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Question-214928

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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.
$${yes},\:{i}\:{think}\:{it}'{s}\:{correct}. \\ $$
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)
$${center}\:{of}\:{rod}\:\left({x},\:{y}\right) \\ $$$${x}=\frac{{L}\mathrm{cos}\:\theta}{\mathrm{2}} \\ $$$${y}=\frac{{L}\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$$\omega=−\frac{{d}\theta}{{dt}} \\ $$$${u}=\frac{{dx}}{{dt}}=\frac{\omega{L}\mathrm{sin}\:\theta}{\mathrm{2}}\:\:\:\:\left(\rightarrow\right) \\ $$$${v}=−\frac{{dy}}{{dt}}=\frac{\omega{L}\mathrm{cos}\:\theta}{\mathrm{2}}\:\:\:\:\left(\downarrow\right) \\ $$$$\frac{{mgL}\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right)}{\mathrm{2}}=\frac{{m}\left({u}^{\mathrm{2}} +{v}^{\mathrm{2}} \right)}{\mathrm{2}}+\frac{\omega^{\mathrm{2}} }{\mathrm{2}}×\frac{{mL}^{\mathrm{2}} }{\mathrm{12}} \\ $$$${gL}\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right)=\frac{\omega^{\mathrm{2}} {L}^{\mathrm{2}} }{\mathrm{4}}+\frac{\omega^{\mathrm{2}} {L}^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\Rightarrow\omega^{\mathrm{2}} =\frac{\mathrm{3}{g}\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right)}{{L}} \\ $$$$\Rightarrow\omega\frac{{d}\omega}{{d}\theta}=−\frac{\mathrm{3}{g}\:\mathrm{cos}\:\theta}{\mathrm{2}{L}} \\ $$$${a}_{{x}} =\frac{{du}}{{dt}}=−\omega\frac{{du}}{{d}\theta}=−\frac{\omega{L}}{\mathrm{2}}\left(\omega\:\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\:\frac{{d}\omega}{{d}\theta}\right) \\ $$$${a}_{{x}} =\frac{\mathrm{3}{g}\:\mathrm{cos}\:\theta\left(\mathrm{3}\:\mathrm{sin}\:\theta−\mathrm{2}\:\mathrm{sin}\:\theta_{\mathrm{0}} \right)}{\mathrm{4}} \\ $$$${N}={ma}_{{x}} \\ $$$$\Rightarrow\frac{{N}}{{mg}}=\frac{\mathrm{3}\:\mathrm{cos}\:\theta\left(\mathrm{3}\:\mathrm{sin}\:\theta−\mathrm{2}\:\mathrm{sin}\:\theta_{\mathrm{0}} \right)}{\mathrm{4}} \\ $$$${N}=\mathrm{0}: \\ $$$$\mathrm{3}\:\mathrm{sin}\:\theta−\mathrm{2}\:\mathrm{sin}\:\theta_{\mathrm{0}} =\mathrm{0}\: \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{\mathrm{2}\:\mathrm{sin}\:\theta_{\mathrm{0}} }{\mathrm{3}} \\ $$
Commented by ajfour last updated on 24/Dec/24
Awesome! I had given up almost. Thank you sir.
$${Awesome}!\:{I}\:{had}\:{given}\:{up}\:{almost}. \\ $$$${Thank}\:{you}\:{sir}. \\ $$

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