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Question Number 214578 by ajfour last updated on 12/Dec/24

Commented by ajfour last updated on 12/Dec/24

$$Ifreleasedasshown,afterwhattime$$$$doestheotherendofseesawhits$$$$theground.(b>a)$$

Answered by mr W last updated on 13/Dec/24

Commented by mr W last updated on 13/Dec/24

$$say\mu =\frac{b}{a},\xi =\frac{h}{a}$$$$e=\frac{a+b}{2}-a=\frac{b-a}{2}=\frac{(\mu -1)a}{2}$$$$I=\frac{m{(a+b)}^2}{12}+{me}^2=\frac{{ma}^2[{(\mu +1)}^2+3{(\mu -1)}^2]}{12}$$$$=\frac{{ma}^2({\mu }^2-\mu +1)}{3}$$$$orI=\frac{m(a^3+b^3)}{3(a+b)}=\frac{{ma}^2({\mu }^3+1)}{3(\mu +1)}$$$$att=0:$$$$\sin \alpha =\frac{h}{a}=\xi \Rightarrow \alpha ={\sin }^{-1}\xi$$$$\theta =-\alpha =-{\sin }^{-1}\xi$$$$att=T:$$$$\sin \beta =\frac{h}{b}=\frac{\xi }{\mu }\Rightarrow \beta ={\sin }^{-1}\frac{\xi }{\mu }$$$$\theta =\beta ={\sin }^{-1}\frac{\xi }{\mu }$$$$att:$$$$\omega =\frac{d\theta }{dt}$$$$mge(\sin \alpha +\sin \theta )=\frac{I{\omega }^2}{2}$$$$mg\left(\frac{\mu -1}{2}\right)a(\xi +\sin \theta )=\frac{{ma}^2({\mu }^2-\mu +1){\omega }^2}{6}$$$$g(\mu -1)(\xi +\sin \theta )=\frac{a({\mu }^2-\mu +1){\omega }^2}{3}$$$${\omega }^2=\frac{3g({\mu }^2-1)(\xi +\sin \theta )}{a({\mu }^3+1)}$$$$\frac{dt}{d\theta }=\frac{1}{\omega }=\sqrt{\frac{a({\mu }^3+1)}{3g({\mu }^2-1)(\xi +\sin \theta )}}$$$$\Rightarrow T=\sqrt{\frac{a({\mu }^3+1)}{3g({\mu }^2-1)}}{\int }_{-{\sin }^{-1}\xi }^{{\sin }^{-1}\frac{\xi }{\mu }}\frac{d\theta }{\sqrt{\xi +\sin \theta }}$$$$$$$$example:$$$$\xi =\frac{h}{a}=\frac{1}{3},\mu =\frac{b}{a}=2$$$$T=\sqrt{\frac{a}{g}}{\int }_{-{\sin }^{-1}\frac{1}{3}}^{{\sin }^{-1}\frac{1}{6}}\frac{d\theta }{\sqrt{\frac{1}{3}+\sin \theta }}$$$$\approx 1.451940097932\sqrt{\frac{a}{g}}$$

Commented by ajfour last updated on 13/Dec/24

$$mg\left(\frac{b-a}{2}\right)\cos \theta =\frac{m}{3}\left(\frac{a^3+b^3}{a+b}\right)\frac{d^2\theta }{{dt}^2}$$$$\frac{d}{dt}\left(\frac{d\theta }{dt}\right)=\frac{\omega d\omega }{d\theta }=\frac{3g}{2}\left(\frac{b^2-a^2}{b^3+a^3}\right)\cos \theta =\lambda \cos \theta$$$$\frac{\omega }{\sqrt{2}}=\sqrt{\lambda }\sqrt{\sin \theta +\sin \alpha }$$$$T=\sqrt{\frac{b^3+a^3}{3g(b^2-a^2)}}{\int }_{-\alpha }^{\beta }\frac{d\theta }{\sqrt{\sin \alpha +\sin \theta }}$$$$$$

Commented by ajfour last updated on 13/Dec/24

$$Thankssir.meticulouslydone.$$

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