CHAPTER 11. Answer to Checkpoint Questions

96 CHAPTER 11 ROTATIONAL MOTION CHAPTER 11 Answer to Checkpoint Questions 1. (b) and (c). (a) and (d) 3. (a) yes; (b) no; (c) yes; (d) yes 4. all tie 5. 1,, 4, 3 6. (a) 1 and 3 tie, 4, then and 5 tie (zero) 7. (a) downward in the gure; (b) less Answer to Questions 1. (a) positive; (b) zero; (c) negative; (d) negative. (a) clockwise; (b) counterclockwise; (c) yes; (d) positive; (e) constant 3. (a) and 3; (b) 1 and 3; (c) 4 4. all tie 5. (a) and (c) 6. (a) c, a, then b and d tie; (b) b, then a and c tie, then d 7. (a) all tie; (b), 3, and then 1 and 4 tie 8. less 9. b, c, a 10. (a) hoop (mass at greatest distance from rotational axis); (b) prism (mass at least distance) 11. less 1. larger 13. 90, then 70 and 110 tie 14. all tie

CHAPTER 11 ROTATIONAL MOTION 97 15. nite angular displacements are not commutative Solutions to Exercises & Problems 1E (a) The angle is 1:8 m1: m 1:50 rad: (b) (1:5 rad)(180 rad) 85:9 : (c) The arc length is l r (0:6 rad)(:40 m) 1:49 m. E (a) The angular velocity is (b) The angular acceleration is! d dt d dt (at + bt3 ct 4 ) a + 3bt 4ct 3 : d! dt d dt (a + 3bt 4ct 3 ) 6bt 1ct : 3E Use! T. (a) For the second hand T 60 s so! 60 s 0:105 rad/s: (b) For the minute hand T 60 min 3600 s so! 3600 s 1:75 10 3 rad/s: (c) For the hour hand T 4 h so! [(4 h)(3600 s h)] 1:45 10 4 rad/s: 4E (a) The time for one revolution is the circumference of the orbit divided by the speed v of the sun: T Rv, where R is the radius of the orbit. Since R :3 10 4 ly (:3 10 4 ly)(9:460 10 1 km/ly) :18 10 17 km, T (:18 1017 km) 50 km/s 5:5 10 15 s : (b) The number of revolutions is the total time t divided by the time T for one revolution: N tt. Convert the total time from years to seconds. The result for the number of revolutions is N (4:5 109 y)(3:16 10 7 s/y) 6 : 5:5 10 15 s

98 CHAPTER 11 ROTATIONAL MOTION 5E (a) The angular velocity is given by At t :0 s and at t 4:0 s (b) The average angular acceleration is! d dt d dt (4:0t 3:0t + t 3 ) 4:0 6:0t + 3t :!(:0 s) 4:0 6:0(:0) + 3(:0) 4:0 rad/s ;!(4:0 s) 4:0 6:0(4:0) + 3(4:0) 8 rad/s :! t 8 rad/s 4:0 rad/s 1 rad/s : 4:0 s :0 s (c) The angular acceleration is d! dt d dt (4:0 6:0t + 3t ) 6:0 + 6t : At t :0 s and at t 4:0 s (:0 s) 6:0 + 6(:0) 6:0 rad/s ; (4:0 s) 6:0 + 6(4:0) 18 rad/s : 6E (a) Evaluate (t) + 4t + t 3 for t 0 to obtain (0) rad. (b) The angular velocity (in rad/s) is given by!(t) ddt 8t + 6t. Evaluate this expression for t 0 to obtain!(0) 0. (c) For t 4:0 s,! 8(4:0) + 6(4:0) 130 rad/s. (d) The angular acceleration (in rad/s ) is given by d!dt 8 + 1t. For t :0 s, 8 + 1(:0) 3 rad/s. (e) The angular acceleration, given by 8 + 1t, depends on the time and so is not constant. 7P (a) The angular speed is given by!(t)! 0 + Z t 0 dt! 0 +! 0 + at 4 bt 3 : Z t 0 (4at 3 3bt )dt

CHAPTER 11 ROTATIONAL MOTION 99 (b) The angular displacement is (t) 0 + Z t 0!(t)dt 0 + Z t 0 +! 0 t + 1 5 at5 1 4 bt4 ; where 0 is the initial angular displacement at t0. 0 (! 0 + at 4 bt 3 )dt 8P The number of spins is n (60 ft)(1800 rev/min)(1 min60 s) (85 mi/h)(579 ft/mi)(1 h3600 s) 14 rev : 9P The time t during which the diver makes.5 revolutions is given by 1 g(t) h, where h 10 m. Thus the average angular speed of the diver is! (:5 rev)( rad/rev) p (10 m)(9:80 m/s ) 11 rad/s : 10P (a) To avoid touching the spokes, the arrow must go through the wheel in not more than t (18) rev :5 rev/s 0:050 s : The minimum speed of the arrow is then v min 0 cm0:050 s 400 cm/s 4:0 m/s: (b) No (as is apparent from the calculation above). 11E (a)! t 3000 rev/min 100 rev/min 9000 rev/min : (1 s)(1 min60 s) (b) The number of revolutions in t 1 s 0:0 min is n! 0 t + 1 t (100 rev/min)(0:0 min) + 1 (100 rev/min )(0:0 min) 4: 10 rev :

300 CHAPTER 11 ROTATIONAL MOTION 1E (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 33:33 rev/min 0:50 min 67 rev/min : (b) Use! 0 t + 1 t (33:33 rev/min)(0:50 min) + 1 ( 66:7 rev/min )(0:50 min) 8:3 rev. 13E (a) The time t is given by t! (10 rad/s)(4:0 ras/s ) 30 s: (b) The angle is 1 t 1 (4:0 rad/s )(30 s) 1:8 10 3 rad: 14E (a) 5:6 m(8:0 10 m) 1:4 10 rad: (b) Use 1 t to obtain t: t r s (1:4 10 rad) 14 s : 1:5 rad/s 15E The angular acceleration of the wheel is! t (0:90 1:00)(50 rev/min)1:0 min 5 rev/min ; so the angular speed at the end of the second minute is! (0:90)(50 rev/min) (5 rev/min )(1:0 min) 00 rev/min : 16E (a)!t 5:0 rad/s 0:0 s 1:5 rad/s : (b)! 0 t + 1 t (5:0 rad/s )(0:0 s) + ( 1:5 rad/s )(0:0 s) 50 rad: (c) The number of revolutions is n rad 50 rad rad 39:8 rev: 17E (a) The angular acceleration satises 5 rad 1 (5:0 s) ; which gives :0 rad/s :

CHAPTER 11 ROTATIONAL MOTION 301 (b) The average angular velocity is! t (c) The instantaneous angular velocity is 5 rad 5:0 s 5:0 rad/s : (d) The additional angle turned is! (:0 rad/s )(5:0 s) 10 rad/s : (10:0 s) (5:0 s) 1 (:0 rad/s )[(10:0 s) (5:0 s) ] 75 rad : 18P The angular acceleration during the rst 0 s is (5:0 rad/s):0 s :5 rad/s : The angular displacement at t 0 s is then 1 1 t 1(:5 rad/s )(0 s) 500 rad: From t 0 s to t 40 s, the angular speed is! (:5 rad/s )(0 s) 50 rad/s: The angular displacement from t 0 s to t 40 s is then (50 rad/s)(40 s 0 s) 1000 rad. The total angle turned is thus 1 + 500 rad + 1000 rad 1500 rad. 19P (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 :00 rad/s. For the interval so 90:0 rad! i t + 1 (t) tt + 1 (t) ; t 1 t 1 t 90:0 rad (:00 rad/s )(3:00 s) (b)! i t (:00 rad/s )(13:5 s) 7:0 rad/s: 3:00 s 13:5 s : 0P 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 t. Solve for! 0 :! 0 1 t t 10 rad 1 (3:0 rad/s )(4:0 s) 4:0 s 4 rad/s : Now use!! 0 + t to nd the time when the wheel was at rest (! 0): t! 0 4 rad/s 3:0 rad/s 8:0 s :

30 CHAPTER 11 ROTATIONAL MOTION That is, the wheel started from rest 8:0 s before the start of the 4:0 s interval. 1P (a) Use!t 1! 0t. The time required is t! 0 (40 rev)( rad/rev) 1:5 rad/s 3:4 10 s : (b)! 0 t ( 1:5 rad/s)(3:4 10 s) 4:5 10 3 rad/s : (c) Let the angular speed p of the wheel be! as it nishes the rst 0 turns. Then!! 0 (), or!! 0 +. Thus the time required is p t 0!! 0! 0 +! 0 p (1:5 rad/s) + ( 4:5 10 3 rad/s )(40 rev)( rad/rev) 1:5 rad/s 4:5 10 3 rad/s 98 s : P (a) The angle through which the line will turn as a function of time t is (t) 0 +! 0 t + 1 t : To maximize (t), set ddt0: d dt d tt0 dt ( 0 +! 0 t + 1 t )! 0 + t 0 0 ; tt0 which gives t 0! 0. Thus! 0 max (t 0 ) 0 + 1!0 (4:7 rad/s) ( 0:5 rad/s ) 44 rad :! 0 (b) Let b max! 0 t b + 1 t b and solve for t b. Use max 44 rad,! 0 4:7 rad/s, and 0:5 rad/s. You should get t b 5:5 s or 3 s. (c) Let 0 10:5 rad! 0 t c + 1 t c and solve for t c. You should get t c :1 s or 40 s.

CHAPTER 11 ROTATIONAL MOTION 303 (d) θ 44rad rad -.1s 5.5s 3s 40s t -10.5rad 3P (a) Use! f! i to calculate :! f! i (15 rev/s) (10 rev/s) (60 rev) 1:0 rev/s : (b) Use!t [(! i +! f )]t. The time required is (c) The time required is t (c) The number of revolutions is! i +! f t 0!0 (60 rev) 15 rev/s + 10 rev/s 4:8 s : 10 rev/s 1:04 rev/s 9:6 s : n 1 t0 1 (1:0 rev/s )(9:6 s) 48 rev : 4P (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 :0 rev/s: (b) The angular acceleration is (10 rev/s :0 rev/s)15 s 0:53 rev/s : The time elapsed is then t! i (:0 rev/s)(0:53 rev/s ) 3:8 s:

304 CHAPTER 11 ROTATIONAL MOTION 5E The magnitude of the acceleration is given by a! 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)( rad/rev)(60 s/min) 3:49 rad/s. Thus a r! (0:15 m)(3:49 rad/s ) 1:8 m/s : The acceleration vector is toward the center of the record. 6E (a)! [(33 + 13) rev/min]( rad/rev)(1:0 min60 s) 3:5 rad/s: (b) v 1 r 1 (3:5 rad/s)(5:9 in.) 1 in./s: (c) v r (3:5 rad/s)(:9 in.) 10 in./s: 7E Use v r!. First convert 50 km/h to m/s: (50 km/h)(1000 m km)(3600 sh) 13:9 m/s. Then! vr (13:9 m/s)(110 m) 0:13 rad/s. 8E (a)! (00 rev/min)( rad/rev)(1:00 min60 s) 0:9 rad/s: (b) v!r (0:9 rad/s)(1:0 m) 1:5 m/s: (c) Use!t to obtain :! t 1000 rev/min 00 rev/min 800 rev/min : (60 s)(1:00 min60 s) (d) Solve from! f! :! f! (1000 rev/min) (00 rev/min) (800 rev/min ) 600 rev : 9E The average angular acceleration is! t v rt 5 m/s 1 m/s (0:75 m)(6: s) 5:6 rad/s :

CHAPTER 11 ROTATIONAL MOTION 305 30E (a)! T (1:0 yr)(365)(4)(3600 s)1:0 yr :0 10 7 rad/s: (b) v r e s (:0 10 7 rad/s)(1:49 10 11 m) 3:0 10 4 m/s: (c) a! r (:0 10 7 rad/s) (1:49 10 11 m) 5:9 10 3 m/s : The direction of a is toward the sun. 31E (a) The longitudinal separation between Helsinki and the explosion site is 10 5 77. Since a time dierence of T 1 da 4 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 T (4 h) 5:1 h : 360 360 (b) Now 10 ( 0 ) 1 so the required time delay would be 1 t T (4 h) 8:1 h : 360 360 3E (a) The angular velocity at t 5:0 s is! d dt d t5:0 s dt (0:30t ) (0:30)(5:0 s) 3:0 rad/s : t5:0 s (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 a t dv dt r r d dt r d dt (0:30t ) (10 m)()(0:30) 6:0 m/s : t5:0 s (d) The radial (centripetal) acceleration is a c! r (3:0 rad/s) (10 m) 90 m/s. 33E (a)! vr (:90 10 4 km/h)(1:00 h3600 s)(3: 10 3 km) :50 10 3 rad/s: (b) a r! r (:50 10 3 rad/s) (3: 10 6 m) 0: m/s : (b) a t dvdt 0.

306 CHAPTER 11 ROTATIONAL MOTION 34E (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 f max s Mg M! 0 R. This gives! 0 p s gr: (b) coin flies off R original circular path with radius R and angular speed ω 0 35P (a) The angular acceleration is! t 0 150 rev/min (: h)(60 min1 h) 1:1 rev/min : (b) The number of rotations is n! 0 t + 1 t (150 rev/min)(: h)(60 min1 h) + 1 ( 1:13 rev/min )[(: h)(60 min1 h)] 9:9 10 3 rev : (c) The tangential linear acceleration is a t r ( 1:1 rev/min )( rad/rev)(1 min 3600 s )(50 cm) 0:99 mm/s : (d) The centripetal acceleration is a c! r [(75 rev/min)( rad/rev)(1 min60 s)] (50 cm) 31 m/s a t : p The net acceleration is then a a c + a t ' a c 31 m/s :

CHAPTER 11 ROTATIONAL MOTION 307 36P (a) a t r (14: rad/s )(:83 cm) 40: cm/s : (b) a! r [(760 rev/min)( rad/rev)(1:00 min60 s)] (:83 cm) :36 10 3 m/s : (c) Use v (!r) as to compute s: s v a [(760 rev/min)( rad/rev)(1:00 min60 s)] (:36 10 3 m/s ) 83: m : 37P (a) The speed as seen by the pilot is v 1!r (00 rev/min)( rad/rev)(1 min60 s)(1:5 m) 3:1 10 m/s : (b) The speed as seen by an observer on the ground is v q (480 km/h) + v 1 p [(480 km/h)(10 3 m/km)(1 h3600 s)] + (3:1 10 m/s) 3:4 10 m/s : 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 500 1:6 10 rad. That time is so the angular velocity of the wheel is t ` c (500 m) :998 10 8 m/s 3:34 10 6 s! t 1:6 10 rad 3:34 10 6 s 3:8 10 3 rad/s : (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 10 3 rad/s) 190 m/s : 39P (a) The linear speed at t 15:0 s is v a t t (0:500 m/s )(15:0 s) 7:50 m/s: The centerpetal acceleration is then a c v r where r 30:0 m, and the net acceleration is

308 CHAPTER 11 ROTATIONAL MOTION s p v a a t + a c a t + r s (7:50 m/s) (0:500 m/s ) + 1:94 m/s : 30:0 m (b) Refer to the gure to the right. Note that a t k v: The angle between v and a is cos 1 at a cos 1 0:500 m/s 1:94 m/s 75:1 : a c a θ a t v 40P (a) The Earth makes one rotation per day and 1 d (4 h)(3600 s/h) 8:64 10 4 s, so the angular velocity of the Earth is ( rad)(8:64 10 4 s) 7:7 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 10 6 m). So its speed is v (R cos 40 )! (6:37 10 6 m)(cos 40 )(7:7 10 5 rad/s) 3:5 10 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 10 6 m)(7:7 10 5 rad/s) 463 m/s. 41P 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 r A A r C C, where A is the angular acceleration of wheel A and C is the angular acceleration of wheel C. Thus C ra r C A 10 cm 5 cm (1:6 rad/s ) 0:64 rad/s : 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 C 10:5 rad/s 0:64 rad/s 16 s :

CHAPTER 11 ROTATIONAL MOTION 309 4P (a) The linear speed of a point on belt 1 is (b) The angular speed of B is v 1! A r A (10 rad/s)(15 cm) 1:5 10 cm/s :! B v 1 r B 1:5 10 cm/s 10 cm (c) The angular speed of B 0 is! B 0! B 15 rad/s: (d) The linear speed of a point on belt is (e) The angular speed of C is 15 rad/s : v! B 0r B 0 (15 rad/s)(5 cm) 75 cm/s :! C v r C 75 cm/s 5 cm 3:0 rad/s : 43P (a) a! r [(33 1 3 rev/min)( rad/rev)(1:0 min60 s)] (6:0 10 m) 0:73 m/s : (b) Use ma f s f s; max s mg to nd s; min : s; min a g 0:73 m/s 9:80 m/s 0:075 : (c) The radial acceleration of the object is a r! r, while the tangential acceleration is a t dvdt rd!dt r. Thus a p a r + a t p (! r) + (r) : If the object is not to slip at any time, we must let f s;max s mg ma max m p (! max r) + (r) ; where! max 33 1 rev 3:49 rad/s: Thus 3 p p (! s;min max r) + (r) (! max r) + (! max rt) g g p (3:49 rad/s)4 + [(3:49 rad/s)0:5 s] (6:0 10 m) 9:80 m/s 0:11 :

310 CHAPTER 11 ROTATIONAL MOTION 44P (a) The angular acceleration is given by d! dt T dt dt : For the pulsar described so dt dt 1:6 10 5 s/y 3:16 10 7 s/y 4:00 10 13 ; (4:00 10 13 ) :3 10 9 rad/s : (0:033 s) 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: t! 0 T ( :3 10 9 rad/s )(0:033 s) 8:3 1010 s : This is about 600 years. (c) The pulsar was born 1996 1054 94 years ago. This is equivalent to (94 y)(3:16 10 7 s/y) :98 10 10 s. It angular velocity was then!! 0 + t T + t 0:033 s + ( :3 10 9 rad/s )( :98 10 10 s) :6 10 rad/s : Its period was T! :6 10 rad/s :4 10 s : 45E The kinetic energy is given by K 1 I!, where I is the rotational inertia and! is the angular velocity. Here! (60 rev/min)( radrev) 60 s/min 63:0 rad/s : Thus I K (4400 J)! (63:0 rad/s) 1:3 kgm :

CHAPTER 11 ROTATIONAL MOTION 311 46P Let the linear and angular speeds of the oxygen molecule (of mass M) be v and!. Then 1 Mv 3 I!, which gives r s M! v 3I (500 m/s) (5:30 10 6 kg) 3(1:94 10 46 kg m ) 6:75 101 rad/s : 47E Since the rotational inertia of a cylinder of mass M and radius R is I 1 MR, the kinetic energy of a cylinder when it rotates with angular velocity! is K 1 I! 1 4 MR!. For the rst cylinder K 1 4 (1:5 kg)(0:5 m) (35 rad/s) 1:1 10 3 J. For the second K 1 4 (1:5 kg)(0:75 m) (35 rad/s) 9:7 10 3 J. 48E (a) The rotational inertia in kgm is I (14000 upm )(1:66 10 7 kg/u)(10 4 m pm ) :3 10 47 kgm : (b) The rotatitional kinetic energy in ev is K 1 I! 1 (:3 10 47 kg m )(4:3 10 1 rad/s) (1 ev1:6 10 19 J) 1:3 10 3 ev 1:3 mev: 49E (a) The rotational inertia is I 1 mr 1 (110 kg) 1:1 m (b) The rotational kinetic energy is :1 10 kgm : K 1 I! 1 (:1 10 kg m )[(1:5 rev/s)( rad/rev)] 1:10 10 4 J : 50E (a) I 0 m` + m(`) + 1 3 (M)(`) 5m` + 8 3 M`:

31 CHAPTER 11 ROTATIONAL MOTION (b) K 1 I 0! 5 m + 4 3 M `! : 51E (a) The rotational inertia of the three blades (each of mass m and length L) is 1 I 3 3 ml ml (40 kg)(5: m) 6:49 10 3 kgm : (b) The rotational kinetic energy is K 1 I! 1 (6:49 103 kgm )[(350 rev/min)(1 min60 s)( rad/rev)] 4:36 10 6 J 4:36 MJ : 5E (a) I 5 M ere 5 (5:98 104 kg)(6:37 10 6 m) 9:71 10 37 kgm : (b) K 1 I! 1 (9:74 1037 kgm ) :57 10 9 J: 86400 s (c) It would be able to supply energy for a time duration of t K np :57 10 9 J (6:4 10 9 )(1:0 10 3 W) 4:0 1016 s 1:3 10 9 y : 53E Use the parallel-axis theorem: I I cm + Mh, where I cm 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 0:0 m 0:30 m. Now I cm M`1 (0:56 kg)(1:0 m) 1 4:67 10 kgm, so I 4:67 10 kgm + (0:56 kg)(0:30 m) 9:7 10 kgm. 54P 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 I cm. According to the parallel-axis theorem I I cm + Mh, where M is the mass of the rigid body and h is the separation between the two parallel axes. Since Mh 0, I cm I Mh I ;

CHAPTER 11 ROTATIONAL MOTION 313 so I cm is the smallest. 55P Divide the rim of the hoop into N small sections. Denote the mass of the n-th one as m i and its separtion from the center of the hoop as r i ( R): Then I NX i1 m i r i NX i1 m i R R N X i1 m i MR : r i R m i Alternatively, we may use the integral approach by taking N! 1 and letting m i! dm: Z Z Z I r dm R dm R dm MR : ring ring ring 56P Use the parallel-axis theorem. According to Table 11-(j) the rotational inertia of a uniform slab about an axis through the center and perpendicular to the large faces is given by I cm M 1 (a + b ) : A parallel axis through a corner is a distance h p (a) + (b) from the center, so I I cm + Mh M 1 (a + b ) + M 4 (a + b ) 1 3 M(a + b ) : 57P (a) I x 4X i1 m i y i [50(:0) + (5)(4:0) + 5( 3:0) + 30(4:0) ] gcm 1:3 10 3 gcm : (b) I y 4X i1 m i x i [50(:0) + (5)(0) + 5(3:0) + 30(:0) ] gcm 5:5 10 gcm :

314 CHAPTER 11 ROTATIONAL MOTION (c) I z 4X i1 m i (x i + y i ) I x + I y 1:3 10 3 gcm + 5:5 10 gcm 1:9 10 gcm 3 : (d) The answer to (c) is A + B. 58P (a) According to Table 11-(c) the rotational inertia of a uniform solid cylinder about its central axis is given by I C 1 MR ; where M is its mass and R is its radius. For a hoop with mass M and radius R H Table 11{ (a) gives I H MR H : If the two bodies have the same mass then they will have the same rotational inertia if R R H, or R H R p. (b) You want the rotational inertia to be given by I Mk, where M is the mass of the arbitrary body and k is the radius of the equivalent hoop. Thus r I k M : 59P (a) The rotational kinetic energy is (b) The operation time is K 1 I! 1 1 MR! 1 4 M(!R) 1 (500 kg)[(00 rad/s)(1:0 m)] 4 4:9 10 7 J : t K P 4:9 107 J 8:0 10 3 W (6: 103 s)(1:0 min60 s) 1:0 10 min : 60E Use rf sin.

CHAPTER 11 ROTATIONAL MOTION 315 (a) (0:15 m)(111 N) sin 30 8:4 Nm: (b) (0:15 m)(111 N) sin 90 17 Nm: (c) (0:15 m)(111 N) sin 180 0: 61E 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/s )(1:5 m) sin 30 4:6 Nm. For the position of the ball shown the torque is counterclockwise. θ mg θ 6E max mgr mgd (70 kg)(9:80 m/s )(0:40 m) 1:4 10 Nm: 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 r 1 F 1 sin 1 is associated with F 1 and a negative torque of magnitude r F sin is associated with F. Both of these are about O. The net torque about O is r 1 F 1 sin 1 r F sin : (b) Substitute the given values to obtain (1:30 m)(4:0 N) sin 75:0 (:15 m)(4:90 N) sin 60:0 3:85 Nm : 64P The net torque is A + B + C F A r A sin A F B r B sin B + F C r C sin C (10 N)(8:0 m)(sin 135 ) (16 N)(4:0 m)(sin 90 ) + (19 N)(3:0 m)(sin 160 ) 1 Nm :

316 CHAPTER 11 ROTATIONAL MOTION 65E I 3:0 Nm 5:0 rad/s 1:8 kg m : 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!! 0 t 6:0 rad/s 0 10 3 s 8: rad/s : (b) If I is the rotational inertia of the diver then the magnitude of the torque acting on her is I (1:0 kgm )(8: rad/s ) 3:38 10 Nm. 67E The magnitude of the net torque exerted on the cylinder of mass m and radius R net F 1 R F R F 3 R 1. The resulting angular acceleration of the cylinder is is net net I MR (F 1R F R F 3 R 1 ) MR [(6:0 N)(0:1 m) (4:0 N)(0:1 m) (:0 N)(0:050 m)] (:0 kg)(0:1 m) 9:7 rad/s ; which is counterclockwise. 68E (a) The rotational inertia is I Ml (1:30 kg)(0:780 m) 0:791 kgm : (b) The torque that must be applied is fl (:3010 N)(0:780 m) 1:7910 Nm: 69E (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 Nm 6:0 rad/s 155 kgm : (b) The rotational inertia of the shell is given by I (3)MR (see Table 11- of the text). This means M 3I R 3(155 kgm ) 64:4 kg : (1:90 m)

CHAPTER 11 ROTATIONAL MOTION 317 70P Use F r I, where satises 1 t : Here 90 and t 30 s: The force needed is then F I r I(t ) r (8:7 104 kgm )(90 )(180 ) (:4 m)(30 s) 1:3 10 N : 71P (a) The angular acceleration is I F (t)r I (b) The angular speed is [(0:50)(3:0) + (0:30)(3:0) ] N(0:10 m) 4: 10 rad/s : t3:0 s 1:0 10 3 kgm!! 0 + Z t 0 dt Z t r I (0:5t + 0:10t 3 ) 0 F (t)r I r I Z t (0:10 m)[(0:5)(3:0) + (0:10)(3:0) 3 ] N 1:0 10 3 kgm 5:0 10 rad/s : 0 (0:50t + 0:30t )dt 7P Let the tension in the cord be T. Then for the wheel of radius R we have net T R I, and for the object of mass M we have Mg sin T Ma: Also, a R. Solve these equations for I: M(g sin I a)r a 0:054 kgm : (:0 kg)[(9:80 m/s )(sin 0 ) :0 m/s ](0:0 m) :0 m/s 73P (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 (MR 5)!t!t; or I! MR! : t 5t

318 CHAPTER 11 ROTATIONAL MOTION Thus for the smaller sphere 1 (1:65 kg)(0:6 m) (317 rad/s) (5)(15:5 s) 0:689 Nm and for the larger one 1 R R 1 (0:689 Nm) 0:854 m 9:84 Nm : 0:6 m (b) The force applied is F R. Thus for the smaller sphere F 1 1 R 1 (0:689 N m)(0:6 m) 3:05 N; and for the larger one F R (9:84 Nm)(0:854 m) 11:5 N: 74P (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 at, so a y (0:750 m) 6:00 10 m/s : t (5:00 s) The lighter block has an acceleration of 6:00 10 m/s, upward. (b) For the heavier block m h g T h m h a, where m h is its mass and T h is the tension in the part of the cord that is attached to it. Thus T h m h (g a) (0:500 kg)(9:8 m/s 6:00 10 m/s ) 4:87 N. (c) For the lighter block m l g T l m l a, so T l m l (g + a) (0:460 kg)(9:8 m/s +6:00 10 m/s ) 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 ar (6:00 10 m/s )(5:00 10 m) 1:0 rad/s. (e) The net torque acting on the pulley is (T h T l )R. Equate this to I and solve for I: I (T h T l )R (4:87 N 4:54 N)(5:00 10 m) 1:38 10 kgm 1:0 rad/s : 75P The angular acceleration for both masses satises mgl which gives mgl 1 I (ml 1 +ml ); g(l l 1 ) l 1 + l (9:80 m/s )(0:80 m 0:0 m) (0:80 m) + (0:0 m) 8:65 rad/s :

CHAPTER 11 ROTATIONAL MOTION 319 The linear acceleration of the left mass is then a 1 l 1 (8:65 rad/s )(0:80 m) 6:9 m/s ; and that of the right mass is a l (8:65 rad/s )(0:0 m) 1:7 m/s : 76P (a) From 1 t we get t. (b) a R Rt : (c) For the lower mass Mg T 1 Ma, which gives T 1 M(g a) M(g Rt ): (d) For the pulley net (T 1 T )R I, which gives T T 1 I R Mg t MR + I R : 77E (a) The speed of v of the mass m after it has descended d 50 cm is given by v ad. where a is calculated in Sample Problem 11-11. Thus v p ad r s (mg)d M + m 4(50 g)(980 cm/s )(50 cm) 400 g + (50 g) (b) The answer is still 1:4 m/s, since it is independent of R. 1:4 10 cm/s 1:4 m/s : 78E P! which is equivalent to 9 ftlb. (100 hp)(746 W/hp) (1800 rev/min)( rad/rev)(1:00 min60 s) 396 Nm ; 79E (a) The work that needs to be done is W K 1 I! 1 MR! 1 (3:0 kg)(1:0 m) [(80 rev/min)( rad/rev)(1 min60 s)] 1:98 10 4 J : (b) The power required is P W t 1:98 104 J 15:0 s 1:3 10 3 W :

30 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. 80E (a) K 1 I! 1 1 m`! 1 m`! : 3 6 (b) Use conservation of energy: K mgh. The center of mass rises by h K mg m`! 6mg `! 6g : 81P (a) The angular speed! of the rotation of the Earth is given by! T, where T 1 da 86400 s. Thus! 86400 s 7:7 10 5 rad/s ; and the angular acceleration required to accelerate the Earth from rest to! in one day is!t. The touque needed is then I I! T (9:71 107 kgm )(7:7 10 5 rad/s) 86400 s 8:17 10 8 Nm ; where we used the data for I obtained in 5E, part (a). (b) As calculated in 5E, part (b), the kinetic energy of the Earth associated with its rotation about its own axis is K :57 10 9 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 P K T :57 109 J 86400 s :97 10 1 kw : 8P Use conservation of energy. The initial energy before falling is E i mg`, 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 E f 1 I! 1 (1 3 m`)! 1 6 mv ; where v!` is the speed of the other end. Thus E i 1 mgl E f 1 6 mv, which gives v p 3g` p 3(9:80 m/s )(1:00 m) 5:4 m/s :

CHAPTER 11 ROTATIONAL MOTION 31 83P Use conservation of mechanical energy. The center of mass is at the midpoint of the cross bar of the H and it drops by `, where ` is the length of any one of the rods. The gravitational potential energy decreases by Mg`, where M is the mass of the body. The initial kinetic energy is zero and the nal kinetic energy may be written 1 I!, where I is the rotational inertia of the body p and! is its angular velocity when it is vertical. Thus 0 Mg` + 1 I! and! Mg`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 (M3)`, where M3 is its mass. The cross bar is a rod that rotates around one end so its contribution is (M3)`3 M`9. The total rotational inertia is I (M`3) + (M`9) 4M`9. The angular velocity is r s r Mg` Mg` 9g! I 4M`9 4` : 84P (a) Use the parallel-axis theorem to nd I: I I cm + Mh 1 MR + Mh 1 (0 kg)(0:10 m) + (0 kg)(0:50 m) 0:15 kgm : (b) Conservation of energy requires that Mgh 1 I!, where! is the angular speed of the cylinder as it passes through the lowest position. Solve for!: r s Mgh (0 kg)(9:80 m/s! )(0:050 m) 11 rad/s : I 0:15 kg m 85P (a) The (centripetal) force exerted on an innitesimal portion of the blade with mass dm located a distance r from the rotational axis is given by df (dm)! r. Thus for the entire blade of mass M and length L the total force is given by F Z df Z! r dm M L Z L 0! r dr M! r L L 0 M! L 1 (110 kg)[(30 rev/min)( rad/rev)(1:00 min60 s)] (7:80 m) 4:8 10 5 N :

3 CHAPTER 11 ROTATIONAL MOTION (b) The torque is! I I t (c) The work done is 1! 3 ML t 1 [(30 rev/min)( rad/rev)(1:00 min60 s) (110 kg)(7:8 m) 3 6:7 s 1:1 10 4 Nm : W K 1 I! 1 1 3 ML! 1 6 (110 kg)(7:80 m) [(30 rev/min)( rad/rev)(1:00 min60 s)] 1:3 10 6 J : 86P Denote the sphere as 1, the pully as, and the hanging mass as 3. From the work-energy theorem (or conservation of mechanical energy) mgh 1 I 1! 1 + 1 I! + 1 mv 3 ; where! 1 v 3 R,! v 3 r, and I I. Solve for v 3 : s s gh v 3 1 + I 1 mr + I mr gh 1 + M3m + Imr : 87P (a) If ` is the length of the chimney then the radial component of the acceleration of the top is given by a r `!, where! is the angular velocity. Use conservation of mechanical energy to nd an expression for! 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 (`) cos. Initially the potential energy is U i Mg(`) and the kinetic energy is zero. Write 1 I! for the kinetic energy when the chimney makes the angle with the vertical. Here I is its rotational inertia. Conservation of energy then leads to Mg` Mg(`) cos + 1 I!, so! (Mg`I)(1 cos ). Thus Mg` a r `! (1 cos ) : I The chimney is rotating about its base, so I Mg`3 and a r 3g(1 cos ).

CHAPTER 11 ROTATIONAL MOTION 33 (b) The tangential component of the acceleration of the chimney top is given by a t `, where is the angular acceleration. Dierentiate! (Mg`I)(1 cos ) with respect to time, replace d!dt with, and replace ddt with! to obtain! (Mg`I)! sin or (Mg`I) sin. Thus a t ` Mg` I sin 3g sin ; where I M`3 was used to obtain the last result. (c) The angle for which a t g is the solution to (3) sin g or sin 3. It is 41:8. 88P (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)( rad/rev)(3:0 s) 7:66 rad/s. (b) Use I, where is the torque and I is the rotational inertia. The contribution of the rod to I is M`1, where M is its mass and ` is its length. The contribution of each ball is m(`), where m is the mass of a ball. The total rotational inertia is M` m` I 1 + 4 (6:40 kg)(1:0 m) 1 + (1:06 kg)(1:0 m) 1:53 kgm : The torque is (1:53 kgm )(7:66 rad/s ) 11:7 Nm : (c) Since the system comes to rest the change in the mechanical energy is simply the negative of the initial kinetic energy K i 1 I! 0 1 (1:53 kgm )[(39:0 rev/s)( rad/rev)] 4:59 10 4 J ; i.e, the amount of mechanical energy dissipated is 4:59 10 4 J. (d) Use! 0 t + 1 t This is 390 64 rev. ()(39:0 rad/s)(3:0 s) + 1 ( 7:66 rad/s )(3:0 s) 390 rad : (e) Only the loss in mechanical energy can still be computed without additional information. It is 4:59 10 4 J no matter how varies with time, as long as the system comes to rest.

34 CHAPTER 11 ROTATIONAL MOTION 89P (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, where 40. Thus its angular acceleration is net I mgl cos (ml 3) 3g cos l 3(9:80 m/s )(cos 40 ) (:0 m) (b) Conservation of energy gives K 1 l I! U mg sin : Thus the angular speed! is 5:6 rad/s :! r mgl sin I r 3g sin l r 3(9:80 m/s )(sin 40 ) :0 m 3:1 rad/s : 90P The work W required is equal to the rotational kinetic energy of the system: W 1 I! 1 [Ma + Ma + M(b a )] 1 M(3a + b )! 1 (0:40 kg)[3(0:30 m) + (0:50 m) ](5:0 rad/s) :6 J : 91P (a) Use conservation of energy. At the top of the hill the kinetic energy is K i 0 and the gravitational potential energy is U i 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 mv + 1 I!, where v is the speed of the car, I is the rotational inertia of the ywheel, and! is its angular velocity. Conservation of energy leads to mgd sin 1 mv + 1 I!. 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 mv + 1 IA v and r mgd sin v m + IA : Use the fact that! 40 rev/s 1:5110 3 rad/s when v 80 km/h : ms to nd the value of A: A!v (1:51 10 3 rad/s)(: m/s) 68:0 rad/m. The rotational inertia of the ywheel (assumed to be a uniform disk) is I 1 MR 1(00 N9:8 m/s )(0:55 m) 3:09 kgm. Then This is 4:1 km/h. v s (800 kg)(9:8 m/s )(1500 m) sin 5 11:7 m/s : 800 kg + (3:09 kgm )(68:0 rad/s)

CHAPTER 11 ROTATIONAL MOTION 35 (b) Dierentiate v mgd sin (m + IA ) with respect to time, replace dvdt with the acceleration a, and replace dddt with v. The result is so Since! Av, Aa Amg sin m + IA va a mgv sin m + IA : mg sin m + IA : (68:0 rad/s)(800 kg)(9:8 m/s ) sin 5 800 kg + (3:09 kgm )(68:0 rad/m) 3:08 rad/s : (c) The energy stored in the ywheel is E 1 I! and the rate at which it is changing is de dt I! IAv (3:09 kgm )(68:0 rad/m)(11:7 m/s)(3:08 rad/s ) 7:57 10 3 W : 9 (a) Rotational kinetic energy is not conserved; (b) For k 0:5, the time to reach the same angular velocity is about 5 s while for k 0:50 it is about 6 s. The nal angular velocities are the same for the two cases, as is the nal rotational kinetic energy. 93 (a) 3:4 10 5 gcm ; (b) :9 10 5 gcm ; (c) 6:3 10 5 gcm ; (d) (1: cm) i + (5:9 cm) j 94 (a) 1:6 m/s ; (b) 4:6 N; (c) 4:9 N