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MEC2401 Engineering Dynamics

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MEC2401 Engineering Dynamics

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Course Code: MEC2401
University: The University Of Queensland

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Country: Australia

Question:
The locomotive A has mass MA = 4.7 Mg andwastravelling in constant velocity of 100 km/h and the Tanker B has mass MB = 1.4 Mg and was travelling in constant velocity of 80 km/h.
 
1. Calculate velocities of the locomotive and the tanker after the collision if the locomotive and the tanker become entangled and move off together after the collision. 2. Calculate velocities of the locomotive and the tanker after the collision in terms of e (0 < e < 1) which is the coefficient of restitution between the locomotive and the tanker.3. Calculate the possible energy losses for Case (1) and Case (2) for the value of e = 0.8.Comment on the severity of collision depending on your calculated values. (10 marks)List all the assumptions clearly for each case.  Answer: Given that the particle has a mass of m=0.6kg Angular velocity w=2.4rad/s ?=320 ?´= 2.4rad/s  r´ =  = =  = 0.7075 m Calculate the linear velocity of the rod r´ = 0.6 sec? ( (tan? )( ?´)) = 0.6sec(32)(tan(32)(2.4) =0.6(1.11791)(0.6248(2.4)) =1.06085m/s Calculate the linear acceleration of the rod  r´´ 0.6sec? (tan?(?´)) tan? +sec?(sec2?´(?) ?´+sec(tan??´´ ) (0.6sec?´×tan2?×?´)+ (sec?×sec2?×?2) +0 = (0.6 ×1.1791× 0.6242 ×2.4) + (1.1791×1.3902×5.76) = (0.6611) + (9.4417) = 10.1028 m/s2 Determine the radial acceleration ar =r´´ - r?2 = 10.1028-(0.7075(2.4)2 =10.1028 – 4.0752 = 6.0276 m/s 2  Angular acceleration a?= r?´´ +2r´´ ? =(0.7075×0 + (2×1.0608×2.4) =5.09184m/s2 Now calculate the force due to the linear acceleration mar= Fr mar=µcos ?- mgcos ? (0.6 )(6.0276)= µ cos 32 – ( 0.6 ×9.81×cos 32) 3.6156=0.8480N-4.9916 0.8480N=8.6081 N= 10.1511N Part 2  The normal force of the slot on particle ma?=Fr (0.6) (5.09184) =F-µsin ?+mgsin ? = 3.055104=F-(10.1511(0.5299)+3.1191 3.055104=F8.4981 F= 11.5532   Velocity of the cart from linear momentum F×t= M1V1+M2V2 48× 5 = 35×u + 29 V 46×5 = ( 35+29)V V=  V= 4.067 m/s Part 3 p×t×r = Iw 48×5×0.14 =mrk2w 48×5×0.14 = 35×0.262 w w= w= 14.20rad/s     Given that mass = MD (mass of drum)= 58kg  and radius of gyration Ko= 0.25 m Mass of block =MB= 18kg The distance travelled by block S1= 3.2 m  α = Angular acceleration of drum and ? = angular velocity of the drum From newton´s 2nd law of motion. For block = ma mBg-T= mBa . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Similarly applying the same law = I. α  Where I is the moment of inertia T.r1 = I α α=  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 We also know that; I= MD  × K02 = 58×(0.25)2           = 3.625 kg-m2 Since pt   M is a part of sting as well as of drum so its acceleration with respect to to string and also with respect to drum is same a= r1× α    Put the value of α in the equation 2 above = a´ =  = T= 111.89a Putting the value of T in the equation 1 MBg= 111.89×a a =  a= 1.36m/s2 Now applying Newton´s 3rd equation of motion V12= U2+2as1  Where U=0 the initial velocity and V is the final velocity V1= V1= 2.95m/s  Break is applied at a distance of 0.28m from a or at outer radius   Is is frictional force N is the normal force to the drum R is the reaction force acting on the handle First we have to obtain the N=R which is acting on the drum Balancing moment at acting on handle about point A P×(xc) =0 P(0.6+0.82)= R(0.6) Is= Due to this reaction force, frictional force is given by Frictional force = µR = 0.5 =   Is=    As part C is moving upward so frictional force will oppose this motion and hence it acts in downward direction. Now Applying Newton´s law for drum T.r1- Is-r2 =Iα T.r1 = Anticlockwise torque Is.r2 = Clockwise torque ¨ T×0.18 – Is ×0.28 = 3.625 α T= And again applying newton´s second law mBg-T=mBa 18×9.81 - Is = 3.625× + 18× a 176.58-1.56Is= 20.14a+18a a= a= 4.63-0.041Is Applying Newton´s 3rd equation  V12= U2+2aS1   Final velocity V=0 2as = (2.95)2 a=  = 1.36m/s2 4.63-0.041Is = 1.36 = Is  P=  79.756 P= 67.399 N Given data ?1= 00 ?4= 1270 AB=310 mm BC=420mm CD=480mm Part 1 Obtaining the distance AC AC= AC= AC= AC=   AC=243.32mm Obtaining the angle ?2 Cos (180- ?2) = Cos (180-?2) = -0.1398  180- ?2 = cos-1 ( -0.1398) - ?2= 98.036-180 + ?2=+81.964 ?2= 81.964 Find the angle ?3 ?3= 81.964+35 ?3= 116.964 Write the equation of from velocity analysis AB ?2Sin ?2+ BC ?3 Sin ?3+ CD ?4 Sin ?4 =0 ( 310× 3.5×sin81.964)+(420× w3×sin 116.964)+(480× w4×sin 127)=0 (1085×0.9901)+(374.342×w3)+(383.345×w4)=0 1074.2585+374.342w3 +383.345w4=0 Through dividing through by 374.342 2.869+w3+1.024w4=0  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  . . 1 ABw2 ?2 +BCCos w3 ?3 +cos w4cos ?4 =0 (310×3.5×cos (81.964)+(420×w3×cos 116.964)+480×w3×cos127)=0 151.677-190.4408w3-288.871w4=0 0.7964-w3-1.5168w4=0  . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .   ... 2 Solving equation 1 and 2 simultaneously 2.869+w3+1.024w4=0 0.7964-w3-1.516w4=0 We obtain   3.6654-0.492w4=0   -0.492w4=-3.6654 W4=    W4= 7.45 rad/seconds Substitute equation 1 in w4 value 2.869+w3+1.024(7.45)=0 W3= -10.4978 rad/s Therefore the angular velocity of the link BC and CD is -10.4978 and 7.45 respectively Writing the equations from accelerations r2α2 sin?2+r2w22 cos ?2 +r3 α3sin ?3+r3w32+r4 α4sin?4+r4w42cos ?4 =0 (310×2.2 ×sin 81.964)(310×3.52 × cos 81.964)+(420×α3×sin116.964)+(420×(-10.4978)2 ×cos 116.964)+(480α4×sin 127)+( 490×7.452 cos 127)=0 765.303+3797.6397+374.342α3- 20987.3053+383.34α4-16033.07439=0 374.342α3+383.345α4-32547.436 ……………………………………………………3 r2α2cos?2-r2w22sin?2+r3α3?3- r3w32sin?3+r4α4cos ?4-r4w42sin?4=0 (310×2.2×cos81.964)-(310×3.52sin81.964) +( 310×2.2 ×cos 81.964)-(420×(-10.4978)2×sin 116.964)+(480×cos 127)-(480×7.452×sin 127)=0  95.34-3760.21-190.44α3-4125.9-288.87α4-21276.6 =0 -190.44α3-288.87α4-66.195 =0 -190.44α3-288.87α4= 66.196 ……………………………………………………….. 4 By solving equation 3 and equation 4 simultaneously we obtain                        α3+73004.22α4=6198333.712 71289.69048α3+108136.17α4=24779.56 α4= 175.725 rad/s2 Substituting α4 value in equation 3 374.342α3+383.345(175.725)= 32547.436 374.342α3= 32547.436-673.3 374.342α3= -34815.86 Α3=-93.005 rad/s2 Therefore, the angular acceleration of the link BC and CD is – 93.005 rad/s2 and 175.725rad/s2 respectively. Find the kinetic energy of the link AB KE= ½ |AB w22  KE= ½  KE= 0.490J/kg Find the kinetic energy of the link BC KE= ½ |BC 32  KE= ½ KE= 8.099J/kg Kinetic energy of the kink CD KE= ½ |AB w22  KE= ½  KE= 10.57J/kg At point C The horizontal component of acceleration, ie  tangential component , at = 5×1.2 = 0.75rad/s radial component , ar=0.5×1.82 horizontal component = 1.62cos 38-0.75cos 52 = 0.814 rad/s2 ( towards the right ) Vertical component = 0.75sin52+1.62sin 38 = 1.58 rad/s2  (towards down)  CASE 1: Given MA= 4.7 mg                  MB= 1.4mg VA=100km/h                             VB=80km /h The angle between the velocities of the two objects is given to be (180-30) = 1200  The Collison hence is a perfect elastic, after the collision there is a single combined body of mass  MA+MB = 6.1 mg The momentum is conserved From the formula of inelastic collision |Vf|2 =  +  + VAVB cos ? =  +  + ×100×80×-0.5   =5936.576+337.11-707.34 =5566.347 Vf= 74.6km/ CASE 2: Initial momentum P1=MaVa+MBVB If = MAVAf+MBVBf Einitial = ½MAVA12 +½ MBVB2 Efinally =½MAVAF2+½MFB2 Conservation of momentum Pinitial =Pfinal MAVA+MBVB= MAVf + MBVBf VAf=  VBf=  VAf=  VAF= 1+  ( = VBF +   ( From geometry; VAF=   ( VBF= VAF= V= 91.73 units VBF= VBF= 99.08 units For case 1 Ei=½ MAVA2+½MBVB2 ½×4.7×1002 + ½1.4×802 = 23500+4480 27980 Units Efinal = ½Vf2×(MA+MB) = ½ ×5566.346V6.1 = 16977.3553 Loss in energy = |Efinal –Einitial| = 27980-16977.3553 = 11002.644 Units Case 2 Einitial = 27980 units Efinal ½ VAF2×MA+½VBF2×MB = 19773.82 +6871.79 = 26645.61 Units Loss in KE = Einitial –Efinal = 27980-26645.61 =1334.39 units Assumption for case 1 The below are some assumptions The amount of time elapse during the collision is very small as compared to the time between the collisions(Kircanski, 2013). There are huge number of particles involved in the collision Assumption for case 2 The volume which is occupied by the particle during collision is negligible as compared to the volume of the container(Holzner, 2014). The particles in collision are always in a random motion.   References Holzner, S. (2014). Physics For Dummies. Hull: John Wiley & Sons. Kircanski, M. (2013). Kinematics and Trajectory Synthesis of Motions. Chicago: Springer Science & Business Media. Free Membership to World's Largest Sample Bank To View this & another 50000+ free samples. Please put your valid email id. E-mail Yes, alert me for offers and important updates Submit  Download Sample Now Earn back the money you have spent on the downloaded sample by uploading a unique assignment/study material/research material you have. After we assess the authenticity of the uploaded content, you will get 100% money back in your wallet within 7 days. UploadUnique Document DocumentUnder Evaluation Get Moneyinto Your Wallet Total 6 pages PAY 4 USD TO DOWNLOAD *The content must not be available online or in our existing Database to qualify as unique. Cite This Work To export a reference to this article please select a referencing stye below: APA MLA Harvard OSCOLA Vancouver My Assignment Help. (2020). Engineering Dynamics. Retrieved from https://myassignmenthelp.com/free-samples/mec2401-engineering-dynamics/random-motion.html. "Engineering Dynamics." My Assignment Help, 2020, https://myassignmenthelp.com/free-samples/mec2401-engineering-dynamics/random-motion.html. My Assignment Help (2020) Engineering Dynamics [Online]. Available from: https://myassignmenthelp.com/free-samples/mec2401-engineering-dynamics/random-motion.html[Accessed 18 December 2021]. My Assignment Help. 'Engineering Dynamics' (My Assignment Help, 2020) accessed 18 December 2021.

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