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Question Number 72198 by mr W last updated on 26/Oct/19

Commented by mr W last updated on 26/Oct/19

$$thesmallballwithmassmandradius$$$$risreleasedatthegivenposition.$$$$findtheperiodoftheswingmotion.$$$$purerollingisassumed.$$

Commented by ajfour last updated on 26/Oct/19

Commented by ajfour last updated on 26/Oct/19

$$\mathrm{\Omega }=\frac{d\theta }{dt}$$$$fr=I\alpha ,\alpha r=\frac{dv}{dt},v=(R-r)\mathrm{\Omega }$$$$mg\cos \theta -f=m\frac{dv}{dt}$$$$\Rightarrow mg\cos \theta -\frac{I\alpha r}{r^2}=m\alpha r$$$$\Rightarrow g\cos \theta =\frac{7}{5}(R-r)\mathrm{\Omega }\left(\frac{d\mathrm{\Omega }}{d\theta }\right)$$$$\Rightarrow {\mathrm{\Omega }}^2=\frac{10g}{7(R-r)}\sin \theta$$$$\Rightarrow {\int }_0^{\pi }\frac{d\theta }{\sqrt{\sin \theta }}=\left(\sqrt{\frac{10g}{7(R-r)}}\right)\frac{T}{2}$$$$\Rightarrow T=\sqrt{\frac{28(R-r)}{10g}}{\int }_0^{\pi }\frac{d\theta }{\sqrt{\sin \theta }}.$$$$\phi =\frac{R\theta }{r}-\theta$$$$\frac{d\phi }{dt}=\frac{(R-r)}{r}\frac{d\theta }{dt}=\frac{(R-r)}{r}\mathrm{\Omega }$$$$\Rightarrow \omega r=v=(R-r)\mathrm{\Omega }$$

Commented by ajfour last updated on 26/Oct/19

Commented by ajfour last updated on 26/Oct/19

$$Fortheroundtrip\theta =2\pi$$$$\phi =\frac{2\pi R}{r}-2\pi$$

Commented by ajfour last updated on 26/Oct/19

Commented by ajfour last updated on 26/Oct/19

$$\phi =\frac{R\theta }{r}-\theta =\frac{(R-r)\theta }{r}$$$$\frac{d\phi }{dt}=\frac{(R-r)}{r}\frac{d\theta }{dt}$$$$ord\phi =\frac{Rd\theta }{r}-d\theta$$

Commented by mr W last updated on 26/Oct/19

$$thanksalotfortheniceexplanation!$$$$nowiseewheretheproblemis.$$

Answered by mr W last updated on 26/Oct/19

Commented by mr W last updated on 26/Oct/19

$$\phi =\frac{R-r}{r}\theta$$$$\omega =\frac{d\phi }{dt}=\frac{R-r}{r}\times \frac{d\theta }{dt}=\frac{R-r}{r}\mathrm{\Omega }with\mathrm{\Omega }=\frac{d\theta }{dt}$$$$\alpha =\frac{d\omega }{dt}=\frac{R-r}{r}\times \frac{d\mathrm{\Omega }}{d\theta }=\frac{R-r}{r}\times \mathrm{\Omega }\frac{d\mathrm{\Omega }}{d\theta }$$$$I_P=I_C+{mr}^2=\frac{2{mr}^2}{5}+{mr}^2=\frac{7{mr}^2}{5}$$$$I_P\alpha =mgr\cos \theta$$$$\frac{7{mr}^2}{5}\times \frac{R-r}{r}\mathrm{\Omega }\frac{d\mathrm{\Omega }}{d\theta }=mgr\cos \theta$$$$\mathrm{\Omega }d\mathrm{\Omega }=\frac{5g}{7R}\cos \theta d\theta$$$${\int }_0^{\mathrm{\Omega }}\mathrm{\Omega }d\mathrm{\Omega }=\frac{5g}{7(R-r)}{\int }_0^{\theta }\cos \theta d\theta$$$$\frac{{\mathrm{\Omega }}^2}{2}=\frac{5g}{7(R-r)}\sin \theta$$$$\Rightarrow \mathrm{\Omega }=\frac{d\theta }{dt}=\sqrt{\frac{10g}{7(R-r)}}\sqrt{\sin \theta }$$$$\Rightarrow dt=\sqrt{\frac{7(R-r)}{10g}}\times \frac{d\theta }{\sqrt{\sin \theta }}$$$$\Rightarrow t=\sqrt{\frac{7(R-r)}{10g}}\times {\int }_0^{\theta }\frac{d\theta }{\sqrt{\sin \theta }}$$$$\Rightarrow T=4\times \sqrt{\frac{7(R-r)}{10g}}\times {\int }_0^{\frac{\pi }{2}}\frac{d\theta }{\sqrt{\sin \theta }}$$$${\int }_0^{\frac{\pi }{2}}\frac{d\theta }{\sqrt{\sin \theta }}\approx 2.6221$$$$\Rightarrow T\approx 4\times \sqrt{\frac{7(R-r)}{10g}}\times 2.6221=8.775\sqrt{\frac{R-r}{g}}$$

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