Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 5

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1 CHAPTE 5 1. The centripetal acceleration is a = v /r = (500 m/s) /( m)(9.80 m/s /g) = 4.5g up.. (a) The centripetal acceleration is a = v /r = (1.35 m/s) /(1.0 m) = 1.5 m/s toward the center. (b) The net horizontal force that produces this acceleration is F net = ma = (5.0 kg)(1.5 m/s ) = 38.0 N toward the center. 3. The centripetal acceleration of the Earth is a = v /r = (¹r/T) /r = 4¹ r/t = 4¹ ( m)/( s) = m/s toward the Sun. The net force that produces this acceleration is F net = m E a = ( kg)( m/s ) = N toward the Sun. This force is the gravitational attraction from the Sun. 4. The force on the discus produces the centripetal acceleration: F = ma = mv /r; 80 N = (.0 kg)v /(1.00 m), which gives v = 1 m/s. 5. We write F = ma from the force diagram for the stationar hanging mass, with down positive: F T = ma = 0; which gives F T =. For the rotating puck, the tension provides the centripetal acceleration, F = Ma : F T = Mv /. When we combine the two equations, we have Mv / =, which gives v = (/M) 1/. Mg puck F T F T 6. For the rotating ball, the tension provides the centripetal acceleration, F = Ma : F T = Mv /. We see that the tension increases if the speed increases, so the maximum tension determines the maximum speed: F Tmax = Mv max /; 60 N = (0.40 kg)v max /(1.3 m), which gives v max = 14 m/s. If there were friction, it would be kinetic opposing the motion of the ball around the circle. Because this is perpendicular to the radius and the tension, it would have no effect. Page 5 1

2 7. If the car does not skid, the friction is static, with F fr ² µ s. This friction force provides the centripetal acceleration. We take a coordinate sstem with the x-axis in the direction of the centripetal acceleration. We write F = ma from the force diagram for the auto: x-component: F fr = ma = mv /; -component: = 0. The speed is maximum when F fr = F fr,max = µ s. When we combine the equations, the mass cancels, and we get µ s g = v max /; (0.80)(9.80 m/s ) = v max /(70 m), which gives v max = 3 m/s. The mass canceled, so the result is independent of the mass. F fr a x 8. At each position we take the positive direction in the direction of the acceleration. (a) At the top of the path, the tension and the weight are downward. We write F = ma from the force diagram for the ball: F T1 + = mv /; F T1 + (0.300 kg)(9.80 m/s ) = (0.300 kg)(4.15 m/s) /(0.850 m), which gives F T1 = 3.14 N. (b) At the bottom of the path, the tension is upward and the weight is downward. We write F = ma from the force diagram for the ball: F T = mv /; F T (0.300 kg)(9.80 m/s ) = (0.300 kg)(4.15 m/s) /(0.850 m), which gives F T = 9.0 N. F T1 F T 9. The friction force provides the centripetal acceleration. We take a coordinate sstem with the x-axis in the direction of the centripetal acceleration. We write F = ma from the force diagram for the auto: x-component: F fr = ma = mv /; -component: = 0. If the car does not skid, the friction is static, with F fr ² µ s. Thus we have mv / ² µ s, or µ s ³ v /g = [(95 km/h)/(3.6 ks/h)] /(9.80 m/s )(85 m). Thus µ s ³ F fr a x 10. The horizontal force on the astronaut produces the centripetal acceleration: F = ma = mv /r; (7.75)(.0 kg)(9.80 m/s ) = (.0 kg)v /(10.0 m), which gives v = 7.6 m/s. The rotation rate is ate = v/¹r = (7.6 m/s)/¹(10.0 m) = rev/s. Note that the results are independent of mass, and thus are the same for all astronauts. Page 5

3 11. The static friction force provides the centripetal acceleration. We write F = ma from the force diagram for the coin: x-component: F fr = mv /; -component: = 0. The highest speed without sliding requires F fr,max = µ s. The maximum speed before sliding is v max = ¹/T min = ¹f max = ¹(0.110 m)(36/min)/(60 min/s) = m/s. Thus we have µ s = mv max / µ s (9.80 m/s ) = (0.415 m/s) /(0.110 m), which gives µ s = Mg F fr x 1. At the top of the trip, both the normal force and the weight are downward. We write F = ma from the force diagram for the passenger: -component: + = mv /. The speed v will be minimum when the normal force is minimum. The normal force can onl push awa from the seat, that is, with our coordinate sstem it must be positive, so min = 0. Thus we have v min = g, or v min = (g) 1/ = [(9.80 m/s )(8.6 m)] 1/ = 9. m/s. 13. At the top of the hill, the normal force is upward and the weight is downward, which we select as the positive direction. (a) We write F = ma from the force diagram for the car: m car g car = mv /; (1000 kg)(9.80 m/s ) car = (1000 kg)(0 m/s) /(100 m), which gives car = N. (b) When we appl a similar analsis to the driver, we have (70 kg)(9.80 m/s ) pass = (70 kg)(0 m/s) /(100 m), which gives pass = N. (c) For the normal force to be equal to zero, we have (1000 kg)(9.80 m/s ) 0 = (1000 kg)v /(100 m), which gives v = 31 m/s (110 km/h or 70 mi/h). 14. To feel weightless the normal force will be zero and the onl force acting on a passenger will be that from gravit, which provides the centripetal acceleration: = mv /, or v = g; v = (9.80 m/s )(50 ft)(0.305 m/ft), which gives v = 8.64 m/s. We find the rotation rate from ate = v/¹ = [(8.64 m/s)/¹(50 ft)(0.305 m/ft)](60 s/min) = 11 rev/min. Page 5 3

4 15. We check the form of a = v /r b using the dimensions of each variable: [a ] = [v/t] = [d/t ] = [L/T ]; [v ] = [(d/t) ] = [(L/T) ] = [L /T ]; [r] = [d] = [L]. Thus we have [v /r] = [L /T ]/[L] = [L/T ], which are the dimensions of a. 16. The masses will have different velocities: v 1 = ¹r 1 /T = ¹r 1 f; v = ¹r /T = ¹r f. We choose the positive direction toward the center of the circle. For each mass we write F r = ma r : m 1 : F T1 F T = m 1 v 1 /r 1 = 4¹ m 1 r 1 f ; m : F T = m v /r = 4¹ m r f. When we use this in the first equation, we get F T1 = F T + 4¹ m 1 r 1 f ; thus F T1 = 4¹ f (m 1 r 1 + m r ); F T = 4¹ f m r. r 1 F T1 r m1 v 1 F T F T m v 17. We convert the speed: (90 km/h)/(3.6 ks/h) = 5 m/s. We take the x-axis in the direction of the centripetal acceleration. We find the speed when there is no need for a friction force. We write F = ma from the force diagram for the car: x-component: 1 sin θ = ma 1 = mv 1 /; -component: 1 cos θ = 0. Combining these, we get v 1 = g tan θ = (9.80 m/s )(70 m) tan 1, which gives v 1 = 1.1 m/s. Because the speed is greater than this, a friction force is required. Because the car will tend to slide up the slope, the friction force will be down the slope. We write F = ma from the force diagram for the car: x-component: sin θ + F fr cos θ = ma = mv /; -component: cos θ F fr sin θ = 0. We eliminate b multipling the x-equation b cos θ, the -equation b sin θ, and subtracting: F fr = m{[(v / ) cos θ ] g sin θ} = (100 kg)({[(5 m/s) /(70 m)] cos 1 } (9.80 m/s ) sin 1 ) = N down the slope. F fr θ θ a x 18. The velocit of the people is v = ¹/T = ¹f = ¹(5.0 m/rev)(0.50 rev/s) = 15.7 m/s. The force that prevents slipping is an upward friction force. The normal force provides the centripetal acceleration. We write F = ma from the force diagram for the person: x-component: = mv /; -component: F fr = 0. F fr Because the friction is static, we have F fr ² µ s, or ² µ s mv /. Thus we have x µ s ³ g/v = (9.80 m/s )(5.0 m)/(15.7 m/s) = 0.0. There is no force pressing the people against the wall. The feel the normal force and thus are appling Page 5 4

5 the reaction to this, which is an outward force on the wall. There is no horizontal force on the people except the normal force. Page 5 5

6 19. The mass moves in a circle of radius r and has a centripetal acceleration. We write F = ma from the force diagram for the mass: x-component: F T cos θ = mv /r; -component: F T sin θ = 0. Combining these, we get rg = v tan θ; (0.600 m)(9.80 m/s ) = (7.54 m/s) tan θ, which gives tan θ = 0.103, or θ = We find the tension from F T = /sin θ = (0.150 kg)(9.80 m/s )/ sin 5.91 = 14.3 N. F T θ x 0. We convert the speeds: (70 km/h)/(3.6 ks/h) = 19.4 m/s; (90 km/h)/(3.6 ks/h) = 5.0 m/s. a At the speed for which the curve is banked perfectl, there is no need for a friction force. We take the x-axis in the direction of the centripetal acceleration. F x fr We write F = ma from the force diagram for the car: θ x-component: 1 sin θ = ma 1 = mv 1 /; θ -component: 1 cos θ = 0. Combining these, we get v 1 = g tan θ. (19.4 m/s) = (80 m)(9.80 m/s )tan θ, which gives tan θ = 0.48, or θ = 5.7. At a higher speed, there is need for a friction force, which will be down the incline. If the automobile does not skid, the friction is static, with F fr ² µ s. We write F = ma from the force diagram for the car: x-component: sin θ + F fr cos θ = ma = mv /; -component: cos θ F fr sin θ = 0. We eliminate F fr b multipling the x-equation b sin θ, the -equation b cos θ, and adding: = m{[(v / ) sin θ ] + g cos θ}. B reversing the trig multipliers and subtracting, we eliminate to get F fr = m{[(v / ) cos θ ] g sin θ}. If the automobile does not skid, the friction is static, with F fr ² µ s : m{[(v / ) cos θ ] g sin θ} ² µ s m{[(v / ) sin θ ] + g cos θ}, or µ s ³ {[(v / ) cos θ ] g sin θ}/{[(v / ) sin θ ] + g cos θ} = [(v /g )] tan θ]/{[(v /g ) tan θ ] + 1}. When we express tan θ in terms of the design speed, we get µ s ³ [(v /g ) (v 1 /g )]/{[(v /g )(v 1 /g )] + 1} = (1/g )(v v 1 )/[(v 1 v /g ) + 1] = [1/(9.80 m/s )(80 m)][(5.0) (19.4 m/s) ]/{[(19.4 m/s)(5.0 m/s)/(9.80 m/s )(80 m)] + 1} = At the bottom of the dive, the normal force is upward, which we select as the positive direction, and the weight is downward. The pilot experiences the upward centripetal acceleration at the bottom of the dive. We find the minimum radius of the circle from the maximum acceleration: a max = v / min ; F Page 5 6

7 (9.0)(9.80 m/s ) = (310 m/s) / min, which gives min = m. Because the pilot is diving verticall, he must begin to pull out at an altitude equal to the minimum radius: 1.1 km. Page 5 7

8 . For the components of the net force we have F tan = ma tan = (1000 kg)(3. m/s ) = N; F = ma = (1000 kg)(1.8 m/s ) = N. 3. We find the constant tangential acceleration from the motion around the turn: v tan = v 0 + a tan (x tan x 0 ) [(30 km/h)/(3.6 ks/h)] = 0 + a tan [¹(00 m) 0], which gives a tan = 6.9 m/s. The centripetal acceleration depends on the speed, so it will increase around the turn. We find the speed at the halfwa point from v 1 = v 0 + a tan (x 1 x 0 ) = 0 + (6.9 m/s )[¹(100 m) 0], which gives v 1 = 6.8 m/s. The radial acceleration is a = v 1 / = (6.8 m/s) /(00 m) = 19.7 m/s. The magnitude of the acceleration is a = (a tan + a ) 1/ = [(6.9 m/s ) + (19.7 m/s ) ] 1/ = 0.7 m/s. On a flat surface, = Mg; and the friction force must provide the acceleration: F fr = ma. With no slipping the friction is static, so we have F fr ² µ s, or Ma ² µ s Mg. Thus we have µ s ³ a/g = (0.7 m/s )/(9.80 m/s ) = (a) We find the speed from the radial component of the acceleration: a = a cos θ = v 1 / ; (1.05 m/s ) cos 3.0 = v 1 /(.70 m), which gives v 1 = 1.3 m/s. (b) Assuming constant tangential acceleration, we find the speed from v = v 1 + a tan t = (1.3 m/s) + (1.05 m/s )(sin 3.0 )(.00 s) = 3.01 m/s. a tan θ a a 5. Because the spacecraft is Earth radii above the surface, it is 3 Earth radii from the center. The gravitational force on the spacecraft is F = GMm/r = ( N m /kg )( kg)(1400 kg)/[3( m)] = N. 6. The acceleration due to gravit on the surface of a planet is g = F/m = GM/. For the Moon we have g Moon = ( N m /kg )( kg)/( m) = 1.6 m/s. 7. The acceleration due to gravit on the surface of a planet is g = F/m = GM/. If we form the ratio of the two accelerations, we have Page 5 8

9 g planet /g Earth = (M planet /M Earth )/( planet / Earth ), or g planet = g Earth (M planet /M Earth )/( planet / Earth ) = (9.80 m/s )(1)/(.5) = 1.6 m/s. Page 5 9

10 8. The acceleration due to gravit on the surface of a planet is g = F/m = GM/. If we form the ratio of the two accelerations, we have g planet /g Earth = (M planet /M Earth )/( planet / Earth ), or g planet = g Earth (M planet /M Earth )/( planet / Earth ) = (9.80 m/s )(.5)/(1) = 4.5 m/s. 9. (a) The mass does not depend on the gravitational force, so it is.10 kg on both. (b) For the weights we have w Earth = Earth = (.10 kg)(9.80 m/s ) = 0.6 N; w planet = planet = (.10 kg)(1.0 m/s ) = 5. N. 30. The acceleration due to gravit at a distance r from the center of the Earth is g = F/m = GM Earth /r. If we form the ratio of the two accelerations for the different distances, we have g h /g surface = [( Earth )/( Earth + h)] = [(6400 km)/(6400 km km)] which gives g h = 0.91g surface. 31. The acceleration due to gravit on the surface of the neutron star is g = F/m = GM/ = ( N m /kg )(5)( kg)/( m) = m/s. 3. The acceleration due to gravit at a distance r from the center of the Earth is g = F/m = GM Earth /r. If we form the ratio of the two accelerations for the different distances, we have g/g surface = ( Earth /) ; 1/10 = [(6400 km)/], which gives = m. 33. The acceleration due to gravit on the surface of the white dwarf star is g = F/m = GM/ = ( N m /kg )( kg)/( m) = m/s. 34. The acceleration due to gravit at a distance r from the center of the Earth is g = F/m = GM Earth /r. If we form the ratio of the two accelerations for the different distances, we have g/g surface = [( Earth )/( Earth + h)] ; (a) g = (9.80 m/s )[(6400 km)/(6400 km + 3. km)] = 9.8 m/s. (b) g = (9.80 m/s )[(6400 km)/(6400 km km)] = 4.3 m/s. Page 5 10

11 35. We choose the coordinate sstem shown in the figure and find the force on the mass in the lower left corner. Because the masses are equal, for the magnitudes of the forces from the other corners we have F 1 = F 3 = Gmm/r 1 = ( N m /kg )(7.5 kg)(7.5 kg)/(0.60 m) = N; F = Gmm/r = ( N m /kg )(7.5 kg)(7.5 kg)/[(0.60 m)/cos 45 ] 1 F 1 L = N. F From the smmetr of the forces we see that the resultant will be along the diagonal. The resultant force is F = F 1 cos 45 + F 4 F 3 = ( N) cos N = N toward center of the square. L 3 x 36. For the magnitude of each attractive gravitational force, we have F V = GM E M V /r V = Gf V M E /r Sun Venus Earth Jupiter Saturn V = ( N m /kg )(0.815)( kg) /[( ) 10 9 m] = N; F J = GM E M J /r J = Gf J M E /r J = ( N m /kg )(318)( kg) /[( ) 10 9 m] = N; F S = GM E M S /r V = Gf S M E /r S = ( N m /kg )(95.1)( kg) /[( ) 10 9 m] = N. The force from Venus is toward the Sun; the forces from Jupiter and Saturn are awa from the Sun. For the net force we have F net = F J + F S F V = N N N = N awa from the Sun. 37. The acceleration due to gravit on the surface of a planet is g = F/m = GM/. If we form the ratio of the two accelerations, we have g Mars /g Earth = (M Mars /M Earth )/( Mars / Earth ) ; 0.38 = [M Mars /( kg)]/(3400 km/6400 km), which gives M Mars = kg. 38. We relate the speed of the Earth to the period of its orbit from v = ¹/T. The gravitational attraction from the Sun must provide the centripetal acceleration for the circular orbit: GM E M S / = M E v / = M E (¹/T) / = M E 4¹ /T, so we have GM S = 4¹ 3 /T ; ( N m /kg )M S = 4¹ ( m) 3 /( s), which gives M S = kg. This is the same as found in Example Page 5 11

12 39. The gravitational attraction must provide the centripetal acceleration for the circular orbit: GM E m/ = mv /, or v = GM E /( E + h) = ( N m /kg )( kg)/( m m), which gives v = m/s. 40. The greater tension will occur when the elevator is accelerating upward, which we take as the positive direction. We write F = ma from the force diagram for the monke: F T cos θ = ma; 0 N (17.0 kg)(9.80 m/s ) = (17.0 kg)a, which gives a = 3.14 m/s upward. Because the rope broke, the tension was at least 0 N, so this is the minimum acceleration. a F T 41. We relate the speed of rotation to the period of rotation from v = ¹/T. For the required centripetal acceleration, we have a = v / = (¹/T) / = 4¹ /T ;!(9.80 m/s ) = 4¹ (16 m)/t, which gives T = 11 s. 4. We relate the speed to the period of revolution from v = ¹/T. For the required centripetal acceleration, we have a = v / = (¹/T) / = 4¹ /T ; 9.80 m/s = 4¹ ( m)/t, which gives T = s (1.41 h). The result is independent of the mass of the satellite. 43. We relate the speed to the period of revolution from v = ¹/T. The required centripetal acceleration is provided b the gravitational attraction: GM M m/ = mv / = m(¹/t) / = m4¹ /T, so we have GM M = 4¹ ( M + h) 3 /T ; ( N m /kg )( kg) = 4¹ ( m m) 3 /T, which gives T = s =.0 h. Page 5 1

13 44. We take the positive direction upward. The spring scale reads the normal force expressed as an effective mass: /g. We write F = ma from the force diagram: = ma, or m effective = /g = m(1 + a/g). + (a) For a constant speed, there is no acceleration, so we have m effective = m(1 + a/g) = m = 58 kg. (b) For a constant speed, there is no acceleration, so we have m effective = m(1 + a/g) = m = 58 kg. (c) For the upward (positive) acceleration, we have m effective = m(1 + a/g) = m( g/g) = 1.33(58 kg) = 77 kg. (d) For the downward (negative) acceleration, we have m effective = m(1 + a/g) = m(1 0.33g/g) = 0.67(58 kg) = 39 kg. (e) In free fall the acceleration is g, so we have m effective = m(1 + a/g) = m(1 g/g) = We relate the orbital speed to the period of revolution from v = ¹/T. The required centripetal acceleration is provided b the gravitational attraction: GM S m/ = mv / = m(¹/t) / = m4¹ /T, so we have GM S = 4¹ 3 /T. For the two extreme orbits we have ( N m /kg )( kg) = 4¹ ( m ) 3 /T inner, which gives T inner = s = 5 h 35 min; ( N m /kg )( kg) = 4¹ ( m ) 3 /T outer, which gives T outer = s = 19 h 50 min. Because the mean rotation period of Saturn is between the two results, with respect to a point on the surface of Saturn, the edges of the rings are moving in opposite directions. 46. The centripetal acceleration has a magnitude of a = v / = (¹/T) / = 4¹ /T = 4¹ (1.0 m)/(1.5 s) = 3.03 m/s. At each position we take the positive direction in the direction of the acceleration. Because the seat swings, the normal force from the seat is upward and the weight is downward. The apparent weight is measured b the normal force. (a) At the top, we write F = ma from the force diagram: F Ttop + = ma, or F Ttop = (1 a /g). For the fractional change we have Fractional change = (F Ttop )/ = a /g = (3.03 m/s )/(9.80 m/s ) = ( 30.9%). (b) At the bottom, we write F = ma from the force diagram: F Tbottom = ma, or F Tbottom = (1 + a /g). For the fractional change we have Fractional change = (F Tbottom )/ = + a /g = + (3.03 m/s )/(9.80 m/s ) = (+ 30.9%). top bottom Page 5 13

14 47. The acceleration due to gravit is g = F grav /m = GM/ = ( N m /kg )( kg)/( m) = 0.8 m/s. We take the positive direction toward the Moon. The apparent weight is measured b the normal force. We write F = ma from the force diagram: + = ma, (a) For a constant velocit, there is no acceleration, so we have + = 0, or = = (70 kg)(0.8 m/s ) = 0 N (toward the Moon). (b) For an acceleration toward the Moon, we have + = ma, or = m(g a) = (70 kg)(0.8 m/s.9 m/s ) = N (awa from the Moon). Mg 48. We determine the period T and radius r of the satellite s orbit, and relate the orbital speed to the period of revolution from v = ¹r/T. We know that the gravitational attraction provides the centripetal acceleration: GM planet m/r = mv /r = m(¹r/t) /r = m4¹ r/t, so we have M planet = 4¹ r 3 /GT. 49. We relate the speed to the period of revolution from v = ¹r/T, where r is the distance to the midpoint. The gravitational attraction provides the centripetal acceleration: Gmm/(r) = mv /r = m(¹r/t) /r = m4¹ r/t, so we have m = 16¹ r 3 /GT = 16¹ ( m) 3 /( N m /kg )[(5.0 r)( s/r)] = kg. 50. (a) We relate the speed of rotation to the period of revolution from v = ¹/T. We know that the gravitational attraction provides the centripetal acceleration: GM planet m/ = mv / = m(¹/t) / = m4¹ /T, so we have M planet = 4¹ 3 /GT. Thus the densit is ρ = M planet /V = [4¹ 3 /GT ]/9¹ 3 = 3¹/GT. (b) For the Earth we have ρ = 3¹/GT = 3¹/( N m /kg )[(90 min)(60 s/min)] = kg/m 3. Note that the densit of iron is kg/m From Kepler s third law, T = 4¹ 3 /GM E, we can relate the periods of the satellite and the Moon: (T/T Moon ) = (/ Moon ) 3 ; (T/7.4 d) = [( m)/( m)] 3, which gives T = das (1.41 h). 5. From Kepler s third law, T = 4¹ 3 /GM S, we can relate the periods of Icarus and the Earth: (T Icarus /T Earth ) = ( Icarus / Earth ) 3 ; (410 d/365 d) = [ Icarus /( m)] 3, which gives Icarus = m. Page 5 14

15 53. From Kepler s third law, T = 4¹ 3 /GM S, we can relate the periods of the Earth and Neptune: (T Neptune /T Earth ) = ( Neptune / Earth ) 3 ; (T Neptune /1 r) = [( m)/( m)] 3, which gives T Neptune = r. 54. We use Kepler s third law, T = 4¹ 3 /GM E, for the motion of the Moon around the Earth: T = 4¹ 3 /GM E ; [(7.4 d)(86,400 s/d)] = 4¹ ( m) 3 /( N m /kg )M E, which gives M E = kg. 55. From Kepler s third law, T = 4¹ 3 /GM S, we can relate the periods of Halle s comet and the Earth to find the mean distance of the comet from the Sun: (T Halle /T Earth ) = ( Halle / Earth ) 3 ; (76 r/1 r) = [ Halle /( m)] 3, which gives Halle = m. If we take the nearest distance to the Sun as zero, the farthest distance is d = Halle = ( m) = m. It is still orbiting the Sun and thus is in the Solar Sstem. The planet nearest it is Pluto. 56. We relate the speed to the period of revolution from v = ¹r/T, where r is the distance to the center of the Milk Wa. The gravitational attraction provides the centripetal acceleration: GM galax M S /r = M S v /r = M S (¹r/T) /r = M S 4¹ r/t, so we have M galax = 4¹ r 3 /GT = 4¹ [(30,000 l)( m/l)] 3 / ( N m /kg )[( r)( s/r)] = kg. The number of stars ( Suns ) is ( kg)/( kg) = From Kepler s third law, T = 4¹ 3 /GM Jupiter, we have M Jupiter = 4¹ 3 /GT. (a) M Jupiter = 4¹ Io3 /GT Io = 4¹ ( m) 3 /( N m /kg )[(1.77 d)(86,400 s/d)] = kg. (b) M Jupiter = 4¹ Europa3 /GT Europa = 4¹ ( m) 3 /( N m /kg )[(3.55 d)(86,400 s/d)] = kg; M Jupiter = 4¹ Ganmede3 /GT Ganmede = 4¹ ( m) 3 /( N m /kg )[(7.16 d)(86,400 s/d)] = kg; M Jupiter = 4¹ Callisto3 /GT Callisto = 4¹ ( m) 3 /( N m /kg )[(16.7 d)(86,400 s/d)] = kg. The results are consistent. 58. From Kepler s third law, T = 4¹ 3 /GM Jupiter, we can relate the distances of the moons: (/ Io ) 3 = (T/T Io ). Thus we have ( Europa / km) 3 = (3.55 d/1.77 d), which gives Europa = km. ( Ganmede / km) 3 = (7.16 d/1.77 d), which gives Ganmede = km. ( Callisto / km) 3 = (16.7 d/1.77 d), which gives Callisto = km. Page 5 15

16 All values agree with the table. Page 5 16

17 59. (a) From Kepler s third law, T = 4¹ 3 /GM S, we can relate the periods of the assumed planet and the Earth: (T planet /T Earth ) = ( planet / Earth ) 3 ; (T planet /1 r) = (3) 3, which gives T planet = 5. r. (b) No, the radius and period are independent of the mass of the orbiting bod. 60. (a) In a short time interval t, the planet will travel a distance vt along its orbit. This distance has been exaggerated on the diagram. Kepler s second law d N d F states that the area swept out b a line from the Sun v N t to the planet in time t is the same anwhere on the Sun orbit. If we take the areas swept out at the nearest and farthest points, as shown on the diagram, and approximate the areas as triangles (which is a good approximation for ver small t), we have!d N (v N t) =!d F (v F t), which gives v N /v F = d F /d N. (b) For the average velocit we have ( = ¹[!(d N + d F )]/T = ¹( m m)/( s) = m/s. From the result for part (a), we have v N /v F = d F /d N = 1.5/1.47 = 1.034, or v N is 3.4% greater than v F. For this small change, we can take each of the extreme velocities to be ± 1.7% from the average. Thus we have v N = 1.017( m/s) = m/s; v F = 0.983( m/s) = m/s. v F t 61. An apparent gravit of one g means that the normal force from the band is, where g = GM E / E. The normal force and the gravitational attraction from the Sun provide the centripetal acceleration: GM S m/ SE + mgm E / E = mv / SE, or v = G[(M S / SE ) + (M E SE / E )] = ( N m /kg )[( kg)/( m) + ( kg)( m)/( m) ], which gives v = m/s. For the period of revolution we have T = ¹ SE /v = [¹( m)/( m/s)]/(86,400 s/da) = 9.0 Earth-das. F grav 6. The acceleration due to gravit at a distance r from the center of the Earth is g = F/m = GM Earth /r. If we form the ratio of the two accelerations for the different distances, we have g/g surface = [ Earth /( Earth + h)] ; 1/ = [( km)/( km + h)], which gives h = km. Page 5 17

18 63. The net force on Tarzan will provide his centripetal acceleration, which we take as the positive direction. We write F = ma from the force diagram for Tarzan: F T = ma = mv /. The maximum speed will require the maximum tension that Tarzan can create: 1400 N (80 kg)(9.80 m/s ) = (80 kg)v /(4.8 m), which gives v = 6.1 m/s. v F T 64. Yes. If the bucket is traveling fast enough at the top of the circle, in addition to the weight of the water a force from the bucket, similar to a normal force, is required to provide the necessar centripetal acceleration to make the water go in the circle. From the force diagram, we write + = ma = mv top /. The minimum speed is that for which the normal force is zero: 0 + = mv top,min /, or v top,min = (g) 1/. 65. We find the speed of the skaters from the period of rotation: v = ¹r/T = ¹(0.80 m)/(3.0 s) = 1.68 m/s. The pull or tension in their arms provides the centripetal acceleration: F T = mv /; = (60.0 kg)(1.68 m/s) /(0.80 m) =.1 10 N. 66. If we consider a person standing on a scale, the apparent weight is measured b the normal force. The person is moving with the rotational speed of the surface of the Earth: v = ¹ E /T = ¹( m)/(86,400 s) = 464 m/s. We take down as positive and write F = ma: + = ma = mv / E, or = effective = mv / E. Thus g effective g = v / E = (464 m/s) /( m) = m/s (0.343% of g). 67. Because the gravitational force is alwas attractive, the two forces will be in opposite directions. If we call the distance from the Earth to the Moon D and let x be the distance from the Earth where the magnitudes of the forces are equal, we have GM M m/(d x) = GM E m/x, which becomes M M x = M E (D x). ( kg)x = ( kg)[( m) x], which gives x = m from Earth s center. Page 5 18

19 68. (a) People will be able to walk on the inside surface farthest from the center, so the normal force can provide the centripetal acceleration. (b) The centripetal acceleration must equal g: g = v / = (¹/T) / = 4¹ /T ; 9.80 m/s = ( m) /T, which gives T = 47.1 s. For the rotation speed we have revolutions/da = (86,400 s/da)/(47.1 s/rev) = rev/da. 69. We take the positive direction upward. The spring scale reads the normal force expressed as an effective mass: /g. We write F = ma from the force diagram: = ma, or m effective = /g = m(1 + a/g); 80 kg = (60 kg)[1 + a/(9.80 m/s )], which gives a = 3.3 m/s upward. The direction is given b the sign of the result At each position we take the positive direction in the direction of the acceleration. (a) The centripetal acceleration is a = v /, so we see that the radius is minimum for a maximum centripetal acceleration: (6.0)(9.80 m/s ) = [(1500 km/h)/(3.6 ks/h)] / min, which gives min = m = 3.0 km. (b) At the bottom of the circle, the normal force is upward and the weight is downward. We write F = ma from the force diagram for the ball: bottom = mv / = m(6.0g); bottom (80 kg)(9.80 m/s ) = (80 kg)(6.0)(9.80 m/s ), which gives bottom = N. (c) At the top of the path, both the normal force and the weight are downward. We write F = ma from the force diagram for the ball: top + = mv /; top + (80 kg)(9.80 m/s ) = (80 kg)(6.0)(9.80 m/s ), which gives top = N. top bottom 71. The acceleration due to gravit on the surface of a planet is g P = F grav /m = GM P /r, so we have M P = g P r /G. Page 5 19

20 7. (a) The attractive gravitational force on the plumb bob is F M = GmM M /D M. Because F = 0, we see from the force diagram: tan θ = F M / = (GmM M /D M )/(mgm E / E ), where we have used GM E / E for g. Thus we have θ = tan 1 (M M E /M E D M ). (b) For the mass of a cone with apex half-angle α, we have M M = ρv = ρ@¹h 3 tan α = ( kg/m 3 )@¹( m) 3 tan 30 = kg. (c) Using the result from part (a) for the angle, we have tan θ = M M E /M E D M = ( kg)( m) /( kg)( m) = 10 5, which gives θ ( ). θ F T F M θ F T F M 73. We convert the speed: (100 km/h)/(3.6 ks/h) = 7.8 m/s. At the speed for which the curve is banked perfectl, a there is no need for a friction force. We take the x-axis in the direction of the centripetal acceleration. We write F = ma from the force diagram for the car: F x fr x-component: 1 sin θ = ma 1 = mv 1 /; θ -component: 1 cos θ = 0. Combining these, we get v θ 1 = g tan θ; (7.8 m/s) = (60 m)(9.80 m/s ) tan θ, which gives tan θ = 1.31, or θ = 5.7. At a higher speed, there is need for a friction force, which will be down the incline to help provide the greater centripetal acceleration. If the automobile does not skid, the friction is static, with F fr ² µ s. At the maximum speed, F fr = µ s. We write F = ma from the force diagram for the car: x-component: sin θ + µ s cos θ = ma = mv max /; -component: cos θ µ s sin θ = 0, or (cos θ µ s sin θ) =. When we eliminate b dividing the equations, we get v max = g[(sin θ + µ s cos θ)/(sin θ µ s cos θ)] = (9.80 m/s )(60 m)[(sin cos 5.7 )/(sin cos 5.7 )], which gives v max = 39.5 m/s = 140 km/h. At a lower speed, there is need for a friction force, which will be up the incline to prevent the car from sliding down the incline. If the automobile does not skid, the friction is static, with F fr ² µ s. At the minimum speed, F fr = µ s. The reversal of the direction of F fr can be incorporated in the above equations b changing the sign of µ s, so we have v min = g[(sin θ µ s cos θ)/(sin θ + µ s cos θ)] = (9.80 m/s )(60 m)[(sin cos 5.7 )/(sin cos 5.7 )], which gives v min = 0.7 m/s = 74 km/h. Thus the range of permissible speeds is 74 km/h < v < 140 km/h. Page 5 0

21 74. We relate the speed to the period from v = ¹/T. To be apparentl weightless, the acceleration of gravit must be the required centripetal acceleration, so we have a = g = v / = (¹/T) / = 4¹ /T ; 9.80 m/s = 4¹ ( m)/t, which gives T = s (1.41 h). 75. (a) The attractive gravitational force between the stars is providing the required centripetal acceleration for the circular motion. (b) We relate the orbital speed to the period of revolution from v = ¹r/T, where r is the distance to the midpoint. The gravitational attraction provides the centripetal acceleration: Gmm/(r) = mv /r = m(¹r/t) /r = m4¹ r/t, so we have m = 16¹ r 3 /GT = 16¹ ( m) 3 /( N m /kg )[(1.6 r)( s/r)] = kg. 76. The chandelier swings out until the tension in the suspension provides the centripetal acceleration, which is the centripetal acceleration of the train. The forces are shown in the diagram. We write F = ma from the force diagram for the chandelier: x-component: F T sin θ = mv /r; -component: F T cos θ = 0. When these equations are combined, we get tan θ = v /rg; tan 17.5 = v /(75 m)(9.80 m/s ), which gives v = 9. m/s. θ F T a x 77. The acceleration due to gravit on the surface of a planet is g P = F grav /m = GM P /. If we form the ratio of the expressions for Jupiter and the Earth, we have g Jupiter /g Earth = (M Jupiter /M Earth )( Earth / Jupiter ) ; g Jupiter /g Earth = [( kg)/( kg)][( m)/( m)], which gives g Jupiter =.56g Earth. This has not taken into account the centripetal acceleration. We ignore the small contribution on Earth. The centripetal acceleration on the equator of Jupiter is a = v / = (¹/T) / = 4¹ /T = 4¹ ( m)/[(595 min)(60 s/min)] =. m/s = 0.g Earth. The centripetal acceleration reduces the effective value of g: g Jupiter = g Jupiter a =.56g Earth 0.g Earth =.3g Earth. 78. The gravitational attraction from the core must provide the centripetal acceleration for the orbiting stars: GM star M core / = M star v /, so we have M core = v /G = (780 m/s) ( m)/( N m /kg ) = kg. If we compare this to our Sun, we get M core /M Sun = ( kg)/( kg) = Page 5 1