Chapter 10 Solutions - PDF Free Download

Chapter 0 Solutions 0. (a) α ω ω i t.0 rad/s 4.00 rad/s 3.00 s θ ω i t + αt (4.00 rad/s )(3.00 s) 8.0 rad 0. (a) ω ω π rad 365 days π rad 7.3 days *0.3 ω i 000 rad/s α 80.0 rad/s day 4 h day 4 h h 3600 s.99 0 7 rad/s h 3600 s.65 0 6 rad/s (a) ω ω i + αt [000 (80.0)(0.0)] 00 rad/s 0 ω i + αt t ω i α 000 80.0 5.0 s 0.4 (a) Let ω h and ω m be the angular speeds of the hour hand and minute hand, so that π rad ω h h π 6 rad/h and ω m π rad/h Then if θ h and θ m are the angular positions of the hour hand and minute hand with respect to the o'clock position, we have θ h ω h t and θ m ω m t For the two hands to coincide, we need θ m θ h + π n, where n is a positive integer. Therefore, we may write ω m t ω h t π n, or πn πn t n ω m ω h π π 6 n h 000 by Harcourt College Publishers. All rights reserved.

Chapter 0 Solutions Construct the following table: n t n (h) time (h:min:s) 0 0.00 :00:00.09 :05:7.8 :0:55 3 3.7 3:6: 4 4.36 4::49 5 5.45 5:7:6 6 6.55 6:3:44 7 7.64 7:38: 8 8.73 8:43:38 9 9.8 9:49:05 0 0.9 0:54:33 Let θ s and ω s be the angular position and angular speed of the second hand, then ω s π rad/min 0π rad/h and θ s ω s t For all three hands to coincide, we need θ s θ m + πk (k is any positive integer) at any of the times given above. That is, we need ω s t n ω m t n πk, or k ω s ω m π t n 8 π n (3)(4)(59)n to be an integer. This is possible only for n 0 or. Therefore, all three hands coincide only when straight up at o'clock. 0.5 ω i 00 rev.00 min.00 min π rad 60.0 s.00 rev 0π 3 rad/s, ω f 0 (a) t ω f ω i α 0 0π/3.00 s 5.4 s θ ω t ω f + ω i t 0π 6 rad/s 0π s 7.4 rad 6 *0.6 ω i 3600 rev/min 3.77 0 rad/s θ 50.0 rev 3.4 0 rad and ω f 0 ω f ω i + αθ 0 (3.77 0 rad/s) + α(3.4 0 rad) α.6 0 rad/s 000 by Harcourt College Publishers. All rights reserved.

Chapter 0 Solutions 3 0.7 (a) θ t 0 5.00 rad ω t 0 dθ 0.0 + 4.00t dt t 0 0.0 rad/s t 0 α t 0 dω 4.00 rad/s dt t 0 θ t 3.00 s 5.00 + 30.0 + 8.0 53.0 rad ω t 3.00 s dθ 0.0 + 4.00t dt t 3.00 s.0 rad/s t 3.00s α t 3.00 s dω 4.00 rad/s dt t 3.00s *0.8 ω 5.00 rev/s 0.0π rad/s We will break the motion into two stages: () an acceleration period and () a deceleration period. While speeding up, θ ω t While slowing down, 0 + 0.0π rad/s (8.00 s) 40.0π rad θ 0.0π rad/s + 0 ω t (.0 s) 60.0π rad So, θ τotal θ + θ 00π rad 50.0 rev 0.9 θ θ i ω i t + α t and ω ω i + α t are two equations in two unknowns ω i and α. ω i ω α t θ θ i (ω α t)t + α t ωt α t 37.0 rev π rad rev (98.0 rad/s)(3.00 s) α (3.00 s) 3 rad 94 rad (4.50 s )α 000 by Harcourt College Publishers. All rights reserved.

Chapter 0 Solutions 5 *0.0 (a) ω θ t rev π rad day 86400 s 7.7 0 5 rad/s t θ ω 07 π rad 7.7 0 5 rad/s 360.57 0 4 s (48 min) *0. Estimate the tire s radius at 0.50 m and miles driven as 0,000 per year. θ s r.00 04 mi 609 m 0.50 m mi 6.44 0 7 rad yr θ 6.44 0 7 rad rev yr π rad rev.0 07 yr or ~0 7 rev yr *0. Main Rotor: v rω (3.80 m) 450 rev π rad min min rev 60 s 79 m/s v sound v 79 m s 343 m/s 0.5v sound Tail Rotor: v rω (0.50 m) 438 rev π rad min min rev 60 s m/s v sound v m s 343 m/s 0.644v sound 0.3 (a) v rω; ω v r 45.0 m/s 50 m 0.80 rad/s a r v r (45.0 m/s) 50 m 8.0 m/s toward the center of track 0.4 v 36.0 km h h 0 3 m 3600 s km 0.0 m/s ω v r 0.0 m/s 0.50 m 40.0 rad/s 000 by Harcourt College Publishers. All rights reserved.

6 Chapter 0 Solutions 0.5 Given r.00 m, α 4.00 rad/s, ω i 0, and θ i 57.3.00 rad (a) ω ω i + α t 0 + α t At t.00 s, ω (4.00 rad /s )(.00 s) 8.00 rad/s v rω (.00 m)(8.00 rad/s) 8.00 m/s a r rω (.00 m)(8.00 rad/s) 64.0 m/s a t rα (.00 m)(4.00 rad/s ) 4.00 m/s The magnitude of the total acceleration is: a a r + a t (64.0 m/s ) + (4.00 m/s ) 64. m/s The direction of the total acceleration vector makes an angle φ with respect to the radius to point P: a t φ tan tan a 4.00 r 64.0 3.58 (c) θ θ i + ω i t + α t (.00 rad) + (4.00 rad/s )(.00 s) 9.00 rad 0.6 (a) ω v r 5.0 m/s.00 m 5.0 rad/s ω f ω i + α( θ) α ω f ω i ( θ) (5.0 rad/s) 0 39.8 rad/s [(.5 rev)(π rad/rev)] (c) t ω α 5.0 rad/s 39.8 rad/s 0.68 s 0.7 (a) s v t (.0 m/s)(9.00 s) 99.0 m θ s r 99.0 m 0.90 m 34 rad 54.3 rev ω v r.0 m/s 0.90 m 75.9 rad/s. rev/s 000 by Harcourt College Publishers. All rights reserved.

Chapter 0 Solutions 7 0.8 K A + U A K p + U p 6.00 9.80 5.00 (6.00) v p + 6.00 9.80.00 a r P v p 58.8 m /s Radial acceleration at P, g a r v p R 9.4 m/s Tangential acceleration at P, a t g 9.80 m/s 0.9 (a) ω πf π rad 00 rev 60.0 rev/s 6 rad/s v ωr (6 rad/s)(3.00 0 m) 3.77 m/s (c) a r ω r (6) (8.00 0 ) 60 m/s.6 km/s (d) s θr ω t r (6 rad/s)(.00 s)(8.00 0 m) 0. m 0.0 Just before it starts to skid, F r ma r f mv r µ s n µ s mg µ s v rg ω r g (ω ω i)r g αθr g a tθ g µ s (.70 m/s )(π/) 9.80 m/s 0.545 *0. (a) x r cos θ (3.00 m) cos (9.00 rad) (3.00 m) cos 56.73 m y r sin θ (3.00 m) sin (9.00 rad) (3.00 m) sin 56.4 m r xi + yj (.73i +.4j) m 56 360 56. This is between 90.0 and 80, so the object is in the second quadrant. The vector r makes an angle of 56 with the positive x-axis or 4.3 with the negative x-axis. 000 by Harcourt College Publishers. All rights reserved.

8 Chapter 0 Solutions (d) The direction of motion (i.e., the direction of the velocity vector is at 56 + 90.0 46 from the positive x axis. The direction of the acceleration vector is at 56 + 80 336 from the positive x axis. m a 56 x (c) v [(4.50 m/s) cos 46 ]i + [(4.50 m/s) sin 46 ]j v (.85i 4.0j) m/s (e) a v r (4.50 m/s) 3.00 m 6.75 m/s directed toward the center or at 336 a (6.75 m/s )(i cos 336 + j sin 336 ) (6.5i.78j) m/s (f) F ma (4.00 kg)[(6.5i.78j) m/s ] (4.6i.j) N *0. When completely rewound, the tape is a hollow cylinder with a difference between the inner and outer radii of ~ cm. Let N represent the number of revolutions through which the driving spindle turns in 30 minutes (and hence the number of layers of tape on the spool). We can determine N from: N θ π ω( t) π ( rad/s)(30 min)(60 s/min) π rad/rev 86 rev Then, thickness ~ cm N ~0 cm 0.3 m 4.00 kg, r y 3.00 m; m.00 kg, r y.00 m; m 3 3.00 kg, r 3 y 3 4.00 m; ω.00 rad/s about the x-axis (a) I x m r + m r + m r (4.00)(3.00) + (.00)(.00) + (3.00)(4.00) I x 9.0 kg m y K R I xω (9.0)(.00) 84 J 4.00 kg y 3.00 m v r ω (3.00)(.00) 6.00 m/s O x v r ω (.00)(.00) 4.00 m/s v 3 r 3 ω (4.00)(.00) 8.00 m/s.00 kg 3.00 kg y -.00 m y -4.00 m 000 by Harcourt College Publishers. All rights reserved.

Chapter 0 Solutions 9 K m v (4.00)(6.00) 7.0 J K m v (.00)(4.00) 6.0 J K 3 m 3v 3 (3.00)(8.00) 96.0 J K K + K + K 3 7.0 + 6.0 + 96.0 84 J I xω 0.4 v 38.0 m/s ω 5 rad/s RATIO Iω mv 5 mr ω mv RATIO 5 (3.80 0 ) (5) (38.0) 60 0.5 (a) I j m j r j y (m) In this case, 3.00 kg 4 3.00 kg r r r 3 r 4 r (3.00 m) + (.00 m) 3.0 m I [ 3.0 m] [3.00 +.00 +.00 + 4.00]kg 0 3 x (m) 43 kg m K R Iω (43 kg m )(6.00 rad/s).00 kg 4.00 kg.57 0 3 J 0.6 The moment of inertia of a thin rod about an axis through one end is I 3 ML. The total rotational kinetic energy is given as K R I hω h + I mω m, with I h m hl h 3 (60.0 kg)(.70 m) 3 46 kg m, and I m m ml m 3 (00 kg)(4.50 m) 3 675 kg m 000 by Harcourt College Publishers. All rights reserved.

0 Chapter 0 Solutions In addition, ω h (π rad) ( h) h 3600 s.45 0 4 rad/s, while ω m (π rad) ( h) h 3600 s.75 0 3 rad/s. Therefore, K R (46)(.45 0 4 ) + (675)(.75 0 3 ).04 0 3 J 0.7 I Mx + m(l x) x di dx Mx m(l x) 0 (for an extremum) ml x M + m M L m d I dx m + M; therefore I is minimum when ml the axis of rotation passes through x M + m x L x which is also the center of mass of the system. The moment of inertia about an axis passing through x is I CM M ml + m m M + m M + m where µ Mm M + m L µl Mm M + m L 0.8 We assume the rods are thin, with radius much less than L. Call the junction of the rods the origin of coordinates, and the axis of rotation the z-axis. For the rod along the y-axis, I 3 ml from the table. For the rod parallel to the z-axis, the parallel-axis theorem gives I mr + m L 4 ml axis of rotation 000 by Harcourt College Publishers. All rights reserved.

Chapter 0 Solutions In the rod along the x-axis, the bit of material between x and x + dx has mass (m/l)dx and is at distance r x + (L/) from the axis of rotation. The total rotational inertia is: I total 3 ml + 4 ml + L/ L/ (x + L /4)(m/L)dx 7 ml + m L x 3 3 L/ L/ + ml 4 x L/ L/ 7 ml + ml + ml 4 ml *0.9 Treat the tire as consisting of three parts. The two sidewalls are each treated as a hollow cylinder of inner radius 6.5 cm, outer radius 30.5 cm, and height 0.635 cm. The tread region is treated as a hollow cylinder of inner radius 30.5 cm, outer radius 33.0 cm, and height 0.0 cm. Use I m(r + R ) for the moment of inertia of a hollow cylinder. Sidewall: m π [(0.305 m) (0.65 m) ] (6.35 0 3 m)(.0 0 3 kg/m 3 ).44 kg I side (.44 kg) [(0.65 m) + (0.305 m) ] 8.68 0 kg m Tread: m π [(0.330 m) (0.305 m) ] (0.00 m)(.0 0 3 kg/m 3 ).0 kg I tread (.0 kg) [(0.330 m) + (0.305 m) ]. kg m Entire Tire: I total I side + I tread (8.68 0 kg m ) +. kg m.8 kg m 0.30 (a) I I CM + MD MR + MR 3 MR I I CM + MD 5 MR + MR 7 5 MR 000 by Harcourt College Publishers. All rights reserved.

Chapter 0 Solutions *0.3 Model your body as a cylinder of mass 60.0 kg and circumference 75.0 cm. Then its radius is 0.750 m π 0.0 m and its moment of inertia is MR (60.0 kg)(0.0 m) 0.43 kg m ~ 0 0 kg m kg m 0.3 τ 0 mg(3r) Tr T Mg sin 45.0 0 Mg sin 45.0 T m T 3g 530g 3g 77 kg 500 kg(g) sin 45.0 (530)(9.80) N 3r r 500 kg m θ 45 000 by Harcourt College Publishers. All rights reserved.

Chapter 0 Solutions 3 0.33 τ (0.00 m)(.0 N) (0.50 m)(9.00 N) (0.50 m)(0.0 N) 0.0 N 3.55 N m The thirty-degree angle is unnecessary information. 30.0 a O.0 N b 9.00 N Goal Solution G: By simply examining the magnitudes of the forces and their respective lever arms, it appears that the wheel will rotate clockwise, and the net torque appears to be about 5 Nm. O: To find the net torque, we simply add the individual torques, remembering to apply the convention that a torque producing clockwise rotation is negative and a counterclockwise torque is positive. A: τ Fd τ (.0 N)(0.00 m) (0.0 N)(0.50 m) (9.00 N)(0.50 m) τ 3.55 N m The minus sign means perpendicularly into the plane of the paper, or it means clockwise. L: The resulting torque has a reasonable magnitude and produces clockwise rotation as expected. Note that the 30 angle was not required for the solution since each force acted perpendicular to its lever arm. The 0-N force is to the right, but its torque is negative that is, clockwise, just like the torque of the downward 9-N force. 0.34 Resolve the 00 N force into components perpendicular to and parallel to the rod, as F par (00 N) cos 57.0 54.5 N.00 m 0.0 37.0 and 0.0 F perp (00 N) sin 57.0 83.9 N 00 N Torque of F par 0 since its line of action passes through the pivot point. Torque of F perp is τ (83.9 N)(.00 m) 68 N m (clockwise) 000 by Harcourt College Publishers. All rights reserved.

4 Chapter 0 Solutions *0.35 The normal force exerted by the ground on each wheel is n mg 4 (500 kg)(9.80 m/s ) 4 The torque of friction can be as large as 3680 N τ max f max r (µ s n)r (0.800)(3680 N)(0.300 m) 88 N m The torque of the axle on the wheel can be equally as large as the light wheel starts to turn without slipping. *0.36 We calculated the maximum torque that can be applied without skidding in Problem 35 to be 88 N m. This same torque is to be applied by the frictional force, f, between the brake pad and the rotor for this wheel. Since the wheel is slipping against the brake pad, we use the coefficient of kinetic friction to calculate the normal force. τ fr (µ k n)r, so n τ µ k r 88 N m (0.500)(0.0 m) 8.0 03 N 8.0 kn 0.37 m 0.750 kg F 0.800 N (a) τ rf (30.0 m)(0.800 N) 4.0 N m α τ I rf mr 4.0 (0.750)(30.0) 0.0356 rad/s (c) a T α r (0.0356)(30.0).07 m/s *0.38 τ 36.0 N m Iα ω f ω i + α t 0.0 rad/s 0 + α(6.00 s) α 0.00 6.00 rad/s.67 rad/s (a) I τ α 36.0 N m.67 rad/s.6 kg m ω f ω i + α t 0 0.0 + α(60.0) α 0.67 rad/s τ Iα (.6 kg m)(0.67 rad/s ) 3.60 N m 000 by Harcourt College Publishers. All rights reserved.

Chapter 0 Solutions 5 (c) Number of revolutions θ θ i + ω i t + α t During first 6.00 s θ (.67)(6.00) 30. rad During next 60.0 s θ 0.0(60.0) (0.67)(60.0) 99 rad θ total (39 rad) rev π rad 0.39 For m : F y ma y +n m g 0 5.4 rev T I, R n m g 9.6 N m f k µ k n 7.06 N T F x ma x 7.06 N + T (.00 kg)a () m For the pulley τ Iα t T R + T R a MR R T + T (0.0 kg) a T + T (5.00 kg)a () n T n f k T T f k n T m g Mg m g 000 by Harcourt College Publishers. All rights reserved.

6 Chapter 0 Solutions For m : +n m g cos θ 0 n 6.00 kg(9.80 m/s ) cos 30.0 50.9 N f k µ k n 8.3 N 8.3 N T + m g sin θ m a 8.3 N T + 9.4 N (6.00 kg)a (3) (a) Add equations () () and (3): 7.06 N 8.3 N + 9.4 N (3.0 kg)a a 4.0 N 3.0 kg 0.309 m/s T.00 kg (0.309 m/s ) + 7.06 N 7.67 N T 7.67 N + 5.00 kg(0.309 m/s ) 9. N 0.40 I mr (00 kg)(0.500 m).5 kg m ω i 50.0 rev/min 5.4 rad/s 0.500 m n 70.0 N α ω f ω i t 0 5.4 rad/s 6.00 s 0.873 rad/s f µn τ Iα (.5 kg m )( 0.873 rad/s ) 0.9 N m The magnitude of the torque is given by fr 0.9 N m, where f is the force of friction. Therefore, f 0.9 N m 0.500 m, and f µ k n yields µ k f n.8 N 70.0 N 0.3 0.4 I MR.80 kg(0.30 m) 0.84 kg m τ Iα (a) F a (4.50 0 m) 0 N(0.30 m) 0.84 kg m (4.50 rad/s ) F a (0.89 N m + 38.4 N m) 4.50 0 m 87 N F b (.80 0 m) 38.4 N m 0.89 N m F b.40 kn 000 by Harcourt College Publishers. All rights reserved.

Chapter 0 Solutions 7 0.4 We assume the rod is thin. For the compound object I 3 M rodl + 5 M ballr + M ball D I 3.0 kg (0.40 m) + 5 0.0 kg(4.00 0 m) + 0.0 kg(0.80 m) I.60 kg m (a) K f + U f K i + U i + E Iω + 0 0 + M rod g(l/) + M ball g(l + R) + 0 (.60 kg m ) ω.0 kg(9.80 m/s )(0.0 m) + 0.0 kg(9.80 m/s )(0.80 m) (.60 kg m ) ω 56.3 J ω 8.38 rad/s (c) v rω (0.80 m)8.38 rad/s.35 m/s (d) v v i + a(y y i ) v 0 + (9.80 m/s )(0.80 m).34 m/s The speed it attains in swinging is greater by.35/.34.0040 times 0.43 Choose the zero gravitational potential energy at the level where the masses pass. K f + U gf K i + U gi + E m v + m v + Iω 0 + m gh i + m gh i + 0 (5.0 + 0.0) v + v (3.00)R 5.0(9.80)(.50) + 0.0(9.80)(.50) R (6.5 kg) v 73.5 J v.36 m/s 000 by Harcourt College Publishers. All rights reserved.

8 Chapter 0 Solutions 0.44 Choose the zero gravitational potential energy at the level where the masses pass. K f + U gf K i + U gi + E m v + m v + Iω 0 + m gh i + m gh i + 0 (m + m ) v + v MR R m g d + m g d m + m + M v (m m )g d v (m m )gd m + m + M 0.45 (a) 50.0 T 50.0 9.80 a F g TR Iα I a R I MR 0.0938 kg m ω 3 kg 50.0 T 5.0 TR I n T T.4 N 0.50 m a 50.0.4 5.0 7.57 m/s v a(y i 0) 9.53 m/s 000 by Harcourt College Publishers. All rights reserved.

Chapter 0 Solutions 9 Use conservation of energy: (K + U) i (K + U) f mgh mv + Iω mgh mv + I v R v m + I R v mgh m + I R (50.0 N)(6.00 m) 5.0 kg + 0.0938 (0.50) 9.53 m/s Goal Solution G: Since the rotational inertia of the reel will slow the fall of the weight, we should expect the downward acceleration to be less than g. If the reel did not rotate, the tension in the string would be equal to the weight of the object; and if the reel disappeared, the tension would be zero. Therefore, T < mg for the given problem. With similar reasoning, the final speed must be less than if the weight were to fall freely: v f < gy m/s O: We can find the acceleration and tension using the rotational form of Newton s second law. The final speed can be found from the kinematics equation stated above and from conservation of energy. Free-body diagrams will greatly assist in analyzing the forces. A: (a) Use τ Iα to find T and a. First find I for the reel, which we assume to be a uniform disk: I MR 3.00 kg (0.50 m) 0.0938 kg m The forces on the reel are shown, including a normal force exerted by its axle. From the diagram, we can see that the tension is the only unbalanced force causing the reel to rotate. T 6.00 m 50.0 N 000 by Harcourt College Publishers. All rights reserved.

0 Chapter 0 Solutions τ Iα becomes n(0) + F g (0) + T(0.50 m) (0.0938 kg m )(a/0.50 m) () where we have applied a t rα to the point of contact between string and reel. The falling weight has mass m F g g 50.0 N 9.80 m/s 5.0 kg For this mass, F y ma y becomes +T 50.0 N (5.0 kg)( a) () Note that since we have defined upwards to be positive, the minus sign shows that its acceleration is downward. We now have our two equations in the unknowns T and a for the two linked objects. Substituting T from equation () into equation (), we have: [50.0 N (5.0 kg)a](0.50 m) 0.0938 kg m a 0.50 m.5 N m (.8 kg m)a (0.375 kg m)a.5 N m a(.65 kg m) or a 7.57 m/s and T 50.0 N 5.0 kg(7.57 m/s ).4 N For the motion of the weight, v f v i + a(y f y i ) 0 + (7.57 m/s )(6.00 m) v f 9.53 m/s The work-energy theorem can take account of multiple objects more easily than Newton's second law. Like your bratty cousins, the work-energy theorem keeps growing between visits. Now it reads: (K + K,rot + U g + U g ) i (K + K,rot + U g + U g ) f 0 + 0 + m gy i + 0 m v f + I ω f + 0 + 0 Now note that ω v r as the string unwinds from the reel. Making substitutions: 50.0 N(6.00 m) (5.0 kg) v f + (0.0938 kg m ) v f 0.50 m 300 N m (5.0 kg) v f + (.50 kg) v f v f (300 N m) 6.60 kg 9.53 m/s L: As we should expect, both methods give the same final speed for the falling object. The acceleration is less than g, and the tension is less than the object s weight as we predicted. Now that we understand the effect of the reel s moment of inertia, this problem solution could be applied to solve other real-world pulley systems with masses that should not be ignored. 000 by Harcourt College Publishers. All rights reserved.

Chapter 0 Solutions 0.46 τ θ Iω (5.0 N m)(5.0 π) (0.30 kg m ) ω ω 90 rad/s 30.3 rev/s 0.47 From conservation of energy, I v r + mv mgh I v r mgh mv I mr gh v 0.48 E Iω 3000 π MR 60.0 P E t.00 04 J/s 6.7 0 6 J t E P 6.7 06 J.00 0 4 67 s 0.3 min J/s 0.49 (a) Find the velocity of the CM (K + U) i (K + U) f 0 + mgr Iω Pivot R ω mgr I mgr 3 mr g v CM R 4g 3R Rg 3 v L v CM 4 Rg 3 (c) v CM mgr m Rg 000 by Harcourt College Publishers. All rights reserved.

Chapter 0 Solutions *0.50 The moment of inertia of the cylinder is I mr (8.6 kg)(.50 m) 9.8 kg m and the angular acceleration of the merry-go-round is found as α τ I (Fr) I (50.0 N)(.50 m) (9.8 kg m ) 0.87 rad/s At t 3.00 s, we find the angular velocity ω ω i + α t ω 0 + (0.87 rad/s )(3.00 s).45 rad/s and K Iω (9.8 kg m )(.45 rad/s) 76 J 0.5 mg l sin θ 3 ml α α 3 g l sin θ a t 3 g l sin θ r Then 3 g l r > g sin θ t θ a t t θ g sinθ g for r > 3 l About /3 the length of the chimney will have a tangential acceleration greater than g sin θ. *0.5 The resistive force on each ball is R DρAv. Here v rω, where r is the radius of each ball s path. The resistive torque on each ball is τ rr, so the total resistive torque on the three ball system is τ total 3rR. The power required to maintain a constant rotation rate is P τ total ω 3rRω. This required power may be written as P τ total ω 3r [DρA(rω) ]ω (3r 3 DAω 3 )ρ With ω π rad rev 0 3 rev min min 60.0 s 000π 30.0 rad/s, P 3(0.00 m) 3 (0.600)(4.00 0 4 m )(000π/30.0 s) 3 ρ or P (0.87 m 5 /s 3 )ρ where ρ is the density of the resisting medium. 000 by Harcourt College Publishers. All rights reserved.

Chapter 0 Solutions 3 (a) In air, ρ.0 kg/m 3, and P (0.87 m 5 /s 3 )(.0 kg/m 3 ) 0.99 N m/s 0.99 W In water, ρ 000 kg/m 3 and P 87 W 0.53 (a) I MR (.00 kg)(7.00 0 m) 4.90 0 3 kg m α τ I 0.600 4.90 0 3 rad/s α ω t t ω α 00 π 60.03 s θ αt ( rad/s)(.03 s) 64.7 rad 0.3 rev 0.54 For a spherical shell 3 dm r 3 [(4πr dr)ρ]r I di 3 (4πr ) r ρ(r)dr I R 0 3 (4πr4 ) 4..6 r R 0 3 kg m 3 dr 3 4π(4. 0 3 ) R5 5 3 4π(.6 0 3 ) R5 6 I 8π 3 (03 ) R 5 4. 5.6 6 M dm R 4πr 4..6 r 0 R 03 dr 4π 0 3 4. 3.6 4 R 3 I MR 8π 3 (03 ) R 5 4. 5.6 6 4π 0 3 R 3 R 4. 3.6 4 3.907.83 0.330 000 by Harcourt College Publishers. All rights reserved.

Chapter 0 Solutions 5 0.55 (a) W K Iω Iω i I(ω ω i ) where I mr (.00 kg)(0.500 m) 8.00 rad 0 4.00 J s t ω 0 α ωr a (8.00 rad/s)(0.500 m).50 m/s.60 s (c) θ θ i + ω i t + αt ; θ i 0; ω i 0 θ αt.50 m/s 0.500 m (.60 s) 6.40 rad s rθ (0.500 m)(6.40 rad) 3.0 m < 4.00 m Yes 0.56 (a) I mr (00 kg)(0.300 m) 9.00 kg m ω (000 rev/min)( min/60 s)(π rad)/ rev 05 rad/s W K Iω (9.00)(04.7) 49.3 kj (c) ω f 500 rev/min min/60 s(π rad/ rev) 5.4 rad/s K f Iω.3 kj W K.3 kj 49.3 kj 37.0 kj *0.57 α 0.0 rad/s (5.00 rad/s 3 )t dω/dt ω dω t [ 0.0 5.00t]dt 0.0t.50t ω 65.0 rad/s 65.0 0 ω dθ dt 65.0 rad/s (0.0 rad/s )t (.50 rad/s 3 )t (a) At t 3.00 s, ω 65.0 rad/s (0.0 rad/s )(3.00 s) (.50 rad/s 3 )(9.00 s ).5 rad/s 000 by Harcourt College Publishers. All rights reserved.

6 Chapter 0 Solutions θ dθ t ω dt t [65.0 rad/s (0.0 rad/s )t (.50 rad/s 3 )t ]dt 0 0 o θ (65.0 rad/s)t (5.00 rad/s )t (0.833 rad/s 3 )t 3 At t 3.00 s, θ (65.0 rad/s)(3.00 s) (5.00 rad/s )9.00 s (0.833 rad/s 3 )7.0 s 3 θ 8 rad 0.58 (a) MK MR, K R MK ML, K L 3 6 (c) MK 5 MR, K R 5 y 0.59 (a) Since only conservative forces act, E 0, so K f + U f K i + U i Iω + 0 0 + mg L, where I 3 ml L ω 3g/L pivot x τ Iα so that in the horizontal, mg L ml 3 α α 3g L (c) a x a r rω L ω 3g a y a t rα α L 3g 4 (d) Using Newton's second law, we have R x ma x 3 mg R y mg ma y or R y 4 mg 000 by Harcourt College Publishers. All rights reserved.

Chapter 0 Solutions 7 0.60 The first drop has a velocity leaving the wheel given by mv i mgh, so v gh (9.80 m/s )(0.540 m) 3.5 m/s The second drop has a velocity given by v gh (9.80 m/s )(0.50 m) 3.6 m/s From ω v r, we find ω v r 3.5 m/s 0.38 m 8.53 rad/s and ω v r 3.6 m/s 0.38 m 8.9 rad/s or α ω ω θ (8.9 rad/s) (8.53 rad/s) 4π 0.3 rad/s 0.6 At the instant it comes off the wheel, the first drop has a velocity v, directed upward. The magnitude of this velocity is found from K i + U gi K f + U gf mv + 0 0 + mgh or v gh and the angular velocity of the wheel at the instant the first drop leaves is ω v R gh R Similarly for the second drop: v gh and ω v R gh R The angular acceleration of the wheel is then α ω ω θ gh /R gh /R (π) g(h h ) πr 0.6 Work done Fs (5.57 N)(0.800 m) 4.46 J A and Work K Iω f Iω i (The last term is zero because the top starts from rest.) F Thus, 4.46 J (4.00 0 4 kg m ) ω f A 000 by Harcourt College Publishers. All rights reserved.

Chapter 0 Solutions 9 0.63 K f Mv f + Iω f : U f Mgh f 0; K i Mv i + Iω i 0 U i (Mgh) i : f µn µmg cos θ; ω v r ; h d sin θ and I mr (a) E E f E i or fd K f + U f K i U i fd Mv f + Iω f Mgh (µmg cos θ)d Mv + (mr /)(v /r )/ Mgd sin θ M + m v Mgd v d 4gd v Mgd sin θ (µmg cos θ)d or (sin θ µ cos θ) (m/) + M M (m + M) (sin θ µ cos θ) / v v i as, v d ad a v d d g M (sin θ µ cos θ) m + M 0.64 (a) E 5 MR (ω ) E 5 (5.98 04 )(6.37 0 6 ) π.57 0 9 J 86400 de dt dt d 5 MR π T 5 MR (π) ( T 3 ) dt dt π 5 MR T dt T dt (.57 0 9 J) 0 0 6 s 86400 s 3.6 0 7 s de dt.63 07 J/day (86400 s/day) 000 by Harcourt College Publishers. All rights reserved.

30 Chapter 0 Solutions 0.65 θ ωt t θ ω (3.0 /360 ) rev 900 rev/60 s v 0.800 m 39 m/s 0.00574 s 0.00574 s v θ 3 ω 0.66 (a) Each spoke counts as a thin rod pivoted at one end. d I MR + n mr 3 By the parallel-axis theorem, I MR + nmr 3 + (M + nm)r MR + 4 nmr 3 *0.67 Every particle in the door could be slid straight down into a high-density rod across its bottom, without changing the particle s distance from the rotation axis of the door. Thus, a rod 0.870 m long with mass 3.0 kg, pivoted about one end, has the same rotational inertia as the door: I 3 ML 3 (3.0 kg)(0.870 m) 5.80 kg m The height of the door is unnecessary data. 000 by Harcourt College Publishers. All rights reserved.

Chapter 0 Solutions 3 0.68 τ f will oppose the torque causing the motion: τ Iα TR τ f τ f TR Iα () M Now find T, I and α in given or known terms and substitute into equation () F y T mg ma then T m(g a) () R/ R/ m y also y v i t + at a y t (3) and α a R y Rt (4) I M R + R 5 8 MR (5) Substituting (), (3), (4) and (5) into () we find τ f m g y t R 5 8 MR y (Rt ) R m g y t 5 My 4 t 0.69 (a) While decelerating, τ f Iα' (0 000 kg m ).00 rev/min (π rad/rev)( min/60 s) 0.0 s τ f 49 N m While accelerating, τ Iα or τ τ f I( ω/ t) τ 49 N m + (0 000 kg m ) 0.00 rev/min (π rad/rev)( min/60 s).0 s τ.6 0 3 N m P τ f ω (49 N m) 0.0 rev min π rad rev min 60 s 439 W ( 0.6 hp) 000 by Harcourt College Publishers. All rights reserved.

3 Chapter 0 Solutions 0.70 (a) W K + U R W K f K i + U f U i 0 mv + Iω mgd sin θ kd m k ω (I + mr ) mgd sin θ + kd θ ω mgd sin θ + kd I + mr ω (0.500 kg)(9.80 m/s )(0.00 m)(sin 37.0 ) + (50.0 N/m)(0.00 m).00 kg m + (0.500 kg)(0.300 m) ω.8 +.00.05 3.04.74 rad/s 0.7 (a) m g T m a.00 m/s T m (g a) (0.0 kg)(9.80.00)m/s 56 N T T T m g sin 37.0 + m a T (5.0 kg)(9.80 sin 37.0 +.00)m/s 8 N 5.0 kg m 37.0 m 0.0 kg (T T )R Iα I a R I (T T )R a (56 8)N(0.50 m).00 m/s.7 kg m Goal Solution G: In earlier problems, we assumed that the tension in a string was the same on either side of a pulley. Here we see that the moment of inertia changes that assumption, but we should still expect the tensions to be similar in magnitude (about the weight of each mass ~50 N), and T > T for the pulley to rotate clockwise as shown. If we knew the mass of the pulley, we could calculate its moment of inertia, but since we only know the acceleration, it is difficult to estimate I. We at least know that I must have units of kgm, and a 50-cm disk probably has a mass less than 0 kg, so I is probably less than 0.3 kgm. O: For each block, we know its mass and acceleration, so we can use Newton s second law to find the net force, and from it the tension. The difference in the two tensions causes the pulley to rotate, so this net torque and the resulting angular acceleration can be used to find the pulley s moment of inertia. 000 by Harcourt College Publishers. All rights reserved.

Chapter 0 Solutions 33 A: (a) Apply F ma to each block to find each string tension. The forces acting on the 5-kg block are its weight, the normal support force from the incline, and T. Taking the positive x axis as directed up the incline, F x ma x yields: (m g) x + T m (+a) (5.0 kg)(9.80 m/s ) sin 37 + T (5.0 kg)(.00 m/s ) T 8 N Similarly for the counterweight, we have F y ma y, or T m g m ( a) T (0.0 kg)(9.80 m/s ) (0.0 kg)(.00 m/s ) So, T 56 N Now for the pulley, τ r(t T ) Iα. We may choose to call clockwise positive. The angular acceleration is α a r.00 m/s 0.50 m 8.00 rad/s τ Iα or (0.50 m)(56 N 8 N) I(8.00 rad/s ) I 9.38 N m 8.00 rad/s.7 kg m L: The tensions are close to the weight of each mass and T > T as expected. However, the moment of inertia for the pulley is about 4 times greater than expected. Unless we made a mistake in solving this problem, our result means that the pulley has a mass of 37.4 kg (about 80 lb), which means that the pulley is probably made of a dense material, like steel. This is certainly not a problem where the mass of the pulley can be ignored since the pulley has more mass than the combination of the two blocks! 0.7 For the board just starting to move, τ Iα mg l cos θ 3 ml α R mg α 3 g l cos θ 000 by Harcourt College Publishers. All rights reserved.

34 Chapter 0 Solutions The tangential acceleration of the end is a t lα 3 g cos θ Its vertical component is a y a t cos θ 3 g cos θ If this is greater than g, the board will pull ahead of the ball in falling: (a) 3 g cos θ g cos θ 3 so cos θ 3 and θ 35.3 When θ 35.3 ( cos θ /3), the cup will land underneath the release-point of the ball if r c l cos θ l cos θ cos θ l 3 cos θ (c) When l.00 m, and θ 35.3 (.00 m) r c 3 /3 0.86 m which is (.00 m 0.86 m) 0.84 m from the moving end 0.73 At t 0, ω 3.50 rad/s ω 0 e 0. Thus, ω 0 3.50 rad/s At t 9.30 s, ω.00 rad/s ω 0 e σ(9.30 s), yielding σ 6.0 0 s (a) α dω dt At t 3.00 s, d(ω 0e σt ) dt ω 0 ( σ)e σt α (3.50 rad/s)( 6.0 0 s )e 3.00(6.0 0 ) 0.76 rad/s θ t ω 0 e σt dt ω 0 0 σ [e σt ] ω 0 σ [ e σt ] At t.50 s, 3.50 rad/s θ (6.0 0 )/s [ e (6.0 0 )(.50) ] 8. rad.9 rev 000 by Harcourt College Publishers. All rights reserved.

36 Chapter 0 Solutions 0.74 Consider the total weight of each hand to act at the center of gravity (mid-point) of that hand. Then the total torque (taking CCW as positive) of these hands about the center of the clock is given by L h τ m h g L m sin θ h m m g sin θ m g (m hl h sin θ h + m m L m sin θ m ) If we take t 0 at o'clock, then the angular positions of the hands at time t are θ h ω h t, where ω h π 6 rad/h and θ m ω m t, where ω m π rad/h Therefore, τ 4.90 m s [(60.0 kg)(.70 m) sin (πt/6) + (00 kg)(4.50 m) sin πt] or τ 794 N m[sin(πt/6) +.78 sin πt], where t is in hours. (a) (i) At 3:00, t 3.00 h, so τ 794 N m [sin(π/) +.78 sin 6π] 794 N m (ii) At 5:5, t 5 h + 5 60 h 5.5 h, and substitution gives: τ 50 N m (iii) At 6:00, τ 0 N m (iv) At 8:0, τ 60 N m (v) At 9:45, τ 940 N m The total torque is zero at those times when sin(πt/6) +.78 sin πt 0 We proceed numerically, to find 0, 0.55955,..., corresponding to the times :00:00 :30:55 :58:9 :3:3 :57:0 :33:5 :56:9 3:33: 3:56:55 4:3:4 4:58:4 5:30:5 6:00:00 6:9:08 7:0:46 7:7:36 8:03:05 8:6:38 9:03:3 9:6:35 0:0:59 0:7:9 :0:4 :9:05 000 by Harcourt College Publishers. All rights reserved.