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Question-216219




Question Number 216219 by mr W last updated on 30/Jan/25
Commented by mr W last updated on 31/Jan/25
a ladder with length L and mass m  is released from rest at the position  as shown. find the time the ladder  takes to fall onto the floor completely.  all surfaces are frictionless.
aladderwithlengthLandmassmisreleasedfromrestatthepositionasshown.findthetimetheladdertakestofallontothefloorcompletely.allsurfacesarefrictionless.
Answered by mahdipoor last updated on 01/Feb/25
1)force diagram :  Ni^�  in  N=(0,2Lcosθ)  Mj^�  in M=(2Lsinθ,0)  −mgj^�  in (Lsinθ,Lcosθ)  2)dynamic diagram :  θ, (ccw is +)  (0,y_N ^(..) ) in N     &   (x_M ^(..) ,0) in M  3)eq :  ΣM=Iθ^(..)    (about CG)  ⇒−NLsinθ+MLcosθ=Iθ^(..)   ΣF=ma   ⇒M−mg=my^(..)   ⇒N=mx^(..)   condition :  (0,y_N ^(..) )=(x^(..) ,y^(..) )+Lθ^(.2) (sinθ,−cosθ)−Lθ^(..) (cosθ,sinθ)  (x_M ^(..) ,0)=(x^(..) ,y^(..) )+Lθ^(.2) (−sinθ,cosθ)+Lθ^(..) (cosθ,sinθ)  4)final eq:  (I/(mL^2 ))θ^(..) +θ^(.2) cos2θ+θ^(..) sin2θ=(g/L)cosθ  ⇒θ(0)=sin^(−1) (((√(L^2 −h^2 ))/L))   , θ^. (θ)=0  ⇒ans=T=θ^(−1) ((π/2))
1)forcediagram:Ni^inN=(0,2Lcosθ)Mj^inM=(2Lsinθ,0)mgj^in(Lsinθ,Lcosθ)2)dynamicdiagram:θ,(ccwis+)(0,y..N)inN&(x..M,0)inM3)eq:ΣM=Iθ..(aboutCG)NLsinθ+MLcosθ=Iθ..ΣF=maMmg=my..N=mx..condition:(0,y..N)=(x..,y..)+Lθ.2(sinθ,cosθ)Lθ..(cosθ,sinθ)(x..M,0)=(x..,y..)+Lθ.2(sinθ,cosθ)+Lθ..(cosθ,sinθ)4)finaleq:ImL2θ..+θ.2cos2θ+θ..sin2θ=gLcosθθ(0)=sin1(L2h2L),θ.(θ)=0ans=T=θ1(π2)
Commented by mr W last updated on 01/Feb/25
thanks!
thanks!
Answered by mr W last updated on 01/Feb/25
Commented by mr W last updated on 05/Feb/25
let (h/L)=η  ω=(dθ/dt)  y=L cos θ  at t=0:   y=h=L cos θ_0  ⇒θ_0 =cos^(−1) η  (1/2)[((mL^2 )/(12))+m((L/2))^2 ]ω^2 =((mg(h−y))/2)  ((L^2 ω^2 )/3)=g(h−L cos θ)  ω^2 =((3g)/L)(η−cos θ)  ⇒ω(dω/dθ)=((3g)/(2L)) sin θ  ⇒ω=(√(((3g)/L)(η−cos θ)))  u_x =((Lω cos θ)/2)  a_x =(L/2)(−ω^2  sin θ+cos θ ((ωdω)/dθ))      =(L/2)[−((3g)/L)(η−cos θ) sin θ+cos θ ((3g)/(2L)) sin θ]       =((3g)/4)(−2η+3 cos θ)sin θ  N=ma_x =((3mg)/4)(−2η+3 cos θ)sin θ  N=0:  ⇒−2η+3 cos θ=0  ⇒cos θ=((2η)/3) ⇒θ_1 =cos^(−1) ((2η)/3)    dt=(√(L/(3g)))(dθ/( (√(η−cos θ))))  ∫_0 ^T_1  dt=(√(L/(3g)))∫_θ_0  ^θ_1  (dθ/( (√(η−cos θ))))  ⇒T_1 =(√(L/(3g)))∫_(cos^(−1) η) ^(cos^(−1) ((2η)/3)) (dθ/( (√(η−cos θ))))  ω_1 =(√(((3g)/L)(η−cos θ_1 )))  u_1 =u_x ∣_(t=T_1 ) =((Lω_1  cos θ_1 )/2)=((cos θ_1 )/2)(√(3gL(η−cos θ_1 )))  upon t=T_1 :  u_x =u_1 =((cos θ_1 )/2)(√(3gL(η−cos θ_1 )))=(η/3)(√(ηgL))  v_y =((Lω sin θ)/2)  (1/2)×((mL^2 ω^2 )/(12))+((m(u_x ^2 +v_y ^2 ))/2)=((mg(h−y))/2)  (1/2)×((mL^2 ω^2 )/(12))+(m/2)×(((η^3 gL)/9)+((L^2 ω^2 sin^2  θ)/4))=((mg(h−L cos θ))/2)  ⇒ω=(√((12g(η−(η^3 /9)−cos θ))/(L(1+3 sin^2  θ))))  (dθ/dt)=(√((12g(η−(η^3 /9)−cos θ))/(L(1+3 sin^2  θ))))  dt=(√((L(1+3 sin^2  θ))/(12g(η−(η^3 /9)−cos θ)))) dθ  ∫_0 ^T_2  dt=(√(L/(3g)))∫_θ_1  ^(π/2) (√((1+3 sin^2  θ)/(4(η−(η^3 /9)−cos θ)))) dθ  ⇒T_2 =(√(L/(3g)))∫_(cos^(−1) ((2η)/3)) ^(π/2) (√((1+3 sin^2  θ)/(4(η−(η^3 /9)−cos θ)))) dθ  total time needed:  T=T_1 +T_2 =λ(√(L/(3g)))  with   λ=∫_(cos^(−1) η) ^(cos^(−1) ((2η)/3)) (dθ/( (√(η−cos θ))))+∫_(cos^(−1) ((2η)/3)) ^(π/2) (√((1+3 sin^2  θ)/(4(η−(η^3 /9)−cos θ)))) dθ  example: η=(h/L)=0.8  To≈1.73205081(√(L/g))
lethL=ηω=dθdty=Lcosθatt=0:y=h=Lcosθ0θ0=cos1η12[mL212+m(L2)2]ω2=mg(hy)2L2ω23=g(hLcosθ)ω2=3gL(ηcosθ)ωdωdθ=3g2Lsinθω=3gL(ηcosθ)ux=Lωcosθ2ax=L2(ω2sinθ+cosθωdωdθ)=L2[3gL(ηcosθ)sinθ+cosθ3g2Lsinθ]=3g4(2η+3cosθ)sinθN=max=3mg4(2η+3cosθ)sinθN=0:2η+3cosθ=0cosθ=2η3θ1=cos12η3dt=L3gdθηcosθ0T1dt=L3gθ0θ1dθηcosθT1=L3gcos1ηcos12η3dθηcosθω1=3gL(ηcosθ1)u1=uxt=T1=Lω1cosθ12=cosθ123gL(ηcosθ1)upont=T1:ux=u1=cosθ123gL(ηcosθ1)=η3ηgLvy=Lωsinθ212×mL2ω212+m(ux2+vy2)2=mg(hy)212×mL2ω212+m2×(η3gL9+L2ω2sin2θ4)=mg(hLcosθ)2ω=12g(ηη39cosθ)L(1+3sin2θ)dθdt=12g(ηη39cosθ)L(1+3sin2θ)dt=L(1+3sin2θ)12g(ηη39cosθ)dθ0T2dt=L3gθ1π21+3sin2θ4(ηη39cosθ)dθT2=L3gcos12η3π21+3sin2θ4(ηη39cosθ)dθtotaltimeneeded:T=T1+T2=λL3gwithλ=cos1ηcos12η3dθηcosθ+cos12η3π21+3sin2θ4(ηη39cosθ)dθexample:η=hL=0.8To1.73205081Lg
Commented by Tawa11 last updated on 01/Feb/25
When will I know book like this.  Weldone sir
WhenwillIknowbooklikethis.Weldonesir

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