Question Number 176660 by mr W last updated on 24/Sep/22
Commented by mr W last updated on 25/Sep/22
$${find}\:{the}\:{time}\:{the}\:{rod}\:{takes}\:{to}\:{reach} \\ $$$${the}\:{ground}.\: \\ $$$${the}\:{coefficient}\:{of}\:{friction}\:{between} \\ $$$${rod}\:{and}\:{wall}\:{is}\:\mu\:{which}\:{is}\:{small}\:{enough} \\ $$$${so}\:{that}\:{the}\:{rod}\:{can}\:{slip}\:{on}\:{the}\:{wall}. \\ $$$${the}\:{friction}\:{between}\:{rod}\:{and}\:{ground} \\ $$$${is}\:{large}\:{enough}\:{so}\:{that}\:{the}\:{rod}\:{doesn}'{t} \\ $$$${slip}\:{on}\:{the}\:{ground}. \\ $$
Answered by mr W last updated on 25/Sep/22
Commented by mr W last updated on 26/Sep/22
$${L}=\sqrt{{b}^{\mathrm{2}} +{h}^{\mathrm{2}} } \\ $$$${I}=\frac{{ML}^{\mathrm{2}} }{\mathrm{3}}=\frac{{M}\left({b}^{\mathrm{2}} +{h}^{\mathrm{2}} \right)}{\mathrm{3}} \\ $$$$\omega=\frac{{d}\theta}{{dt}} \\ $$$$\alpha=\frac{{d}^{\mathrm{2}} \theta}{{dt}^{\mathrm{2}} }=\frac{{d}\omega}{{dt}}=\omega\frac{{d}\omega}{{d}\theta} \\ $$$${b}\:\mathrm{tan}\:\varphi={h}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\varphi=\mathrm{tan}^{−\mathrm{1}} \left(\frac{{h}\:\mathrm{sin}\:\theta}{{b}}\right) \\ $$$$\frac{{d}\varphi}{{dt}}=\frac{\frac{{h}\:\mathrm{cos}\:\theta}{{b}}}{\mathrm{1}+\left(\frac{{h}\:\mathrm{sin}\:\theta}{{b}}\right)^{\mathrm{2}} }×\omega \\ $$$$\frac{{d}^{\mathrm{2}} \varphi}{{dt}^{\mathrm{2}} }=\frac{\frac{{h}\:\mathrm{cos}\:\theta}{{b}}}{\mathrm{1}+\left(\frac{{h}\:\mathrm{sin}\:\theta}{{b}}\right)^{\mathrm{2}} }×\omega×\frac{{d}\omega}{{d}\theta}+\omega^{\mathrm{2}} ×\left[−\frac{\left(\frac{{h}\:\mathrm{cos}\:\theta}{{b}}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\left(\frac{{h}\:\mathrm{sin}\:\theta}{{b}}\right)^{\mathrm{2}} \right)^{\mathrm{2}} }−\frac{\frac{{h}\:\mathrm{sin}\:\theta}{{b}}}{\mathrm{1}+\left(\frac{{h}\:\mathrm{sin}\:\theta}{{b}}\right)^{\mathrm{2}} }\right] \\ $$$$\frac{{d}^{\mathrm{2}} \varphi}{{dt}^{\mathrm{2}} }=\frac{\frac{{h}\:\mathrm{cos}\:\theta}{{b}}}{\mathrm{1}+\left(\frac{{h}\:\mathrm{sin}\:\theta}{{b}}\right)^{\mathrm{2}} }×\omega×\frac{{d}\omega}{{d}\theta}−\frac{\omega^{\mathrm{2}} }{\mathrm{1}+\left(\frac{{h}\:\mathrm{sin}\:\theta}{{b}}\right)^{\mathrm{2}} }×\left[\frac{\left(\frac{{h}\:\mathrm{cos}\:\theta}{{b}}\right)^{\mathrm{2}} }{\mathrm{1}+\left(\frac{{h}\:\mathrm{sin}\:\theta}{{b}}\right)^{\mathrm{2}} }+\frac{{h}\:\mathrm{sin}\:\theta}{{b}}\right] \\ $$$${I}\frac{{d}^{\mathrm{2}} \varphi}{{dt}^{\mathrm{2}} }={N}\:{h}\:\mathrm{sin}\:\theta \\ $$$${N}=\frac{{M}\left({b}^{\mathrm{2}} +{h}^{\mathrm{2}} \right)}{\mathrm{3}{h}\:\mathrm{sin}\:\theta\:\left(\mathrm{1}+\left(\frac{{h}\:\mathrm{sin}\:\theta}{{b}}\right)^{\mathrm{2}} \right)}\left\{\frac{{h}\:\mathrm{cos}\:\theta}{{b}}×\omega×\frac{{d}\omega}{{d}\theta}−\omega^{\mathrm{2}} \left[\frac{\left(\frac{{h}\:\mathrm{cos}\:\theta}{{b}}\right)^{\mathrm{2}} }{\mathrm{1}+\left(\frac{{h}\:\mathrm{sin}\:\theta}{{b}}\right)^{\mathrm{2}} }+\frac{{h}\:\mathrm{sin}\:\theta}{{b}}\right]\right\} \\ $$$${I}\frac{{d}^{\mathrm{2}} \theta}{{dt}^{\mathrm{2}} }=−{fh}+\frac{{Mgh}\:\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$${I}\frac{{d}^{\mathrm{2}} \theta}{{dt}^{\mathrm{2}} }=−\mu{Nh}+\frac{{Mgh}\:\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$$\omega\frac{{d}\omega}{{d}\theta}=−\frac{\mu}{\mathrm{sin}\:\theta\:\left(\mathrm{1}+\left(\frac{{h}\:\mathrm{sin}\:\theta}{{b}}\right)^{\mathrm{2}} \right)}\left\{\frac{{h}\:\mathrm{cos}\:\theta}{{b}}×\omega×\frac{{d}\omega}{{d}\theta}−\omega^{\mathrm{2}} \left[\frac{\left(\frac{{h}\:\mathrm{cos}\:\theta}{{b}}\right)^{\mathrm{2}} }{\mathrm{1}+\left(\frac{{h}\:\mathrm{sin}\:\theta}{{b}}\right)^{\mathrm{2}} }+\frac{{h}\:\mathrm{sin}\:\theta}{{b}}\right]\right\}+\frac{\mathrm{3}{gh}\:\mathrm{sin}\:\theta}{\mathrm{2}\left({b}^{\mathrm{2}} +{h}^{\mathrm{2}} \right)} \\ $$$${let}\:\lambda=\frac{{h}}{{b}} \\ $$$$\omega\frac{{d}\omega}{{d}\theta}=−\frac{\mu}{\mathrm{sin}\:\theta\:\left(\mathrm{1}+\lambda^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta\right)}\left\{\lambda\:\mathrm{cos}\:\theta×\omega×\frac{{d}\omega}{{d}\theta}−\omega^{\mathrm{2}} \left(\frac{\lambda^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{1}+\lambda^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}+\lambda\:\mathrm{sin}\:\theta\right)\right\}+\frac{\mathrm{3}{g}\lambda\:\mathrm{sin}\:\theta}{\mathrm{2}{b}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\lambda}+\frac{\mu\:\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta\:\left(\mathrm{1}+\lambda^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta\right)}\right)\frac{{d}\omega^{\mathrm{2}} }{{d}\theta}−\frac{\mu}{\mathrm{sin}\:\theta\:\left(\mathrm{1}+\lambda^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta\right)}\left(\frac{\lambda\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{1}+\lambda^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}+\mathrm{sin}\:\theta\right)\omega^{\mathrm{2}} −\frac{\mathrm{3}{g}\:\mathrm{sin}\:\theta}{\mathrm{2}{b}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}=\mathrm{0} \\ $$$$\frac{{d}\omega^{\mathrm{2}} }{{d}\theta}−\frac{\mathrm{2}\mu\lambda\left(\lambda\mathrm{cos}^{\mathrm{2}} \:\theta+\lambda^{\mathrm{2}} \mathrm{sin}^{\mathrm{3}} \:\theta+\mathrm{sin}\:\theta\right)}{\left(\mathrm{sin}\:\theta\:\left(\mathrm{1}+\lambda^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta\right)+\mu\lambda\:\mathrm{cos}\:\theta\right)\left(\mathrm{1}+\lambda^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta\right)}\omega^{\mathrm{2}} −\frac{\mathrm{3}{g}\lambda\:\mathrm{sin}^{\mathrm{2}} \:\theta\:\left(\mathrm{1}+\lambda^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta\right)}{{b}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)\left(\mathrm{sin}\:\theta\:\left(\mathrm{1}+\lambda^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta\right)+\mu\lambda\:\mathrm{cos}\:\theta\right)}=\mathrm{0} \\ $$$${let}\:\Phi=\omega^{\mathrm{2}} \\ $$$$\frac{{d}\Phi}{{d}\theta}−\frac{\mathrm{2}\mu\lambda\left(\lambda\mathrm{cos}^{\mathrm{2}} \:\theta+\lambda^{\mathrm{2}} \mathrm{sin}^{\mathrm{3}} \:\theta+\mathrm{sin}\:\theta\right)}{\left[\mathrm{sin}\:\theta\:\left(\mathrm{1}+\lambda^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta\right)+\mu\lambda\:\mathrm{cos}\:\theta\right]\left(\mathrm{1}+\lambda^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta\right)}\Phi−\frac{\mathrm{3}{g}\lambda\:\mathrm{sin}^{\mathrm{2}} \:\theta\:\left(\mathrm{1}+\lambda^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta\right)}{{b}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)\left[\mathrm{sin}\:\theta\:\left(\mathrm{1}+\lambda^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta\right)+\mu\lambda\:\mathrm{cos}\:\theta\right]}=\mathrm{0} \\ $$$${with}\:{p}\left(\theta\right)=\frac{\mathrm{2}\mu\lambda\left(\lambda\mathrm{cos}^{\mathrm{2}} \:\theta+\lambda^{\mathrm{2}} \mathrm{sin}^{\mathrm{3}} \:\theta+\mathrm{sin}\:\theta\right)}{\left[\mathrm{sin}\:\theta\:\left(\mathrm{1}+\lambda^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta\right)+\mu\lambda\:\mathrm{cos}\:\theta\right]\left(\mathrm{1}+\lambda^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:{q}\left(\theta\right)=\frac{\mathrm{3}{g}\lambda\:\mathrm{sin}^{\mathrm{2}} \:\theta\:\left(\mathrm{1}+\lambda^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta\right)}{{b}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)\left[\mathrm{sin}\:\theta\:\left(\mathrm{1}+\lambda^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta\right)+\mu\lambda\:\mathrm{cos}\:\theta\right]} \\ $$$$\Rightarrow\frac{{d}\Phi}{{d}\theta}−{p}\left(\theta\right)\Phi−{q}\left(\theta\right)=\mathrm{0} \\ $$$${we}\:{get}\:\Phi=\Phi\left(\theta\right) \\ $$$$\Rightarrow\omega=\sqrt{\Phi\left(\theta\right)} \\ $$$$\Rightarrow\frac{{d}\theta}{{dt}}=\sqrt{\Phi\left(\theta\right)} \\ $$$$\Rightarrow{dt}=\frac{{d}\theta}{\:\sqrt{\Phi\left(\theta\right)}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{{T}} {dt}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\theta}{\:\sqrt{\Phi\left(\theta\right)}} \\ $$$$\Rightarrow{T}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\theta}{\:\sqrt{\Phi\left(\theta\right)}} \\ $$
Commented by Tawa11 last updated on 25/Sep/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$