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Question Number 216219 by mr W last updated on 30/Jan/25

Commented by mr W last updated on 31/Jan/25

$$aladderwithlengthLandmassm$$$$isreleasedfromrestattheposition$$$$asshown.findthetimetheladder$$$$takestofallontothefloorcompletely.$$$$allsurfacesarefrictionless.$$

Answered by mahdipoor last updated on 01/Feb/25

$$1)forcediagram:$$$$N\hat{\mathrm{i}}inN=(0,2Lcos\theta )$$$$M\hat{\mathrm{j}}inM=(2Lsin\theta ,0)$$$$-mg\hat{\mathrm{j}}in(Lsin\theta ,Lcos\theta )$$$$2)dynamicdiagram:$$$$\theta ,(ccwis+)$$$$(0,{\stackrel{..}{y}}_N)inN\mathrm{\&}({\stackrel{..}{x}}_M,0)inM$$$$3)eq:$$$$\mathrm{\Sigma }M=I\stackrel{..}{\theta }\left(aboutCG\right)$$$$\Rightarrow -NLsin\theta +MLcos\theta =I\stackrel{..}{\theta }$$$$\mathrm{\Sigma }F=ma$$$$\Rightarrow M-mg=m\stackrel{..}{y}$$$$\Rightarrow N=m\stackrel{..}{x}$$$$condition:$$$$(0,{\stackrel{..}{y}}_N)=(\stackrel{..}{x},\stackrel{..}{y})+L\stackrel{.2}{\theta }(sin\theta ,-cos\theta )-L\stackrel{..}{\theta }(cos\theta ,sin\theta )$$$$({\stackrel{..}{x}}_M,0)=(\stackrel{..}{x},\stackrel{..}{y})+L\stackrel{.2}{\theta }(-sin\theta ,cos\theta )+L\stackrel{..}{\theta }(cos\theta ,sin\theta )$$$$4)finaleq:$$$$\frac{I}{{mL}^2}\stackrel{..}{\theta }+\stackrel{.2}{\theta }cos2\theta +\stackrel{..}{\theta }sin2\theta =\frac{g}{L}cos\theta$$$$\Rightarrow \theta \left(0\right)={sin}^{-1}\left(\frac{\sqrt{L^2-h^2}}{L}\right),\stackrel{.}{\theta }\left(\theta \right)=0$$$$\Rightarrow ans=T={\theta }^{-1}\left(\frac{\pi }{2}\right)$$

Commented by mr W last updated on 01/Feb/25

$$thanks!$$

Answered by mr W last updated on 01/Feb/25

Commented by mr W last updated on 05/Feb/25

$$let\frac{h}{L}=\eta$$$$\omega =\frac{d\theta }{dt}$$$$y=L\cos \theta$$$$att=0:$$$$y=h=L\cos {\theta }_0\Rightarrow {\theta }_0={\cos }^{-1}\eta$$$$\frac{1}{2}[\frac{{mL}^2}{12}+m{\left(\frac{L}{2}\right)}^2]{\omega }^2=\frac{mg(h-y)}{2}$$$$\frac{L^2{\omega }^2}{3}=g(h-L\cos \theta )$$$${\omega }^2=\frac{3g}{L}(\eta -\cos \theta )$$$$\Rightarrow \omega \frac{d\omega }{d\theta }=\frac{3g}{2L}\sin \theta$$$$\Rightarrow \omega =\sqrt{\frac{3g}{L}(\eta -\cos \theta )}$$$$u_x=\frac{L\omega \cos \theta }{2}$$$$a_x=\frac{L}{2}(-{\omega }^2\sin \theta +\cos \theta \frac{\omega d\omega }{d\theta })$$$$=\frac{L}{2}[-\frac{3g}{L}(\eta -\cos \theta )\sin \theta +\cos \theta \frac{3g}{2L}\sin \theta ]$$$$=\frac{3g}{4}(-2\eta +3\cos \theta )\sin \theta$$$$N={ma}_x=\frac{3mg}{4}(-2\eta +3\cos \theta )\sin \theta$$$$N=0:$$$$\Rightarrow -2\eta +3\cos \theta =0$$$$\Rightarrow \cos \theta =\frac{2\eta }{3}\Rightarrow {\theta }_1={\cos }^{-1}\frac{2\eta }{3}$$$$$$$$dt=\sqrt{\frac{L}{3g}}\frac{d\theta }{\sqrt{\eta -\cos \theta }}$$$${\int }_0^{T_1}dt=\sqrt{\frac{L}{3g}}{\int }_{{\theta }_0}^{{\theta }_1}\frac{d\theta }{\sqrt{\eta -\cos \theta }}$$$$\Rightarrow T_1=\sqrt{\frac{L}{3g}}{\int }_{{\cos }^{-1}\eta }^{{\cos }^{-1}\frac{2\eta }{3}}\frac{d\theta }{\sqrt{\eta -\cos \theta }}$$$${\omega }_1=\sqrt{\frac{3g}{L}(\eta -\cos {\theta }_1)}$$$$u_1=u_x{\mid }_{t=T_1}=\frac{L{\omega }_1\cos {\theta }_1}{2}=\frac{\cos {\theta }_1}{2}\sqrt{3gL(\eta -\cos {\theta }_1)}$$$$upont=T_1:$$$$u_x=u_1=\frac{\cos {\theta }_1}{2}\sqrt{3gL(\eta -\cos {\theta }_1)}=\frac{\eta }{3}\sqrt{\eta gL}$$$$v_y=\frac{L\omega \sin \theta }{2}$$$$\frac{1}{2}\times \frac{{mL}^2{\omega }^2}{12}+\frac{m(u_x^2+v_y^2)}{2}=\frac{mg(h-y)}{2}$$$$\frac{1}{2}\times \frac{{mL}^2{\omega }^2}{12}+\frac{m}{2}\times (\frac{{\eta }^{3}gL}{9}+\frac{L^2{\omega }^2{\sin }^2\theta }{4})=\frac{mg(h-L\cos \theta )}{2}$$$$\Rightarrow \omega =\sqrt{\frac{12g(\eta -\frac{{\eta }^3}{9}-\cos \theta )}{L(1+3{\sin }^2\theta )}}$$$$\frac{d\theta }{dt}=\sqrt{\frac{12g(\eta -\frac{{\eta }^3}{9}-\cos \theta )}{L(1+3{\sin }^2\theta )}}$$$$dt=\sqrt{\frac{L(1+3{\sin }^2\theta )}{12g(\eta -\frac{{\eta }^3}{9}-\cos \theta )}}d\theta$$$${\int }_0^{T_2}dt=\sqrt{\frac{L}{3g}}{\int }_{{\theta }_1}^{\frac{\pi }{2}}\sqrt{\frac{1+3{\sin }^2\theta }{4(\eta -\frac{{\eta }^3}{9}-\cos \theta )}}d\theta$$$$\Rightarrow T_2=\sqrt{\frac{L}{3g}}{\int }_{{\cos }^{-1}\frac{2\eta }{3}}^{\frac{\pi }{2}}\sqrt{\frac{1+3{\sin }^2\theta }{4(\eta -\frac{{\eta }^3}{9}-\cos \theta )}}d\theta$$$$totaltimeneeded:$$$$T=T_1+T_2=\lambda \sqrt{\frac{L}{3g}}$$$$with$$$$\lambda ={\int }_{{\cos }^{-1}\eta }^{{\cos }^{-1}\frac{2\eta }{3}}\frac{d\theta }{\sqrt{\eta -\cos \theta }}+{\int }_{{\cos }^{-1}\frac{2\eta }{3}}^{\frac{\pi }{2}}\sqrt{\frac{1+3{\sin }^2\theta }{4(\eta -\frac{{\eta }^3}{9}-\cos \theta )}}d\theta$$$$example:\eta =\frac{h}{L}=0.8$$$$To\approx 1.73205081\sqrt{\frac{L}{g}}$$

Commented by Tawa11 last updated on 01/Feb/25

$$\mathrm{When}\mathrm{will}\mathrm{I}\mathrm{know}\mathrm{book}\mathrm{like}\mathrm{this}.$$$$\mathrm{Weldone}\mathrm{sir}$$

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