Peason Physics Level 0 Unit III Cicula Motion, Wok, and Enegy: Chapte 5 Solutions Student Book page 4 Concept Check 1. he axis of otation is though the cente of the Fisbee diected staight up and down.. A typical bicycle has six axes of otation: the axis fo each pedal of the bike (); the axis though which the pedal ams otate, which is attached to the lage geas at the font of the bike (1); the axes fo each wheel of the bike (); and the axis though which the handle bas otate when the bike is steeed (1). Student Book page 47 Concept Check he foce that acts as the centipetal foce is fiction. Fiction holds the pebble in the tead of the tie and povides the necessay foce to keep the pebble moving in its cicula path. 5.1 Check and Reflect Knowledge 1. Students answes will vay. Examples of objects that move with unifom cicula motion could include: he popelle of an aicaft tuning with a constant speed. he wheel of a ca tuning with a constant speed. he motion of the planets aound the Sun, o moons aound a planet. (his motion can be consideed unifom until section 5..) Any cicula motion whee the speed is unifom. Examples of non-unifom cicula motion could include: he popelle of an aicaft that is slowing down o speeding up. he wheel of a ca that is slowing down o speeding up. Any cicula motion whee the speed is not constant.. he centipetal foce acting in each case is: (a) fiction between the wheels and the oad (static fiction) (b) tension in the ope (c) foce of gavity between the Moon and Eath Applications. he speed vaies with the squae oot of the adius. 4. he speed is the magnitude of the velocity. If the cicula motion is unifom, the speed will be constant. he velocity is continually changing as the object s diection changes. Peason Physics Solutions Unit III Chapte 5 1 Copyight 009 Peason Education Canada
Extensions 5. If the wheels wee oval, the bike would expeience acceleation and deceleation each tun. he motion of the bike would not be unifom. 6. When the pebble comes in contact with the gound, it is not moving elative to the gound. his is why it does not easily dislodge. o get it to dislodge, thee ae two stategies: Stop the wheel fom tuning while the pebble is in contact with the gound. his has the effect of making the pebble move elative to the gound, and fiction might wench it loose. Speed up so that the fictional foce holding the pebble in the tead of the wheel is insufficient to povide the centipetal foce necessay to keep it moving in a cicula path. he pebble will fly off at a tangent. Example 5.1 Pactice Poblems 1. 00 ev f min 00 ev 1 min min 60 s 5.00 ev s 5.00 Hz he fequency of the popelle is 5.00 Hz.. f = 40 Hz 60s pm f min ev 60 s (40 ) s min.4 10 pm Student Book page 50 he otational fequency of the moto is.4 10 pm.. 4 f 6.0 10 pm 4 ev 1min 6.0 10 min 60s ev 1.0 10 s 1.0 10 Hz 1 f 1 1.0 10 Hz 1.0 10 s Peason Physics Solutions Unit III Chapte 5 Copyight 009 Peason Education Canada
he centifuge s fequency is 1.0 10 Hz and its peiod is 1.0 10 - s. Student Book page 51 Example 5. Pactice Poblems 1. Given v = 61.0 km/h = 0.50 m Requied peiod of the aceca s wheels () he speed of the oute edge of the wheel is the same as the aceca. o detemine the peiod, manipulate the equation fo the velocity aound a cicle to solve fo. Convet the ca s speed to appopiate SI units befoe using it in the equation. 61.0 km 1 h 1000 m v h 600 s 1 km 7.50 m/s v v 0.50 m m 7.50 s 0.00 s Paaphase It takes 0.00 s fo the ties of the aceca to make one evolution.. Given = 16.1 km = 1.61 10 4 m f = 716 Hz Requied speed at the pulsa s equato (v) Any point on the equato is at a adius of 16.1 km fom the cente of the pulsa. he speed can be detemined by conveting the fequency to peiod and using equation 1. Peason Physics Solutions Unit III Chapte 5 Copyight 009 Peason Education Canada
1 f 1 716 Hz 0.00197 s v 4 (1.6110 m) 0.00197 s 7 7.410 m/s Paaphase he speed at the pulsa s equato is 7.4 10 7 m/s. Student Book page 55 Example 5. Pactice Poblems 1. Given D = 8.0 cm = 0.80 m, so = 0.140 m = 0.110 s Requied centipetal acceleation (a c ) Fist, detemine the speed at the oute edge of the Fisbee. hen use the equation fo centipetal acceleation. v 0.140 m 0.110 s 7.997 m/s v ac m 7.997 s 0.140 m 4.57 10 m/s Paaphase he centipetal acceleation at the oute edge of the Fisbee is 4.57 10 m/s.. Given ac 15.0 m/s =.00 cm = 0.000 m Requied speed at the edge of the top (v) Peason Physics Solutions Unit III Chapte 5 4 Copyight 009 Peason Education Canada
Detemine the speed by manipulating the equation fo centipetal acceleation. v ac v a c m 15.0 0.000 m s 1.94 m/s Paaphase he speed at the edge of the top is 1.94 m/s.. Given D = 14.0 m ac 57.0 m/s Requied peiod () Fist, calculate the adius of the blade. hen detemine the speed of the oto blades so that you can find the peiod using the equation fo centipetal acceleation. D 14.0 m D 14.0 m 7.00 m v ac v a c m 57.0 (7.00 m) s 1.0 m/s v = v (7.00 m) m 1.0 s 0.1 s Paaphase he peiod of the helicopte blade is 0.1 s. Peason Physics Solutions Unit III Chapte 5 5 Copyight 009 Peason Education Canada
Student Book page 56 Example 5.4 Pactice Poblems 1. Given m = 7.50 kg v = 65.9 m/s = 7.7 cm = 0.77 m Requied centipetal foce on the blade (F c ) Fo this question, it is best to assume that the entie mass of the fan blade is at the cente of the blade. his is often efeed to as the cente of mass. You can detemine the centipetal foce on the blade most accuately by using the adius fom the cente of otation to the cente of mass. Fc m 7.50 kg65.9 s 0.77 m 6 1.610 N Paaphase he centipetal foce on the blade is 1.6 10 6 N.. Fc F c v m 0.660 N0.0 m 0.001 kg 8.5 m/s he speed of the wheel is 8.5 m/s. Example 5.5 Pactice Poblems Student Book page 59 1. Given m = 100 kg = 7.17 m μ s = 0.80 Requied speed of the hockey playe as he begins to slip (v) Peason Physics Solutions Unit III Chapte 5 6 Copyight 009 Peason Education Canada
he foce of fiction povides the centipetal foce. Stat with this equality and solve fo speed. Fc Ff sfn FN mg s mg v g s m 0.807.17 m9.81 s 7.5 m/s Paaphase he speed of the hockey playe is 7.5 m/s.. Given = 100 m m = 100 kg v = 95.0 km/h Requied coefficient of static fiction (μ s ) Convet the speed of the ca to appopiate SI units. he foce of fiction is the centipetal foce, so you can wite the equation Fc Ff, and then solve fo μ s. Mass will cancel and is not needed. Peason Physics Solutions Unit III Chapte 5 7 Copyight 009 Peason Education Canada
95.0 km 1h 1000m v h 600 s 1km 6.4 m/s m sg v sg v s g m 6.4 s m (100 m) 9.81 s 0.710 Paaphase he coefficient of static fiction is 0.710.. Given m = 600.0 g = 0.6000 kg v =.0 m/s μ s = 0.90 Requied adius () he foce of fiction is the centipetal foce, so Fc Ff. Solve fo the adius. Peason Physics Solutions Unit III Chapte 5 8 Copyight 009 Peason Education Canada
Fc Ff s mg v sg m.0 s m 0.909.81 s 1.0 m Paaphase he adius of the ca s tun is 1.0 m. Student Book page 61 Concept Check he centipetal foce exeted on the Moon is the foce of gavity between Eath and the Moon. his can be expessed as follows: F F c g Moon Eath Moon m Gm m Gm Eath he quantity is constant. heefoe, if the velocity of the Moon wee to incease, its obital adius would have to be educed (smalle obit). Similaly, if the velocity of the Moon wee to decease, its obital adius would have to incease (lage obit) to maintain the same centipetal foce. Student Book page 6 Example 5.6 Pactice Poblems 1. Given = 15.0 cm = 0.150 m g = 9.81 m/s Requied speed (v) he minimum speed the ca can have without falling off is detemined by Fc Fg. By equating the two foces, you can solve fo the minimum speed. Peason Physics Solutions Unit III Chapte 5 9 Copyight 009 Peason Education Canada
F c F g mg v g v g m (0.150 m) 9.81 s 1.1 m/s Paaphase he minimum speed equied fo the ca to move aound the loop without falling off is 1.1 m/s.. Given v = 0.0 m/s Requied maximum adius of the loop () Detemine the adius by equating Fc Fg. he mass is not necessay as it will cancel out. F c F g mg v g m 0.0 s m 9.81 s 40.8 m Paaphase he maximum adius that the olle coaste loop can have is 40.8 m. Student Book page 6 Concept Check 1. he ope is most likely to beak when the bucket is at the bottom of the swing, because this is whee the tension on the ope is the geatest. At the top of the swing, the tension is the least.. he centipetal foce does not depend on an object s position. It is always the same as long as the speed does not change. It is given by: m Fc Peason Physics Solutions Unit III Chapte 5 10 Copyight 009 Peason Education Canada
Student Book page 64 Example 5.7 Pactice Poblems Note: Fo Pactice Poblems 1 and, the bucket is moving with unifom cicula motion. he magnitude of the centipetal foce is constant in all positions. 1. Given m = 1.5 kg = 0.75 m v =.00 m/s g = 9.81 m/s Requied tension on the ope at position A (top of swing) he centipetal foce is a net foce that, in this case, is the sum of the tension and foce of gavity. Use the same efeence coodinates as in the example. ake up to be positive and down to be negative. At the top of the bucket s swing, the tension and foce of gavity ae both down. When you substitute them into the equation fo centipetal foce, they will be negative. Fc F Fg Fc F Fg F F F c g mg m (1.5 kg).00 s m (1.5 kg) 9.81 0.75 m s 18.0 N14.7 N.N Paaphase he tension on the ope at position A is. N [down].. Given m = 1.5 kg = 0.75 m v =.00 m/s g = 9.81 m/s Peason Physics Solutions Unit III Chapte 5 11 Copyight 009 Peason Education Canada
Requied tension on the ope at position B (paallel to the gound) In this position, the tension is pependicula to the foce of gavity and they ae independent vectos. he foce of gavity does not affect the tension. he tension is the centipetal foce. F F Fc Fc m (1.5 kg).00 s 0.75m 18 N 18 N [left] Paaphase he tension on the ope in position B is 18 N [left].. Given m = 0.98 kg = 0.40 m F = 79.0 N [down] g = 9.81 m/s Requied speed of the ock (v) o detemine the speed of the ock, you need to know the centipetal foce. At the top of the swing, the centipetal foce is the sum of the tension and the foce of gavity, which ae both downwad. Daw a diagam to show the efeence coodinates, whee up is positive and down is negative. Once you detemine the centipetal foce, find the speed of the ock by manipulating the equation and solving fo v. Peason Physics Solutions Unit III Chapte 5 1 Copyight 009 Peason Education Canada
F F F F F F c g c g m 79.0 N (0.98 kg) 9.81 s 79.0 N 9.61 N 88.6 N It is not necessay to keep the negative sign because you ae solving fo speed, which is a scala quantity. Fc F c v m (88.6 N)(0.40 m) 0.98 kg 6.0 m/s Paaphase he speed of the ock at the top of its cicula path is 6.0 m/s. Student Book page 67 Example 5.8 Pactice Poblems 1. ac 4 f ac f 4 m 9.81 s 4 0.0 m 0.0910 Hz 9.10 10 Hz Fo the space station to simulate Eath s gavity, it would have to spin at a fequency of 9.10 10 Hz.. Given m = 454.0 g = 0.4540 kg = 1.50 m f = 150.0 pm Requied centipetal foce (F c ) Convet fequency into SI units and substitute it into the equation. Peason Physics Solutions Unit III Chapte 5 1 Copyight 009 Peason Education Canada
150.0 ev 1 min f min 60 s.5 Hz F 4 mf c 4 0.4540 kg 1.50 m.5 Hz 1.68 10 N Paaphase he centipetal foce acting on the mass is 1.68 10 N. Student Book page 68 5. Check and Reflect Knowledge 1. When the ca makes one-quate of a tun ight, it will be headed east and the centipetal foce will be diected south.. he Moon expeiences centipetal acceleation because it is moving moe o less in a cicula path. he centipetal foce is povided by the gavitational attaction between Eath and the Moon.. Students answes may vay. Possible answes include: inceasing the speed deceasing the adius inceasing the fequency deceasing the peiod 4. (a) he foce of gavity does not vay as the object moves. (b) he tension vaies thoughout the object s motion. he tension is least at the top of the cicle and geatest at the bottom. 5. While making a hoizontal tun, the aiplane is tilted towad the cente of the cicula path. Pessue on the aileons and udde makes the plane tun. he centipetal foce esults fom the hoizontal component of the lift foce on the plane. he vetical component keeps the plane aibone. Applications 6. v v 0.5m m 15.0 s 0. s he peiod of otation of the wheels is 0. s. Peason Physics Solutions Unit III Chapte 5 14 Copyight 009 Peason Education Canada
7. v 1.0 m 0.0400 s 1.8810 m/s he speed of the oute tip of the popelle blade is 1.88 10 m/s. 8. Given m = 1500 kg = 100.0 m μ s = 0.70 Requied maximum speed with which the ca can make the tun without skidding off the cuve (v) he foce of fiction is the centipetal foce. Use Ff Fc, and solve fo speed. Ff Fc F F F s ( mg) v g s N N g s m 0.70100.0 m9.81 s 6 m/s Paaphase he maximum speed that the ca can have to make the tun is 6 m/s. Peason Physics Solutions Unit III Chapte 5 15 Copyight 009 Peason Education Canada
9. 100.0 km 1h 1000 m v h 600s 1km 7.78 m/s v ac m 7.78 s 90.0 m 8.57 m/s he centipetal acceleation of the ca is 8.57 m/s. 10. Given = 8.9 m f = 5 pm Requied magnitude of the centipetal acceleation (a c ) magnitude of the centipetal acceleation in compaison to gavity (g = 9.81 m/s ) Fist convet fequency to SI units and detemine the magnitude of the centipetal acceleation of the am. hen detemine how many times geate it is than gavity by using a simple atio. ev 1 min f 5 min 60 s 0.58 Hz ac 4 f 4 (8.9 m)(0.58 Hz) 1. 10 m/s he atio of this acceleation to gavity is: m 1. 10 s 1 m 9.81 s Paaphase he astonaut expeiences an acceleation of 1. 10 m/s, which is 1 times geate than the acceleation of gavity. 11. Given D = 0.0 cm = 0.00 m Requied minimum speed necessay to successfully move though the loop (v) Peason Physics Solutions Unit III Chapte 5 16 Copyight 009 Peason Education Canada
At the top of the loop, the minimum speed occus when the centipetal foce is the foce of gavity only, Fc Fg. Use this equality to solve fo the speed. Remembe to convet the diamete to adius. D 0.00 m 0.150 m Fc Fg mg v g m 0.150 m9.81 s 1.1 m/s Paaphase he minimum speed that the toy aceca must have to go aound the loop is 1.1 m/s. 1. 4 h 600 s 86400 s 4 ac 6 4 6.810 m 86400 s 0.07 m/s he centipetal acceleation of a peson standing at the equato is 0.07 m/s. 1. Given = 0.40 m m = 0.010 g = 0.000010 kg 4 F f 4.4 10 N Requied fequency of the tie s otation that will fling the ant off the tie (f) he centipetal foce is caused by the ant holding onto the tie (foce of fiction). he faste the tie otates, the moe foce the ant must apply to stay on. Since the maximum foce the Peason Physics Solutions Unit III Chapte 5 17 Copyight 009 Peason Education Canada
4 ant can apply is 4.4 10 N, you need to detemine the fequency that ceates this foce. Any highe fequency will cause the ant to fly off. Remembe to convet the mass to the appopiate SI units. Fc F Fc 4 mf Fc f 4 m f 4 4.410 N 4 4.410 N 4 (0.000010 kg)(0.40 m) 1.7 Hz Paaphase he ant will fly off the tie if the fequency of the wheel exceeds 1.7 Hz. Extensions 14. Given fequency of pulley 1, f 1 = 00.0 pm Requied fequency of the lage pulley (f ) Since both pulleys ae connected by a belt along thei oute im, the speed of the two pulleys at thei oute im is the same. Wite the equality v 1 = v and solve using the equation fo velocity, v. f1 00.0 pm 00.0 ev 1 min. Hz min 60 s 1 v f f v v 1 1 1 f f f 1 1 f f f 1 1 (0.10m)(.Hz) 0.5m 1. Hz Paaphase he fequency of the lage pulley is 1. Hz o 80 pm. Peason Physics Solutions Unit III Chapte 5 18 Copyight 009 Peason Education Canada
15. he ca that has the lage tuning adius will have the advantage because it has to move faste in the tun in ode to stay beside the othe ca. As it comes out of the tun, it will be tavelling faste than the othe ca. Example 5.9 Pactice Poblems 1. J KJ whee K 1 a /AU Student Book page 7 a J 1 (5.0 AU) AU 11.87 a he obital peiod of Jupite is 11.87 Eath yeas.. Given p = 90 55 d Requied mean obital adius ( p ) Use Keple s constant elating peiod and adius. o keep the same units, convet Pluto s peiod to yeas. 90 55 d p 47.9 a d 65.4 a P KP whee K 1a /AU P K P (47.9a) a 1 AU 9.465AU Paaphase he mean obital adius of Pluto is 9.465 AU. Peason Physics Solutions Unit III Chapte 5 19 Copyight 009 Peason Education Canada
. d Kd whee K 1 a /AU d K d a 1 (45.0 AU) AU 0 a he obital peiod of the debis is 0 Eath yeas. Student Book page 74 Concept Check 1. he obit would be pefectly cicula. Students could daw an ellipse with a semi-majo axis equal to the semi-mino axis to see fo themselves.. he Moon and Callisto do not obit the sun. he Moon obits Eath, and Callisto obits Jupite. Keple s thid law applies only to objects obiting the same focus.. Yes, Keple s constant fo ou sola system could be used in this new sola system because the planet has the same obital peiod and obital adius as Eath. 4. Astonomes could pick one of the six planets at andom and use its obital peiod and adius to detemine a new Keple s constant that applies to that sola system. he new constant would then apply to all the planets in that system. 5. Planets obit stas with a peiod that depends on thei obital adius as shown by Keple s thid law: a Ka, whee a is any planet. Points moving on a otating disk as shown in Figue 5.4 on page 65 in the Student Book all have the same peiod egadless of thei obital adius. Example 5.10 Pactice Poblems 1. D D D D 9.74d 1.10 d 8.774 10 m 16.0 d itan s obital peiod is 16.0 days. Student Book page 75 Peason Physics Solutions Unit III Chapte 5 0 Copyight 009 Peason Education Canada
. Given CH 147d Requied aveage obital adius of Cassini-Huygens ( CH ) o detemine the aveage obital adius of the Cassini-Huygens pobe, you cannot use Keple s constant as it only applies to planets obiting the Sun. Instead use the moon ethys and Keple s thid law. CH CH CH CH 8 (.947 10 m) (147d) (1.887 d) 9 5.8 10 m 6 5.8 10 km Paaphase he aveage obital adius of the Cassini-Huygens pobe is 5.8 10 6 km.. he neaest moon is Callisto. X C X C X C X C 9 9.810 m 16.689 d 9 1.8810 m 186 d he obital peiod of moon X is 186 days. Peason Physics Solutions Unit III Chapte 5 1 Copyight 009 Peason Education Canada
Example 5.11 Pactice Poblems 1. Fc Fg N Gmm N Sun GmSun v GmSun v 11 Nm 0 6.67 10 1.99 10 kg kg 1 4.50 10 m 5.4 10 m/s Neptune s obital speed is 5.4 10 m/s.. F F c g M GmMmU m v GmU v GmU v 11 Nm 5 6.67 10 8.68 10 kg kg m 6.6810 s 8 1.0 10 m he adius of Mianda s obit is 1.0 10 8 m. Example 5.1 Pactice Poblems Student Book page 78 Student Book page 79 1. Given height of the space station above Eath, altitude = 59. km Requied obital speed of the space station (v) Peason Physics Solutions Unit III Chapte 5 Copyight 009 Peason Education Canada
he space station is an atificial satellite that is elatively nea to Eath. Add Eath s adius to its altitude so you can detemine the space station s aveage speed. altitude 59. km 5.5910 m Distance fom the cente of Eath: 6 5 adius of Eath altitude 6.810 m.5910 m 6 6.7910 m Fc Fg missv GmISSmEath GmEath v GmEath v 11 Nm 4 6.67 10 (5.97 10 kg) kg 6 6.7910 m 7.69 10 m/s Paaphase he obital speed of the Intenational Space Station is 7.69 10 m/s.. Given height of Chanda above Eath, altitude = 114 59 km Requied obital peiod of Chanda () Chanda is an atificial satellite that is elatively nea to Eath. Add Eath s adius to its altitude so you can detemine Chanda s obital adius. altitude 114 59km 8 1.145 910 m Distance fom the cente of Eath: 6 8 adius of Eath altitude 6.810 m 1.145 910 m 8 1.09 7 10 m Peason Physics Solutions Unit III Chapte 5 Copyight 009 Peason Education Canada
F F c g Chanda Chanda Eath m v Gm m GmEath v GmEath v 11 Nm 4 6.67 10 (5.97 10 kg) kg 8 1.09 710 m 1.814 10 m/s v v 8 (1.09 7 10 m ) m 1.814 10 s 5 4.19 10 s Paaphase he obital peiod of Chanda is 4.19 10 5 s. Student Book page 81 Concept Check 1. It occued to Newton that the foce of gavity may not be confined to Eath but could also exist out in space. He easoned that the gavitational attaction between Eath and the Moon was esponsible fo the centipetal foce, and theefoe the Moon s obit.. Newton knew that gavity existed only between objects that have mass. He also knew it was the foce of gavity that was esponsible fo the centipetal foce on planets. If he wee told ou sola system is obiting the cente of ou galaxy, then he would assume that thee is a lage mass thee to ceate the gavitational, and hence centipetal foce.. By equating the foce of gavity with the centipetal foce, Newton was able to substitute his equation fo the foce of gavity to detemine the mass of an object being obited. He used it to detemine the mass of Eath using the Moon s obital adius and peiod. 5. Check and Reflect Student Book page 86 Knowledge 1. An astonomical unit is the length of the semi-majo axis of Eath s obit. his is the same as the mean obital adius of Eath. Peason Physics Solutions Unit III Chapte 5 4 Copyight 009 Peason Education Canada
. he obital adius of a planet is not constant because planetay obits ae ellipses. An ellipse is an elongated cicle with a adius that is not constant.. he lage the eccenticity, the moe elongated (o squashed) the cicle. heefoe, an eccenticity of 0.9 is vey elongated. (A comet might have an eccenticity like this.) 4. A planet s obital velocity is geatest at its peihelion, whee it is closest in its obit to the Sun. Its slowest velocity occus when it is fathest fom the Sun, at its aphelion. 5. Fo Keple s thid law to be valid, the two satellites that ae equated must obit the same focus. 6. Keple s constant fo moons obiting the same planet is not the same as the constant fo planets obiting the Sun. he constant is diffeent because the moons ae obiting a diffeent focus. 7. he Sun does expeience obital petubation. It wobbles in its obit because the combined mass of the planets causes it to evolve aound thei common cente of mass. Applications 8. he elationship between a planet s obital peiod and its adius is given by Keple s thid law: K o. Gaphing this elation yields: he gaph of a planet s peiod as a function of its obital adius will look like the gaph above. 9. K whee K 1 a /AU V V V V K (0.615 a) 1 a AU 0.7AU he obital adius of Venus is 0.7 AU. 10. Detemine the speed of Sedna fom its obital adius using equation 1. Fist convet the adius fom astonomical units to metes. 11 1.49 10 m 479.5 AU 1 AU 1 7.14510 m he foce of gavity between the Sun and Sedna is the centipetal foce: Peason Physics Solutions Unit III Chapte 5 5 Copyight 009 Peason Education Canada
Fc Fg msednav GmSednamSun GmSun v GmSun v 11 Nm 0 6.67 10 (1.99 10 kg) kg 1 7.145 10 m 1.6 10 m/s he aveage obital speed of Sedna is 1.6 10 m/s. 11. E I E I E I E I 8 6.71010 m 1.769 d 8 4.0 10 m.547 d Euopa s obital peiod is.547 d. 1. Given 8 M 1.855 10 m M 0.94 d Requied aveage obital speed (v) Fist convet the obital peiod to SI units befoe substituting them into the equation. 86 400s M 0.94 d d 4 8.1910 s v 8 1.85510 m 4 8.1910 s 4 1.410 m/s Paaphase he aveage obital speed of Mimas is 1.4 10 4 m/s. Peason Physics Solutions Unit III Chapte 5 6 Copyight 009 Peason Education Canada
1. 86 400 s 4.7 d d 7 1.941410 s F F 4 c g mvenus GmVenusmSun Venus m 4 G Sun Venus 0 1.98 10 kg 11 4 1.0810 m 1.941410 s 6.67 10 7 11 Nm s he mass of the Sun is 1.98 10 0 kg. Extensions 14. Students answes may vay. hey might suggest that satellites can eseve some fuel so that when they wea out o ae no longe useful, they can fie thei ockets to put them on a eenty path into Eath s atmosphee whee they will bun up. Convesely, the satellites could push themselves to a highe obit whee they would be out of the way. 15. Gaph of Obital Velocity vs. Obital Radius fo an Atificial Satellite Obiting Eath Chapte 5 Review Student Book pages 88 89 Knowledge 1. he diection of centipetal acceleation is towad the cente of the cicle.. (a) he centipetal foce is poduced by wate against the boat s hull. (b) he centipetal foce is poduced by ai against the wings. (c) he centipetal foce is poduced by Eath s gavitational pull on the satellite. Peason Physics Solutions Unit III Chapte 5 7 Copyight 009 Peason Education Canada
. he tug you hand feels as you swing an object in a cicle on the end of a ope is the eaction foce of the ope on you hand. he foce you hand exets on the ope is the centipetal foce. 4. 5. A spinning tie stetches because the paticles of the tie continually ty to move in a staight line. he centipetal foce is povided to the pats of the tie though the elatively elastic mateial of the tie itself. he eaction foce of the centipetal foce then visibly stetches the less igid pats of the tie. 6. he cable is likely to beak at the bottom of its cicula path. In this position, the tension in the cable is the geatest, as it alone is poviding the centipetal foce in addition to balancing the weight of the object. At the top of the cicula path, the centipetal foce is povided by the foce of gavity and tension in the cable (both ae in the same diection). 7. A motocycle making a tun expeiences the same centipetal foce a ca does. his is the foce of fiction exeted on the ties by the oad. 8. A tuck needs a lage tuning adius than a ca because (1) its taile would cut the cone if the cuve is too shap and () a tuck is highe and theefoe much easie to tip ove. As fa as fiction is concened, the mass of the vehicle is not significant. 9. A planet moves fastest in its obit when it is neaest the Sun, at its peihelion. 10. he semi-majo axis of an elliptical obit epesents the mean obital adius of the planet. 4 11. (a) Equation 14 is meath ; m Eath epesents the mass of Eath, and Moon epesents Moon G the obital peiod of Eath s Moon. Equation 14 is pesented with these vaiables to show how Newton found the mass of Eath. (b) Equation 14 can be used in the most geneal case whee m is the mass of the object being obited, and is the peiod of its satellite. 1. Keple s second law states that a planet, in its obit aound the Sun, will sweep out equal aeas in equal times. o do this, the planet must move faste when it is close to the Sun than when it is fathe away. 1. Keple s constant (K) is 1 a /AU and is based on the obit of Eath aound the Sun. So accoding to Keple s thid law, it is only applicable to othe celestial bodies obiting the same focus (the Sun). It is possible to detemine Keple s constant fo any set of bodies obiting the same focus, but the values will be diffeent. Applications 14. ac 4 f if f is doubled: f f a c 4 f 4 4f If the fequency of a spinning object is doubled, the centipetal acceleation becomes fou times geate. Peason Physics Solutions Unit III Chapte 5 8 Copyight 009 Peason Education Canada
15. It must be eleased at the point whee it is moving vetically upwad. his occus when the ope is in the hoizontal position. 16. Fc Fg mg v g his is the minimum equied speed and mass does not play a ole in whethe people can go though the highest point in a vetical loop in a olle coaste ide. 17. v 5.0 m 8.0 s 1.0 10 m/s he speed of the eagle is.0 10 1 m/s. 18. Fc m 80.0 kg7.0 s 5.0 m 7.8 10 N he centipetal foce exeted on the passenge is 7.8 10 N. 19. Given = 00.0 m Requied minimum speed the glide must fly (v) he elation Fc Fg descibes the minimum speed. Peason Physics Solutions Unit III Chapte 5 9 Copyight 009 Peason Education Canada
F c F g mg v g m (00.0 m)(9.81 ) s 44. m/s Paaphase he minimum speed the glide must fly to make a pefectly cicula vetical loop is 44. m/s. 0. Given m = 800.0 g = 0.8000 kg = 60.0 cm = 0.600 m f =.0 Hz Requied magnitude of the tension on the ope at the top of the cicle ( F ) At the top of the swing, the foce of gavity and tension ae both acting down. ake down to be negative and up to be positive. he centipetal foce is the sum of these two foces given F F F. Manipulate this equation to solve fo tension. by c g Fc Fg F F Fc Fg Solve fo F g and F c sepaately, then substitute them into the equation. Since both foces ae acting down, they will be negative. Fc 4 mf 4 (0.600 m)(0.8000 kg)(.0 Hz) 75.799 N F mg g m (0.8000 kg) 9.81 s 7.848N F ( 75.799 N) ( 7.848 N) 68 N Paaphase he magnitude of the tension on the ope is 68 N. Peason Physics Solutions Unit III Chapte 5 0 Copyight 009 Peason Education Canada
1. Given = 40.0 m m = 1600.0 kg μ = 0.500 v 0.0 km/h Requied detemine if the ca skids off the oad o detemine if the ca skids off the oad, calculate the maximum speed at which the ca can ound the cuve without skidding. Compae it with the speed the ca is tavelling (0.0 km/h). o detemine the maximum possible speed, use the equality F c F g. F F c g FN whee FN Fg ( mg) v g m (0.500)(40.0m) 9.81 s 14.01m/s 14.01 m 600 s 1km s h 1000 m 50.4 km/h Paaphase he maximum possible speed that the ca can go aound the tun without skidding is 50.4 km/h. he ca is tavelling at 0.0 km/h, so it won t skid.. Given D = 5.4 cm = 0.54 m f = 750 pm Requied magnitude of the centipetal acceleation at the edge of the blade ( F c ) Detemine the centipetal acceleation fom the fequency of otation and the adius. Remembe to convet both the fequency and adius to appopiate SI units fist. D 5.4 cm o 0.54 m D 0.54 m 0.17 m Peason Physics Solutions Unit III Chapte 5 1 Copyight 009 Peason Education Canada
750 ev 1 min f min 60 s 1.5 Hz ac 4 f 4 (0.17 m)(1.5 Hz) 7.8 10 m/s Paaphase he magnitude of the centipetal acceleation at the edge of the blade is 7.8 10 m/s.. ac 4 f ac f 4 4 m 8.8810 s 4 0.90 m 49.99 Hz 60 s 49.99 Hz min.010 pm he fequency of the popelle is.0 10 pm. 4. Given v = 107 000 km/h Requied mathematical veification that the speed of Eath in its obit aound the Sun is 107 000 km/h Detemine the speed of Eath fom the Sun s mass and fom Eath s mean obital adius using equation 1. Convet the speed to km/h to compae it with the value given. Fc Fg meathv GmEathmSun GmSun v GmSun v 11 Nm 0 6.67 10 (1.99 10 kg) kg.9747 10 m/s 11 1.50 10 m 4 Peason Physics Solutions Unit III Chapte 5 Copyight 009 Peason Education Canada
Now convet this speed to km/h: 4.9747 10 m 600 s 1 km s h 1000 m 107 089 km/h ~107 000 km/h Paaphase he aveage obital speed of Eath is ~107 000 km/h. 5. v ac v a c m 6.87 (5.0 m) s 1.1 m/s he speed of the hubcap is 1.1 m/s. 6. (a) Given 8 e.14 10 s e 0.0 cm 0.00 m 1 m e 9.11 10 kg Requied speed of the electon ( v ) Detemine the speed of the electon fom its obital peiod and adius. v (0.00 m) -8.14 10 s 7 6.00 10 m/s Paaphase he speed of the electon is 6.00 10 7 m/s. (b) Given 8 e.14 10 s e 0.0 cm 0.00 m 1 me 9.1110 kg 7 6.0010 m/s fom pat (a) Requied centipetal acceleation of the electon ( a c ) Detemine the centipetal acceleation of the electon using equation 5. Peason Physics Solutions Unit III Chapte 5 Copyight 009 Peason Education Canada
v ac 7 m 6.0010 s 0.00 m 16 1.0 10 m/s Paaphase he centipetal acceleation of the electon is 1.0 10 16 m/s. 7. K H H H H K 76.5 a a 1 AU 18.0AU he mean obital adius of Halley s comet is 18.0 AU. 8. Given 400.0 d p p 11 1.0 10 m Requied mass of the sta being obited ( m Sta ) Newton s vesion of Keple s thid law detemines the mass of a body being obited. Make sue to use the appopiate SI units. 86400 s p 400.0 d d 7.45610 s F F 4 c g m p Gmm p Sta p m m 4 Sta p G 4 p Sta Gp 0 1.09 10 kg 11 4 1.0 10 m 11 Nm 7 6.67 10.456 10 s kg Peason Physics Solutions Unit III Chapte 5 4 Copyight 009 Peason Education Canada
Paaphase he mass of the sta is 1.09 10 0 kg. 9. (a) v 0.710 m m 14 s 18 1.1610 s 10.6610 a he peiod of ou sola system in its obit aound the black hole is 1.16 10 18 s o.66 10 10 a. (b) Given 18 S 1.156 10 s fom pat (a) 0 B.7 10 m Requied mass of the black hole (m B ) Since ou Sun is obiting the black hole at the cente of ou galaxy, use Newton s vesion of Keple s thid law to detemine the black hole s mass. Fc Fg 4 msun GmSunmB Sun 4 mb G Sun 0 4.710 m 11 Nm 18 6.67 10 1.15610 s kg 6 5.1810 kg Paaphase he mass of the black hole is 5.18 10 6 kg. (c) v ac m 14 s 0.7 10 m 15 6.7110 m/s he centipetal acceleation of ou sola system caused by the black hole is 6.71 10 15 m/s. Peason Physics Solutions Unit III Chapte 5 5 Copyight 009 Peason Education Canada
0. (a) Given 6 E 6.8 10 m Requied speed of the cannon ball ( v ) Detemine the speed by teating the cannonball as a satellite and equating Fc Fg. F c F g mg v g v g whee is the adius of Eath 6 m (6.810 m) 9.81 s 7.9110 m/s Paaphase he cannonball must tavel at a speed of 7.91 10 m/s to obit Eath at the suface. (b) he mass of the cannonball is not elated to its obital speed. Eath v Eath v 6 (6.810 m ) m 7.9110 s 5.07 10 s he time it takes the cannonball to obit Eath once is independent of its mass, and is 5.07 10 s. Peason Physics Solutions Unit III Chapte 5 6 Copyight 009 Peason Education Canada
1. 7. d 6.5910 s 4 Eath ac 8 4.84410 m 6.5910 s.710 m/s Fc mac m 7.510 kg.710 s 0.0010 N he centipetal acceleation of the Moon in obit aound Eath is.710 m/s. he Moon expeiences a centipetal foce of.00 10 0 N.. Given 7 G 6.0 10 m Requied Galatea s obital peiod ( G ) Use the obital peiod and adius of Neptune s moon, Neeid, to find the obital peiod of Galatea using Keple s thid law. 60.14 d 5.51 10 m N 9 N G N G N N G G N 7 60.14 d 6.010 m 9 5.5110 m 0.40 d 0.40 d 4 86 400 s d.7110 s Paaphase he obital peiod of Galatea is 0.40 d o.71 10 4 s. Peason Physics Solutions Unit III Chapte 5 7 Copyight 009 Peason Education Canada
. Given 5 m 8.68 10 kg U 8 1.909 10 m Requied obital speed of Aiel ( v ) Fc Fg A GmAmU GmU v GmU v 11 Nm 5 6.67 10 (8.68 10 kg) kg 8 1.90910 m 5.5110 m/s Paaphase he obital speed of Aiel is 5.51 10 m/s. 4. (a) Given 4 4.46 x 10 s 10.8 x 10 m Requied mass of sta ( m Sta ) Fc Fg 4 m p Gmm p Sta 4 msta G 10 4.810 m 4 11 Nm 4.4610 s 6.67 10 kg 4.0110 kg Paaphase he mass of the sta being obited is 4.01 10 kg. (b) Given 4 A 4.46 x 10 s 10.8 x 10 m A Peason Physics Solutions Unit III Chapte 5 8 Copyight 009 Peason Education Canada
Requied obital adius of second planet ( B ) A B A B A B B A 10 6.810 m 6.1910 s 4 4.4610 s 11 6.8 10 m he obital adius of the second planet is 6.8 10 11 m. Extensions 5. Students answes may vay. wo misconceptions wee: (i) Centipetal foce acts adially outwad. (ii) Centipetal foce is a foce unto itself. 6. A peson standing at the equato is moving aound in a cicle that has a adius of 6.8 10 6 m and a peiod of 4 h. o do this, the peson needs a centipetal foce that is povided by the foce of gavity. Only the left-ove pat of the gavitational foce is measued as tue weight. A peson standing on one of the poles is standing on Eath s axis of otation. he peson s peiod is the same: 4 h, but the adius of the cicula path is zeo. heefoe, the centipetal foce is zeo. he total amount of gavitational foce is measued as tue weight. Peason Physics Solutions Unit III Chapte 5 9 Copyight 009 Peason Education Canada