : t (60 s)(1:00 2 (d) Solve from !f !2 = 2: 2 (200 2 !2 !2 = f 2 = (1000 rev/min)rev/min2rev/min) = 600 rev : 2(800 ) The average angular acceleration is 25 m/s = ! = rvt = (0:75 m)(612 m/s = 5:6 rad/s2 : t :2 s=2) 29E CHAPTER 11 ROTATIONAL MOTION 305 (a) 30E (b) v = re s = (2:0 10 7 rad/s)(1:49 1011 m) = 3:0 104 m/s: (c) a = !2 r = (2:0 10 7 rad/s)2 (1:49 1011 m) = 5:9 10 3 m/s2 : The direction of a is toward the sun. (a) The longitudinal separation between Helsinki and the explosion site is = 102 25 = 77 . Since a time dierence of T = 1 da = 24 h corresponds to a longitudinal dierence of 360 , the asteroid would have hit Helsinki should it have arrived a time t later, where 77 t = 360 T = 360 (24 h) = 5:1 h : 31E 2 ! = 2 = (1:0 yr)(365)(24)(3600 s)=1:0 yr = 2:0 10 7 rad/s: T (b) Now = 102 ( 20 ) = 122 so the required time delay would be 122 t = 360 T = 360 (24 h) = 8:1 h : (a) The angular velocity at t = 5:0 s is d = dt (0:30t2 ) = 2(0:30)(5:0 s) = 3:0 rad/s : ! = d dt t=5:0 s t=5:0 s 32E (b) The linear speed at t = 5:0 s is v = !r = (3:0 rad/s)(10 m) = 30 m/s: (c) The tangential acceleration at t = 5:0 s is 2 d2 = (10 m)(2)(0:30) = 6:0 m/s2 : at = dv = r = r d 2 = r dt2 (0:30t2 ) dt dt t=5:0 s (d) The radial (centripetal) acceleration is ac = !2 r = (3:0 rad/s)2 (10 m) = 90 m/s2 . (a) ! = v=r = (2:90 104 km/h)(1:00 h=3600 s)=(3:22 103 km) = 2:50 10 3 rad/s: (b) ar = !2 r = (2:50 10 3 rad/s)2 (3:22 106 m) = 20:2 m/s2 : (b) at = dv=dt = 0. 33E 306 CHAPTER 11 ROTATIONAL MOTION (a) As the angular speed is increased to !0 , the force of friction f , which provides the centripetal force for the coin (of mass M ), has reached its maximum value: f = fmax = p 2 s Mg = M!0 R. This gives !0 = s g=R: (b) 34E coin flies off R original circular path with radius R and angular speed 0 (a) The angular acceleration is 0 150 = ! = (2:2 h)(60rev/min = 1:1 rev/min2 : t min=1 h) (b) The number of rotations is 35P n = !0 t + 1 t2 2 1 = (150 rev/min)(2:2 h)(60 min=1 h) + 2 ( 1:13 rev/min2 )[(2:2 h)(60 min=1 h)]2 = 9:9 103 rev : (c) The tangential linear acceleration is at = r = ( 1:1 rev/min2 )(2 rad/rev)(1 min2 =3600 s2 )(50 cm) = 0:99 mm/s2 : (d) The centripetal acceleration is ac = !2 r = [(75 rev/min)(2 rad/rev)(1 min=60 s)]2 (50 cm) = 31 m/s2 at : The net acceleration is then a = a2 + a2 ' ac = 31 m/s2 : t c p CHAPTER 11 ROTATIONAL MOTION 307 (a) at = r = (14:2 rad/s2 )(2:83 cm) = 40:2 cm/s2 : (b) a = !2 r = [(2760 rev/min)(2 rad/rev)(1:00 min=60 s)]2 (2:83 cm) = 2:36 103 m/s2 : (c) Use v2 = (!r)2 = 2as to compute s: 36P v2 = [(2760 rev/min)(2 rad/rev)(1:00 min=60 s)]2 = 83:2 m : s = 2a 2(2:36 103 m/s2 ) 37P (a) The speed as seen by the pilot is v1 = !r = (200 rev/min)(2 rad/rev)(1 min=60 s)(1:5 m) = 3:1 102 m/s : (b) The speed as seen by an observer on the ground is 2 v = (480 km/h)2 + v1 = [(480 km/h)(103 m/km)(1 h=3600 s)]2 + (3:1 102 m/s)2 = 3:4 102 m/s : q p 38P (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 10 2 rad. That time is 2(500 m) t = 2c` = 2:998 108 m/s = 3:34 10 6 s so the angular velocity of the wheel is 10 2 rad ! = = 1:326 10 6 s = 3:8 103 rad/s : t :34 (b) If r is the radius of the wheel the linear speed of a point on its rim is v = r! = (0:05 m)(3:8 103 rad/s) = 190 m/s : 39P (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 centerpetal acceleration is then ac = v2 =r where r = 30:0 m, and the net acceleration is 308 CHAPTER 11 s s ROTATIONAL MOTION 2 2 + a2 = a2 + v a = at c t r p 2 2 2 )2 + (7:50 m/s) = (0:500 m/s 30:0 m 2 = 1:94 m/s2 : (b) Refer to the gure to the right. Note that at k v: The angle between v and a is = cos at a 2 1 0:500 m/s = cos 1:94 m/s2 = 75:1 : 1 40P ac at a v (a) The Earth makes one rotation per day and 1 d = (24 h)(3600 s/h) = 8:64 104 s, so the angular velocity of the Earth is (2 rad)=(8:64 104 s) = 7:27 10 5 rad/s. (b) Use v = r!, where r is the radius of its orbit. A point on the Earth at a latitude of 40 goes around a circle of radius r = R cos 40 , where R is the radius of the Earth (6:37 106 m). So its speed is v = (R cos 40 )! = (6:37 106 m)(cos 40 )(7:27 10 5 rad/s) = 3:5 102 m/s. (c) At the equator the value of ! is the same but the latitude is 0 and the speed is v = R! = (6:37 106 m)(7:27 10 5 rad/s) = 463 m/s. 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 rA A = rC C , where A is the angular acceleration of wheel A and C is the angular acceleration of wheel C . Thus 41P r 10 cm C = rA A = 25 cm (1:6 rad/s2 ) = 0:64 rad/s2 : C Since the angular velocity of wheel C is given by !C = C t, the time for it to reach an angular velocity of 100 rev/min (= 10:5 rad/s) is ! t = C = 10:5 rad/s2 = 16 s : C 0:64 rad/s CHAPTER 11 ROTATIONAL MOTION 309 (a) The linear speed of a point on belt 1 is 42P v1 = !A rA = (10 rad/s)(15 cm) = 1:5 102 cm/s : (b) The angular speed of B is 2 v !B = r 1 = 1:5 10 cm/s = 15 rad/s : 10 cm B (c) The angular speed of B 0 is !B = !B = 15 rad/s: (d) The linear speed of a point on belt 2 is 0 v2 = !B rB = (15 rad/s)(5 cm) = 75 cm/s : 0 0 (e) The angular speed of C is v !C = r 2 = 75 cm/s = 3:0 rad/s : 25 cm C 43P (a) a = !2 r = [(33 1 rev/min)(2 rad/rev)(1:0 min=60 s)]2 (6:0 10 2 m) = 0:73 m/s2 : 3 (b) Use ma = fs fs; max = s mg to nd s; min : 0 73 m/s2 s; min = a = 9::80 m/s2 = 0:075 : g 2 (c) The radial acceleration of the object is ar = !pr, while the tangential acceleration is p at = dv=dt = rd!=dt = r. Thus a = a2 + a2 = (!2 r)2 + (r)2 : If the object is not to t r slip at any time, we must let 2 fs;max = s mg mamax = m (!max r)2 + (r)2 ; p where !max = 33 1 rev = 3:49 rad/s: Thus 3 s;min = = p (3:49 rad/s)4 + [(3:49 rad/s)=0:25 s]2 (6:0 10 2 m) 9:80 m/s2 = 0:11 : p 2 (!max r)2 + (r)2 = g p 2 (!max r)2 + (!max r=t)2 g 310 CHAPTER 11 ROTATIONAL MOTION 44P (a) The angular acceleration is given by 2 = d! = T dT : 2 dt dt For the pulsar described dT = 1:26 10 5 s/y = 4:00 10 dt 3:16 107 s/y so 13 ; 2 = (0:033 s)2 (4:00 10 13 ) = 2:3 10 9 rad/s2 : The minus sign indicates that the angular acceleration is opposite the angular velocity and the pulsar is slowing down. (b) Solve ! = !0 + t for the time t when ! = 0: 2 2 t = !0 = T = = 8:3 1010 s : 9 rad/s2 )(0:033 s) ( 2:3 10 This is about 2600 years. (c) The pulsar was born 1996 1054 = 942 years ago. This is equivalent to (942 y)(3:16 107 s/y) = 2:98 1010 s. It angular velocity was then 2 ! = !0 + t = 2 + t = 0:033 s + ( 2:3 10 9 rad/s2 )( 2:98 1010 s) = 2:6 102 rad/s : T Its period was 2 T = 2! = 2:6 102 rad/s = 2:4 10 2 s : 45E The kinetic energy is given by K = 1 I!2 , where I is the rotational inertia and ! is the 2 angular velocity. Here ! = (602 rev/min)(2 rad=rev) = 63:0 rad/s : 60 s/min Thus 2(24400 J) I = 2K = (63:0 rad/s)2 = 12:3 kg m2 : 2 ! CHAPTER 11 ROTATIONAL MOTION 311 Let the linear and angular speeds of the oxygen molecule (of mass M ) be v and !. Then 1 Mv 2 = 3 I! 2 , which gives 2 2 26 kg) ! = v M = (500 m/s) 3(1(5:30 1046 kg m2 ) = 6:75 1012 rad/s : 3I :94 10 r s 46P 47E Since the rotational inertia of a cylinder of mass M and radius R is I = 1 MR2 , the kinetic 2 energy of a cylinder when it rotates with angular velocity ! is K = 1 I!2 = 1 MR2 !2 . 2 4 For the rst cylinder K = 1 (1:25 kg)(0:25 m)2 (235 rad/s)2 = 1:1 103 J. For the second 4 K = 1 (1:25 kg)(0:75 m)2 (235 rad/s)2 = 9:7 103 J. 4 (a) The rotational inertia in kg m2 is 48E I = (14000 u pm2 )(1:66 10 27 kg/u)(10 24 m2 = pm2 ) = 2:3 10 47 kg m2 : (b) The rotatitional kinetic energy in eV is 1 K = 1 I!2 = 2 (2:3 10 47 kg m2 )(4:3 1012 rad/s)2 (1 eV=1:6 10 2 = 1:3 10 3 eV = 1:3 meV: 49E 19 J) (a) The rotational inertia is 1 I = 1 mR2 = 2 (1210 kg) 1:21 m 2 2 2 = 2:21 102 kg m2 : (b) The rotational kinetic energy is 1 K = 1 I!2 = 2 (2:21 102 kg m2 )[(1:52 rev/s)(2 rad/rev)]2 2 = 1:10 104 J : 50E 8 (a) I0 = m`2 + m(2`)2 + 1 (2M )(2`)2 = 5m`2 + 3 M`2 : 3 312 CHAPTER 11 ROTATIONAL MOTION 1 5 (b) K = 2 I0 !2 = 2 m + 4 M `2 !2 : 3 (a) The rotational inertia of the three blades (each of mass m and length L) is 1 I = 3 3 mL2 = mL2 = (240 kg)(5:2 m)2 = 6:49 103 kg m2 : (b) The rotational kinetic energy is 51E K = 1 I!2 = 1 (6:49 103 kg m2 )[(350 rev/min)(1 min=60 s)(2 rad/rev)]2 2 2 = 4:36 106 J = 4:36 MJ : 52E 2 2 (a) I = 2 Me Re = 5 (5:98 1024 kg)(6:37 106 m)2 = 9:71 1037 kg m2 : 5 1 I!2 = 1 (9:74 1037 kg m2 ) 2 2 = 2:57 1029 J: (b) K = 2 2 86400 s (c) It would be able to supply energy for a time duration of :57 29 K t = nP = (6:4 2109 )(110 J 3 W) = 4:0 1016 s = 1:3 109 y : :0 10 53E Use the parallel-axis theorem: I = Icm + Mh2 , where Icm is the rotational inertia about a parallel axis through the center of mass, M is the mass, and h is the distance between the two axes. In this case the axis through the center of mass is at the 0:50 m mark so h = 0:50 m 0:20 m = 0:30 m. Now Icm = M`2 =12 = (0:56 kg)(1:0 m)2 =12 = 4:67 10 2 kgm2 , so I = 4:67 10 2 kg m2 + (0:56 kg)(0:30 m)2 = 9:7 10 2 kg m2 . Let the moment of inertia of a rigid body about a certain axis be I . Now consider another axis, which is parallel to the existing one and passes through the center of mass of the rigid body. The moment of inertia about the new axis is Icm . According to the parallel-axis theorem I = Icm + Mh2 , where M is the mass of the rigid body and h is the separation between the two parallel axes. Since Mh2 0, 54P Icm = I Mh2 I ; CHAPTER 11 ROTATIONAL MOTION 313 so Icm is the smallest. Divide the rim of the hoop into N small sections. Denote the mass of the n-th one as mi and its separtion from the center of the hoop as ri (= R): Then 55P mi ri =R I= X X mi ri2 = mi R2 i=1 i=1 N N X = R2 mi = MR2 : i=1 N Alternatively, we may use the integral approach by taking N ! 1 and letting mi ! dm: I= Z ring r2 dm = Z ring R2 dm = R2 Z ring dm = MR2 : 56P Use the parallel-axis theorem. According to Table 11-2(j) the rotational inertia of a uniform slab about an axis through the center and perpendicular to the large faces is given by Icm = M (a2 + b2 ) : 12 A parallel axis through a corner is a distance h = (a=2)2 + (b=2)2 from the center, so 1 I = Icm + Mh2 = M (a2 + b2 ) + M (a2 + b2 ) = 3 M (a2 + b2 ) : 12 4 57P p (a) Ix = (b) X 4 i=1 mi yi2 = [50(2:0)2 + (25)(4:0)2 + 25( 3:0)2 + 30(4:0)2 ] g cm2 = 1:3 103 g cm2 : Iy = X 4 i=1 mi x2 = [50(2:0)2 + (25)(0)2 + 25(3:0)2 + 30(2:0)2 ] g cm2 = 5:5 102 g cm2 : i 314 CHAPTER 11 ROTATIONAL MOTION (c) Iz = X 4 i=1 mi (x2 + yi2 ) = Ix + Iy = 1:3 103 g cm2 + 5:5 102 g cm2 = 1:9 102 g cm3 : i (d) The answer to (c) is A + B . 58P where M is its mass and R is its radius. For a hoop with mass M and radius RH Table 11{ 2(a) gives 2 IH = MRH : If the two bodies have the same mass then they will have the same rotational inertia if p 2 R2 =2 = RH , or RH = R= 2. (b) You want the rotational inertia to be given by I = Mk2 , where M is the mass of the arbitrary body and k is the radius of the equivalent hoop. Thus (a) According to Table 11-2(c) the rotational inertia of a uniform solid cylinder about its central axis is given by 1 IC = 2 MR2 ; I k= M: 59P r (a) The rotational kinetic energy is 1 1 1 1 K = 2 I!2 = 2 2 MR2 !2 = 4 M (!R)2 1 = 4 (500 kg)[(200 rad/s)(1:0 m)]2 = 4:9 107 J : (b) The operation time is 10 J t = K = 84::09 103 W = (6:2 103 s)(1:0 min=60 s) = 1:0 102 min : P 7 Use = rF sin . 60E CHAPTER 11 ROTATIONAL MOTION 315 (a) = (0:152 m)(111 N) sin 30 = 8:4 N m: (b) = (0:152 m)(111 N) sin 90 = 17 N m: (c) = (0:152 m)(111 N) sin 180 = 0: 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 , so if ` is the length of the rod then the torque associated with this force has magnitude = mg` sin = (0:75 kg)(9:8 m/s2 )(1:25 m) sin 30 = 4:6 N m. For the position of the ball shown the torque is counterclockwise. 62E 61E mg max = mgr = mgd=2 = (70 kg)(9:80 m/s2 )(0:40 m=2) = 1:4 102 N m: 63P (a) Take a torque that tends to cause a counterclockwise rotation from rest to be positive and a torque that tends to cause a clockwise rotation from rest to be negative. Thus a positive torque of magnitude r1 F1 sin 1 is associated with F1 and a negative torque of magnitude r2 F2 sin 2 is associated with F2 . Both of these are about O. The net torque about O is = r1 F1 sin 1 r2 F2 sin 2 : (b) Substitute the given values to obtain = (1:30 m)(4:20 N) sin 75:0 (2:15 m)(4:90 N) sin 60:0 = 3:85 N m : 64P The net torque is = A + B + C = FA rA sin A FB rB sin B + FC rC sin C = (10 N)(8:0 m)(sin 135 ) (16 N)(4:0 m)(sin 90 ) + (19 N)(3:0 m)(sin 160 ) = 12 N m : 316 CHAPTER 11 ROTATIONAL MOTION 65E 32: N m I = = 25:00rad/s2 = 1:28 kg m2 : 66E (a) Use the kinematic equation ! = !0 + t, where !0 is the initial angular velocity, ! is the nal angular velocity, is the angular acceleration, and t is the time. This gives 6: = ! t !0 = 22020 rad/s s = 28:2 rad/s2 : 10 3 (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. The magnitude of the net torque exerted on the cylinder of mass m and radius R2 is net = F1 R2 F2 R2 F3 R1 . The resulting angular acceleration of the cylinder is 67E net F2 R = net = MR2 =2 = 2(F1 R2 MR22 F3 R1 ) I 2 2 (4: N)(0 12 m) = 2[(6:0 N)(0:12 m) (2:00kg)(0::12 m)2 (2:0 N)(0:050 m)] = 9:7 rad/s2 ; which is counterclockwise. (a) The rotational inertia is I = Ml2 = (1:30 kg)(0:780 m)2 = 0:791 kg m2 : (b) The torque that must be applied is = fl = (2:30 10 2 N)(0:780 m) = 1:79 10 2 Nm: 69E 68E (a) Use = I, where is the net torque acting on the shell, I is the rotational inertia of the shell, and is its angular acceleration. This gives I = = 960 N m 2 = 155 kg m2 : 6:20 rad/s (b) The rotational inertia of the shell is given by I = (2=3)MR2 (see Table 11-2 of the text). This means kg m2 M = 23I2 = 3(15590 m)2 ) = 64:4 kg : R 2(1: CHAPTER 11 ROTATIONAL MOTION 317 Use = Fr = I, where satis es = 1 t2 : Here = 90 and t = 30 s: The force needed 2 is then 2 4 m2 )(90 F = I = I (2=t ) = 2(8:7 10(2kgm)(30 s)2)(=180 ) = 1:3 102 N : r r :4 70P (a) The angular acceleration is (0:30)(3:0)2 = = F (It)r = [(0:50)(3:0) :+ 10 3 kg m]2N (0:10 m) = 4:2 102 rad/s2 : I 10 t=3:0 s 71P (b) The angular speed is ! = !0 + dt = 0 0 r (0:25t2 + 0:10t3 ) =I 2 3 = (0:10 m)[(01::25)(3:0) 3 + (0:10)(3:0) ] N 0 10 kg m2 = 5:0 102 rad/s : Z t Z t F (t)r r Z t (0:50t + 0:30t2 )dt I =I 0 Let the tension in the cord be T . Then for the wheel of radius R we have net = TR = I, and for the object of mass M we have Mg sin T = Ma: Also, a = R. Solve these equations for I : 72P M (g sin a)R2 = (2:0 kg)[(9:80 m/s2 )(sin 20 ) 2:0 m/s2 ](0:20 m)2 I= a 2:0 m/s2 = 0:054 kg m2 : (a) The torque required to bring a uniform sphere of radius r and mass M from rest to an angular velocity ! in a time duration t is given by = =I = =(2MR2 =5) = !=t = !=t; or 2 = I! = 2MR ! : t 5t 73P 318 CHAPTER 11 ROTATIONAL MOTION Thus for the smaller sphere 1 = 2(1:65 kg)(0:226 m)s)(317 rad/s) = 0:689 N m (5)(15:5 and for the larger one 2 R 2 = 1 R2 1 2 = (0:689 N m) 0::854 m 0 226 m 2 = 9:84 N m : (b) The force applied is F = =R. Thus for the smaller sphere F1 = 1 =R1 = (0:689 N m)=(0:226 m) = 3:05 N; and for the larger one F2 = 2 =R2 = (9:84 Nm)=(0:854 m) = 11:5 N: (a) Use constant acceleration kinematics. If down is taken to be positive and a is the acceleration of the heavier block, then its coordinate is given by y = 1 at2 , so 2 750 y a = 22 = 2(0::00 s)m) = 6:00 10 2 m/s2 : 2 t (5 The lighter block has an acceleration of 6:00 10 2 m/s2 , upward. (b) For the heavier block mh g Th = mh a, where mh is its mass and Th is the tension in the part of the cord that is attached to it. Thus Th = mh (g a) = (0:500 kg)(9:8 m/s2 6:00 10 2 m/s2 ) = 4:87 N. (c) For the lighter block ml g Tl = ml a, so Tl = ml (g + a) = (0:460 kg)(9:8 m/s2 +6:00 10 2 m/s2 ) = 4:54 N. (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=R = (6:00 10 2 m/s2 )=(5:00 10 2 m) = 1:20 rad/s2 . (e) The net torque acting on the pulley is = (Th Tl )R. Equate this to I and solve for I: 2 I = (Th Tl )R = (4:87 N 4:54 N)(5:00 10 m) = 1:38 10 2 kg m2 : 1:20 rad/s2 75P 74P 2 2 The angular acceleration for both masses satis es = mgl2 mgl1 = I = (ml1 + ml2 ); which gives 2 )(0 80 m 0:20 = g(ll22+ ll21 ) = (9:80 m/sm)2 :+ (0:20 m)2 m) = 8:65 rad/s2 : (0:80 1 2 CHAPTER 11 ROTATIONAL MOTION 319 The linear acceleration of the left mass is then a1 = l1 = (8:65 rad/s2 )(0:80 m) = 6:9 m/s2 ; and that of the right mass is a2 = l2 = (8:65 rad/s2 )(0:20 m) = 1:7 m/s2 : (a) From = 1 t2 we get = 2=t2 . 2 (b) a = R = 2R=t2 : (c) For the lower mass Mg T1 = Ma, which gives T1 = M (g a) = M (g 2R=t2 ): (d) For the pulley net = (T1 T2 )R = I, which gives 76P T2 = T1 I = Mg 2 MR + I : R t2 R (a) The speed of v of the mass m after it has descended d = 50 cm is given by v2 = 2ad. where a is calculated in Sample Problem 11-11. Thus 77E v = 2ad = p r 2(2mg)d = M + 2m s 4(50 g)(980 cm/s2 )(50 cm) = 1:4 102 cm/s = 1:4 m/s : 400 g + 2(50 g) (b) The answer is still 1:4 m/s, since it is independent of R. 78E which is equivalent to 292 ft lb. 79E (100 hp)(746 W/hp) = P = (1800 rev/min)(2 rad/rev)(1:00 min=60 s) = 396 N m ; ! (a) The work that needs to be done is 1 W = K = 2 I!2 = 1 MR2 !2 2 1 = 2 (32:0 kg)(1:20 m)2 [(280 rev/min)(2 rad/rev)(1 min=60 s)]2 = 1:98 104 J : 4 P = W = 1:98 010 J = 1:32 103 W : t 15: s (b) The power required is 320 CHAPTER 11 ROTATIONAL MOTION Note that the work done on the wheel (and, consequently, the power delivered to the wheel) are negative, which tend to slow down its rotation. 1 1 1 (a) K = 1 I!2 = 2 3 m`2 !2 = 6 m`2 !2 : 2 (b) Use conservation of energy: K = mgh. The center of mass rises by 2 2 2 2 K h = mg = m` ! = ` 6! : 6mg g 80E and the angular acceleration required to accelerate the Earth from rest to ! in one day is = !=T . The touque needed is then 27 2 27 5 = I = I! = (9:71 10 kg m )(7:s 10 rad/s) = 8:17 1028 N m ; T 86400 where we used the data for I obtained in 52E, part (a). (a) The angular speed ! of the rotation of the Earth is given by ! = 2=T , where T = 1 da = 86400 s. Thus 2 ! = 86400 s = 7:27 10 5 rad/s ; 81P (b) As calculated in 52E, part (b), the kinetic energy of the Earth associated with its rotation about its own axis is K = 2:57 1029 J. Thus this much energy would be needed to bring it from rest into rotation at the present angular speed. (c) The power needed is 29 P = K = 2:57 10s J = 2:97 1021 kW : T 86400 Use conservation of energy. The initial energy before falling is Ei = mg`=2, where m is the mass of the stick and ` is the length. As the stick rotates about one end and falls onto the table, the nal energy is 1 1 Ef = 2 I!2 = 2 ( 1 m`2 )!2 = 1 mv2 ; 3 6 where v = !` is the speed of the other end. Thus Ei = 1 mgl = Ef = 1 mv2 , which gives 2 6 82P v = 3g` = 3(9:80 m/s2 )(1:00 m) = 5:42 m/s : p p CHAPTER 11 ROTATIONAL MOTION 321 Use conservation of mechanical energy. The center of mass is at the midpoint of the cross bar of the H and it drops by `=2, where ` is the length of any one of the rods. The gravitational potential energy decreases by Mg`=2, where M is the mass of the body. The initial kinetic energy is zero and the nal kinetic energy may be written 1 I!2 , where I is 2 the rotational inertia of the body and ! is its angular velocity when it is vertical. Thus p 0 = Mg`=2 + 1 I!2 and ! = Mg`=I . Since the rods are thin the one along the axis 2 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)`2 , where M=3 is its mass. The cross bar is a rod that rotates around one end so its contribution is (M=3)`2 =3 = M`2 =9. The total rotational inertia is I = (M`2 =3) + (M`2 =9) = 4M`2 =9. The angular velocity is 83P Mg` ! = Mg` = 4M`2 =9 = 9g : I 4` r s r (a) Use the parallel-axis theorem to nd I : 1 I = Icm + Mh2 = 2 MR2 + Mh2 1 = 2 (20 kg)(0:10 m)2 + (20 kg)(0:50 m)2 = 0:15 kg m2 : 84P (b) Conservation of energy requires that Mgh = 1 I!2 , where ! is the angular speed of the 2 cylinder as it passes through the lowest position. Solve for !: != r 2Mgh = s I 2(20 kg)(9:80 m/s2 )(0:050 m) = 11 rad/s : 0:15 kg m2 (a) The (centripetal) force exerted on an in nitesimal portion of the blade with mass dm located a distance r from the rotational axis is given by dF = (dm)!2 r. Thus for the entire blade of mass M and length L the total force is given by 85P M Z L !2 r dr = M!2 r2 L = M!2 L F = dF = L 0 2L 0 2 = 1 (110 kg)[(320 rev/min)(2 rad/rev)(1:00 min=60 s)]2 (7:80 m) 2 = 4:8 105 N : Z Z !2 r dm = 322 CHAPTER 11 ROTATIONAL MOTION (b) The torque is 1 = I = I ! = 3 ML2 ! t t = 1 (110 kg)(7:8 m)2 [(320 rev/min)(2 rad/rev)(1:00 min=60 s) 3 6:7 s 4 Nm : = 1:1 10 (c) The work done is 1 I!2 = 1 1 ML2 !2 W = K = 2 2 3 1 = 6 (110 kg)(7:80 m)2 [(320 rev/min)(2 rad/rev)(1:00 min=60 s)]2 = 1:3 106 J : 86P Denote the sphere as 1, the pully as 2, and the hanging mass as 3. From the work-energy theorem (or conservation of mechanical energy) 1 2 1 2 1 2 mgh = 2 I1 !1 + 2 I2 !2 + 2 mv3 ; where !1 = v3 =R, !2 = v3 =r, and I2 = I . Solve for v3 : 2 2gh v3 = 1 + I =mRgh+ I =mr2 = 1 + 2M=3m + I=mr2 : 2 1 2 87P s s (a) If ` is the length of the chimney then the radial component of the acceleration of the top is given by ar = `!2 , where ! is the angular velocity. Use conservation of mechanical energy to nd 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 = (`=2) cos . Initially the potential energy is Ui = Mg(`=2) and the kinetic energy is zero. Write 1 I!2 for the kinetic energy when the chimney makes 2 the angle with the vertical. Here I is its rotational inertia. Conservation of energy then leads to Mg`=2 = Mg(`=2) cos + 1 I!2 , so !2 = (Mg`=I )(1 cos ). Thus 2 Mg`2 (1 cos ) : ar I The chimney is rotating about its base, so I = Mg`2 =3 and ar = 3g(1 cos ). = `!2 = CHAPTER 11 ROTATIONAL MOTION 323 (b) The tangential component of the acceleration of the chimney top is given by at = `, where is the angular acceleration. Dierentiate !2 = (Mg`=I )(1 cos ) with respect to time, replace d!=dt with , and replace d=dt with ! to obtain 2! = (Mg`=I )! sin or = (Mg`=2I ) sin . Thus at = ` = Mg` sin = 32g sin ; 2I where I = M`2 =3 was used to obtain the last result. (c) The angle for which at = g is the solution to (3=2) sin = g or sin = 2=3. It is = 41:8 . (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, or = !0 =t = (39:0 rev/s)(2 rad/rev)=(32:0 s) = 7:66 rad/s2 . (b) Use = I, where is the torque and I is the rotational inertia. The contribution of the rod to I is M`2 =12, where M is its mass and ` is its length. The contribution of each ball is m(`=2)2 , where m is the mass of a ball. The total rotational inertia is 2 2 2 2 I = M` + 2 m` = (6:40 kg)(1:20 m) + (1:06 kg)(1:20 m) = 1:53 kg m2 : 12 4 12 2 88P 2 The torque is = (1:53 kg m2 )(7:66 rad/s2 ) = 11:7 N m : 2 Ki = 1 I!0 = 1 (1:53 kg m2 )[(39:0 rev/s)(2 rad/rev)]2 2 2 4 J; = 4:59 10 (c) Since the system comes to rest the change in the mechanical energy is simply the negative of the initial kinetic energy i.e, the amount of mechanical energy dissipated is 4:59 104 J. (d) Use = !0 t + 1 t2 2 1 = (2)(39:0 rad/s)(32:0 s) + 2 ( 7:66 rad/s2 )(32:0 s)2 = 3920 rad : This is 3920=2 = 624 rev. (e) Only the loss in mechanical 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. 324 CHAPTER 11 ROTATIONAL MOTION (b) Conservation of energy gives (a) Let the mass of the rod be m and its length be l. The net torque exerted on the rod at the instant it is released is net = mgl cos =2, where = 40 . Thus its angular acceleration is mgl m/s2 = net = 2(mlcos3) = 3g cos = 3(9:80 2(2:0 )(cos 40 ) = 5:6 rad/s2 : 2= I 2l m) 89P l K = 1 I!2 = U = mg 2 sin : 2 Thus the angular speed ! is r r r mgl sin = 3g sin = 3(9:80 m/s2 )(sin 40 ) = 3:1 rad/s : != I l 2:0 m The work W required is equal to the rotational kinetic energy of the system: 1 1 W = 1 I!2 = 2 [2Ma2 + 2Ma2 + M (b2 a2 )] = 2 M (3a2 + b2 )!2 2 1 = 2 (0:40 kg)[3(0:30 m)2 + (0:50 m)2 ](5:0 rad/s)2 = 2:6 J : 90P (a) Use conservation of energy. At the top of the hill the kinetic energy is Ki = 0 and the gravitational potential energy is Ui = mgh = mgd sin , where m is the total mass of the car, d is the length of the hill, and is its slope. At the bottom of the hill the kinetic energy is 1 mv2 + 1 I!2 , where v is the speed of the car, I is the rotational inertia of the ywheel, 2 2 and ! is its angular velocity. Conservation of energy leads to mgd sin = 1 mv2 + 1 I!2 . 2 2 Since the ywheel is geared to the driveshaft the velocity of the car and the angular velocity of the ywheel are proportional to each other. Write ! = Av, where A is the constant of proportionality. Thus mgd sin = 1 mv2 + 1 IA2 v2 and 2 2 r mgd sin v = 2m + IA2 : Use the fact that ! = 240 rev/s = 1:51 103 rad/s when v = 80 km/h = 22:2 m=s to nd the value of A: A = !=v = (1:51 103 rad/s)=(22:2 m/s) = 68:0 rad/m. The rotational inertia of the ywheel (assumed to be a uniform disk) is I = 1 MR2 = 1 (200 N=9:8 m/s2 )(0:55 m)2 2 2 = 3:09 kg m2 . Then s 91P v= This is 42:1 km/h. 2(800 kg)(9:8 m/s2 )(1500 m) sin 5 = 11:7 m/s : 800 kg + (3:09 kg m2 )(68:0 rad/s)2 CHAPTER 11 ROTATIONAL MOTION 325 (b) Dierentiate v2 = 2mgd sin =(m + IA2 ) with respect to time, replace dv=dt with the acceleration a, and replace dd=dt with v. The result is so Since ! = Av, mgv sin 2va = 2m + IA2 : mg a = m +sin 2 : IA Amg sin = (68:0 rad/s)(800 kg)(9:8 m/s2 ) sin 5 = 3:08 rad/s2 : = Aa = m + IA2 800 kg + (3:09 kg m2 )(68:0 rad/m)2 (c) The energy stored in the ywheel is E = 1 I!2 and the rate at which it is changing is 2 dE = I! = IAv dt = (3:09 kg m2 )(68:0 rad/m)(11:7 m/s)(3:08 rad/s2 ) = 7:57 103 W : 92 (a) Rotational kinetic energy is not conserved; (b) For k = 0:25, the time to reach the same angular velocity is about 52 s while for k = 0:50 it is about 26 s. The nal angular velocities are the same for the two cases, as is the nal rotational kinetic energy. 93 (a) 3:4 105 g cm2 ; (b) 2:9 105 g cm2 ; (c) 6:3 105 g cm2 ; (d) (1:2 cm) i + (5:9 cm) j 94 (a) 1:6 m/s2 ; (b) 4:6 N; (c) 4:9 N ...
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This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.

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