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Question Number 214928 by ajfour last updated on 24/Dec/24

Commented by ajfour last updated on 24/Dec/24

https://youtu.be/_ME_QSE6MIo?si=JxKs4xa21FSBKJSp Do you think the analysis is all correct here?

Commented by mr W last updated on 24/Dec/24

yes, i think it′s correct.

yes,ithinkitscorrect.

Answered by mr W last updated on 24/Dec/24

center of rod (x, y)  x=((Lcos θ)/2)  y=((Lsin θ)/2)  ω=−(dθ/dt)  u=(dx/dt)=((ωLsin θ)/2)    (→)  v=−(dy/dt)=((ωLcos θ)/2)    (↓)  ((mgL(sin θ_0 −sin θ))/2)=((m(u^2 +v^2 ))/2)+(ω^2 /2)×((mL^2 )/(12))  gL(sin θ_0 −sin θ)=((ω^2 L^2 )/4)+((ω^2 L^2 )/(12))  ⇒ω^2 =((3g(sin θ_0 −sin θ))/L)  ⇒ω(dω/dθ)=−((3g cos θ)/(2L))  a_x =(du/dt)=−ω(du/dθ)=−((ωL)/2)(ω cos θ+sin θ (dω/dθ))  a_x =((3g cos θ(3 sin θ−2 sin θ_0 ))/4)  N=ma_x   ⇒(N/(mg))=((3 cos θ(3 sin θ−2 sin θ_0 ))/4)  N=0:  3 sin θ−2 sin θ_0 =0   ⇒sin θ=((2 sin θ_0 )/3)

centerofrod(x,y)x=Lcosθ2y=Lsinθ2ω=dθdtu=dxdt=ωLsinθ2()v=dydt=ωLcosθ2()mgL(sinθ0sinθ)2=m(u2+v2)2+ω22×mL212gL(sinθ0sinθ)=ω2L24+ω2L212ω2=3g(sinθ0sinθ)Lωdωdθ=3gcosθ2Lax=dudt=ωdudθ=ωL2(ωcosθ+sinθdωdθ)ax=3gcosθ(3sinθ2sinθ0)4N=maxNmg=3cosθ(3sinθ2sinθ0)4N=0:3sinθ2sinθ0=0sinθ=2sinθ03

Commented by ajfour last updated on 24/Dec/24

Awesome! I had given up almost.  Thank you sir.

Awesome!Ihadgivenupalmost.Thankyousir.

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