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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.
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