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Question Number 144420 by ajfour last updated on 25/Jun/21

Commented by ajfour last updated on 25/Jun/21

Two cylinders one on top of  another are released and  motion initiated with a mild  disturbance. If all surfaces  are frictionless find the  speeds u and v of the cylinders  just when they loose contact  with each other.

Twocylindersoneontopofanotherarereleasedandmotioninitiatedwithamilddisturbance.Ifallsurfacesarefrictionlessfindthespeedsuandvofthecylindersjustwhentheyloosecontactwitheachother.

Answered by ajfour last updated on 25/Jun/21

Commented by ajfour last updated on 25/Jun/21

energy conserva^n :   ....(i)  mg(2R)(1−cos θ)  =(1/2)m{v^2 +(u_r cos θ−v)^2 +u_r ^2 sin^2 θ}  momentum conserva^n :   mv=m(u_r cos θ−v)      ⇒   u_r cos θ=2v    ...(ii)  m(centripetal acc.)=F_r   ((mu_r ^2 )/(2R))=mgcos θ     ....(iii)  ⇒ 2gR=(u_r ^2 /(cos θ))    ....(A)  now  using  (ii)  (v^2 /u_r ^2 )=((cos^2 θ)/4)      .....(B)  using all (A),(B) in ..(i)  ((2(1−cos θ))/(cos θ))=((cos^2 θ)/2)+1−cos^2 θ  ⇒  4−4cos θ=−cos^3 θ+2cos θ  ⇒  cos^3 θ−6cos θ+4=0  ⇒ cos θ=(√3)−1  now  v^2 =((u_r ^2 cos^2 θ)/4)=((gRcos^3 θ)/2)  v^2 =((gR((√3)−1)^3 )/2)=gR(3(√3)−5)   v=(√(gR(3(√3)−5)))

energyconservan:....(i)mg(2R)(1cosθ)=12m{v2+(urcosθv)2+ur2sin2θ}momentumconservan:mv=m(urcosθv)urcosθ=2v...(ii)m(centripetalacc.)=Frmur22R=mgcosθ....(iii)2gR=ur2cosθ....(A)nowusing(ii)v2ur2=cos2θ4.....(B)usingall(A),(B)in..(i)2(1cosθ)cosθ=cos2θ2+1cos2θ44cosθ=cos3θ+2cosθcos3θ6cosθ+4=0cosθ=31nowv2=ur2cos2θ4=gRcos3θ2v2=gR(31)32=gR(335)v=gR(335)

Answered by mr W last updated on 25/Jun/21

Commented by mr W last updated on 25/Jun/21

x_B =x_A −2R sin θ  y_B =R+2R cos θ  ω=(dθ/dt)  v=(dx_A /dt)  a=(dv/dt)  u_(Bx) =(dx_B /dt)=v−2R cos θ ω=−v  ⇒v=R cos θ ω  u_(By) =(dy_B /dt)=−2R sin θ ω  (1/2)m(v^2 +u_(Bx) ^2 +u_(By) ^2 )=mg2R(1−cos θ)  v^2 +(v−2R cos θ ω)^2 +(−2R sin θ ω)^2 =4gR(1−cos θ)  R^2  cos^2  θ ω^2 +R^2  cos^2  θ ω^2 +4R^2  sin^2  θ ω^2 =4gR(1−cos θ)  (1+sin^2  θ)ω^2 =2g(1−cos θ)  ⇒ω^2 =((2g(1−cos θ))/(R(1+sin^2  θ)))  ⇒ω=(√((2g(1−cos θ))/(R(2−cos^2  θ))))  2ω(dω/dθ)=((2g)/R)[((sin θ)/(1+sin^2  θ))−((2sin θ cos θ(1−cos θ))/((1+sin^2  θ)^2 ))]  ω(dω/dθ)=((g sin θ(cos^2  θ−2 cos θ+2))/(R(1+sin^2  θ)^2 ))  a=(dv/dt)=Rω(−sin θ ω+cos θ (dω/dθ))    ma=Nsin θ  N=0 ⇒a=0  −sin θ ω+cos θ (dω/dθ)=0  tan θ ω^2 =ω (dω/dθ)=((g sin θ(cos^2  θ−2 cos θ+2))/(R(1+sin^2  θ)^2 ))  tan θ ((2g(1−cos θ))/(R(1+sin^2  θ)))=((g sin θ(cos^2  θ−2 cos θ+2))/((1+sin^2  θ)^2 ))  ((2(1−cos θ))/(cos θ))=((cos^2  θ−2 cos θ+2)/(1+sin^2  θ))  ((2−2cos θ)/(cos θ))=((cos^2  θ−2 cos θ+2)/(2−cos^2  θ))  cos^3  θ−6 cos θ+4=0  (cos θ−2)(cos^2  θ+2 cos θ−2)=0  ⇒cos θ=(√3)−1  ⇒θ=cos^(−1) ((√3)−1)≈42.9°  v= (√((2gR(1−cos θ)cos^2  θ)/(2−cos^2  θ)))  ⇒v=(2−(√3)) (√(((√3)+1)gR))≈0.443(√(gR))

xB=xA2RsinθyB=R+2Rcosθω=dθdtv=dxAdta=dvdtuBx=dxBdt=v2Rcosθω=vv=RcosθωuBy=dyBdt=2Rsinθω12m(v2+uBx2+uBy2)=mg2R(1cosθ)v2+(v2Rcosθω)2+(2Rsinθω)2=4gR(1cosθ)R2cos2θω2+R2cos2θω2+4R2sin2θω2=4gR(1cosθ)(1+sin2θ)ω2=2g(1cosθ)ω2=2g(1cosθ)R(1+sin2θ)ω=2g(1cosθ)R(2cos2θ)2ωdωdθ=2gR[sinθ1+sin2θ2sinθcosθ(1cosθ)(1+sin2θ)2]ωdωdθ=gsinθ(cos2θ2cosθ+2)R(1+sin2θ)2a=dvdt=Rω(sinθω+cosθdωdθ)ma=NsinθN=0a=0sinθω+cosθdωdθ=0tanθω2=ωdωdθ=gsinθ(cos2θ2cosθ+2)R(1+sin2θ)2tanθ2g(1cosθ)R(1+sin2θ)=gsinθ(cos2θ2cosθ+2)(1+sin2θ)22(1cosθ)cosθ=cos2θ2cosθ+21+sin2θ22cosθcosθ=cos2θ2cosθ+22cos2θcos3θ6cosθ+4=0(cosθ2)(cos2θ+2cosθ2)=0cosθ=31θ=cos1(31)42.9°v=2gR(1cosθ)cos2θ2cos2θv=(23)(3+1)gR0.443gR

Commented by ajfour last updated on 25/Jun/21

Thanks Sir, right answer!

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