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Question Number 82806 by ajfour last updated on 24/Feb/20

Commented by ajfour last updated on 24/Feb/20

$$\mathrm{If}\mathrm{released}\mathrm{as}\mathrm{shown}\mathrm{by}\mathrm{dashed}$$$$\mathrm{lines},\mathrm{subsequently}\mathrm{find}\theta \left(\mathrm{x}\right).$$$$\mathrm{Assume}\mathrm{length}\mathrm{of}\mathrm{rod}\mathrm{L}.$$

Answered by mr W last updated on 24/Feb/20

Commented by mr W last updated on 24/Feb/20

$$a=g\sin \theta$$$$N=m(g\cos \theta -\alpha x)$$$$Mg\frac{L\cos \theta }{2}+Nx=\frac{{ML}^2}{3}\alpha$$$$Mg\frac{L\cos \theta }{2}+m(g\cos \theta -\alpha x)x=\frac{{ML}^2}{3}\alpha$$$$(\frac{ML}{2}+mx)g\cos \theta =(\frac{{ML}^2}{3}+{mx}^2)\alpha$$$$\alpha =\frac{d^2\theta }{{dt}^2}=\frac{g(\frac{ML}{2}+mx)\cos \theta }{\frac{{ML}^2}{3}+{mx}^2}$$$$a=\frac{d^{2}x}{{dt}^2}=g\sin \theta$$$$x\left(0\right)=c\left(markedasaindiagram\right)$$$$v\left(0\right)=x^{\prime }\left(0\right)=0$$$$\theta \left(0\right)=0$$$$\omega \left(0\right)={\theta }^{\prime }\left(0\right)=0$$$$let\mu =\frac{m}{M},\xi =\frac{x}{L}$$$$\frac{d^2\theta }{{dt}^2}=\frac{g(\frac{1}{2}+\mu \xi )\cos \theta }{L(\frac{1}{3}+\mu {\xi }^2)}...\left(i\right)$$$$\frac{d^2\xi }{{dt}^2}=\frac{g}{L}\sin \theta ...\left(ii\right)$$$$$$$$wesolve\left(i\right)and\left(ii\right)toget\xi \left(t\right)and$$$$\theta \left(t\right)\left(theoretically\right)$$$$......$$

Commented by ajfour last updated on 24/Feb/20

$$\mathrm{I}=\frac{{\mathrm{ML}}^2}{3}$$$$\tau =\mathrm{Mg}\left(\frac{\mathrm{L}}{2}\right)\cos \theta +\mathrm{mgrcos}\theta$$$$-\mathrm{Nr}=\mathrm{I}\alpha .......\left(\mathrm{i}\right)$$$$\mathrm{gsin}\theta =\frac{\mathrm{vdv}}{\mathrm{dr}}-{\omega }^2\mathrm{r}+\alpha \mathrm{r}....\left(\mathrm{ii}\right)$$$$\mathrm{let}\overline{\mathrm{r}}={\mathrm{re}}^{\mathrm{i}\theta }$$$$\overline{\mathrm{v}}=\frac{\mathrm{dr}}{\mathrm{dt}}{\mathrm{e}}^{\mathrm{i}\theta }+\mathrm{r}\frac{\mathrm{d}}{\mathrm{d}\theta }\left({\mathrm{e}}^{\mathrm{i}\theta }\right)\left(\frac{\mathrm{d}\theta }{\mathrm{dt}}\right)$$$$\overline{\mathrm{a}}=\frac{{\mathrm{d}}^2\mathrm{r}}{{\mathrm{dt}}^2}{\mathrm{e}}^{\mathrm{i}\theta }+2\left(\frac{\mathrm{dr}}{\mathrm{dt}}\right)\left(\frac{\mathrm{d}\theta }{\mathrm{dt}}\right)\frac{\mathrm{d}}{\mathrm{d}\theta }\left({\mathrm{e}}^{\mathrm{i}\theta }\right)$$$$+\mathrm{r}{\left(\frac{\mathrm{d}\theta }{\mathrm{dt}}\right)}^2\frac{{\mathrm{d}}^2}{\mathrm{d}{\theta }^2}\left({\mathrm{e}}^{\mathrm{i}\theta }\right)+\mathrm{r}\left(\frac{{\mathrm{d}}^2\theta }{{\mathrm{dt}}^2}\right){\mathrm{e}}^{\mathrm{i}\theta }$$$$\Rightarrow {\mathrm{a}}_{\theta }=2\omega \mathrm{v},{\mathrm{a}}_{\mathrm{r}}=\frac{{\mathrm{d}}^2\mathrm{r}}{{\mathrm{dt}}^2}-{\omega }^2\mathrm{r}+\alpha \mathrm{r}$$$$\mathrm{mgcos}\theta -\mathrm{N}=2\mathrm{m}\omega \mathrm{v}.....\left(\mathrm{iii}\right)$$$$\mathrm{Now}\mathrm{using}\left(\mathrm{iii}\right)\mathrm{in}\left(\mathrm{i}\right):$$$$\mathrm{Mg}\left(\frac{\mathrm{L}}{2}\right)\cos \theta +\mathrm{mgrcos}\theta$$$$+2\mathrm{m}\omega \mathrm{vr}-\mathrm{mgrcos}\theta =\mathrm{I}\alpha ...\left(1\right)$$$$\Rightarrow \mathrm{Mg}\left(\frac{\mathrm{L}}{2}\right)\cos \theta +2\mathrm{m}\omega \mathrm{vr}=\mathrm{I}\alpha$$$$\mathrm{differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{t}$$$$-\mathrm{Mg}\left(\frac{\mathrm{L}}{2}\right)\left(\frac{\mathrm{d}\theta }{\mathrm{dt}}\right)\sin \theta$$$$+2\mathrm{m}\alpha \mathrm{vr}+2\mathrm{m}\omega \mathrm{r}\left(\frac{\mathrm{dv}}{\mathrm{dt}}\right)+2\mathrm{m}\omega {\mathrm{v}}^2=\mathrm{I}\left(\frac{\mathrm{d}\alpha }{\mathrm{dt}}\right)$$$$........$$$$$$

Commented by mr W last updated on 24/Feb/20

$$youareright,thankssir!$$$$ifoundnowaytosolvetheequations.$$

Commented by ajfour last updated on 24/Feb/20

$$\mathrm{sir}\mathrm{in}\mathrm{the}{3}^{\mathrm{rd}}\mathrm{line},\mathrm{should}\mathrm{be}$$$$\mathrm{Nx}\mathrm{instead}\mathrm{of}\mathrm{Nxcos}\theta .$$

Commented by mr W last updated on 24/Feb/20

$$\frac{d\xi }{dt}=\omega \frac{d\xi }{d\theta }$$$$\frac{d^2\xi }{{dt}^2}=\alpha \frac{d\xi }{d\theta }+{\omega }^2\frac{d^2\xi }{d{\theta }^2}$$$$\frac{g(\frac{1}{2}+\mu \xi )\cos \theta }{L(\frac{1}{3}+\mu {\xi }^2)}\times \frac{d\xi }{d\theta }+{\omega }^2\frac{d^2\xi }{d{\theta }^2}=\frac{g}{L}\sin \theta$$$$\omega \frac{d\omega }{d\theta }=\frac{g(\frac{1}{2}+\mu \xi )\cos \theta }{L(\frac{1}{3}+\mu {\xi }^2)}$$$$$$$${\omega }^2=2{\int }_0^{\theta }\frac{g(\frac{1}{2}+\mu \xi )\cos \theta }{L(\frac{1}{3}+\mu {\xi }^2)}d\theta$$$$\Rightarrow 2{\int }_0^{\theta }\frac{(\frac{1}{2}+\mu \xi )\cos \theta }{(\frac{1}{3}+\mu {\xi }^2)}d\theta \times \frac{d^2\xi }{d{\theta }^2}+\frac{(\frac{1}{2}+\mu \xi )\cos \theta }{(\frac{1}{3}+\mu {\xi }^2)}\times \frac{d\xi }{d\theta }=\sin \theta$$

Commented by ajfour last updated on 24/Feb/20

$$\mathrm{Thanks}\mathrm{sir},\mathrm{bringing}\mathrm{it}\mathrm{as}\mathrm{far},\mathrm{but}$$$$\mathrm{let}\mathrm{us}\mathrm{rather}\mathrm{focus}\mathrm{on}\mathrm{something}$$$$\mathrm{smoother}..$$

Commented by mr W last updated on 24/Feb/20

$$yessir!$$$${what}^{\prime }saboutQ81968?canyoupost$$$$yoursolutionsir?$$

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