Question and Answers Forum
Question Number 129795 by ajfour last updated on 19/Jan/21
Commented by ajfour last updated on 19/Jan/21
Answered by ajfour last updated on 23/Jan/21
Commented by ajfour last updated on 23/Jan/21
![From work-energy balance mg((L/2))[sin 2φ_0 −sin 2φ] =(1/2)m(v_(cm) ^2 )+(1/2)I_(cm) ω^2 ucos 2φ−v=−Rcosec^2 φ((dφ/dt)) OP =p=Rcot φ (dp/dt)=−(Rcosec^2 φ)(dφ/dt)=u ⇒ (dφ/dt)=−(u/R)sin^2 φ usin 2φ=ωRcot φ v_(cm) ^2 =v_(x,cm) ^2 +v_(y,cm) ^2 v_x =(d/dt)(Rcot φ−(L/2)cos 2φ) v_y =(d/dt)((L/2)sin 2φ) v_(cm) ^2 =(Lsin 2φ−Rcosec^2 φ)^2 ((dφ/dt))^2 +(Lcos 2φ)^2 ((dφ/dt))^2 ω=((usin 2φ)/(Rcot φ))=−(((cosec^2 φsin 2φ)/(cot φ)))(dφ/dt) −−−−−−−−−−−−−−− gL(sin 2φ_0 −sin 2φ) =((dφ/dt))^2 [(I_(cm) /m)(((cosec^2 φsin 2φ)/(cot φ)))^2 +(Lsin 2φ−Rcosec^2 φ)^2 +(Lcos 2φ)^2 ] ⇒ ((dφ/dt))^2 =((gL(sin 2φ_0 −sin 2φ))/((I_(cm) /m)(((cosec^2 φsin 2φ)/(cot φ)))^2 +(Lsin 2φ−Rcosec^2 φ)^2 +(Lcos 2φ)^2 )) ω^2 =((4gL(sin 2φ_0 −sin 2φ_1 ))/(((4L^2 )/3)+R^2 cosec^4 φ−4RLcot φ)) ω^2 =((12gλ(sin α−sin θ))/(R(4λ^2 +3cosec^4 (θ/2)−12λcot (θ/2)))) say φ=φ_1 =tan^(−1) ((R/L)) Now phase II starts: ...............................](Q130237.png)
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Question and Answers Forum | ||
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Question Number 129795 by ajfour last updated on 19/Jan/21 | ||
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Commented by ajfour last updated on 19/Jan/21 | ||
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Answered by ajfour last updated on 23/Jan/21 | ||
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Commented by ajfour last updated on 23/Jan/21 | ||
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