X3.17a 2050PHYS-205(17) (Kaldon-20619) Name _________S O L U T I O N_______WMU-Summer I 2006 Exam 3A - 100,000 points + 20,000 ¶ points Don’t Forget Your Paper!06/15/2006•Rev.5State Any Assumptions You Need To Make – Show All Work – Circle Any Final AnswersUse Your Time Wisely – Work on What You Can – Be Sure to Write Down EquationsFeel Free to Ask Any Questions¶2a ¶2b ¶2c ¶2eEXAM 3 [FORM - A]PHYS-2050 (KALDON-17)SUMMER I 2006WMUShouldn’t the game Twisterbe called “Torquer”?Physics 2050 / Exam 3 [Form-A] Summer 2006 Page 2Roving the Red Planet (35,000 points)1.) (a) Three summers ago NASA sent the first of two rover probes toMars – Spirit and Opportunity are still at work! To get to Mars, a400. kg probe first has to leave the Earth. On Earth the probe weighs3920 N. What would this 400. kg probe weigh on Mars ( mass = 6.42× 1023 kg ; radius = 3.37 × 106 m )? G = 6.67 × 10 -11 N·m²/kg²FGMmrkg kgmNGNmkg==´´´=-×211 23626 67 10 6 42 10 400337 10150822....dichbgchPhysics 2050 / Exam 3 [Form-A] Summer 2006 Page 3To launch a rocket to Mars, an intrepid NASA scientist/engineer first has to get to work in his car. Theparking deck has a mass of 11,400 kg and is 20.0 m wide. Support Piers 1 and 2 are located 3.00 m fromthe ends. The 1950 kg car is 6.00 m from the left and our 1320 N scientist/engineer is 3.00 meters fromthe center-of-mass of the car as shown. Find the support forces (b) F1 and (c) F2. To get full credit, youmust include the F.B.D. and the F.R.D.F1 is 0 meters from pivot.Car is 3.00 meters from pivot.Man is 6.00 meters from pivot.Deck c.m. is 7.00 meters from pivot.F2 is 14.0 meters from pivot.wmg kg ms NwNwmg kgms NFFFw w wFmw mw mw mFmw mw mw mFNm Ncar carmandeck decky car man deckcar man deckcar man deck== ==== ==+- - - ==- - - ==+ +=+åå1950 9 81 19 130132011 400 9 81 11 800014 0 3 00 6 00 7 00 014 0 3 00 6 00 7 0019 130 3 00 1320 6 002212222bgchbgchbgbg bg bgbgbg bg bgbgbgbg./ ,,./,.. . ... . .,. .tmNmmNFw w w FNN N NNcar man deckbgb gbg+==+ + -=++-=11 800 7 0014 010 57019 130 1320 11 800 10 57021 68012,..,,,,,Physics 2050 / Exam 3 [Form-A] Summer 2006 Page 4(d) Our scientist/engineer thinks he’s got a powerful car – it’s got 245 hp (182,800 W). If he wants to gethis car from rest up to 60.0 mph (26.8 m/s), how much time would this take, assuming he can use all thatpower?WKKK mvkg m s JPWttWPJWfi==-====== =D1221221950 26 8 700 300700 300182 8003831bgb g./ ,,,.sec(e) When our scientist/engineer gets to his destination, he pulls on the parking brake with a force of 127N. The parking brake has a 1.00 m length of cable ( Y = 10.0×1010 N/m² , cross-section A = 9.00 × 10-6m² ). How much does the cable stretch?YFALLFLALLFLAYNmmNmmm====´´==´--DDD00062 10 24127 100900 10 100 100 0001411 1411 10bgbgchc h.../..Physics 2050 / Exam 3 [Form-A] Summer 2006 Page 5A Notch Above the Usual (30,000 points)2.) ¶(a) A plate of mass m has sides of 4a and 2b. Find the center of masscoordinate xcm by integrating xMxdmcm=z1, using the x- and y-axes asshown.Mass on the left side should be twice the mass of the rightside, since it has twice the area.left M b M b dm dxright M b M b dm dxxMxdmMxdmMxdmMxdxMxdxMxdxMxdxMxMxMbMbbMbMbMbMbbbLL LRR Rcmleft rightLbRbbLbRbbLbRbbLR://;://;//llllllllllll== === === +=+=+=+=+-=+-=zz zzzzz231302022022222232132211 11122242242bgbgchbgbgch3226365608333+= + = =bbbbb. ORAab ababMAMabdm dA dxdyxMxdmMxdxdyMxdxdy xdxdyMxdx dy xdx dyMxaxaMbabbaMbabcmaaabbbaaabbbbbb=+========+LNMOQP=+LNMOQP=+LNMMOQPP=+-LNMOQP=+zzzzzzzzzz()()()()426611242224422243220432004320202222222sssssssssbg bgbg bgbgaMab abMabMMababbbbgLNMOQP=+==FHGIKJ==ss235165560 83332222.Physics 2050 / Exam 3 [Form-A] Summer 2006 Page 6¶(b) A plate of mass m has sides of a and b. Find the moment of inertia I of theplate about the y-axis as shown, by integrating Irdm=z2 .llllll==®== ==LNMOQP=FHGIKJ=FHGGIKJJ=zz zMadm dx r xIrdm xdx xdxxaMaaMaaaa;;220203033132333ch¶(c) A torque rt to tighten a bolt consists of a force being applied at a distance from the axis of rotation.As the bolt gets tighter, it gets harder and harder to turn the bolt, so the torque as a function of angle isgiven by t = C q² , where C is some constant with appropriate units. If the total work done by applyingthis torque through two complete revolutions is 1500. J, then find C.Wd CdCCradCrad JCJradJradORNmrad==== -====×zzt qqqqppp2043043333233406615 1500150066152 267 2 267bgejchch......Those quasi-units are slippery!(d) In the diagram at right, the mass m1 is moving down. Does rt on the pulley point up,down, left, right, in, out ?Rotation is counter-clockwise, so using RHR, thethumb points OUT.The Torque vector rt points OUT of the page.Physics 2050 / Exam 3 [Form-A] Summer 2006 Page 7¶(e) An solid disk of mass 71.3 kg and radius 1.21 m has a motion that follows the following equation.Find the angular acceleration, a, at time t = 1.00 sec.w () (. /) (. / ) (. / )t rad s rad s rad s=+ -2 00 2 00 2 00232ttwaqwaa() (. /) (. / ) (. / )(. /) (. / ) (. / )(. / ) (. / )(. / ) (. / )(. / ) (. / ) . sec./. /.sect rad s rad s rad sddtddtddtrad s rad s rad srad s rad srads radsrads radsrad s radt=+ -===+-=+ -=-=-=-=200 200 200200 200 2000200 2200200 400200 400 100200 4000232222322323100232ttttttbgsrad s222000=- ./If It’s Green and Can Be Mowed, It’s a Lawn (35,000 points)3.) Dr. Phil’s 1994 Sears self-propelled gas lawnmower has a 5.55 hp (4140 W)Eager One engine. Theclaim in an ad is that “the cutting bar strikes each blade of grass five times as the mower moves forward atnormal walking speed (1.00 m/s)”. This works out to the blade spinning 2.00 revolutions in 0.700 sec.(a) What is the angular velocity w of the blade? Given that the blade has a length L = 0.700 m, what is thelinear speed vblade at the edge of the blade? Note this is not the same speed v of the mower going forward.wpwpp========×=20010040700 2 03500350 414004398..sec/sec.; /../sec./sec./revsradDmrD mvr m radmradmsbgb gPhysics 2050 / Exam 3 [Form-A] Summer 2006 Page 8(b) The ad continues, “to meet Federal safety standards, the blade brake will stop the blade in onerevolution”. Find the angular acceleration a needed to stop the blade. If you didn’t get an answer to (a)use w0 = 5.00 rad/sec.ww aqqwapap wawpppp202002020222220222244441257=+ -=+=-=-=-=- =-bgbgbgbgradradradradradrad rad/sec/sec . /sec(c) The 2.00 kg blade can be modeled as a solid rod or bar. Find