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Question Number 31584 by Tinkutara last updated on 10/Mar/18

Answered by mrW2 last updated on 10/Mar/18

x_A =l cos θ v_A =(dx_A /dt)=−l sin θ ω ⇒ω=−(v_A /(l sin θ)) a_A =(dv_A /dt)=−l sin θ α−l cos θ ω^2 ⇒α=((−a_A −l cos θ ω^2 )/(l sin θ))=−(1/(l sin θ))(a_A + ((cos θ v_A ^2 )/(l sin^2 θ))) y_B =l sin θ v_B =l cos θ ω a_B =l cos θ α−l sin θ ω^2 =−((cos θ)/(sin θ))(a_A + ((cos θ v_A ^2 )/(l sin^2 θ)))−(v_A ^2 /(l sin θ)) =−cot θ a_A −(v_A ^2 /l)(((cos^2 θ)/(sin^3 θ))+(1/(sin θ))) ⇒a_B =−(cot θ a_A +(v_A ^2 /(l sin^3 θ)))

xA=lcosθvA=dxAdt=−lsinθω⇒ω=−vAlsinθaA=dvAdt=−lsinθα−lcosθω2⇒α=−aA−lcosθω2lsinθ=−1lsinθ(aA+cosθvA2lsin2θ)yB=lsinθvB=lcosθωaB=lcosθα−lsinθω2=−cosθsinθ(aA+cosθvA2lsin2θ)−vA2lsinθ=−cotθaA−vA2l(cos2θsin3θ+1sinθ)⇒aB=−(cotθaA+vA2lsin3θ)

Commented by Tinkutara last updated on 10/Mar/18

Commented by Tinkutara last updated on 10/Mar/18

Do I need to further do vector sum of these? And how to find ω and Iα?

DoIneedtofurtherdovectorsumofthese?AndhowtofindωandIα?

Commented by Tinkutara last updated on 11/Mar/18

Please see it's my book solution but I can't understand that.

Commented by mrW2 last updated on 11/Mar/18

i can not help you sir. i personally won′t do in that way as the book.

icannothelpyousir.ipersonallywon′tdointhatwayasthebook.

Commented by Tinkutara last updated on 11/Mar/18

I am not asking you sir to solve that way. I am just asking that the book is correct to say that the final acceleration is the vector sum of that 3 so we need to find that? I highly appreciate your method!

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