Question and Answers Forum

All Questions      Topic List

Mechanics Questions

Previous in All Question      Next in All Question      

Previous in Mechanics      Next in Mechanics      

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)))

$${x}_{{A}} ={l}\:\mathrm{cos}\:\theta \\ $$$$ \\ $$$${v}_{{A}} =\frac{{dx}_{{A}} }{{dt}}=−{l}\:\mathrm{sin}\:\theta\:\omega \\ $$$$\Rightarrow\omega=−\frac{{v}_{{A}} }{{l}\:\mathrm{sin}\:\theta} \\ $$$${a}_{{A}} =\frac{{dv}_{{A}} }{{dt}}=−{l}\:\mathrm{sin}\:\theta\:\alpha−{l}\:\mathrm{cos}\:\theta\:\omega^{\mathrm{2}} \\ $$$$\Rightarrow\alpha=\frac{−{a}_{{A}} −{l}\:\mathrm{cos}\:\theta\:\omega^{\mathrm{2}} }{{l}\:\mathrm{sin}\:\theta}=−\frac{\mathrm{1}}{{l}\:\mathrm{sin}\:\theta}\left({a}_{{A}} +\:\frac{\mathrm{cos}\:\theta\:{v}_{{A}} ^{\mathrm{2}} }{{l}\:\mathrm{sin}^{\mathrm{2}} \:\theta}\right) \\ $$$$ \\ $$$${y}_{{B}} ={l}\:\mathrm{sin}\:\theta \\ $$$${v}_{{B}} ={l}\:\mathrm{cos}\:\theta\:\omega \\ $$$${a}_{{B}} ={l}\:\mathrm{cos}\:\theta\:\alpha−{l}\:\mathrm{sin}\:\theta\:\omega^{\mathrm{2}} =−\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}\left({a}_{{A}} +\:\frac{\mathrm{cos}\:\theta\:{v}_{{A}} ^{\mathrm{2}} }{{l}\:\mathrm{sin}^{\mathrm{2}} \:\theta}\right)−\frac{{v}_{{A}} ^{\mathrm{2}} }{{l}\:\mathrm{sin}\:\theta} \\ $$$$=−\mathrm{cot}\:\theta\:{a}_{{A}} −\frac{{v}_{{A}} ^{\mathrm{2}} }{{l}}\left(\frac{\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{sin}^{\mathrm{3}} \:\theta}+\frac{\mathrm{1}}{\mathrm{sin}\:\theta}\right) \\ $$$$\Rightarrow{a}_{{B}} =−\left(\mathrm{cot}\:\theta\:{a}_{{A}} +\frac{{v}_{{A}} ^{\mathrm{2}} }{{l}\:\mathrm{sin}^{\mathrm{3}} \:\theta}\right) \\ $$

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?

$${Do}\:{I}\:{need}\:{to}\:{further}\:{do}\:{vector}\:{sum} \\ $$$${of}\:{these}?\:{And}\:{how}\:{to}\:{find}\:\omega\:{and}\:{I}\alpha? \\ $$

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.

$${i}\:{can}\:{not}\:{help}\:{you}\:{sir}.\:{i}\:{personally} \\ $$$${won}'{t}\:{do}\:{in}\:{that}\:{way}\:{as}\:{the}\:{book}. \\ $$

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!

Terms of Service

Privacy Policy

Contact: info@tinkutara.com