Laws of Motion; Circular Motion

Transcription

1 Pactice Test: This test coves Newton s Laws of Motion, foces, coefficients of fiction, fee-body diagams, and centipetal foce. Pat I. Multiple Choice 3m 2m m Engine C B A 1. A locomotive engine of unknown mass pulls a seies of aiload cas of vaying mass: the fist ca has mass m, the second ca has mass 2m, and the last ca has mass 3m. The cas ae connected by links A, B, and C, as shown. Which link expeiences the geatest foce as the tain acceleates to the ight? a. A b. B c. C d. Which link depends on the mass of the engine. e. A, B, and C all expeience the same foce. FNomal Fapplied Ffiction Fg 2. The fee-body diagam shows all foces acting on a box suppoted by a hoizontal suface, whee the length of each foce vecto is popotional to its magnitude. Which statement below is coect? a. The box is acceleating downwads because the foce of gavity is geate than the nomal foce. b. The box is acceleating to the ight, but not upwads. c. The box is acceleating upwads, but not to the ight. d. The box is acceleating upwads and to the ight. e. None of the statements above is coect. 3. A 0.50-kg object moves along the x-axis accoding to the function x = 4t 3 + 2t 1, whee x is in metes and t is in seconds. What is the magnitude of the net foce acting on the object at time t = 2.0s? a. 50 N b. 25 N c. 46 N d. 48 N e. 24 N

2 Pactice Test: m=1.0 kg h = 60cm L = 100cm 4. To detemine the coefficient of fiction between a block of mass 1.0kg and a 100cm long suface, an expeimente places the block on the suface and begins lifting one end. The block just begins to slip when the end of the suface has been lifted 60cm above the hoizontal. The static coefficient of fiction between the block and the suface is most nealy a b c d e A D ω R B Bb C 5. A lage Feis wheel at an amusement pak has fou seats, located 90 fom each othe and at a distance R fom the axis. Each seat is attached to the wheel by a stong axle. As the Feis wheel otates with a constant angula velocity ω, the seats move past positions A, B, C, and D as shown. At which position does a seat s axle apply the geatest foce to the seat? a. A b. B c. C d. D e. The axles applies the same foce to the seat at all fou positions.

3 Pactice Test: Pat II. Fee Response Top view Pespective view of ea of ca (velocity into the page) v = 14.0m/s = 50.0m v = 14.0m/s 6. A 500-kg ace ca is taveling at a constant speed of 14.0 m/s as it tavels along a flat oad that tuns with a adius of 50.0m. a. Daw a fee-body diagam fo the ca as it negotiates the ight-tuning cuve. b. What is the magnitude of the centipetal foce equied fo the ca to tavel though the tun? c. The coefficient of static fiction between the ties and the oad is Show that the ca will be able to make this tun.

4 Pactice Test: d. What is the maximum velocity that the ca can have, and still make the tun without slipping off the oad? e. Now enginees want to edesign the cuve so that no fiction at all is equied to stay on the oad. How high should they bank the 50.0-mete adius tun so that the ca will be able to tavel though it at 14.0 m/s with no lateal fiction equied fo the ca to make the tun.

5 Pactice Test: µ > 0 µ = 0 m3 = 2.0 kg m2 = 4.0 kg m1 = 2.0 kg 7. Blocks m 1 = 2.0 kg and m 2 = 4.0 kg ae connected by a thin, light cod which is daped ove a light pulley so that mass m 1 is hanging ove the edge of the pulley as shown. The suface between m 2 and the table is essentially fictionless, but thee is fiction between m 2 and m 3, which has a mass of 2.0 kg and is esting on top of m 2. a. Block m 2 is initially held so that it doesn t move. What is the Tension in the cod attached to m 1? b. Block m 2 is now eleased, and it acceleates so that m 3 does not slip, and emains in place atop m 2. i. What is the acceleation of mass m 2? ii. Daw a fee-body diagam of mass m 2, with vecto aows oiginating at the location whee the foce is applied.

6 Pactice Test: iii. What is the Tension in the cod attached to m 1 now as the system acceleates? iv. What is the minimum static coefficient of fiction that can exist between m 2 and m 3 based on this situation? Explain you easoning. v. If the coefficient of static fiction between m 2 and m 3 is 0.50, what is the maximum mass that m 1 can have so that m 3 will acceleate without sliding?

7 8. A billiad ball (mass m = kg) is attached to a light sting that is 0.50 metes long and swung so that it tavels in a hoizontal, cicula path of adius 0.40 m, as shown. a. On the diagam, daw a fee-body diagam of the foces acting on the billiad ball. Pactice Test: 0.50 m b. Calculate the foce of tension in the sting as the ball swings in a hoizontal cicle. c. Detemine the magnitude of the centipetal acceleation of the ball as it tavels in the hoizontal cicle. d. Calculate the peiod T (time fo one evolution) of the ball s motion.

8 Pactice Test: In a diffeent expeiment, the same ball with the same length of sting is now swung in a vetical cicle of adius 0.50 m. e. If the ball is tavelling at 3.00 m/s at the bottom of the vetical, cicula path, what is the tension in the sting at that moment? Include a fee-body diagam as pat of you solution. f. If the ball is tavelling at 3.00 m/s at the top of the vetical, cicula path, what is the tension in the sting at that moment? Include a fee-body diagam as pat of you solution.

9 Pactice Test: g. If the ball is tavelling at 3.00 m/s at the moment when the sting makes an angle of 45 fom the vetical as shown, calculate the tension in the sting. 45 h. The sting can handle a maximum foce of 10.0 Newtons befoe it beaks. What is the maximum speed the ball can have at the bottom of its path befoe the sting beaks?

10 Pactice Test: 9. A ping-pong ball has a mass of 2.7 g and a diamete of 40mm so that its coss-sectional aea is about m 2. The ball is eleased fom the top of a tall cliff at time t = 0, and as it falls though the ai, expeiences a dag foce R = 1 2 DρAv 2, whee D is the dag coefficient (0.5 fo this ping-pong ball), ρ is the density of ai (129 kg/m 3 ), and v is the velocity. a. Daw a fee-body diagam fo the ping-pong ball: i. Just afte it has been eleased ii. Afte it has fallen some distance but befoe it has eached teminal velocity iii. Afte the ball has eached teminal velocity b. Use Newton s Second Law to detemine the ball s acceleation as a function of velocity. c. Detemine the teminal velocity of this ping-pong ball.

11 Pactice Test: d. Develop, but do not solve, a diffeential equation that could be used to detemine the velocity of the ball as a function of time. e. Sketch a gaph of the ball s velocity as a function of time, including the time at which the ball eaches teminal velocity. +v 0 tteminal t - v

12 Pactice Test Solutions: 1. The coect answe is a. Link A is esponsible fo pulling the entie mass of the tain (m + 2m + 3m = 6m total) to the ight. Link B only needs to pull 5m, and Link C only 3m. A moe quantitative analysis, although not equied fo finding the answe hee, might include detemining the net acceleation of the tain as a function of the Foce of the engine and the total mass of the tain: F net = ma F engine = (m 1 + m 2 + m 3 )a F engine a = (m + 2m + 3m) = F engine 6m A fee-body analysis on the 3m ca, then, would detemine that the foce acting on that ca was: " F link = ma = (3m) F % engine $ ' = 1 # 6m & 2 F engine Simila analyses fo ca 2m and ca 1m eveal that link A expeiences a foce of F engine that is geate than the foces on the othe links. FNomal 3m Fgavity C Flink 2. The coect answe is b. The box has a net foce in the positive-x diection, but the foces in the y- diection ae balanced. 3. The coect answe is e. The acceleation of the object is detemined by using a = d 2 x 2, as follows: dt v = dx dt v = d dt (4t 3 + 2t 1) =12t a = dv dt a = d dt (12t 2 + 2) = 24t Substitute in t = 2.0s to get a = 48m/s 2. Use F net = ma to get F net = 24 N.

13 Pactice Test Solutions: 4. The coect answe is b. The amp can be thought of as the hypotenuse of a ight tiangle, with a coesponding ight tiangle as pat of the fee-body diagam fo the block. Fpependicula = 8.0N Fg = 10N Fpaallel = 6.0N The foce of fiction when the block just begins to slip equal the foce F paallel, and the nomal foce F Nomal equals the foce F pependicula. The coefficient of fiction, then, can be calculated: µ = F fiction F Nomal µ = 6.0N 8.0N = The coect answe is c. The axles need to suppot the seats against the foce of gavity, and fo a nonotating Feis wheel, the foce would be the same at each position. Fo a otating wheel, howeve, a centipetal foce is equied to keep the seats moving in a cicle. At position C, the axle needs not only to suppot the seat, but also povide additional foce to keep it acceleating centipetally (moving in a cicle). Quantitatively: F centipetal = mv 2 +F axle F gavity = mv 2 F axle = mv 2 + mg Faxle Fgavity Fcentipetal 6. a. The fee-body diagam fo the ca needs to take into account all the foces acting on the ca. All vectos should have labels. Note that the foce of fiction acts centipetally towad the cente of the cicula motion, and it should not be labeled F centipetal.. FNomal Ffiction b. Fo the ca to be able to make it aound the tun, it needs a centipetal foce of F c = mv 2 FWeight (500kg)(14m /s)2 = =1960N 50m This will obviously be supplied by the foce of fiction between the oad and the ties. c. If the static (non-slipping) coefficient of fiction between the ties and the oad is 0.78, we can detemine the maximum amount of centipetal foce that that fiction will supply: F fiction = µf Nomal F fiction = µmg = (0.78)(500kg)(9.8m /s 2 ) = 3820N

14 Pactice Test Solutions: Because this foce of fiction is geate than the centipetal foce we need (3820N > 1960N), the ca will easily make the tun. d. With that 3820 N maximum fiction foce available to us, the maximum speed the ca can have to negotiate this tun would be: F c = mv 2 F fiction = 3820N = (500kg)v 2 50m v =19.5m /s e. We need to bank the tun so that the hoizontal component of the Nomal foce is what supplies the centipetal foce that keeps the ca moving in a hoizontal FNomal cicle. Notice that we have chosen not to tilt ou x-y axes (as we sometimes do fo inclined planes), because the centipetal foce is diected hoizontally towad the cente of the cicle. Theefoe ou calculations will be easie if we keep standad x-y oientations on ou axes. The analysis: y : F net = ma; F Nomal y F gavity = 0 F Nomal y = F Nomal cosθ = mg x : F net = ma; F Nomal x = mv 2 F Nomal sinθ = mv 2 Combine the x and y equations to get: F Nomal sinθ = mv 2 2 v, so tanθ = F Nomal cosθ = mg g $ v 2 ' θ = tan 1 & ) = 21.8 % g( FWeight

15 Pactice Test Solutions: 7. a. The Tension in the cod may be detemined by dawing a fee-body diagam and examining the foces acting on m 1 : b. F y = ma = 0 F gavity = 0 = F gavity = mg = (2kg)(9.8m /s 2 ) =19.6N i. Once the block m 2 is eleased, the foce of gavity acting on m 1 begins to acceleate the entie system. Although we could do an independent analysis of each individual mass, it s easie in this case simply to sum all the foces that ae acting on the entie mass of the system: F net x = ma a system = F net x m = 1 g m system (m 1 + m 2 + m 3 ) = 19.6 = 2.45m /s2 ( ) FTension Fgavity ii. The fee-body diagam fo m 2 has to include all foces acting on the block. Two impotant details to conside: the Weight of m 3 on top of m 2 applies a foce down on that block, which esults in a lage F Nomal pushing up on m 2. Also, note that thee is a foce of fiction between m 2 and m 3 it is this foce of fiction acting to the ight on m 2 that causes that block to acceleate off to the ight with the system. The logical esult of Fg (m3) Fgavity that, howeve, is that m 3 expeiences that same foce of fiction in the opposite diection, to the left. That fiction woks opposite the foce of Tension, and keeps m 2 fom acceleating as quickly as it would othewise. iii. The Tension in m 1 is easily calculated using a new fee-body diagam fo that block: F y = ma F g = ma = F g ma = mg ma = m(g a) = (2kg)( ) =14.7N whee we ve chosen to make the down diection positive in ou equations. iv. Because m 3 is acceleating to the ight at 2.45 m/s 2, we can detemine the foce of fiction acting on it: = ma F x F fiction = ma = (2kg)(2.45m /s 2 ) = 4.90N Ffiction FNomal FTension We know the Nomal foce acting on m 3, so we can get the minimum coefficient of static fiction as follows: µ = F fiction = 4.90 F Nomal mg = N = 0.25 If thee s a geate coefficient of fiction between these two sufaces ie. if the contact between

16 Pactice Test Solutions: them is moe sticky that s fine. We don t eally have any moe we can say about that. But we do know that because m 3 is still stuck unde the cuent cicumstances, the coefficient of fiction has to be at least v. Fist let s figue out what the maximum acceleation fo that block can be based on fiction: F fiction = µf Nomal = µmg = (0.5)(2)(9.8) = 9.8N F m3 = F fiction = ma a = F fiction m = 9.8N = 4.9m /s 2 2 Now let s look at the system again to see what foces we can apply and NOT exceed an acceleation of 4.9 m/s 2 : F x = ma F g = ma; m 1 g = m total a m 1 g = (m 1 + m 2 + m 3 )(4.9) m 1 (g 4.9) = (m 2 + m 3 )(4.9) m 1 = ( 2 + 4)4.9 = 6kg 4.9 Moe than 6kg fo m 1 and we ll acceleate the system too quickly, causing m 3 to beak loose and stat sliding. 8. a. b. Using the fee-body diagam, and knowing that the billiad ball is not acceleating vetically, we can wite F y = ma y = 0 y F gavity = 0 sinθ = mg We can detemine θ by using the dimensions of the sting and the cicle adius: cosθ = 0.40m 0.50m " θ = cos m % $ ' = 36.9 # 0.50m & Solving fo tension: sin36.9 = (0.15kg)(9.8) = 2.45N Fgavity FTension c. Centipetal acceleation can be detemined using a c = v2, but we don t know the velocity of the ball. Pehaps we can detemine acceleation using a foce analysis. Consideing the x-diection:

17 Pactice Test Solutions: F x = ma x, o F c = ma c x = ma c a c = x m = cosθ m a c = (2.45N)cos(36.9 ) (0.15kg) =13.1m / s 2 d. The peiod T fo one evolution is calculated using T = distance speed = 2π. We haven t yet detemined v v, but now that we know centipetal acceleation, we can get it. Using that elationship: a c = v2, so v = T = 2π v a c = 2π = 2π a c a = 2π 0.40m 13.1m / s 2 =1.10s e. At the bottom of its path, the Foce of tension and the Foce of gavity ae acting in opposite diections. Taking up to the positive diection: F c F g + F Fgavity g (0.15kg)(3.00m / s)2 = + (0.15kg)(9.8m / s 2 ) = 4.17N 0.50m This is much geate than the weight of the billiad ball by itself, just 1.47 N. That s because the Tension in the sting has to suppot the weight of the ball, and acceleate the ball towad the middle of the cicle so that it will tavel in that cicula path. f. At the top of its path, the Foce of tension and the Foce of gavity ae both acting downwads towad the cente of the cicle. Taking down to be the positive diection: F c + F g FTension F g (0.15kg)(3.00m / s)2 = (0.15kg)(9.8m / s 2 ) =1.23N 0.50m It takes some amount of foce diected downwad to keep the ball moving in a cicle at this speed, but gavity helps to supply some of that downwad foce, so the foce equied fom the tension in

18 Pactice Test Solutions: the sting is much less than it was befoe. g. To solve this poblem, we need to think about whethe o not we should tilt ou axes to solve the poblem. We typically want to do that if the diection of acceleation in the poblem has been tilted. Is that the case hee? In which diection is the ball acceleating? Given that the ball is acceleating centipetally (towad the cente of the cicle), we choose to tilt ou axes as indicated. +y Fgavity FTension +x With these axes in mind, we can see that we ll have to split the foce of gavity up into components, and then use F net = ma to calculate the foce of tension. F c F gavity adial + F gavity adial + mgcosθ (3.0m / s)2 = (0.15kg) + (0.15kg)(9.8m / s 2 )cos45 = 3.74N (0.5m) FTension Fgavity h. We can use an analysis simila to what we did in pat (e), although hee we e solving fo velocity instead of tension: F c F g v = ( ) m F g = (0.5m)(10N (0.15kg)(9.8m / s2 )) 0.15kg = 5.33m / s

19 Pactice Test Solutions: 9. a. i. Just afte it has been eleased ii. Afte it has fallen some distance but befoe it has eached teminal velocity iii. Afte the ball has eached teminal velocity Fai Fai Fgavity Fgavity Fgavity Some instuctos pefe that students daw thei vectos with the beginning of the vecto located at the point of application fo that Foce. Thus, Foce of gavity acts on the cente of mass, and oiginates at the cente of the ball, while the foce of ai fiction is acting at the suface of the ball. The labels on the vectos ae impotant, of couse, and thei elative lengths should be to scale.

20 Pactice Test Solutions: b. The ball s acceleation as a function of velocity is detemined using Newton s 2nd Law and the dag function given in the poblem statement: F net = ma F g R = ma mg 1 2 DρAv2 = ma a = g DρAv2 2m (0.5)(129)(1.26e 3)v2 a = 9.8 2(0.0027) = v 2 c. The teminal velocity of the ball will occu when ai esistance R is equal to the Weight of the ball: F net = ma F g R = 0 mg = 1 2 DρAv 2 v = 2mg DρA = = 0.807m /s m d. We can develop this integal by using the function that s been given to us, along with Newton s Second Law. Note, also, that acceleation is the deivative of velocity, so: F net = ma F g R = ma mg 1 2 DρAv 2 = ma = m dv dt dv dt = g DρA 2m v 2 e. The gaph of the ping-pong ball s velocity should begin with 0 velocity, then incease in speed in the downwad (negative velocity) diection with a negative acceleation (slope) that deceases ove time to appoach the constant (teminal) velocity. +v 0 tteminal t - v