# deen's ch1 solution.pdf - Solution Chapter 1 1.1 Falling...

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Solution: Chapter 11.1 Falling BodyBody of massmfalls a distancehin timeth, with gravity the only force(a) Dimensional analysis for initial velocity=0Quantities:th,m,g,h[th]=T,[m]=M,[g]=LT2,[h]=Lmcannot be involved because nothing else contains the dimension M.3 quantities2 dimensions=1 dimensionless group.LetN=thgahb. Then[th]=T=[gahb]=(LT2)aLb=La+bT2aSolve for exponents:a= −12,b=12N=thgh1/2=const.K(b) Dimensional analysis for initial velocity=UQuantities (omittingmagain):th,g,h,U.[U]=LT14 quantities2 dimensions=2 dimensionless groups.LetN1=thgh1/2as in part (a).LetN2=Ugchdbe the second group.[U]=LT1=[gchd]=(LT2)cLd=Lc+dT2cc+d=1and2c= −1givesc=12=d.N2=U(gh)1/2andN1=f(N2).(c) ActualthforU0.
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Solution: Chapter 1z(t)=instantaneous position, measured downward from initial positiond2zdt2=g,z(0)=0,dzdt(0)=Udzdt(t)=gt+Adzdt(0)=A=Uz(t)=gt22+Ut+B,z(0)=B=0z(t)=gt22+Utz(th)=h=gt2h2+Utht2h+2Ugth2hg=0th=2Ug±4U2g2+8hg2Need “+” root becauseth>0.th= −Ug+U2g2+2hgN1=thgh12= −U(gh)12+U2gh+2N1=N22+2N2For part (a), whereN2=0,N1=K=2.For part (b), whereN2>0,N1=f(N2) andf(x)=x2+2x.1.2 Pendulumθ(t)=angle relative to vertical,θ(0)=θ0h(t)=bob height above its lowest pointtp=periodm=bob massR=rod length(a) Dimensional analysisQuantities:tp,R,m,g[tp]=T,[R]=L,[m]=M,[g]=LT2mcannot be involved because nothing else has the dimension M.3 quantities2 dimensions (L, T)=1 dimensionless group
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Solution: Chapter 1LetN=tpRagb.[tp]=T=[Ragb]=La(LT2)b=La+bT2ba+b=0 and2b=1b= −12,a=12N=tpgR1/2=const.K(b) Expression fortpFrom the geometry:cosθ=RhR=1hRddt(cosθ)= −sinθdθdt= −1Rdhdtdhdt=Rsinθdθdt=Rθdθdtfor smallθ.From conservation of energy with no drag or pivot friction:kinetic energy+potential energy=constantv22+gh=gh0v(t)=bob speed,v(0)=0,h(0)=h0h=h0v22gdhdt= −12g2vdvdt= −vgdvdtEquate expressions fordh/dtand setv= ±R dθ/dt:Rθdθdt= −1g±Rdθdt±Rd2θdt= −R2gdθdtd2θdt2d2θdt2= −gRθ,θ(0)=θ0,dθdt(0)=0θ(t)=Acos[(g/R)1/2t]+Bsin[(g/R)1/2t]dθdt(0)=0B=0,θ(0)=θ0A=θ0θ(t)=θ0cos[(g/R)1/2t]For one period, argument=2πandgR1/2tp=2πtp=2πRg1/2orN=tpgR1/2=2π=K
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Solution: Chapter 11.3 Salad DressingDimensions of the quantities involded:[V]=L3[μ]=ML1T1[U]=LT1[γ]=MT2[D]=L5 quantities3 independent dimensions (M, L, T)=2 groupsThe simplest way to form a group containingVisN1=VD3The second group will need to haveU,μ, γ, and possibly againD. From Table 1.6, sucha group is the Capillary number:Ca=μUγ.(Dis not needed again after all.) The general relationship will beN1=F(Ca)where the form of the functionFwould need to be found from other information (pre-sumably, a fit to experimental data).If the availability of Ca had not been noticed, a groupN2could have been found usingN2=UaμbγcDd(LT1)a(ML1T1)b(MT2)cLd=1 (=M0L0T0)Mb+cLab+dTab2c=1
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