1) Consider a particle moving with constant speed that experiences no net force. What path must this particle be taking?

Transcription

1 Chapte 5 Test Cicula Motion and Gavitation 1) Conside a paticle moving with constant speed that expeiences no net foce. What path must this paticle be taking? A) It is moving in a paabola. B) It is moving in a staight line. C) It is moving in a cicle. D) None of the above is definitely tue all of the time. If a paticle expeiences no net foce thee will be no acceleation and so no change in velocity (eithe speed o diection) and so must be taveling in a staight line. Answe C (10-15 seconds) ) When an object expeiences unifom cicula motion, the diection of the acceleation is A) if thee is unifom cicula motion thee is no acceleation. B) is diected towad the cente of the cicula path. C) in the opposite diection of the velocity vecto. D) in the same diection as the velocity vecto. E) is diected away fom the cente of the cicula path. We showed a mathematical agument duing class that duing unifom cicula motion, the diection of the acceleation (diection of the change in velocity) must be towad the cente of the cicle. It can also be easoned this must be the case because the net foce pushing o pulling the object into cicula motion is also towad the cente of the cicle. Theefoe, this is also pependicula to the velocity vecto, which is diected at a tangent to the cicle. Answe B (10-15 seconds). 4) Conside a paticle moving with constant speed that expeiences a constant net foce that is always pependicula to its velocity. What path must this paticle be taking? A) It is moving in a cicle. B) It is moving in a paabola. C) It is moving in a staight line. D) None of the above is definitely tue all of the time. You know that a foce diected at an angle to motion will cause a paticle to tavel a cuved path. If this foce is constantly diected pependiculaly to the motion of the object we have shown though a mathematical agument in class that the path will be a cicle. Answe A (10-15 seconds) 5) Which of the following could povide the foce that is always pependicula to the motion of the paticle descibed in question 4? I II III IV Gavity. Tension in a cod. Static fiction. A nomal foce. A) I and II B) I, II, and IV C) I, II, III, and IV D) II and III All of the above foces could povide the net foce diected pependiculaly to the motion of the paticle to keep it in cicula motion. Answe C (15-0 seconds) 6) An object moves in a cicula path at a constant speed. Compae the diection of the object's velocity and acceleation vectos. A) Vectos ae paallel velocity vecto is tangent to cicle; acceleation vecto is tangent to cicle 1

2 B) Vectos ae paallel velocity vecto points to cente of cicle, acceleation vecto points to cente of cicle C) Vectos ae pependicula velocity vecto is tangent to cicle; acceleation vecto points to cente of cicle D) Vectos ae pependicula--velocity vecto points to cente of cicle, acceleation vecto is tangent to cicle E) The question is meaningless as the acceleation is zeo. We have shown though mathematical analysis that in unifom cicula motion, the acceleation vecto is diected towad the cente of the cicle while the velocity vecto is diected tangent to the cicle, and so, the vectos ae pependicula. Answe C (15-0 seconds) 7) Conside a paticle moving with constant speed such that its acceleation of constant magnitude is always pependicula to its velocity. A) It is moving in a cicle. B) It is moving in a paabola. C) It is moving in a staight line. D) None of the above is definitely tue all of the time. Again, we have shown mathematically that when a paticle is moving at a constant speed and thee is a constant net foce exeted pependicula to the diection of motion the paticle will tavel in a cicula path. Answe A (15-0 seconds) 8) What type of acceleation does an object moving with constant speed in a cicula path expeience? A) Thee is no acceleation if an object moves with constant speed. B) Fee fall. C) Constant acceleation. D) linea acceleation. E) centipetal acceleation. We have consistently called the acceleation of an object towad the cente of a cicle in unifom cicula motion, centipetal acceleation. Answe E (15-0 seconds) 9) What foce is needed to make an object move in a cicle? A) centifugal foce B) static fiction C) centipetal foce D) weight E) tension Although seveal of the mentioned foces could cause an object to move in a cicle, all of them would have to be a centipetal foce, which is the best answe. Answe C (15-0 seconds) 10) When an object expeiences unifom cicula motion, the diection of the net foce is A) is diected towad the cente of the cicula path. B) in the same diection as the motion of the object. C) in the opposite diection of the motion of the object. D) is diected away fom the cente of the cicula path. E) thee is no net foce because thee is no acceleation. As the diection of a net foce has to be in the diection of acceleation, and the diection of acceleation in unifom cicula motion is towad the cente of the cicle, the foce in unifom cicula motion, which we call the centipetal foce, must be diected towad the cente of the cicle. Note when we talk about the foce on an object holding it in cicula motion, we ae not talking

3 about the net foce, (F) R. We ae talking just about what eve foce is pushing/pulling the object into cicula motion. If thee ae othe foces opposing this foce, you must detemine the net foce befoe you can calculate the centipetal acceleation. 11) A olle coaste ca is on a tack that foms a cicula loop in the vetical plane. If the ca is to just maintain contact with tack at the top of the loop, what is the minimum value fo its centipetal acceleation at this point? A) 0.5g downwad B) 0 g C) g upwad D) g downwad E) g upwad Recall that the foces that contibute to keeping the ca in cicula motion at the top of the tack ae foce due to gavity and the nomal foce of the tack on the ca. The foce due to gavity is always going to be the same, but if the ca is just to maintain contact, the nomal foce will be 0N and the foce due to gavity will be supplying all of the centipetal foce. Theefoe, as ma R F G, a R must equal g in a downwad diection. Answe D (0-45 seconds). 1) A olle coaste ca (mass M) is on a tack that foms a cicula loop (adius ) in the vetical plane. If the ca is to just maintain contact with the tack at the top of the loop, what is the minimum value fo its speed at that point (emembe, a value to the ½ powe is the squae oot of that value) (also emembe, you just stated the value fo a in the same situation in question 11)? A) (g)1/ B) (g)1/ C) g D) (0.5g)1/ Fo question 11 you detemined the centipetal acceleation fo this scenaio must be g in the downwad diection. To find v, just isolate v in the centipetal acceleation fomula and use g fo a. a R v v a R g (g)1/ Answe B (45-60 seconds) (Recognize that if you wee just given this question without being given question 11, you would have to eason out both the acceleation and velocity!) 1) A pilot executes a vetical dive, then follows a semi-cicula ac until it is going staight up. Just as the plane is at its lowest point, the foce on him is A) moe than mg, and pointing down. B) less than mg, and pointing down. C) less than mg, and pointing up. D) moe than mg, and pointing up. As the plane is at its lowest point, it is being pulled into a cicula path by a foce that is diected towad the cente of the cicle. This is the foce we ae asking fo (while the net foce would be used in detemining the centipetal acceleation, we ae just asking about the foce diected towad the cente of

4 the cicle). In ode to accomplish this, the foce towad the cente of the cicle must be geate than the foce in the downwad diection, which is mg. If you need to, daw a fee body diagam and wite a net foce equation in the adial diection to pove this to youself. The foce in the upwad diection would actually be povided by the nomal foce fom the seat. Answe D (45-60 seconds) 14) A coin of mass m ests on a tuntable a distance fom the axis of otation. The tuntable otates with a fequency of f. What is the minimum coefficient of static fiction between the tuntable and the coin if the coin is not to slip? A) (4π )/g B) (4πf)/g C) (4πf)/g D) (4πf)/g The foce of static fiction is supplying the foce that keeps the coin taveling in a cicula path that is, it is supplying the centipetal (adial) foce. The velocity can incease until the point of maximum fiction, at which point fiction cannot hold and the coin will slip. So, we can equate the foce of static fiction with centipetal foce and solve fo (isolate) u s. Because all of the answes have 4π in them we have used the fom of centipetal acceleation that equies conveting velocity to π / T and then used 1/f in place of T. Answe B (90-10 seconds) F fs µ s F N ma R µ s ( mg) m v µ s ( mg) m 4π µ T s ( mg) m 4π µ s 4π f g µ 1 ( f ) s ( m ) m 4π 15) A ca goes aound a cuve of adius at a constant speed v. What is the diection of the net foce on the ca? You know that fo the ca to tavel in a cicula path the diection of the net foce needs to be in the diection towad the cente of the cuve. Answe D (10-15 seconds) A) towad the font of the ca B) towad the back of the ca C) away fom the cuve's cente D) towad the cuve's cente 16) A ca goes aound a cuve of adius at a constant speed v. Then it goes aound the same cuve at one thid of the oiginal speed. What is the centipetal foce on the ca as it goes aound the cuve fo the second time, compaed to the fist time? As centipetal foce is popotional to the squae of the velocity, if the velocity is deceased by 1/, the foce will be 1/9 as big. Answe B (15-0 seconds) Wite out the v fomula to visualize this if you need to m v if m A) one thid as big B) one ninth as big C) nine times as big D) thee times as big then ) A ca goes aound a cuve of adius at a constant speed v. Then it goes aound a cuve of adius at speed 4v. What is the centipetal foce on the ca as it goes aound the second cuve, compaed to the fist? f g 4

5 As centipetal foce is popotional to the squae of the velocity, multiplying the velocity by 4 will multiply the foce by 16. Howeve, it is invesely popotional to the adius, so doubling the adius at the same time will decease this by a facto of. Theefoe, foce will be eight times as big. Answe B (15-0 seconds) Wite out the fomula to visualize this if you need to. m v if m 4v A) fou times as big B) eight times as big C) nine halves as big D) thee halves as big then ) A ca of mass m goes aound a banked cuve of adius with speed v. If the oad is fictionless due to ice, the ca can still negotiate the cuve if the hoizontal component of the nomal foce on the ca fom the oad is equal in magnitude to Fo a banked cuve, fiction is not equied to help keep the ca in cicula motion if the x component of the nomal foce is equal to the centipetal foce, which equals mv /. A) mv/. B) mg. C) mg/. D) tan[v/(g)]. 19) Two hoizontal cuves on a bobsled un ae banked at the same angle, but one has twice the adius of the othe. The safe speed (no fiction needed to stay on the un) fo the smalle adius cuve is v. What is the safe speed on the lage adius cuve? You should emembe o know how to deive the banking fomula (see note to quantitative poblems). Reaanging this fo velocity you get: v tanθg. Theefoe, v is popotional to the squae oot of. If you double this will multiply v x Answe C (45-60 seconds) A) appoximately 0.707v B) v C) appoximately 1.41v D) 0.5v 0) The banking angle in a tun on the Olympic bobsled tack is not constant, but inceases upwad fom the hoizontal. Coming aound a tun, the bobsled team will intentionally "climb the wall," then go lowe coming out of the tun. Why do they do this? The geate the banking angle, the geate the x component of the nomal foce, and the geate the coesponding safe velocity. Doing this allows them to take the tun at a highe velocity. Then, coming out of the tun, as they ae highe up on the cuve, they can acceleate thei tangential velocity (speed along thei path) as the will be moving downhill. A) to give the team bette contol, because they ae able to see ahead of the tun B) to pevent the bobsled fom tuning ove C) to take the tun at a faste speed D) to educe the g-foce on them Why does this wok? see above statement 1) Is it possible fo an object moving aound a cicula path to have both centipetal and tangential acceleation? Yes-a dive on a cicula ace tack acceleates both tangentially and centipetally even though thee is tangential acceleation the tack constains the dive to a cicle. Answe B (15-0 seconds) A) No, because then the path would not be a cicle. 5

6 B) Yes, this is possible if the speed is changing. C) No, an object can only have one o the othe at any given time. D) Yes, this is possible if the speed is constant. ) The gavitational foce between two objects is popotional to As the masses of both objects ae multiplied in the numeato of the univesal law of gavity fomula, F G is popotional to the poduct of the masses of the objects. Answe C (10-15 seconds) Wite out the fomula to visualize this if you need to: F G G m 1 m A) the squae of the distance between the two objects. B) the distance between the two objects. C) the poduct of the two objects. D) the squae of the poduct of the two objects. ) The gavitational foce between two objects is invesely popotional to As the squae of the distance between the two objects is in the denominato of the univesal law of gavity fomula, F G is invesely popotional to the squae of the distance between the two objects. Answe D (10-15 seconds) A) the poduct of the two objects. B) the distance between the two objects. C) the squae of the poduct of the two objects. D) the squae of the distance between the two objects. 4) Two objects attact each othe gavitationally. If the distance between thei centes is cut in half, the gavitational foce As the foce due to gavity is invesely popotional to the squae of the distance, if distance is halved the gavitational foce will be inceased by 4 times. Answe B (15-0 seconds) Wite out the fomula to visualize this if you need to: F G Gm m 1 Gm m 1 4 F G A) is cut to one fouth. B) quaduples C) is cut in half. D) doubles. 5) Two objects, with masses m1 and m, ae oiginally a distance apat. The gavitational foce between them has magnitude F. The second object has its mass changed to m, and the distance is changed to /4. What is the magnitude of the new gavitational foce? As foce is popotional to the poduct of the masses, if one mass is doubled, the foce will double. Howeve, as it is invesely popotional to the squae of the distances, if the distance is educed to ¼ of its oiginal value, this will futhe multiply the foce by a facto of 16. In total, the foce will incease by. Answe A (15-0 seconds) distance, if distance is halved the gavitational foce will be inceased by 4 times. Answe B (15-0 seconds) Wite out the fomula to visualize this if you need to: F G Gm m 1 Gm m 1 F G A) F B) F/ C) F/16 D) 16F 4 6

7 6) Two objects, with masses m1 and m, ae oiginally a distance apat. The magnitude of the gavitational foce between them is F. The masses ae changed to m1 and m, and the distance is changed to 4. What is the magnitude of the new gavitational foce? As foce is popotional to the poduct of the masses, if both masses ae doubled, the foce will quaduple. Howeve, as it is invesely popotional to the squae of the distances, if the distance is multiplied by fou, this will decease the foce by a facto of 16. This means that oveall, the foce will be deceased by ¼. Answe B (0-45 seconds) Wite out the fomula to visualize this if you need to: F G Gm m 1 G m m 1 4 A) F/16 B) F/4 C) 16F D) 4F 1 4 F G 8) The acceleation of gavity on the Moon is one-sixth what it is on Eath. An object of mass 48 kg is taken to the Moon. What is its mass thee? Be caeful, this is a tick question. The mass would be the same no matte whee the object was. Answe B (10-15 seconds) A) 48 N A) 8 kg B) 48 kg D) 8 N 9) As a ocket moves away fom the Eath's suface, the ocket's weight As weight is the same quantity as the foce due to gavity, as an object gets futhe away fom the gavitational body the weight deceases. A) deceases. B) emains the same. C) inceases. D) depends on how fast it is moving. 0) A spaceship is taveling to the Moon. At what point is it beyond the pull of Eath's gavity? Accoding to the univesal law of gavitation, thee will neve be a distance between two objects at which the gavitational foce of attaction is 0 N (although thee is cetainly a pactical limit). Answe D (10-15 seconds) A) when it gets above the atmosphee B) when it is half-way thee C) when it is close to the Moon than it is to Eath D) It is neve beyond the pull of Eath's gavity. 1) Suppose a satellite wee obiting the Eath just above the suface. What is its centipetal acceleation? As the acceleation due to gavity is poviding the centipetal foce, the centipetal acceleation at the suface of the eath would be g. Answe D (10-15 seconds) A) lage than g B) Impossible to say without knowing the mass. C) smalle than g D) equal to g ) A hypothetical planet has a mass of half that of the Eath and a adius of twice that of the Eath. What is the acceleation due to gavity on the planet in tems of g, the acceleation due to gavity at the Eath? 7

8 Halving the mass would decease g to ½ g. Doubling the adius would futhe decease g by a facto of 1/4 so in total, g on this planet would be g/8. Answe C (15-0 seconds). Wite out the fomula to visualize this if you need to: g Gm planet G 1 m planet A) g/ B) g/4 C) g/8 D) g 1 8 g ) The acceleation of gavity on the Moon is one-sixth what it is on Eath. The adius of the Moon is one-fouth that of the Eath. What is the Moon's mass compaed to the Eath's? If you eaange the fomula fo gavitational foce at the suface of a gavitational body to isolate mass we see that deceasing g by 1/6 would also decease the mass by 1/6. Howeve, deceasing the adius by ¼ would decease g futhe by a facto of 1/16. In total, mass would less by a facto of 1/96. Answe D (0-45 seconds) Wite out the fomula to visualize this if you need to: g Gm E m E g A) 1/6 B) 1/16 C) 1/4 D) 1/96 1 G g G 1 96 m E 4) Two planets have the same suface gavity, but planet B has twice the adius of planet A. If planet A has mass m, what is the mass of planet B? If you eaange the fomula fo gavitational foce at the suface of a gavitational body to isolate mass we see that having the same suface gavity would have no effect on the calculation of mass. Howeve, doubling the planet s adius would equie that the planet s mass incease by 4 times. So, oveall, the planet s mass would be 4 times geate. Answe B (0-45 seconds). Wite out the fomula to visualize this if you need to: g Gm A m A g G g G A) 0.707m B) 4m C) m D) 1.41m 4 m A 5) Two planets have the same suface gavity, but planet B has twice the mass of planet A. If planet A has adius, what is the adius of planet B? If you eaange the fomula fo gavitational foce at the suface of a gavitational body to isolate adius: g Gm E Gm E g (although we don t cae about G in this case because it is a constant), having the same gavity will not affect the adius calculation, but doubling the mass will have the effect of equiing that the adius be inceased by the squae oot of giving 1.41 Answe C (0-45 seconds) A) B) C) 1.41 D) 4 8

9 6) Conside a small satellite moving in a cicula obit (adius ) about a spheical planet (mass M). Which expession gives this satellite's obital velocity? Make sue you know how to deive the fomula fo the obital velocity (see notes unde quantitative poblems). Knowing that v seconds) A) (GM/)1/ B) GM/ C) (GM/)1/ D) v GM/ Gm E, you can see that the answe is (GM/) 1/. Answe A (0-45 7) Satellite A has twice the mass of satellite B, and otates in the same obit. Compae the two satellite's speeds. Based on the obital velocity fomula, you can see that obital speed depends only on as the adius is the same, the speed must also be the same. Answe D (10-15 seconds) A) The speed of B is twice the speed of A. B) The speed of B is half the speed of A. C) The speed of B is one-fouth the speed of A. D) The speed of B is equal to the speed of A. 8) Who was the fist peson to ealize that the planets move in elliptical paths aound the Sun? Keple Answe C (10-15 seconds) A) Bahe B) Einstein C) Keple D) Copenicus 9) The speed of Halley's Comet, while taveling in its elliptical obit aound the Sun, Although when we solve poblems elated to obits we petend that the obits ae cicula, we know that in fact they ae ellipses, and as such, as a planet gets close to a gavitational body it must speed up. Answe C (10-15 seconds) A) is constant. B) is zeo at two points in the obit. C) inceases as it neas the Sun. D) deceases as it neas the Sun. 40) Let the aveage obital adius of a planet be. Let the obital peiod be T. What quantity is constant fo all planets obiting the Sun? Accoding to Keple s thid law, late poven mathematically by Newton, the atio of the squae of the peiod to the cube of the obital adius (o vice vesa) is constant fo all planets of a planetay system. Answe C (0-45 seconds) A) T/R B) T/R C) T/R D) T/R 9

10 41) A new planet is discoveed to obit aound a sta in a neaby galaxy, with the same obital diamete as the Eath aound ou Sun. If that sta has 9 times the mass of ou Sun, what will the peiod of evolution of that new planet be, compaed to the Eath's obital peiod? Fom the Keple s thid law fomula fo a single obiting body, isolate T: T Gm 4π T 4π Gm. Fom this you can see that if the mass is 4 times geate, T will be affected by 1 ove the squae oot of 4, o 1/ so the new obital peiod will be ½ as much. Answe C (45-60 seconds) A) one-thid as much B) six times as much C) one-half as much D) twice as much 4) The aveage distance fom the Eath to the Sun is defined as one "astonomical unit" (AU). An asteoid obits the Sun in one-thid of a yea. What is the asteoid's aveage distance fom the Sun? You will need to use Kepple s d law fo two gavitational bodies (see note at beginning of quantitative poblems.) You aleady have that the atio of the obital peiods of the asteoid to the eath is 1:. Isolate fo the asteoid and solve. Answe B ( seconds) A T E A T E A) 0.19 AU B) 0.48 AU C).1 AU D) 5. AU A T A T A E 1 1 ( ) (1.0AU ) (1.0 AU ) 1 A.48 AU Quantitative Poblems ************************************************************************************** Notes You will not eceive the banking fomula on the test. Eithe you need to emembe it o be able to deive it fom the centipetal foce fomula and nomal foce calculation: Fo a banked cuve, in ode fo fiction not to be necessay, the x component of the nomal foce must supply all of the centipetal foce that keeps the ca taveling in a cicula path theefoe: F Nx m v You know that F Nx F N sinθ, but F N does not equal mg in this case, no does it equal the opposite of the y component of gavity as we use it in the case of motion down an incline. But, we know that the y component of the nomal foce ( F Ny ) is the opposite of mg (- mg) and we also know that F Ny F N cosθ. So, we can say that F N mg cosθ. We can then substitute this in fo F N in the F F sinθ fomula. Nx N F Nx F N sinθ mg sinθ tanθmg cosθ 10

11 Then we can equate tanθmg m v and solve fo tanθ : tanθ v g We can use this fomula to find banking angle, safe velocity, o adius of a cetain banking angle. θ tan 1 v g v g tanθ v g tanθ ********************************************************************************************** Make sue you undestand that at the suface of a gavitational body, F G mg Gm m 1 g Gm gavitational body, whee is the adius of the gavitational body. ********************************************************************************************** You will not eceive Kepple s thid law on the test. Eithe you need to know Kepple s thid law fo one obiting body T Gm 4π, o two obiting bodies A T B A T B, o you need to know how to deive them by setting the centipetal foce fomula equal to the univesal law of gavity fomula. F G m v G m m 1 4π Gm T T Gm 4π fo one obiting body. Fo two obiting bodies, ecognize that the atio of the obital adius cubed to peiod squaed must be the equal fo both planets obiting because both atios will equal Gm 4π A T B A T B, and so, ********************************************************************************************** You will also need to eithe emembe o be able to deive a fomula fo calculating obital velocity. F G m v G m m 1 E v Gm E ********************************************************************************************** 1) An object moves with a constant speed of 0 m/s on a cicula tack of adius 140 m. What is the acceleation of the object? The motion descibed is unifom constant motion and acceleation is detemined by using the basic centipetal acceleation equation. Plug in the values fo velocity and adius and solve. a R v (0 m / s) 140 m.857 m / s.6 m / s ) The maximum speed aound a level cuve is 0.0 km/h. What is the maximum speed aound a cuve with twice the adius? (Assume all othe factos emain unchanged.) If you take the centipetal acceleation equation and isolate velocity, you find that velocity is popotional to the squae oot of the adius. Theefoe, if you double the adius, the velocity will incease by the poduct of the squae oot of. 11

12 a R v v a R v a R (0.0 km / h) 4.4 km / h ) What is the centipetal acceleation of a point on the peimete of a bicycle wheel of diamete 70 cm when the bike is moving 8.0 m/s? The motion descibed is unifom constant motion and acceleation is detemined by using the basic centipetal acceleation equation. Plug in the values fo velocity and adius and solve (note, adius is.5 cm). a R v (8.0 m / s).5 m 18.9 m / s 180 m / s 4) A point on a wheel otating at 5.00 ev/s is located 0.00 m fom the axis. What is the centipetal acceleation? The motion descibed is unifom constant motion and acceleation is detemined by using the centipetal acceleation equation. Howeve, you discove that you have not been given velocity. Theefoe, you need to find the velocity using the cicumfeence of the cicle divided by the peiod. You then find out that you haven t been given the peiod, but the fequency, which you can convet to the peiod by taking its invese you can do this in one calculation as follows: π ( T ) a R v 4π 4π (4π )( f ) (4π )(.00m)(5.00 ev T 1 s ) 197.9m / s 197m / s f 5) A motocycle has a mass of 50 kg. It goes aound a 1.7 m adius tun at 96.5 km/h. What is the centipetal foce on the motocycle? You know ma R and a R v in and solve. ma R m v (50 kg). You have been given m, v (convet to m/s) and. Plug the values 96.5 km/h (.6 ) 1.7m 111 N 1000 N 6) A 0.50-kg mass is attached to the end of a 1.0-m sting. The system is whiled in a hoizontal cicula path. If the maximum tension that the sting can withstand is 50 N. What is the maximum speed of the mass if the sting is not to beak? Recognize that the foce due to tension is the centipetal foce so you need to find a the velocity at which the centipetal foce equals 50 N. You need to show the centipetal foce fomula, isolate v, plug in the values and solve. F T ma R m v v m (50N )(1.0m) 6.46 m / s 6 m / s.50kg 7) A jet plane flying 600 m/s expeiences an acceleation of 4g when pulling out of the dive. What is the adius of cuvatue of the loop in which the plane is flying? If an aiplane is diving, in ode to pull out of the dive it must follow a cicula path fom to get fom a downwad vetical motion to a hoizontal motion. In ode to be pulled into this cicula motion the pilot manipulates the wings such that ai esistance and lift foces cause a centipetal foce which acceleates the plane towad the cente of the cicle. (Remembe though, that thee would also be a foce due to gavity opposing this, so it is the net foce between these two foces the centipetal foce that causes the centipetal acceleation) The acceleation and velocity of the motion have been given. Isolate in the centipetal acceleation equation, plug the values in and solve. a R v v (600 m / s) 9184 m 9000 m a R (4)(9.8 m / s ) 1

13 8) A pilot makes an outside vetical loop (in which the cente of the loop is beneath him) of adius 00 m. At the top of his loop he is pushing down on his seat with only one-half of his nomal weight. How fast is he going? At the top of the loop if feels to the pilot as if he only weighs half as much as nomal. Keep in mind, howeve, that it is still this weight (foce due to gavity) which is the centipetal foce that is keeping him in cicula motion. So the centipetal foce will equal ½ times the foce due to gavity. This can be set equal to m v whee mass will cancel out. v can then be isolated, values plugged in, and solved fo. ma R m ( 1 g) m v v ( 1 g)() (4.9m / s )(00m) 15. m / s 10 m / s 9) The maximum foce a pilot can stand is about seven times his weight. What is the minimum adius of cuvatue that a jet plane's pilot, pulling out of a vetical dive, can toleate at a speed of 50 m/s? This is a ticky question. You have been told that the foce on the pilot is about 7 mg. This will be the nomal foce povided by the pilot s seat. Howeve, thee will also be an opposite foce povided by the foce due to gavity. So, the net foce poviding the centipetal acceleation is what you need to solve the poblem. Once you ealize this, you can isolate in the centipetal foce equation and plug this net and 50 m/s fo the velocity to get: (F) R F G + F N 1mg + 7mg 6 m g m v v (50m / s) 6g (6)(9.8 m / s ) 106 m 1100 m 10) A ca taveling 0 m/s ounds an 80-m adius hoizontal cuve with the ties on the vege of slipping. How fast can this ca ound a second cuve of adius 0 m? (Assume the same coefficient of fiction between the ca's ties and each oad suface.) If the coefficient of fiction is the same fo both cuves, the foce of static fiction keeping the ca fom sliding will be the same fo both cuves, which means the maximum centipetal acceleation fo both cuves must also be the same. We can detemine a R fo the fist cuve, so we can also use this fo the second cuve. We can then isolate and solve fo v in the second cuve given this a R and the adius of 0 m. (Note, once you ecognize that a R is the same fo both, you could also eason that if you quaduple the adius you would only double the velocity (a R v /) F fstatic and F fstatic fo cuve1 F fstatic fo cuve a R fo cuve1 a R fo cuve a R fo cuve1 v (0 m / s) 80 m 5 m / s v fo cuve a R (5 m / s )(0 m) 40 m / s 11) A ca is negotiating a flat cuve of adius 50 m with a speed of 0 m/s. The centipetal foce povided by fiction is N. What is the mass of the ca? Remembe that the foce of fiction is poviding the centipetal foce keeping the ca in cicula motion, and centipetal foce equals mass times centipetal acceleation. You have been given all of the infomation isolate mass, plug in values and solve. ma R m v F m F f (50 m)(1. x 104 N ) 1500 kg f v (0 m / s) 1) A ca goes aound a flat cuve of adius 50 m at a speed of 14 m/s. What must be the minimum coefficient of fiction between the ties and the oad fo the ca to make the tun? The point at which ties ae on the vege of slipping is the point at which the maximum foce of fiction that is, the static fiction, equals the foce equied to keep the ca in a cicula path fo a cetain velocity and a cetain adius. You can set the foce of static fiction equal to the centipetal foce 1

14 equied. Remembe that F f µ s F N µ s ( mg). This will allow you to cancel mass and solve fo the coefficient of fiction. m v F fs µ s F N µ s ( mg) m v µ s ( m g) µ s v g (14 m / s) (50 m)( 9.8 m / s ).4 1) What minimum banking angle is equied fo an Olympic bobsled to negotiate a 100-m adius tun at 5 m/s without skidding? (Ignoe fiction.) Remembe that we deived a fomula to detemine the banking angle at that point whee fiction did not apply. θ tan 1 v g tan 1 (5 m / s) (100 m)(9.8 m / s ) ) A hoizontal cuve on a bobsled un is banked at a 45 angle. When a bobsled ounds this cuve at the cuve's safe speed (no fiction needed to stay on the un), what is its centipetal acceleation? Being able to deive that banking angle fomula, you ae also awae that tan θ v g and so a R v g tanθ (9.8 m / s )(tan45 0 ) 9.8 m / s o 1.0 g 15) A fictionless cuve of adius 100 m, banked at an angle of 45, may be safely negotiated at a speed of Knowing that tan θ v you can eaange the fomula to isolate v: g v g tanθ (100 m)(9.8 m / s )(tan45 0) m / s 100 m / s 16) A 175-kg ball on the end of a sting is evolving unifomly in a hoizontal cicle of adius m. The ball makes.00 evolutions in a second. (a) Detemine the speed of the ball. v π T π(.500 m) 1 f π 1.00 ev/s (b) Detemine the ball's centipetal acceleation. ( ) π 6.8 m / s a R v 6.8 m / s 78.9 m / s.500 m (c) Detemine the foce a peson must exet on opposite end of the sting. ma R ( 175 kg)(78.9 m / s ) 1.8 x 10 4 N 17) Stating fom est in the pit aea, a ace ca acceleates at a unifom ate to a speed of 45 m/s in 15 s, moving on a cicula tack of adius 500 m. (a) Calculate the tangential acceleation. Remembe that you calculate the tangential acceleation in the same way you calculate any acceleation along a path. 14

15 a Δv Δt 45m / s.0 m / s 15s (b) Calculate the adial acceleation when the instantaneous speed is equal to 0 m/s. Just plug values in the centipetal acceleation fomula to solve this: v (0m / s) 500 m 1.8 m / s 18) The hydogen atom consists of a poton of mass kg and an obiting electon of mass kg. In one of its obits, the electon is m fom the poton. What is the mutual attactive foce between the electon and poton? Just plug in values to the univesal gavity fomula hee: F G Gm 1 m (6.67 x )(1.67 x 10 7 kg)(9.11 x 10 1 kg).6 x 10 ( 5. x m) 47 N 19) What is the gavitational foce on a 70-kg peson standing on the Eath, due to the Moon? The mass of the Moon is kg and the distance to the Moon is m. Just plug in values to the univesal gavity fomula hee: F G Gm 1 m (6.67 x )(7.6 x 10 kg)(70kg).00 N (.8 x 10 8 m) 0) Fo a spacecaft going fom the Eath towad the Sun, at what distance fom the Eath will the gavitational foces due to the Sun and the Eath cancel? Eath's mass: Me kg the Sun's mass: Ms kg Eath-Sun distance: m Remembe that we said in class, the appoach to this poblem is to ecognize that at the point the gavitational foces fom the sun and eath cancel, the foce of gavity fom the on the spacecaft has to equal the foce of gavity fom the eath. We can theefoe set the fomulas equal to each othe and solve fo the distance fom the eath, ecognizing that the distance fom the sun would then be eath-sun minus distance fom the eath. F G sun on caft G m caft m S ( ) E S E G m caft m E E E ( ) m E m S E S E E ( ) E S E m E m S E ( ) m E E S E E (1+ m E m S ) m E m E S m E S S m E m S E S E m E m S E m E m S E S (1+ m E m S ) (5.98 x 10 4 kg) 1.99 x 10 0 kg (1.5 x 1011 m) 1+ (5.98 x 104 kg) 1.99 x 10 0 kg.6 x 10 8 m 1) The mass of the Moon is kg and its mean adius is km. What is the acceleation due to gavity at the suface of the Moon? 15

16 You know at the suface of the moon, F G mg Gm 1 m find g at the suface of the moon. F G mg Gm 1 m g Gm moon g Gm moon (6.67 x )(7.4 x 10 kg) (1.75 x 10 6 m) 1.6 m / s. Just plug in the values to ) An astonaut goes out fo a "space-walk" at a distance above the Eath equal to the adius of the Eath. What is he acceleation due to gavity? You know at the suface of the eath F G m g G m m 1 g Gm E. Theefoe, if is doubled, acceleation due to gavity must decease by ¼. The astonaut s acceleation due to gavity is ¼ g. ) The adius of the Eath is R. At what distance above the Eath's suface will the acceleation of gavity be 4.9 m/s? You know at the suface of the eath F G m g G m m 1 g Gm E. As G and m E ae constants, if you decease g by ½, this means that the value fo is going to need to be doubled. To do this, will need to be multiplied by the squae oot of. This would give a new adius of 1.41 R. Howeve, the question asks fo the distance above the suface of the eath, so the answe would be.41 R. 4) A satellite encicles Mas at a distance above its suface equal to times the adius of Mas. The acceleation of gavity of the satellite, as compaed to the acceleation of gavity on the suface of Mas, is At a distance times the adius of Mas above the suface, the satellite is 4 times the adius away fom the gavitation cente. As gavity deceases by the squae of the adius this means that the acceleation due to gavity on the satellite will be (1/4) 1/16 that of gavity at the suface. 5) An object weighs 4 N on the suface of the Eath. The Eath has adius. If the object is aised to a height of above the Eath's suface, what is its weight? As the foce due to gavity deceases by the squae of the adius, the foce due to gavity at above the suface ( + 4) will be (1/4) 1/16 that at the suface of the eath. (4 N )(1/ 16) 7 N 6) Duing a luna eclipse, the Moon, Eath, and Sun all lie on the same line, with the Eath between the Moon and the Sun. The Moon has a mass of kg; the Eath has a mass of kg; and the Sun has a mass of kg. The sepaation between the Moon and the Eath is given by m; the sepaation between the Eath and the Sun is given by m. (a) Calculate the foce exeted on the Eath by the Moon. Use the law of gavitation to find this foce: F G Gm M m E (6.67 x )(7.6 x 10 kg)(5.98 x 10 4 kg) 1.99 x 10 (.84 x 10 8 m) 0 N (b) Calculate the foce exeted on the Eath by the Sun. Use the law of gavitation to find this foce: F G Gm E m S (6.67 x )(5.98 x 10 4 kg)(1.99 x 10 0 kg).50 x 10 ( x m) N (c) Calculate the net foce exeted on the Eath by the Moon and the Sun. 16

17 If the moon exets a foce of 1.99 x 10 0 N on the eath and the sun exets a foce of.50 x 10 N in the opposite diection, the net foce is given by F F MonE + F SonE 1.99 x 10 0 N ( ) + (.50 x 10 N ).48 x 10 N towad the sun 7) A satellite is in a low cicula obit about the Eath (i.e., it just skims the suface of the Eath). What is the speed of the satellite? (The mean adius of the Eath is m.) You will need to use the fomula fo velocity of an obiting body. Plug in values and solve. v Gm E (6.67 x )(5.98 x 10 4 kg) (6.8 x 10 6 m) 7900 m / s 8) A satellite is in a low cicula obit about the Eath (i.e., it just skims the suface of the Eath). How long does it take to make one evolution aound the Eath? (The mean adius of the Eath is m.) You will need to use the one obiting body vesion of Kepple s d Law. Isolate peiod, plug in the values and solve. T Sat 4π Sat Gm E 4π (6.8 x 10 6 m) 5070 s 84.5 min (6.67 x )(5.98 x 10 4 kg) 9) A satellite is in cicula obit 0 km above the suface of the Eath. It is obseved to have a peiod of 89 min. What is the mass of the Eath? (The mean adius of the Eath is m.) You will need to use the one obiting body vesion of Kepple s d Law. Isolate mass, plug in the values and solve. You need to convet minutes to seconds and km to metes and you need to add the 0 km to the adius of the eath. Sat T Gm E Sat m 4π E 4π Sat GT Sat 4π (6.61 x 10 6 m) (6.67 x )(89 min x 60s / min) 6.0 x 104 kg 0) Euopa, a moon of Jupite, has an obital diamete of m, and a peiod of.55 days. What is the mass of Jupite? You will need to use the one obiting body vesion of Kepple s d Law. Isolate mass, plug in the values and solve. Be caeful this poblem gives you an obital diamete, not an obital adius. E T Gm J m E 4π J 4π E GT E 4π (6.7 x 10 8 m) (6.67 x )(.55 d x 4 x 60 x 60) 1.89 x 107 kg 1) Two moons obit a planet in nealy cicula obits. Moon A has obital adius, and moon B has obital adius 4. Moon A takes 0 days to complete one obit. How long does it take moon B to complete an obit? You will need to use Kepple s d law fo two gavitation bodies (see note at beginning of quantitative poblems.) Isolate T fo moon B, plug in known values and solve you aleady know that the atio of the obital adii is 4:1. A T B A T B T B B T 4 A A 1 (0d) T B 4 ( ) (0d) 160d ) The planet Jupite is m fom the Sun. How long does it take fo Jupite to obit once about the Sun? (The distance fom the Eath to the Sun is m.) You will need to use Kepple s d law fo two gavitation bodies (see note at beginning of 17

18 quantitative poblems.) Isolate T fo Jupite, plug in known values and solve. (You know that the obital peiod of the eath is 1 yea. J T E J T E T J J T 7.78 x 1011 m E E 1.50 x m (1y) T J 7.78 x m 1.50 x m (1y) 11.8 y ) It takes the planet Jupite 1 yeas to obit the Sun once. What is the aveage distance fom Jupite to the Sun? (The distance fom the Eath to the Sun is m.) You will need to use Kepple s d law fo two gavitational bodies (see note at beginning of quantitative poblems.) You aleady have that the atio of the peiods is 1 yeas to 1 yea, and you have been given the aveage obital adius fo the eath. Isolate fo Jupite, plug in known values and solve. J T E J T E 7.9 x m J T J T 1y E E 1y (1.5 x m) J 1y 1y (1.5 x m) 18