a +3bt 2 4ct 3) =6bt 12ct 2. dt 2. (a) The second hand of the smoothly running watch turns through 2π radians during 60 s. Thus,

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1 1. a) Eq leads to ω d at + bt 3 ct 4) a +3bt 4ct 3. dt b) And Eq gives α d a +3bt 4ct 3) 6bt 1ct. dt. a) The second hand of the smoothly running watch turns through π radians during 60 s. Thus, ω π rad/s. 60 b) The minute hand of the smoothly running watch turns through π radians during 3600 s. Thus, ω π rad/s. c) The hour hand of the smoothly running 1-hour watch turns through π radians during 4300 s. Thus, ω π rad/s. 3. a) The time for one revolution is the circumference of the orbit divided by the speed v of the Sun: T πr/v, where R is the radius of the orbit. We convert the radius: R ly ) km/ly ) km where the ly km conversion can be found in Appendix D or figured from basics knowing the speed of light). Therefore, we obtain T π km ) s. 50 km/s b) The number of revolutions N is the total time t divided by the time T for one revolution; that is, N t/t. We convert the total time from years to seconds and obtain y ) s/y ) N s 4. If we make the units explicit, the function is θ 4.0rad/s)t 3.0 rad/s ) t rad/s 3) t 3 but generally we will proceed as shown in the problem letting these units be understood. Also, in our manipulations we will generally not display the coefficients with their proper number of significant figures.

2 a) Eq leads to ω d 4t 3t + t 3) 4 6t +3t. dt Evaluating this at t s yields ω 4.0 rad/s. b) Evaluating the expression in part a) at t 4 s gives ω 4 8 rad/s. c) Consequently, Eq gives α avg ω 4 ω 4 1 rad/s. d) And Eq gives α dω dt d 4 6t +3t ) 6+6t. dt Evaluating this at t s produces α 6.0 rad/s. e) Evaluating the expression in part d) at t 4 s yields α 4 18 rad/s. We note that our answer for α avg does turn out to be the arithmetic average of α and α 4 but point out that this will not always be the case. 5. If we make the units explicit, the function is θ rad + 4 rad/s ) t + rad/s 3) t 3 but in some places we will proceed as indicated in the problem by letting these units be understood. a) We evaluate the function θ at t 0 to obtain θ 0 rad. b) The angular velocity as a function of time is given by Eq. 11-6: ω dθ dt 8 rad/s ) t + 6 rad/s 3) t which we evaluate at t 0 to obtain ω 0 0. c) For t 4 s, the function found in the previous part is ω 4 8)4) + 6)4) 18 rad/s. If we round this to two figures, we obtain ω rad/s. d) The angular acceleration as a function of time is given by Eq. 11-8: α dω dt 8 rad/s + 1 rad/s 3) t which yields α 8 + 1)) 3 rad/s at t s. e) The angular acceleration, given by the function obtained in the previous part, depends on time; it is not constant.

3 6. a) To avoid touching the spokes, the arrow must go through the wheel in not more than 1/8 rev t.5 rev/s s. The minimum speed of the arrow is then 0 cm v min 400 cm/s 4.0 m/s s b) No there is no dependence on radial position in the above computation. 7. Applying Eq. -15 to the vertical axis with +y downward) we obtain the free-fall time: y v 0y t ) gt t 1.4 s. 9.8 Thus, by Eq. 11-5, the magnitude of the average angular velocity is ω avg.5)π) 11 rad/s a) We assume the sense of rotation is positive. Applying Eq. 11-1, we obtain ω ω 0 + αt α / rev/min. b) And Eq gives θ 1 ω 0 + ω) t 1 ) ) 60 which yields θ 40 rev. 9. We assume the sense of initial rotation is positive. Then, with ω 0 > 0 and ω 0 since it stops at time t), our angular acceleration is negative-valued. a) The angular acceleration is constant, so we can apply Eq ω ω 0 + αt). To obtain the requested units, we have t 30/ min. Thus, rev/min α 66.7 rev/min min b) We use Eq : θ ω 0 t + 1 αt 33.33)0.50) )0.50) 8.3 rev.

4 10. We assume the sense of initial rotation is positive. Then, with ω rad/s and ω 0 since it stops at time t), our angular acceleration deceleration ) will be negative-valued: α 4.0 rad/s. a) We apply Eq to obtain t. b) And Eq gives ω ω 0 + αt t s. θ 1 ω 0 + ω) t ) 30) which yields θ 1800 rad. Alternatively, Eq could be used if it is desired to only use the given information as opposed to using the result from part a)) in obtaining θ. If using the result of part a) is acceptable, then any angular equation in Table 11-1 except Eq. 11-1) can be used to find θ. 11. We apply Eq twice, assuming the sense of rotation is positive. We have ω > 0 and α < 0. Since the angular velocity at t 1 min is ω )50) 5 rev/min, we have 5 50 ω 1 ω 0 + αt α 5 rev/min. 1 Next, between t 1 min and t min we have the interval t 1 min. Consequently, the angular velocity at t min is ω ω 1 + α t 5 + 5)1) 00 rev/min. 1. We assume the sense of rotation is positive, which since it starts from rest) means all quantities angular displacements, accelerations, etc.) are positive-valued. a) The angular acceleration satisfies Eq : 5 rad 1 α5.0s) α.0 rad/s. b) The average angular velocity is given by Eq. 11-5: ω avg θ t 5 rad 5.0s 5.0 rad/s. c) Using Eq. 11-1, the instantaneous angular velocity at t 5.0 sis ω.0 rad/s ) 5.0 s) 10 rad/s.

5 d) According to Eq , the angular displacement at t 10 s is θ ω αt )10) 100 rad. Thus, the displacement between t 5 s and t 10 s is θ rad. 13. We take t 0 at the start of the interval and take the sense of rotation as positive. Then at the end of the t 4.0s interval, the angular displacement is θ ω 0 t + 1 αt. We solve for the angular velocity at the start of the interval: ω 0 θ 1 αt t 10 rad rad/s ) 4.0s) 4.0s 4 rad/s. We now use ω ω 0 + αt Eq. 11-1) to find the time when the wheel is at rest: t ω 0 α 4 rad/s 8.0 s. 3.0 rad/s That is, the wheel started from rest 8.0 s before the start of the described 4.0 s interval. 14. The wheel starts turning from rest ω 0 0)att 0, and accelerates uniformly at α.00 rad/s. Between t 1 and t it turns through θ 90.0 rad, where t t 1 t 3.00 s. a) We use Eq with a slight change in notation) to describe the motion for t 1 t t : θ ω 1 t + 1 α t) ω 1 θ t α t which we plug into Eq. 11-1, set up to describe the motion during 0 t t 1 : ω 1 ω 0 + αt 1 θ t α t αt )3.00).00)t 1 yielding t s. b) Plugging into our expression for ω 1 in previous part) we obtain ω 1 θ t α t )3.00) 7.0 rad/s.

6 15. The problem has implicitly) specified the positive sense of rotation. The angular acceleration of magnitude 0.5 rad/s in the negative direction is assumed to be constant over a large time interval, including negative values for t). a) We specify θ max with the condition ω 0 this is when the wheel reverses from positive rotation to rotation in the negative direction). We obtain θ max using Eq : θ max ω o α rad. 0.5) b) We find values for t 1 when the angular displacement relative to its orientation at t 0)isθ 1 rad or.09 rad if we wish to keep track of accurate values in all intermediate steps and only round off on the final answers). Using Eq and the quadratic formula, we have θ 1 ω o t αt 1 t 1 ω o ± ωo +θ 1 α α which yields the two roots 5.5 s and 3 s. c) We find values for t when the angular displacement relative to its orientation at t 0)isθ 10.5 rad. Using Eq and the quadratic formula, we have θ ω o t + 1 αt t ω o ± ωo +θ α α which yields the two roots.1 s and 40 s. d) With radians and seconds understood, the graph of θ versus t is shown below with the points found in the previous parts indicated as small circles).

7 40 θ t The wheel starts turning from rest ω 0 0)att 0, and accelerates uniformly at α>0, which makes our choice for positive sense of rotation. At t 1 its angular velocity is ω rev/s, and at t its angular velocity is ω +15 rev/s. Between t 1 and t it turns through θ 60 rev, where t t 1 t. a) We find α using Eq : ω ω1 +α θ α ) which yields α 1.04 rev/s which we round off to 1.0 rev/s. b) We find t using Eq : θ 1 ω 1 + ω ) t t 60) 4.8 s c) We obtain t 1 using Eq. 11-1: ω 1 ω 0 + αt 1 t s d) Any equation in Table 11-1 involving θ can be used to find θ 1 the angular displacement during 0 t t 1 ); we select Eq ω1 ω0 +αθ 1 θ rev. 1.04)

8 17. The wheel has angular velocity ω rad/s rev/s at t 0, and has constant value of angular acceleration α<0, which indicates our choice for positive sense of rotation. At t 1 its angular displacement relative to its orientation at t 0)isθ 1 +0 rev, and at t its angular displacement is θ +40 rev and its angular velocity is ω 0. a) We obtain t using Eq : θ 1 ω 0 + ω ) t t 40) 0.39 which yields t 335 s which we round off to t 340 s. b) Any equation in Table 11-1 involving α can be used to find the angular acceleration; we select Eq θ ω t 1 αt α 40) 335 which yields α rev/s which we convert to α rad/s. c) Using θ 1 ω 0 t αt 1 Eq ) and the quadratic formula, we have t 1 ω 0 ± ω 0 +θ 1α α 0.39 ± ) ) which yields two positive roots: 98 s and 57 s. Since the question makes sense only if t 1 <t we conclude the correct result is t 1 98s. 18. The wheel starts turning from rest ω 0 0)att 0, and accelerates uniformly at α +4.0 rad/s, which makes our choice for positive sense of rotation. At t 1 its angular displacement relative to its orientation at t 0)isθ 1, and at t its angular velocity is θ, where θ θ 1 θ 80 rad. Also, t t 1 t 4.0 s. a) We find the angular velocity at t 1 using Eq set up to describe the interval t 1 t t ). θ ω 1 t + 1 α t) ω )4.0) 4.0 which yields ω 1 1 rad/s. b) We obtain t 1 using Eq. 11-1: ω 1 ω 0 + αt 1 t s. 4.0

9 19. The magnitude of the acceleration is given by a ω r Eq. 11-3) where r is the distance from the center of rotation and ω is the angular velocity. We convert the given angular velocity to rad/s: ω rev/min)π rad/rev) 60 s/min 3.49 rad/s. Therefore, a 3.49 rad/s ) 0.15 m) 1.8 m/s. The acceleration vector is toward the center of the record. 0. a) We obtain ω b) Using Eq , we have rev/min)π rad/rev) 60 s/min v rω 15)3.49) 5 cm/s rad/s. c) Similarly, when r 7.4 cmwefindv rω 6 cm/s. The goal of this exercise to observe what is and is not the same at different locations on a body in rotational motion ω is the same, v is not), as well as to emphasize the importance of radians when working with equations such as Eq With v /3600) 13.9 m/s, Eq leads to ω v r rad/s a) We obtain ω 00 rev/min)π rad/rev) 60 s/min b) With r 1.0/ 0.60 m, Eq leads to 0.9 rad/s. v rω 0.60)0.9) 1.6 m/s. c) With t 1 min, ω 1000 rev/min and ω o 00 rev/min, Eq gives α ω ω o 800 rev/min. t d) With the same values used in part c), Eq becomes θ 1 ω o + ω) t ) 1) 600 rev.

10 3. a) Using Eq. 11-6, the angular velocity at t 5.0s is ω dθ dt d 0.30t ) 0.30)5.0) 3.0 rad/s. t5.0 dt t5.0 b) Eq gives the linear speed at t 5.0s: v ωr 3.0 rad/s)10 m) 30 m/s. c) The angular acceleration is, from Eq. 11-8, α dω dt d dt 0.60t) 0.60 rad/s. Then, the tangential acceleration at t 5.0s is, using Eq. 11-, a t rα 10 m) 0.60 rad/s ) 6.0 m/s. d) The radial centripetal) acceleration is given by Eq. 11-3: a r ω r 3.0 rad/s) 10 m) 90 m/s. 4. a) Converting from hours to seconds, we find the angular velocity assuming it is positive) from Eq : ω v r km/h ) ) 1.00 h 3600 s rad/s. km b) The radial or centripetal) acceleration is computed according to Eq. 11-3: a r ω r rad/s ) m ) 0. m/s. c) Assuming the angular velocity is constant, then the angular acceleration and the tangential acceleration vanish, since α dω dt 0 and a t rα a) In the time light takes to go from the wheel to the mirror and back again, the wheel turns through an angle of θ π/ rad. That time is t l c 500 m) m/s s so the angular velocity of the wheel is ω θ t rad rad/s. s

11 b) If r is the radius of the wheel, the linear speed of a point on its rim is v ωr rad/s ) 0.05 m) 190 m/s. 6. a) The angular acceleration is α ω t rev/min. h)60 min/1h) 1.14 rev/min. b) Using Eq with t.)60) 13 min, the number of revolutions is θ ω 0 t + 1 αt 150 rev/min)13 min) rev. c) With r 500 mm, the tangential acceleration is a t αr 1.14 rev/min ) 13 min) 1.14 rev/min ) ) ) π rad 1 min 500 mm) 1 rev 60 s which yields a t 0.99 mm/s. d) With r 0.50 m, the radial or centripetal) acceleration is given by Eq. 11-3: a r ω r )) π rad/rev 75 rev/min) 0.50 m) 1 min/60 s which yields a r 31 in SI units and is seen to be much bigger than a t. Consequently, the magnitude of the acceleration is a a r + a t a r 31m/s. 7. a) Earth makes one rotation per day and 1 d is 4 h)3600 s/h) s, so the angular speed of Earth is ω π rad s rad/s. b) We use v ωr, where r is the radius of its orbit. A point on Earth at a latitude of 40 moves along a circular path of radius r R cos 40, where R is the radius of Earth m). Therefore, its speed is v ω R cos 40 ) rad/s ) m ) cos m/s.

12 c) At the equator and all other points on Earth) the value of ω is the same rad/s). d) The latitude is 0 and the speed is v ωr rad/s ) m ) 463 m/s. 8. a) The tangential acceleration, using Eq. 11-, is a t αr 14. rad/s ).83 cm) 40. cm/s. b) In rad/s, the angular velocity is ω 760)π/60) 89, so a r ω r 89 rad/s) m) m/s. c) The angular displacement is, using Eq , θ ω α ) rad. Then, using Eq. 11-1, the distance traveled is s rθ m) rad ) 83. m. 9. 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 α A r A α C r C, where α A is the angular acceleration of wheel A and α C is the angular acceleration of wheel C. Thus, α C ra r C ) α A 10 cm 5 cm ) 1.6 rad/s )0.64 rad/s. Since the angular speed of wheel C is given by ω C α C t, the time for it to reach an angular speed of ω 100 rev/min 10.5 rad/s starting from rest is t ω C α C 10.5 rad/s 0.64 rad/s 16s. 30. The function θ ξe βt where ξ 0.40 rad and β s 1 is describing the angular coordinate of a line which is marked in such a way that all points on it have the same value of angle at a given time) on the object. Taking derivatives with respect to time leads to dθ dt ξβeβt and d θ dt ξβ e βt. a) Using Eq. 11-, we have a t αr d θ dt r 6.4 cm/s.

13 b) Using Eq. 11-3, we have a r ω r ) dθ r.6 cm/s. dt 31. a) A complete revolution is an angular displacement of θ π rad, so the angular velocity in rad/s is given by ω θ/t π/t. The angular acceleration is given by α dω dt π dt T dt. For the pulsar described in the problem, we have Therefore, α dt dt s/y s/y π s) ) ) rad/s. The negative sign indicates that the angular acceleration is opposite the angular velocity and the pulsar is slowing down. b) We solve ω ω 0 + αt for the time t when ω 0: t ω 0 α π αt π rad/s )0.033 s) s. This is about 600 years. c) The pulsar was born years ago. This is equivalent to 938 y) s/y) s. Its angular velocity at that time was ω ω 0 +αt π T +αt π s rad/s ) s) 58 rad/s. Its period was T π ω π 58 rad/s.4 10 s. 3. a) The angular speed in rad/s is ω 33 1 ) ) π rad/rev 3 rev/min 3.49 rad/s. 60 s/min Consequently, the radial centripetal) acceleration is using Eq. 11-3) a ω r 3.49 rad/s) m ) 0.73 m/s.

14 b) Using Ch. 6 methods, we have ma f s f s, max µ s mg, which is used to obtain the minimum allowable) coefficient of friction: µ s, min a g c) The radial acceleration of the object is a r ω r, while the tangential acceleration is a t αr. Thus a a r + a t ω r) +αr) r ω 4 + α. If the object is not to slip at any time, we require f s,max µ s mg ma max mr ω 4 max + α. Thus, since α ω/t from Eq. 11-1), we find µ s,min r ωmax 4 + α g r ωmax 4 +ω max /t) g 0.060) /0.5) The kinetic energy in J) is given by K 1 Iω, where I is the rotational inertia in kg m ) and ω is the angular velocity in rad/s). We have 60 rev/min)π rad/rev) ω 60 s/min Consequently, the rotational inertia is 63.0 rad/s. I K 4400 J) ω 63.0 rad/s) 1.3 kg m. 34. The translational kinetic energy of the molecule is K t 1 mv ) 500) J. With I kg m, we employ Eq. 11-7: K r 3 K t 1 Iω ) 3 which leads to ω rad/s.

15 35. Since the rotational inertia of a cylinder is I 1 MR Table 11-c)), its rotational kinetic energy is K 1 Iω 1 4 MR ω. For the first cylinder, we have K )0.5) 35) J. For the second cylinder, we obtain K )0.75) 35) J. 36. a) Using Table 11-c), the rotational inertia is I 1 mr 1 ) 1.1 m 110 kg) 1 kg m. b) The rotational kinetic energy is, by Eq. 11-7, K 1 Iω kg m ) 1.5 rev/s)π rad/rev)) J. 37. The particles are treated point-like in the sense that Eq yields their rotational inertia, and the rotational inertia for the rods is figured using Table 11-e) and the parallel-axis theorem Eq. 11-9). a) With subscript 1 standing for the rod nearest the axis and 4 for the particle farthest from it, we have I I 1 + I + I 3 + I 4 ) ) Md + M d 8 3 Md +5md. + md Md + M ) ) 3 d + md) b) Using Eq. 11-7, we have K 1 Iω 4 3 Md + 5 ) md ω. 38. 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) kg m.

16 b) The rotational kinetic energy is K 1 Iω kg m ) 350 rev/min) π rad/rev 60 s/min J 4.36 MJ. )) 39. We use the parallel axis theorem: I I com + Mh, where I com is the rotational inertia about the center of mass see Table 11-d)), M is the mass, and h is the distance between the center of mass and the chosen rotation axis. The center of mass is at the center of the meter stick, which implies h 0.50 m 0.0 m 0.30 m. We find I com 1 1 ML kg)1.0m) kg m. Consequently, the parallel axis theorem yields I kg m kg)0.30 m) kg m. 40. a) We show the figure with its axis of rotation the thin horizontal line). We note that each mass is r 1.0 m from the axis. Therefore, using Eq. 11-6, we obtain I m i r i kg)1.0 m) kg m. b) In this case, the two masses nearest the axis are r 1.0 m away from it, but the two furthest from the axis are r m from it. Here, then, Eq leads to I m i r i 0.50 kg)1.0m )+0.50 kg)5.0m )6.0kg m. c) Now, two masses are on the axis with r 0) and the other two are a distance r m away. Now we obtain I.0 kg m.

17 41. We use the parallel-axis theorem. According to Table 11-i), the rotational inertia of a uniform slab about an axis through the center and perpendicular to the large faces is given by I com M 1 a + b ). A parallel axis through the corner is a distance h a/) +b/) from the center. Therefore, I I com + Mh M a + b ) + M 1 4 a + b ) M 3 a + b ). 4. a) We apply Eq. 11-6: 4 I x m i yi 50.0) +5)4.0) ) ) g cm. i1 b) For rotation about the y axis we obtain 4 I y m i x i 50.0) +5)0) +53.0) +30.0) g cm. i1 c) And about the z axis, we find using the fact that the distance from the z axis is x + y ) 4 I z m i x i +yi )I x +I y g cm. i1 d) Clearly, the answer to part c) is A + B. 43. a) According to Table 11-, the rotational inertia formulas for the cylinder radius R) and the hoop radius r) are given by I C 1 MR and I H Mr. Since the two bodies have the same mass, then they will have the same rotational inertia if R /R H,orR H R/. b) We require the rotational inertia to be written as I Mk, where M is the mass of the given body and k is the radius of the equivalent hoop. It follows directly that k I/M. 44. a) Using Table 11-c) and Eq. 11-7, the rotational kinetic energy is K 1 Iω 1 ) 1 MR ω kg)00π rad/s) 1.0m) J.

18 b) We solve P K/t where P is the average power) for the operating time t. t K P J W s which we rewrite as t 100 min. 45. Two forces act on the ball, the force of the rod and the force of gravity. No torque about the pivot point is associated with the force of the rod since that force is along the line from the pivot point to the ball. As can be seen from the diagram, the component of the force of gravity that is.. perpendicular to the rod. is mg sin θ. If l is the θ.. length of the rod, then.. the torque associated θ. with this force has magnitude τ mgl sin θ )9.8)1.5) sin 30.. θ 4.6N m. For the position shown, the torque is... m g. counterclockwise. 46. We compute the torques using τ rf sin φ. τ a 0.15 m)111 N) sin N m τ b 0.15 m)111 N) sin 90 17N m τ c 0.15 m)111 N) sin a) We take a torque that tends to cause a counterclockwise rotation from rest to be positive and a torque tending to cause a clockwise rotation to be negative. Thus, a positive torque of magnitude r 1 F 1 sin θ 1 is associated with F 1 and a negative torque of magnitude r F sin θ is associated with F. The net torque is consequently τ r 1 F 1 sin θ 1 r F sin θ. b) Substituting the given values, we obtain τ 1.30 m)4.0 N) sin m)4.90 N) sin N m. 48. The net torque is τ τ A + τ B + τ C F A r A sin φ A F B r B sin φ B + F C r C sin φ C 10)8.0) sin )4.0) sin )3.0) sin N m.

19 49. a) We use the kinematic equation ω ω 0 + αt, where ω 0 is the initial angular velocity, ω is the final angular velocity, α is the angular acceleration, and t is the time. This gives α ω ω 0 t 6.0 rad/s s 8. rad/s. b) If I is the rotational inertia of the diver, then the magnitude of the torque acting on her is τ Iα 1.0kg m ) 8. rad/s ) N m. 50. The rotational inertia is found from Eq I τ α kg m a) We use τ Iα, where τ is the net torque acting on the shell, I is the rotational inertia of the shell, and α is its angular acceleration. Therefore, I τ α 960 N m 6.0 rad/s 155 kg m. b) The rotational inertia of the shell is given by I /3)MR see Table 11- of the text). This implies M 3I R 3155 kg m ) 1.90 m) 64.4kg. 5. According to the sign conventions used in the book, the magnitude of the net torque exerted on the cylinder of mass m and radius R is τ net F 1 R F R F 3 R N)0.1 m) 4.0 N)0.1 m).0 N)0.05 m) 71 N m. The resulting angular acceleration of the cylinder with I 1 MR according to Table 11-c)) is α τ net I 71 N m 1.0 kg)0.1 m) 9.7 rad/s and is counterclockwise which is the positive sense of rotation).

20 53. We use τ Fr Iα, where α satisfies θ 1 αt Eq ). Here θ 90 π rad and t 30 s. The force needed is consequently F Iα r I θ/t ) r ) π/)/30 ) N. 54. With rightward positive for the block and clockwise negative for the wheel as is conventional), then we note that the tangential acceleration of the wheel is of opposite sign from the block s acceleration which we simply denote as a); that is, a t a. Applying Newton s second law to the block leads to P T ma where m.0 kg. Applying Newton s second law for rotation) to the wheel leads to TR Iα where I kg m. Noting that Rα a t a, we multiply this equation by R and obtain TR Ia T a I R. Adding this to the above equation for the block) leads to P m + IR ) a. Thus, a 0.9 m/s and therefore α 4.6 rad/s, where the negative sign should not be mistaken for a deceleration it simply indicates the clockwise sense to the motion). 55. a) We 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 m) t 5.00 s) m/s. The lighter block has an acceleration of m/s upward. b) Newton s second law for the heavier block is m h g T h m h a, where m h is its mass and T h is the tension force on the block. Thus, T h m h g a) kg) 9.8m/s m/s ) 4.87 N. c) Newton s second law for the lighter block is m l g T l m l a, where T l is the tension force on the block. Thus, T l m l g+a) kg) 9.8m/s m/s ) 4.54 N.

21 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 α a R m/s m 1.0 rad/s. e) The net torque acting on the pulley is τ T h T l )R. Equating this to Iα we solve for the rotational inertia: I T h T l )R α 4.87 N 4.54 N) m) 1.0 rad/s kg m. 56. Since the force acts tangentially at r 0.10 m, the angular acceleration presumed positive) is α τ I Fr ) 0.5t +0.3t I 0.10) t +30t in SI units rad/s ). a) At t 3 s, the above expression becomes α 40 rad/s. b) We integrate the above expression, noting that ω o 0, to obtain the angular speed at t 3s: 3 ω αdt 3 5t +10t 3) rad/s With counterclockwise positive, the angular acceleration α for both masses satisfies τ mgl 1 mgl Iα ml 1+mL )α, by combining Eq with Eq and Eq Therefore, using SI units, α g L 1 L ) L 1 + L 0 9.8) ) rad/s where the negative sign indicates the system starts turning in the clockwise sense. The magnitude of the acceleration vector involves no radial component yet) since it is evaluated at t 0 when the instantaneous velocity is zero. Thus, for the two masses, we apply Eq. 11- and obtain the respective answers for parts a) and b): a 1 α L rad/s ) 0.80 m) 6.9 m/s a α L 8.65 rad/s ) 0.0 m) 1.7 m/s.

22 58. a) The speed of v of the mass m after it has descended d 50cmis given by v ad Eq. -16) where a is calculated as in Sample Problem 11-7 except that here we choose +y downward so a>0). Thus, using g 980 cm/s,wehave v mg)d ad M +m 450)980)50) cm/s ) b) The answer is still cm/s 1.4m/s, since it is independent of R. 59. With ω 1800)π/60) rad/s, we apply Eq : which yields τ 396 N m. P τω τ W rad/s 60. The initial angular speed is ω 80)π/60) 9.3 rad/s. We use Eq for the work and Eq. 7-4 for the average power. a) Since the rotational inertia is Table 11-a)) I 3)1.) 46.1 kg m, the work done is W K 0 1 Iω )9.3) which yields W J. b) The average power in absolute value) is therefore P W t W. 61. a) We apply Eq. 11-7: K 1 Iω 1 ) 1 3 ml ω 1 6 ml ω. b) Simple conservation of mechanical energy leads to K mgh. Consequently, the center of mass rises by h K mg ml ω 6mg L ω 6g. 6. a) The angular speed ω associated with Earth s spin is ω π/t, where T s one day). Thus ω π s rad/s

23 and the angular acceleration α required to accelerate the Earth from rest to ω in one day is α ω/t. The torque needed is then τ Iα Iω ) T ) N m where we used I 5 MR ) ) 5 for Earth s rotational inertia. b) Using the values from part a), the kinetic energy of the Earth associated with its rotation about its own axis is K 1 Iω J. This is how much energy would need to be supplied to bring it starting from rest) to the current angular speed. c) The associated power is P K T J W s 63. We use l to denote the length of the stick. Since its center of mass is l/ from either end, its initial potential energy is 1 mgl, where m is its mass. Its initial kinetic energy is zero. Its final potential energy is zero, and its final kinetic energy is 1 Iω, where I is its rotational inertia about an axis passing through one end of the stick and ω is the angular velocity just before it hits the floor. Conservation of energy yields 1 mgl 1 mgl Iω ω. I The free end of the stick is a distance l from the rotation axis, so its speed as it hits the floor is from Eq ) mgl 3 v ωl. I Using Table 11- and the parallel-axis theorem, the rotational inertial is I 1 3 ml,so v 3gl 3 9.8m/s ) 1.00 m) 5.4 m/s. 64. a) We use the parallel-axis theorem to find the rotational inertia: I I com + Mh 1 MR + Mh 1 0 kg)0.10 m) + 0 kg)0.50 m) 0.15 kg m.

24 b) Conservation of energy requires that Mgh 1 Iω, where ω is the angular speed of the cylinder as it passes through the lowest position. Therefore, Mgh 0)9.8)0.050) ω I rad/s. 65. We use conservation of mechanical energy. The center of mass is at the midpoint of the cross bar of the H and it drops by L/, where L is the length of any one of the rods. The gravitational potential energy decreases by MgL/, where M is the mass of the body. The initial kinetic energy is zero and the final kinetic energy may be written 1 Iω, where I is the rotational inertia of the body and ω is its angular velocity when it is vertical. Thus 0 MgL/+ 1 Iω ω MgL/I. Since the rods are thin the one along the axis of rotation does not contribute to the rotational inertia. All points on the other leg are the same distance from the axis of rotation, so that leg contributes M/3)L, where M/3 is its mass. The cross bar is a rod that rotates around one end, so its contribution is M/3)L /3ML /9. The total rotational inertia is I ML /3)+ML /9) 4ML /9. Consequently, the angular velocity is MgL ω I MgL 4ML /9 9g 4L. 66. From Table 11-, the rotational inertia of the spherical shell is MR /3, so the kinetic energy after the object has descended distance h) is K 1 ) 3 MR ωsphere + 1 Iω pulley + 1 mv. Since it started from rest, then this energy must be equal in the absence of friction) to the potential energy mgh with which the system started. We substitute v/r for the pulley s angular speed and v/r for that of the sphere and solve for v. v 1 m + 1 mgh I r + M 3 gh 1+I/mr )+M/3m) 67. a) We use conservation of mechanical energy to find an expression for ω as a function of the angle θ 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 θ with the vertical,

25 h H/) cos θ. Initially the potential energy is U i MgH/) and the kinetic energy is zero. The kinetic energy is 1 Iω when the chimney makes the angle θ with the vertical, where I is its rotational inertia about its bottom edge. Conservation of energy then leads to MgH/MgH/) cos θ + 1 Iω ω MgH/I)1 cos θ). The rotational inertia of the chimney about its base is I MH /3 found using Table 11-e) with the parallel axis theorem). Thus 3g ω 1 cos θ). H b) The radial component of the acceleration of the chimney top is given by a r Hω,soa r 3g1 cos θ). c) The tangential component of the acceleration of the chimney top is given by a t Hα, where α is the angular acceleration. We are unable to use Table 11-1 since the acceleration is not uniform. Hence, we differentiate ω 3g/H)1 cos θ) with respect to time, replacing dω/dt with α, and dθ/dt with ω, and obtain dω dt ωα 3g/H)ω sin θ α 3g/H) sin θ. Consequently, a t Hα 3g sin θ. d) The angle θ at which a t g is the solution to 3g sin θ /3 and we obtain θ sin θ g. Thus, 68. a) The longitudinal separation between Helsinki and the explosion site is θ The spin of the earth is constant at ω 1 rev 1day h so that an angular displacement of θ corresponds to a time interval of ) 4 h t 77 ) h. b) Now θ 10 0 ) 1 so the required time shift would be ) 4 h t 1 ) h. 69. Analyzing the forces tending to drag the M 514 kg stone down the oak beam, we find F Mgsin θ + µ s cos θ)

26 where µ s 0. static friction is assumed to be at its maximum value) and the incline angle θ for the oak beam is sin 1 3.9/10) 3 but the incline angle for the spruce log is the complement of that). We note that the component of the weight of the workers N of them) which is perpendicular to the spruce log is Nmgcos90 θ) Nmgsin θ, where m 85 kg. The corresponding torque is therefore Nmglsin θ where l m see figure). This must at least) equal the magnitude of torque due to F, so with r 0.7 m,wehave Mgrsin θ + µ s cos θ) Ngmlsin θ. This expression yields N 17 for the number of workers. 70. a) We apply Eq , using the subscript J for the Jeep. ω v J r J 114 km/h km which yields 1140 rad/h or dividing by 3600) 0.3 rad/s for the value of the angular speed ω. b) Since the cheetah has the same angular speed, we again apply Eq , using the subscript c for the cheetah. v c r c ω 9 m)1140 rad/h) which yields m/h or 105 km/h for the cheetah s speed. 71. The Hint given in the problem would make the computation in part a) very straightforward without doing the integration as we show here), but we present this further level of detail in case that hint is not obvious or simply in case one wishes to see how the calculus supports our intuition. a) The centripetal) force exerted on an infinitesimal portion of the blade with mass dm located a distance r from the rotational axis is Newton s second law) df dm)ω r, where dm can be written as M/L)dr and the angular speed is ω 30)π/60) 33.5 rad/s. Thus for the entire blade of mass M and length L the total force is given by F df ω rdm L M ω rdr L 0 Mω r L L Mω L kg)33.5 rad/s) 7.80 m) N.

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