Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 23390 by ajfour last updated on 29/Oct/17

Commented by ajfour last updated on 31/Oct/17

$$\left(a\right)Ifreleasedinthisorientation$$$$atorigin,inwhattimecanthis$$$$onespokeringreturnbackto$$$$origin,assumingitstartsrolling$$$$onitsown,andtherearenonet$$$$energylosses,andit{doesn}^{\prime }tfall$$$$aside.$$

Commented by ajfour last updated on 31/Oct/17

$$\left(b\right)Alsofindtheangularacceleration$$$$\alpha \left(\theta \right)using{Newton}^{\prime }slaws,thatis,$$$$withoutusingenergymethods.$$

Answered by mrW1 last updated on 30/Oct/17

Commented by mrW1 last updated on 31/Oct/17

$$\mathrm{Point}1=\mathrm{COM}\mathrm{of}\mathrm{ring}\mathrm{M}$$$$\mathrm{Point}2=\mathrm{COM}\mathrm{of}\mathrm{rod}\left(\mathrm{spoke}\right)\mathrm{m}$$$${\mathrm{x}}_1=\mathrm{R}\theta$$$${\mathrm{y}}_1=\mathrm{R}$$$${\mathrm{x}}_2={\mathrm{x}}_1+\frac{\mathrm{R}}{2}\sin \theta =\frac{\mathrm{R}}{2}(2\theta +\sin \theta )$$$${\mathrm{y}}_2={\mathrm{y}}_1+\frac{\mathrm{R}}{2}\cos \theta =\frac{\mathrm{R}}{2}(2+\cos \theta )$$$$\omega =\frac{\mathrm{d}\theta }{\mathrm{dt}}$$$${\mathrm{v}}_{\mathrm{x}1}=\frac{{\mathrm{dx}}_1}{\mathrm{dt}}=\mathrm{R}\frac{\mathrm{d}\theta }{\mathrm{dt}}=\mathrm{R}\omega$$$${\mathrm{v}}_{\mathrm{y}1}=\frac{{\mathrm{dy}}_1}{\mathrm{dt}}=0$$$${\mathrm{v}}_{\mathrm{x}2}=\frac{{\mathrm{dx}}_2}{\mathrm{dt}}=\frac{\mathrm{R}}{2}(2+\cos \theta )\omega$$$${\mathrm{v}}_{\mathrm{y}2}=\frac{{\mathrm{dy}}_2}{\mathrm{dt}}=-\frac{\mathrm{R}}{2}\sin \theta \omega$$$$$$$${\mathrm{KE}}_1=\frac{1}{2}\mathrm{M}({\mathrm{v}}_{\mathrm{x}1}^2+{\mathrm{v}}_{\mathrm{y}1}^2)+\frac{1}{2}{\mathrm{I}}_1{\omega }^2$$$$=\frac{1}{2}{\mathrm{MR}}^2{\omega }^2+\frac{1}{2}{\mathrm{MR}}^2{\omega }^2={\mathrm{MR}}^2{\omega }^2$$$${\mathrm{KE}}_2=\frac{1}{2}\mathrm{m}({\mathrm{v}}_{\mathrm{x}2}^2+{\mathrm{v}}_{\mathrm{y}2}^2)+\frac{1}{2}{\mathrm{I}}_2{\omega }^2$$$$=\frac{{\mathrm{mR}}^2{\omega }^2}{8}(5+4\cos \theta )+\frac{{\mathrm{mR}}^2{\omega }^2}{24}$$$$=\frac{{\mathrm{mR}}^2{\omega }^2}{6}(4+3\cos \theta )$$$${\mathrm{KE}}_1+{\mathrm{KE}}_2=\frac{\mathrm{mgR}}{2}(1-\cos \theta )$$$${\mathrm{MR}}^2{\omega }^2+\frac{{\mathrm{mR}}^2{\omega }^2}{6}(4+3\cos \theta )=\frac{\mathrm{mgR}}{2}(1-\cos \theta )$$$$[6\mathrm{M}+4\mathrm{m}+3\mathrm{mcos}\theta ]\mathrm{R}{\omega }^2=3\mathrm{mg}(1-\cos \theta )$$$${\omega }^2=\frac{\mathrm{g}}{\mathrm{R}}\times \frac{1-\cos \theta }{\frac{6\mathrm{M}+4\mathrm{m}}{3\mathrm{m}}+\cos \theta }=\frac{\mathrm{g}}{\mathrm{R}}\times \frac{2{\sin }^2\frac{\theta }{2}}{\frac{6\mathrm{M}+4\mathrm{m}}{3\mathrm{m}}+1-{2\sin }^2\frac{\theta }{2}}$$$${\omega }^2=\frac{\mathrm{g}}{\mathrm{R}}\times \frac{{\sin }^2\frac{\theta }{2}}{\frac{6\mathrm{M}+7\mathrm{m}}{6\mathrm{m}}-{\sin }^2\frac{\theta }{2}}$$$${\omega }^2=\frac{\mathrm{g}}{\mathrm{R}}\times \frac{{\sin }^2\frac{\theta }{2}}{{\mathrm{a}}^2-{\sin }^2\frac{\theta }{2}},\mathrm{with}{\mathrm{a}}^2=\frac{6\mathrm{M}+7\mathrm{m}}{6\mathrm{m}}=\frac{\mathrm{M}}{\mathrm{m}}+\frac{7}{6}>1$$$$\Rightarrow \omega =\sqrt{\frac{\mathrm{g}}{\mathrm{R}}}\times \frac{\sin \frac{\theta }{2}}{\sqrt{{\mathrm{a}}^2-{\sin }^2\frac{\theta }{2}}}=\frac{\mathrm{d}\theta }{\mathrm{dt}}$$$$\Rightarrow \frac{\sqrt{{\mathrm{a}}^2-{\sin }^2\frac{\theta }{2}}}{\sin \frac{\theta }{2}}\mathrm{d}\theta =\sqrt{\frac{\mathrm{g}}{\mathrm{R}}}\mathrm{dt}$$$$\Rightarrow \sqrt{\frac{\mathrm{g}}{\mathrm{R}}}{\int }_0^{\mathrm{t}}\mathrm{dt}={\int }_0^{\theta }\frac{\sqrt{{\mathrm{a}}^2-{\sin }^2\frac{\theta }{2}}}{\sin \frac{\theta }{2}}\mathrm{d}\theta$$$$\Rightarrow \mathrm{t}=\sqrt{\frac{\mathrm{R}}{\mathrm{g}}}{\int }_0^{\theta }\frac{\sqrt{{\mathrm{a}}^2-{\sin }^2\frac{\theta }{2}}}{\sin \frac{\theta }{2}}\mathrm{d}\theta$$$$\mathrm{t}=2\sqrt{\frac{\mathrm{R}}{\mathrm{g}}}{\int }_0^{\theta }\frac{\sqrt{{\mathrm{a}}^2-{\sin }^2\frac{\theta }{2}}}{\sin \frac{\theta }{2}}\mathrm{d}\frac{\theta }{2}$$$$\mathrm{t}=2\sqrt{\frac{\mathrm{R}}{\mathrm{g}}}{\int }_0^{\phi }\frac{\sqrt{{\mathrm{a}}^2-{\sin }^2\phi }}{\sin \phi }\mathrm{d}\phi ,\mathrm{with}\phi =\frac{\theta }{2}$$$$$$$$\mathrm{the}\mathrm{time}\mathrm{which}\mathrm{is}\mathrm{needed}\mathrm{for}\mathrm{a}\mathrm{cycle}\mathrm{is}$$$$\mathrm{T}=2\sqrt{\frac{\mathrm{R}}{\mathrm{g}}}{\int }_0^{\pi }\frac{\sqrt{{\mathrm{a}}^2-{\sin }^2\phi }}{\sin \phi }\mathrm{d}\phi$$$$$$$$\mathrm{F}\left(\phi \right)={\int }^{}\frac{\sqrt{{\mathrm{a}}^2-{\sin }^2\phi }}{\sin \phi }\mathrm{d}\phi$$$$=-\frac{1}{2}[\mathrm{a}\times \ln \mid \frac{\sqrt{{\mathrm{a}}^2-{\sin }^2\phi }+\mathrm{a}\mid \cos \phi \mid }{\sqrt{{\mathrm{a}}^2-{\sin }^2\phi }-\mathrm{a}\mid \cos \phi \mid }\mid +\ln \mid \frac{\sqrt{{\mathrm{a}}^2-{\sin }^2\phi }-\mid \cos \phi \mid }{\sqrt{{\mathrm{a}}^2-{\sin }^2\phi }+\mid \cos \phi \mid }\mid ]+\mathrm{C}$$$$=-\frac{1}{2}[\mathrm{a}\times \ln \mid \frac{\sqrt{({\mathrm{a}}^2-1){\tan }^2\phi +{\mathrm{a}}^2}+\mathrm{a}}{\sqrt{({\mathrm{a}}^2-1){\tan }^2\phi +{\mathrm{a}}^2}-\mathrm{a}}\mid +\ln \mid \frac{\sqrt{({\mathrm{a}}^2-1){\tan }^2\phi +{\mathrm{a}}^2}-1}{\sqrt{({\mathrm{a}}^2-1){\tan }^2\phi +{\mathrm{a}}^2}+1}\mid ]+\mathrm{C}$$$$......$$$$\mathrm{I}\mathrm{can}\mathrm{not}\mathrm{continue}\mathrm{due}\mathrm{to}\mathrm{problems}\mathrm{with}$$$$\mathrm{this}\mathrm{integral}\mathrm{at}\phi =0\mathrm{and}\phi =\pi .$$$$$$$$\mathrm{what}\mathrm{is}\mathrm{wrong}\mathrm{above}???$$

Commented by ajfour last updated on 31/Oct/17

$$takelimit\phi \to 0,and\phi \to \pi$$$$whatwasthetecqniqueto$$$$evaluatetheIntegral,Sir?$$

Commented by mrW1 last updated on 31/Oct/17

$$\mathrm{is}\mathrm{there}\mathrm{a}\mathrm{simple}\mathrm{solution}?$$

Commented by mrW1 last updated on 31/Oct/17

$$\mathrm{but}\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{limit}\mathrm{for}\phi \to 0\mathrm{and}\pi .$$$$\mathrm{I}\mathrm{am}\mathrm{simply}\mathrm{confused}.$$

Commented by mrW1 last updated on 31/Oct/17

$$\mathrm{let}\mathrm{us}\mathrm{take}\mathrm{a}=2,$$$$\underset{\phi \to 0}{\lim ln}\mid \frac{\sqrt{3+{\cos }^2\phi }+2\cos \phi }{\sqrt{3+{\cos }^2\phi }-2\cos \phi }\mid \to \mathrm{\infty }$$$$\underset{\phi \to \pi }{\lim ln}\mid \frac{\sqrt{3+{\cos }^2\phi }+2\cos \phi }{\sqrt{3+{\cos }^2\phi }-2\cos \phi }\mid \to \mathrm{\infty }$$

Commented by ajfour last updated on 31/Oct/17

$${\int }_0^{\pi }\frac{\sqrt{a^2-\sin {}^2\phi }}{\sin \phi }d\phi =2{\int }_0^{\pi /2}\frac{\sqrt{a^2-\cos {}^2\phi }}{\cos \phi }d\phi$$$$whataboutthisintegral,Sir?$$$$butpleaseexplainhowthe$$$$F\left(\phi \right)isintegrated,maybeican$$$$seeifthetwologarithmsmerge.$$

Commented by mrW1 last updated on 31/Oct/17

$$\mathrm{substitute}\mathrm{u}=\tan \mathrm{x}$$$$\Rightarrow \int \frac{\sqrt{{\mathrm{a}}^2+({\mathrm{a}}^2-1){\mathrm{u}}^2}}{\mathrm{u}({\mathrm{u}}^2+1)}\mathrm{du}$$$$\mathrm{substitute}\mathrm{v}=\sqrt{{\mathrm{a}}^2+({\mathrm{a}}^2-1){\mathrm{u}}^2}$$$$\Rightarrow ({\mathrm{a}}^2-1)\int \frac{{\mathrm{v}}^2}{({\mathrm{v}}^2-1)({\mathrm{v}}^2-{\mathrm{a}}^2)}\mathrm{dv}$$$$\Rightarrow {\mathrm{a}}^2\int \frac{\mathrm{dv}}{{\mathrm{v}}^2-{\mathrm{a}}^2}+\frac{1}{2}\int \frac{\mathrm{dv}}{\mathrm{v}+1}-\frac{1}{2}\int \frac{\mathrm{dv}}{\mathrm{v}-1}$$$$=......$$

Commented by ajfour last updated on 31/Oct/17

$$Thankyouimmenselysir.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com