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205 Chapter Phys 5 Solutions 5.3 The Earth has a radius of 6.4 106 m ad completes one revolution about its axis in 24 h. (a) Find the speed of a point on the equator. (b) Find the speed of New York City. This is a kinematics of uniform circular motion problem. Given: Radius of the Earth: Period of rotation: rE = 6.4 106 m T = 24h = 86, 400 s (a) We want to nd the tangential speed of a point at the equator. Use Equation 5.1. vEq = 2rE T vEq = 2 6.4 106 m = 465 m/s 86, 400 s (b) New York City has a latitude of 40.7 so the radius of its orbit will not be the full radius of the Earth (see the gure for Problem 5.9 below). The radius of its orbit will be: r = rE cos 40.7 = 4.85 106 m. vNY = 2r T vNY = 2 4.85 106 m = 353 m/s 86, 400 s 5.6 Consider the motion of the hand of a mechanical clock. If the minute hand of the clock has a length of 6.0 cm, what is the centripetal acceleration of a point at the end of the hand? This is a kinematics of uniform circular motion problem. Given: Period Radius: T = 60 min = 3600 s r = 0.060 m We want to nd the centripetal acceleration, ac . Use Equations 5.5 and 5.1. ac = v2 1 = r r 2r T 2 ac = 4 2 r T2 This last equation is useful to use for problems like this one when you are given the period & radius and want the centripetal acceleration. 4 2 (0.060 m) 2 ac = = 1.8 107 m/s 2 (3600 s) 5.9 Consider points on the Earths surface as sketched in the gure. Because of the Earths rotation, these points undergo uniform circular motion. Compute the centripetal acceleration of (a) a point at the equator, and (b) at latitude of 30 . This is a kinematics of uniform circular motion problem. Given: Radius of the Earth: Period of rotation: rE = 6.4 106 m T = 24h = 86, 400 s !"#$ (a) We want to nd the centripetal acceleration, ac , of a point at the equator. Use the equation from the previous problem. ac = 4 2 r T2 ac = 4 2 6.4 106 m (86, 400 s) 2 = 0.034 m/s 2 (b) At a latitude of 30 the radius of orbit will be: r = rE cos 30 = 5.54 106 m. ac = 4 2 r T2 ac = 4 2 5.54 106 m (86, 400 s) 2 = 0.029 m/s 2 1 5.12 When a ghter pilot makes a quick turn, he experiences a centripetal acceleration. When this acceleration is greater than about 8 g , the pilot will usually lose consciousness (black out). consider a pilot ying at a speed of 900 m/s who wants to make a sharp turn. What is the minimum radius of curvature she can take without blacking out? This is a kinematics of uniform circular motion problem. (a) Given: Maximum centripetal acceleration: Tangential speed: ac = 8 g = 78.4 m/s v = 900 m/s 2 We want to nd the minimum radius of curvature which corresponds to the maximum centripetal acceleration. Use Equation 5.5. ac = v2 r r= v2 ac r= (900 m/s) 78.4 m/s 2 2 = 1.0 104 m The minimum radius is roughly 10 km. 5.14 The Daytona 500 stock car race is held on a track that is approximately 2.5 mi long, and the turns are banked at an angle of 31 . It is currently possible for cars to travel through the turns at a speed of 180 mi/h. Assuming these cars are on the verge of slipping into the outer wall of the racetrack (because they are racing!), nd the coecient of static friction between the tires and the track. This is a dynamics of uniform circular motion problem. Given: Track length: Tangential velocity: Bank angle: L = 2.5 mi = 4025 m v = 180 mi/h = 80.5 m/s = 31 y ! FN We want to nd the coecient of static friction, s , assuming the friction force is at a maximum at this speed. A diagram will help (see right). In order to nd the ! required centripetal acceleration, we need to know the radius of curvature of the banked turns. If we assume a circular track, the radius would be L/2 = 0.398 mi. ! The actual track shape is much more oval so Ill estimate r = 0.20 mi = 322 m. The centripetal acceleration is: ! F fr ! Fg x !"#$ !! ! ! v2 (80.5 m/s) 2 ac = = = 20.1 m/s = 2.05g r 322 m Applying Newtons 2nd Law: 2 ! x component: FN sin Ff r cos = mac F = mac y component: FN cos Ff r sin mg = 0 If the static friction force is at a maximum, we have Ff r = s FN , and this gives us: FN sin + s FN cos = mac FN (sin + s cos ) = mac FN cos s FN sin = mg FN (cos s sin ) = mg This gives us two equations with three unknowns (FN , m, and s ). We want to solve this for s . In order to isolate s , Im going to divide the equation on the left by the equation on the right: FN (sin + s cos ) mac = FN (cos s sin ) mg (sin + s cos ) ac = (cos s sin ) g sin +s cos = (cos s sin ) ac g Both the mass and the normal force cancel out and Im left with one equation for s . Isolate s on one side: s (cos + 2.05 sin ) = 2.05 cos sin s = 2.05 cos 31 sin 31 = 0.65 cos 31 + 2.05 sin 31 Your answer will vary depending on your estimate of the radius of curvature. 2 5.16 Consider the motion of a rock tied to a string of length 0.50 m. The string is spun so that te rock travels in a vertical circle as shown in gure. The mass of the rock is 1.5 kg, and it is twirling at constant speed with a period of 0.33 s. (a) Draw free-body diagrams for the rock when it is at the top and when it is at the bottom of the circle. Your value should include the tension in the string ! but the value of FT is not yet known. (b) What is the total force on the rock directed towards the center of the circle? (c) Find the tension in the string when the rock is at the top and when it is at ! the bottom of the circle. This is a dynamics of uniform circular motion problem. Given: Radius: Period: Mass: r = 0.50 m T = 0.33 s m = 1.5 kg Calculate: Centripetal acceleration: Centripetal force: m = 1.5 kg r = 0.50 m ac = 4T 2r = 181 m/s mac = 272 N 2 2 (a) Free body diagrams for the rock at the top and at the bottom of the circle. m3 FT m1 m2 F Free-body diagram at the bottom: Free body v0 diagram at the top: g FT mg m v0 FT mg FT mg !"#$ F N Fyou (b) If the rock is in uniform circular motion, then the force directed towards the center of the circle is always Ff,max mac = 272 N. mg h (c) At the top both FT and Fg contribute to the centripetal force. At the bottom, FT has to overcome Fg to T h T provide the centripetal force. Fleg T Fleg T 2 2 Top: FT + mg = mac FT = mac mg m FT = (1.5 kg) 181 m/s 9.80 m/s = 257 N Bottom: FT mg = mac FT = mac + mg FT = (1.5 kg) 181 m/s + 9.80 m/s 2 2 m = 286 N 5.20 A roller F coasterF track is designed so F that the car travels upside down on a certain FS FS FS FS FS FS FS FS S S S portion of the track as shown in the gure. What is the minimum speed the roller coaster can have without falling from the track? Assume FN track has a radius of the curvature of 30 m. FS FS FS FS FS FS FS FS FS FS FS This is a dynamics of uniform circular motion problem. Given: Radius of curvature: r = 30 m FS FS ! v FS FS FS FS r FS FS FS FS FS FS ! ! We want to nd the minimum speed that keeps the roller coaster car on the track. At high speeds, the normal force from the track on the car keeps the car in circular motion. At slower speeds, the normal force becomes smaller. At the minimum speed, gravity provides all of the centripetal acceleration at the top. At slower speeds, the car would fall. F = ma v= gr v= mg = mac 2 g= v2 r (9.80 m/s )(30 m) = 17 m/s 3 ! v 5.26 Spin out! An interesting amusement park activity involves a cylindrical room that ! spins about a vertical axis (see gure). Participants in thevride are in contact r with the wall of the room, and the circular motion of the room results in a normal force from the wall on the riders. When the room spins suciently fast, the oor is retracted and the frictional force from the wall keeps the people stuck to the ! wall. Assume the room has a radius of 2.0 m and the coecient of static friction ! between the people and the wallis s = 0.50. r ! r (a) Draw pictures showing the motion of a rider. Give both a side view and a top view. (b) What are all the forces acting on the rider? Add them to your pictures in part (a). Then draw a free-body diagram for a rider. (c) What are the components of all forces directed towards the center of the circle (the radial direction)? (d) Apply Newtons second law along both the vertical and the radial directions. Find the minimum rotation ! rate for which the riders do not slip down the wall. ! v ! This is a dynamics of uniform circular motion problem. ! Ffr ! FN ! ! ! ! Fg ! (a) and (b) View from the side: ! FN ! F fr ! Ffr View from the top: ! Fg ! ! FN ! Ffr ! Fg ! Ffr ! F mg ! FN ! The forces on the person are the normal and frictional forces from the wall, and the force of Earths gravity. r ! ! ! (c) The only one with a component towards the center is the normal force from the wall. ! (d) We want to nd the minimum rotation rate to stay stuck to the wall. This will be when the friction force ! is at a maximum, Ff r = s FN . Given: Apply Newtons second law to the person. F = ma FN = mac Ff r mg = 0 Radius: Coecient of friction: r = 2.0 m s = 0.50 ! ! F ! ! ! ! ! FN ! ! ! Using the fact that the static friction force is at a maximum, Ff r = s FN = s mac , we have: s mac mg = 0 s 4 2 r T2 =g T= 4 2 rs g ! v r T= 4 2 (2.0 m) (0.50) 9.80 m/s 2 = 2.0 s ! ! So the room must complete a minimum of a half rotation every second order in for the person to stay stuck. 5.28 A rock of mass m = 2.5 kg is tied to the end of a string of length L = 1.2 m. The other end of the string is fastened to a ceiling, and the rock is set into motion so that it travels in a horizontal circle of radius r = 0.70 m as sketched in the gure. (a) Draw a picture showing the motion of the rock. Give both a top view and a side view. (b) What are all the forces acting on the rock? Add them to your pictures in part (a). Then draw a free-body diagram for the rock. (c) Find the vertical and horizontal components of all of the forces on the rock. (d) Apply Newtons second law in both the vertical (y ) and the horizontal directions. What is the acceleration along y ? Find the tension in the string. ! 4 m r ! ! F fr ! Fg ! FN fr m! ! ! FN is a dynamics of uniform circular motion problem. Fg r This ! ! ! ! ! FN ! ! ! Fg ! Fg ! ! FT ! ! FT ! Fg ! ! ! (a) and (b) View from ! side: the ! ! Fg ! FT ! Fg ! FT !! ! View from the top: ! ! ! The only forces on the rock are the Earths gravitational force and the tension force from the string. (c) The components of the ! gravitational force: Fgx = 0 and Fgy = mg = 24.5 N Well nd the components of the tension force, FT x = FT cos and FT y = FT sin , when we apply Newtons second law to the rock. ! ! ! (d) Applying Newtons second law to the rock. F = ma FT cos = mac FT sin mg = 0 The acceleration along y is zero as long as it is in uniform circular motion that is horizontal. We can nd the angle knowing that the length of the string (the hypotenuse) is 1.2 m and the horizontal component is 0.70 m. cos = 0.70 m 1.2 m = cos1 (0.5833) = 54.3 Now Ill use the vertical component of Newtons 2nd law to nd the tension. FT sin = mg FT = mg sin FT = (2.5 kg)(9.80 m/s ) = 30.2 N m = 1.5 kg sin 0.50 r =54.3 m 2 5.30 Consider a Ferris wheel in which the chairs hang down from the main wheel via a cable. The cable is 2.0 m long, and the radius of the wheel is 12 m (see gure). When a chair is in the orientation shown in the gure (the 3 ! ! oclock position), the cable attached to the chair makes an angle of = 20 with the vertical. Find the speed of the chair. This is a dynamics of uniform circular motion problem. At this position, the tension provides the centripetal force and balances gravity. Given: Radius: r = 12 m mv 2 FT cos = mg r Divide the horizontal equation by the vertical equation (to get rid of the tension): F = ma FT sin = FT sin mv 2 /r = FT cos mg v= 2 !! tan = v2 gr v g r tan (9.80 m/s )(12 m) tan 20 v = 6.5 m/s 5.34 NASA has built centrifuges to enable astronauts to train in conditions in which the acceleration is very large. The device in Figure P5.34 shows one of these human centrifuges. If the device has a radius of 8.0 m and attains accelerations as large as 5.0 g , what is the rotation rate? This is a dynamics of uniform circular motion problem. Given: Radius: Centripetal acceleration: 5 r = 8.0 m 2 ac = 5.0 g = 49.0 m/s We want to nd the rotation rate. Ill nd the period then take the inverse. ac = 4 2 r T2 T= 4 2 r ac T= 4 2 (8.0 m) 49.0 m/s 2 T = 2.54 s So the wheel rotates at 1/T = 0.39 rev/s. 5.35 Two small objects of mass 20 kg and 30 kg are a distance 1.5 m apart. What is the gravitational force of one of these objects on the other? This is a Newtons Law of Gravitation problem. Given: Separation: Masses: r = 1.5 m m1 = 20 kg, m2 = 30 kg We simply need to use Equation 5.16 to nd the magnitude of the force. FG = Gm1 m2 r2 FG = (6.67 1011 N m2 /kg2 )(20 kg)(30 kg) (1.5 m) 2 FG = 1.8 108 N 5.38 Travel and lose pounds! Your apparent weight is the force you feel on the bottoms of your feet when you are standing. Due to the Earths rotation, your apparent weight is slightly more when you are at the South Pole than when you are at the equator. What is the ratio of your apparent weight at these two locations? Carry three signicant gures in your calculation. This is a dynamics of uniform circular motion problem. At the equator, the radius of orbit is the Earths radius, r = 6/4 106 m. The normal force does not quite balance the gravitational force at the equator. 4 2 mr T2 At the pole, the radius of the orbit has gone to zero and the normal force balances your weight. F = ma mg FNequator = mac FNequator = mg F = ma mg FNpole = 0 FNpole = mg Taking the ratio of these two normal forces: FNpole mg = FNequator mg 4 2 mr/T 2 FNpole 1 = FNequator 1 4 2 r/gT 2 The change is going to be small so I will calculate the factor by itself: 4 2 (6.4 106 m) 4 2 mr = 2 2 = 0.00345 gT 2 (9.80 m/s ) (86, 400 s) Using this, the ratio of normal forces is: FNpole = 1.003 FNequator So your weight at the pole will be 0.3% larger than at the equator. 5.46 Suppose the density of the Earth was somehow reduced from its actual value to 1000 kg/m3 (the density of water). Find the value of g , the acceleration due to gravity, on this new planet. Assume the radius does not change. This is a Newtons Law of Gravitation problem. Radius: r = 6.4 106 m 3 Density: = 1000 kg/m We need to nd the new mass to nd the new value of g using Equation 5.21. Given: Mnew = V = 43 r 3 = 1.10 1024 kg g= GMnew 2 = 1.79 m/s r2 6 5.48 In Section 5.4, we showed that the radius of a geosynchronous orbit about the Earth is 4.2 107 m, compared with the radius of the Earth, which is 6.4 106 m. By what factor is the force of gravity smaller when you are in geosynchronous orbit than when you are on the Earths surface? This is a Newtons Law of Gravitation problem. Given: Radii: rE = 6.4 106 m, rgs = 4.2 107 m Use Newtons Law of Gravitation to nd the factor relating the forces. 2 GmE m/rgs FGgs = 2 FG GmE m/rE 2 FGgs rE =2 FG rgs FGgs = FG 6.4 106 m 4.2 107 m 2 = 0.023 5.50 The region of the solar system between Mars and Jupiter contains many asteroids that orbit the Sun. Consider an asteroid in a circular orbit of radius 5.0 1011 m. Find the period of the orbit. This is a Keplers laws problems. Given: Radius of orbit: r = 5.0 1011 m Ill use the Earth as our reference object which is also orbiting the Sun. (Ill use ES to indicate the period and radius of the Earths orbit around the Sun. ) r3 T2 =3 2 TES rES T = TE r rE 3/2 T = (1.00 yr) 5.0 1011 m 1.5 1011 m 3/2 T = 6.09 yr The orbit takes a little over six years or about 2220 days. 5.52 In recent years, a number of nearby stars have been found to posses planets. Suppose the orbital radius of such a planet is found to be 4.0 1011 m, with a period of 1100 days. Find the mass of the star. This is a Keplers laws problems. r = 0.50 m = 0.50 m Period=of orbit: r m = 1.5 kgm 1.5 kg Given: Radius of orbit: r = 4.0 1011 m T = 1100 days = 9.50 107 s We want to nd the mass of the star that this planet is orbiting. We can use Equation 5.26 with the mass of the star replacing the mass of the Sun. ! ! ! ! T2 = 4 2 GMstar r3 Mstar = 4 2 r3 GT 2 Mstar = 4.2 1030 kg Mstar = (6.67 1011 4 2 (4.0 1011 m)3 N m2 /kg2 )(9.50 107 s)2 !! !! This star is more than twice as massive as our Sun. 5.54 In our derivation of Keplers laws, we assumed the only force on a planet is due to the Sun. In a real solar system, however, the gravitational forces from the other planets can sometimes be important. Calculate the gravitational force of Jupiter on the Earth and compare it to to the magnitude of the force from Sun. Do the calculations for the cases when Jupiter is both closest to and farthest from Earth (see gure). !"#$%&'( )*'%+( !"#$%&'( )*'%+( ,"-( ,"-( )*'%+( )*'%+( ,"-( !"#$%&'( ,"-( !"#$%&'( 7 This is a Newtons Law of Gravitation problem. Given: Radii of orbit: Masses: rES = 1.50 1011 m, rJS = 7.78 1011 m ME = 5.98 1024 kg, MJ = 1900 1024 kg, MS = 1.99 1030 kg We want to nd the gravitational force of Jupiter on the Earth and compare it to the gravitational force of the Sun on the Earth. Ive used Table 5.1 on page 147 to ll in the relevant information for the Earth, Jupiter and Sun. The gure above is not to scale but does show the orientation of the Earth and Jupiter in theirs orbits when they are closest and farthest apart. rnear = rJS rES = 6.28 1011 m rfar = rJS + rES = 9.28 1011 m We just need to calculate three gravitational forces: (1) FG , the Sun on the Earth; (2) FGnear , Jupiter on the Earth when closest; and (3) FGfar , Jupiter on the Earth when farthest apart. FG = (6.67 1011 N m2 /kg2 )(1.99 1030 kg)(5.98 1024 kg) GmS mE = 2 2 rES (1.50 1011 m) (6.67 1011 N m2 /kg2 )(1.90 1027 kg)(5.98 1024 kg) GmJ mE = 2 2 rnear (6.28 1011 m) GmJ mE (6.67 1011 N m2 /kg2 )(1.90 1027 kg)(5.98 1024 kg) = 2 2 rfar (9.28 1011 m) FG = 3.53 1022 N FGnear = 1.92 1018 N FGfar = 8.80 1017 N FGnear = FGfar = When the Earth and Jupiter are closest, the gravitational force from Jupiter on the Earth relative to that from the Sun is: FGnear = 5.44 105 FG When the Earth and Jupiter are farthest apart, the gravitational force from Jupiter on the Earth relative to that from the Sun is: FGfar = 2.49 105 FG While the gravitational force from Jupiter on the Earth may seem large ( 1018 N), it is still very small compared to the gravitational force from the Sun on the Earth. 8 ... View Full Document