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SolutionsManual© toaccompany System Dynamics,FourthEdition by William

PalmIII UniversityofRhodeIsland

SolutionstoProblemsinChapterOne

1.1 ( ) ===332.296.6 Wmg lb.

1.2 ===/100/9.8110.19mWg kg. ( ) ==1000.224822.48 W lb. ( ) == 10.190.068520.698 m slug.

1.3 ( )( ) =+=505/120.304815.37 d m.

1.4 ( )( ) == 31000.304891.44 d m

1.5 ( ) ==1003.281328.1 d ft

1.6 ( ) == 503600/528034.0909 d mph

1.7 ( ) ==1000.621462.14 v mph

1.8 ( )  ==  3 1/601.3411012.43 n ,orapproximately12bulbs.

1.9 ( ) −= 57032/921.1 C

1.10 ( ) += 930/53286 F

1.11 ( ) ==ωπ30002/60314.16 rad/sec.Period === πω 2/60/30001/50 P sec.

1.12 =ω 5rad/sec.Period === πωπ 2/2/51.257 P sec.Frequency === 1/5/2fP π 0.796 Hz.

1.13 Speed ( ) == 405280/360058.6667 ft/sec.Frequency ==58.6667/301.9556 times persecond.

1.14 = 0.005sin6, xt ( ) == 0.0056cos60.03cos6. xtt Velocityamplitudeis0.03m/s. ( ) =−=−60.03sin60.18sin6. xtt Accelerationamplitudeis0.18m/s2.Displacement, velocityandaccelerationallhavethesamefrequency.

1.15 Physicalconsiderationsrequirethemodeltopassthroughtheorigin,soweseeka modeloftheform = fkx Aplotofthedatashowsthatagoodlinedrawnbyeyeisgiven by = 0.2.fx Soweestimate k tobe0.2lb/in.

1.16 Thescriptfileis

x=[0:0.01:1];

subplot(2,2,1)

plot(x,sin(x),x,x),xlabel(’x(radians)’),ylabel(’xandsin(x)’),...

gtext(’ x ’),gtext(’sin(x)’)

subplot(2,2,2)

plot(x,sin(x)-x),xlabel(’ x(radians)’),ylabel(’Error:sin(x)-x ’)

subplot(2,2,3)

plot(x,100*(sin(x)-x)./sin(x)),xlabel(’x(radians)’),...

ylabel(’PercentError’),grid

Theplotsareshowninthefigure.

Figure:forProblem1.16.

Fromthethirdplotwecanseethattheapproximation  sin xx isaccuratetowithin5% if  ||0.5 x radians.

1.17 Forθnear /4,π

Forθnear 3/4, π

1.18 Forθnear /3,π

Forθnear 2/3, π

1.19 For h near25,

1.20 For r near5,

For r near10,

1.21 For h near16, ( ) ( ) ( )

and ( )  0 fh if −16. h

1.22 Constructastraightlinethepassesthroughthetwoendpointsat = 0 p and = 900. p At = 0, p ( ) = 00. f At = 900, p ( ) == 9000.0029000.06. f Thisstraightlineis ( ) == 0.061 90015,000 fppp

c)Let = .vx

d)Let = .vx

From(1.4.4),

From(1.4.2),

1.25 Let ( ) ==− 0 0.hbv

Thus

Also,since = ,ha

1.26 Werequirethat ( ) =− ff htv .From(1.4.3), ( ) =−+ α 2 00 2 t htgvth

Equation(1.4.4)staysthesame,sothethruststaysthesameas(1.4.6): ( ) =+ α 1 fmg

1.27 a.

Find a and .b

Solve(1)and(2)for a and :b

Thus f t isfreetobechosen. b.1.

LLC.

1.28 Thetotalinertiais =++=++=0.0050.0020.0080.015. mSL IIII

FromFigure1.5.4 weobtaintherequiredangularaccelerationfromtheslopeofthevelocitycurve.Thisgives

Thustherequiredtorqueis

Sothemaximumrequiredtorqueis30N m.

Thermsaveragetorqueiscalculatedasfollows:

1.29 Theequationofmotionisnow

Sothemaximumrequiredtorqueis33andthermstorqueis

1.30 Thetotalloaddisplacementisgivenas = θπ 10 f rad.Wealsohave =0.5 f t sanda slewtimeof −= 21 0.4 tt s.Usingasymmetricalvelocityprofilerequiresthat = 1 0.05 t and = 2 0.45. t Thetotalinertiais =++=3433 610510107.510 I

FromTable1.5.1 wehave ==ωπ10/0.469.813 max rad/sor ( ) = ωπ 60/2666.667 max rpm.Sothe loadwillrotateatthatspeedfor0.4s.

Wearegiventhat = 2 dT N m.Sothemaximumtorquerequiredis = 12.4720 max T N m andthermsaveragetorqueis = 5.0924 rms T N m.

1.31 Thetotalloaddisplacementisgivenas = θπ 4 f rad.Wealsohave =1 f t sandaslew timeof −= 21 0.6 tt s.Usingasymmetricalvelocityprofilerequiresthat = 1 0.2 t and = 2 0.8. t Thetotalinertiais =++=2433 1061021011.610 I

FromTable1.5.1 wehave ==ωπ4/0.815.7083 max rad/sor ( ) = ωπ 60/2150 max rpm.Sotheload willrotateatthatspeedfor0.6s.

Wearegiventhat = 0.5 dT N m.Sothemaximumtorquerequiredis = 1.4111 max T N m andthermsaveragetorqueis = 0.7629 rms T N m.

1.32 Theequivalentinertiaoftheloadmassis,from(1.5.11),

Themassofthescrewis ( )

kg r mVρπ andtheinertiaofthescrewis

Sothetotalsysteminertiais

Fromtheequationofmotion,therequiredtorqueis = . TIω Thelineardisplacement x of themassisrelatedtotherotationangleθofthescrewby = /2. xLθπ Thusthelinear acceleration x isrelatedtotheangularacceleration = θωby = 2/. xLωπ Thegivenlinearaccelerationis ( ) =−==0.30/0.221.5 x m/s2.Thus ( ) == 2 21.5 1178.1 0.008rad/s

torque

1.33

>>x=2;y=5;(y*x^3)/(x-y)

ans= -13.3333

>>3*x/(2*y)

ans= 0.6000 >>(3.2)*x*y

ans= 32 >>x^5/(x^5-1)

ans= 1.0323

1.34

>>x=-7-5j;y=4+3j; >>x+y

ans= -3.0000-2.0000i >>x*y

ans= -13.0000-41.0000i >>x/y

ans= -1.7200+0.0400i

1.35

>>(3+6j)/(-7-9j)

ans=

-0.5769-0.1154i

>>(5+4j)/(5-4j)

ans=

0.2195+0.9756i >>(3/2)*j

ans= 0.0000+1.5000i >>3/(2j)

ans= 0.0000-1.5000i

1.36

>>x=5+8j;y=-6+7j; v>>u=x+y

v-1.0000+15.0000i >>v=x*y

v= -86.0000-13.0000i >>w=x/y

0.3059-0.9765i >>z=exp(x)

-2.1594e+01+1.4683e+02i >>r=sqrt(y)

r= 1.2688+2.7586i >>s=x*y^2

s= 6.0700e+02-5.2400e+02i

1.37

>>exp(2)

ans= 7.3891 >>log10(2)

ans= 0.3010 >>log(2)

ans= 0.6931

1.38 >>cos(pi/3)

ans= 0.5000 >>cosd(80)

ans= 0.1736 >>acos(0.7)

ans= 0.7954 >>acosd(0.6)

ans= 53.1301

1.39

>>atan(2) ans= 1.1071

>>atan(100) ans= 1.5608

Theanswersfor(c),(d),and(e)areindegrees.

>>atan2d(3,2)

ans= 56.3099 >>atan2d(3,-2)

ans= 123.6901

>>atan2d(-3,2)

ans= -56.3099

1.40

>>x=1:0.2:5; >>y=7*sin(4*x); >>length(y)

ans= 21 >>y(3)

ans= -4.4189

1.41

roots([13,182,-184,2503])

ans=

-15.6850+0.0000i

0.8425+3.4008i

0.8425-3.4008i

1.42

roots([70,24,-10,20])

ans=

-0.8771+0.0000i

0.2671+0.5044i

0.2671-0.5044i

1.43

>>poly([-2+5j,-2-5j,-7])

ans= 11157203

Thisgivesthepolynomial

Tocheck, >>roots(ans)

ans=

-7.0000+0.0000i

-2.0000+5.0000i

-2.0000-5.0000i

1.44

>>x=linspace(0,2,300);u=2*log10(60*x+1);v=3*cos(6*x);

>>plot(x,u,x,v,’–’),xlabel(’Miles’),ylabel(’Mph’),gtext(’u’),gtext(’v’)

1.45

>>h=polyval([0.0125,-5,500],t);

>>plot(t,h),xlabel(’Time(s)’),ylabel(’Height(m’))

1.46 Thegivenparametersare ,m ,g 0, h and 0. v FollowExample1.4.1andprogram equations(1.4.3)forα and(1.4.2)for . f t

1.47

t1=0:0.001:0.5; om1=300*t1; t2=0.501:0.001:2.5; om2=150*ones(size(t2)); t3=2.501:0.001:3; om3=-300*(t3-2.5)+150; t=[t1,t2,t3]; om=[om1,om2,om3]; plot(t,om)

1.48 Hereisafragmentoftheprogram.Youmustcreatecodetoinputthefollowing parameters:θf , dT , 1t , 2t , f t ,and .I

om max=theta f/t2

alpha max=theta f/(t1*t2)

T max=I*theta f/(t1*t2)+Td

T rms=sqrt(2*I^2*theta f^2/(tf*t1*t2^2)+Td^2)

FollowExample1.5.5.Theequationsare: