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Question Number 71611 by ajfour last updated on 17/Oct/19

Commented by ajfour last updated on 17/Oct/19

$$Findtimeittakesthecylinder$$$$tounwindandslideandhitthe$$$$ground.$$

Commented by ajfour last updated on 18/Oct/19

Answered by Tanmay chaudhury last updated on 18/Oct/19

$$mg-Tcos\alpha =ma$$$$Tsin\alpha =N$$$$T.r=I\left(\frac{a}{r}\right)\to T=\frac{Ia}{r^2}$$$$mg-\frac{Iacos\alpha }{r^2}=ma$$$$a(\frac{Icos\alpha }{r^2}+m)=mg$$$$a=\frac{mg}{\frac{Icos\alpha }{r^2}+m}\to h=\frac{1}{2}{at}^2\to t=\sqrt{\frac{2h}{a}}$$$$t=\sqrt{\frac{2h}{mg}\times (m+\frac{Icos\alpha }{r^2})}$$$$t=\sqrt{\frac{2h}{mg}\times (m+\frac{{mr}^{2}cos\alpha }{2r^2})}[I=\frac{{mr}^2}{2}]$$$$t=\sqrt{\frac{2h}{mg}\times m(1+\frac{r^{2}cos\alpha }{2r^2})}$$$$t=\sqrt{\frac{2h}{g}\times (1+\frac{cos\alpha }{2})}$$$$$$$$\mathit{p}\mathit{l}\mathit{s}c\mathit{h}\mathit{e}\mathit{c}\mathit{k}\mathit{i}\mathit{s}\mathit{i}\mathit{t}\mathit{c}\mathit{o}\mathit{r}\mathit{r}\mathit{e}\mathit{c}\mathit{t}...$$$$$$

Commented by Tanmay chaudhury last updated on 18/Oct/19

$$dofrictionpresent...plsclarify...becauseihave$$$$ignoredfriction$$

Commented by ajfour last updated on 19/Oct/19

$$frictionless,yes,but\alpha goes$$$$changingsir.$$

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

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

$$iuse\beta toreplace\alpha inthequestion$$$$AC=h+\frac{r}{\tan \frac{\beta }{2}}=s+\frac{r}{\tan \frac{\theta }{2}}=k$$$$\tan \frac{\theta }{2}=\frac{r}{k-s}$$$$\cos \theta =\frac{1-{\left(\frac{r}{k-s}\right)}^2}{1+{\left(\frac{r}{k-s}\right)}^2}=1-\frac{2r^2}{{(k-s)}^2+r^2}$$$$mg-T\cos \theta =ma$$$$Tr=I\alpha$$$$I=\frac{{mr}^2}{2}$$$$\alpha =\frac{a}{r}$$$$Tr=\frac{{mr}^2}{2}\times \frac{a}{r}$$$$T=\frac{ma}{2}$$$$mg-\frac{ma\cos \theta }{2}=ma$$$$\Rightarrow a=\frac{2g}{2+\cos \theta }=\frac{2g}{3-\frac{2r^2}{{(k-s)}^2+r^2}}=\frac{2g[{(k-s)}^2+r^2]}{3{(k-s)}^2+r^2}$$$$\Rightarrow -v\frac{dv}{ds}=2g\times \frac{{(k-s)}^2+r^2}{3{(k-s)}^2+r^2}$$$$\Rightarrow {\int }_0^{v}vdv=-2g{\int }_h^s\frac{{(k-s)}^2+r^2}{3{(k-s)}^2+r^2}ds$$$$\Rightarrow \frac{v^2}{2}=2g{\int }_h^s\frac{{(k-s)}^2+r^2}{3{(k-s)}^2+r^2}d(k-s)$$$$\Rightarrow \frac{v^2}{4g}={\int }_{k-h}^{k-s}\frac{u^2+r^2}{3u^2+r^2}du$$$$\Rightarrow \frac{v^2}{4g}=\frac{1}{3}{[\frac{2r}{\sqrt{3}}{\tan }^{-1}\frac{\sqrt{3}u}{r}+u]}_{k-h}^{k-s}$$$$\Rightarrow \frac{v^2}{4g}=\frac{1}{3}[\frac{2r}{\sqrt{3}}({\tan }^{-1}\frac{\sqrt{3}}{\tan \frac{\theta }{2}}-{\tan }^{-1}\frac{\sqrt{3}}{\tan \frac{\beta }{2}})+\frac{r}{\tan \frac{\theta }{2}}-\frac{r}{\tan \frac{\beta }{2}}]$$$$\Rightarrow \frac{3v^2}{4gr}=\frac{2}{\sqrt{3}}({\tan }^{-1}\frac{\sqrt{3}}{\tan \frac{\theta }{2}}-{\tan }^{-1}\frac{\sqrt{3}}{\tan \frac{\beta }{2}})+\frac{1}{\tan \frac{\theta }{2}}-\frac{1}{\tan \frac{\beta }{2}}$$$$\Rightarrow v=\frac{2\sqrt{gr}}{\sqrt{3}}\sqrt{\frac{2}{\sqrt{3}}({\tan }^{-1}\frac{\sqrt{3}}{\tan \frac{\theta }{2}}-{\tan }^{-1}\frac{\sqrt{3}}{\tan \frac{\beta }{2}})+\frac{1}{\tan \frac{\theta }{2}}-\frac{1}{\tan \frac{\beta }{2}}}$$$$\Rightarrow \frac{ds}{dt}=\frac{2\sqrt{gr}}{\sqrt{3}}\sqrt{\frac{2}{\sqrt{3}}({\tan }^{-1}\frac{\sqrt{3}(k-s)}{r}-{\tan }^{-1}\frac{\sqrt{3}}{\tan \frac{\beta }{2}})+\frac{k-s}{r}-\frac{1}{\tan \frac{\beta }{2}}}$$$$\Rightarrow \frac{2\sqrt{gr}}{\sqrt{3}}{\int }_0^{t}dt={\int }_h^r\frac{ds}{\sqrt{\frac{2}{\sqrt{3}}({\tan }^{-1}\frac{\sqrt{3}(k-s)}{r}-{\tan }^{-1}\frac{\sqrt{3}}{\tan \frac{\beta }{2}})+\frac{k-s}{r}-\frac{1}{\tan \frac{\beta }{2}}}}$$$$\Rightarrow t=\frac{\sqrt{3}}{2\sqrt{gr}}{\int }_h^r\frac{ds}{\sqrt{\frac{2}{\sqrt{3}}({\tan }^{-1}\frac{\sqrt{3}(k-s)}{r}-{\tan }^{-1}\frac{\sqrt{3}}{\tan \frac{\beta }{2}})+\frac{k-s}{r}-\frac{1}{\tan \frac{\beta }{2}}}}$$$$......$$

Commented by ajfour last updated on 19/Oct/19

Commented by ajfour last updated on 19/Oct/19

$$Sir,itseemstomethat$$$$v\cos \theta =\omega r,pleasecomment...$$

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

$$ithoughtexactlysoatthebegining,$$$$thenichangedmymind.{let}^{\prime }ssay$$$$thelengthofwireisAD=l,$$$$l+s=k=constant$$$$thespeedofunwinding\omega r=\frac{dl}{dt}=-\frac{ds}{dt}=v.$$$$actuallyiamstillconfusedwhatis$$$$reallytrue.$$

Commented by ajfour last updated on 19/Oct/19

$$samehereSir,confusedsince$$$$lastnight..$$

Commented by mr W last updated on 20/Oct/19

Commented by mr W last updated on 20/Oct/19

$$ihaveclarified:$$$$letAB=s$$$$v=\frac{ds}{dt}$$$$rotationofcylinderd\phi =(\theta +d\theta )+\varphi -\theta =\frac{ds}{r}+d\theta$$$$with\varphi =\frac{ds}{r}=rotationduetounwinding$$$$\omega =\frac{d\phi }{dt}=\frac{v}{r}+\frac{d\theta }{dt}=\frac{v}{r}+v\frac{d\theta }{ds}$$$$\tan \frac{\theta }{2}=\frac{r}{s}$$$$\frac{d\theta }{2{\cos }^2\frac{\theta }{2}}=-\frac{rds}{s^2}=-{\tan }^2\frac{\theta }{2}\frac{ds}{r}$$$$\Rightarrow \frac{d\theta }{ds}=-\frac{1}{r}2{\sin }^2\frac{\theta }{2}$$$$\Rightarrow \omega =\frac{v}{r}+v\frac{d\theta }{ds}=\frac{v}{r}(1-2{\sin }^2\frac{\theta }{2})=\frac{v\cos \theta }{r}$$$$\Rightarrow r\omega =v\cos \theta$$

Commented by mr W last updated on 20/Oct/19

$$newattempt:$$$$s_0=\frac{r}{\tan \frac{\beta }{2}}$$$$s_1=\frac{r}{\tan \frac{\beta }{2}}+h-r$$$$mg(s-s_0)=\frac{1}{2}{mv}^2+\frac{1}{2}I{\omega }^2$$$$g(s-s_0)=\frac{1}{2}{v}^2+\frac{r^2}{4}{\omega }^2$$$$g(s-s_0)=\frac{1}{4}{v}^2(2+{\cos }^2\theta )$$$$\tan \frac{\theta }{2}=\frac{r}{s}$$$$\cos \theta =\frac{1-{\left(\frac{r}{s}\right)}^2}{1+{\left(\frac{r}{s}\right)}^2}=\frac{s^2-r^2}{s^2+r^2}$$$$g(s-s_0)=\frac{1}{4}{v}^2[2+{\left(\frac{s^2-r^2}{s^2+r^2}\right)}^2]$$$$\Rightarrow v=2\sqrt{\frac{g(s-s_0)}{2+{\left(\frac{s^2-r^2}{s^2+r^2}\right)}^2}}$$$$\Rightarrow \frac{ds}{dt}=2\sqrt{g}\sqrt{\frac{s-s_0}{2+{\left(\frac{s^2-r^2}{s^2+r^2}\right)}^2}}$$$$\Rightarrow t=\frac{1}{2\sqrt{g}}{\int }_{s_0}^{s_1}\sqrt{\frac{2+{\left(\frac{s^2-r^2}{s^2+r^2}\right)}^2}{s-s_0}}ds$$

Commented by ajfour last updated on 21/Oct/19

$$Thankssir,icantfindmuch$$$$timetoattemptmyselfsir,i$$$$shallhowevertryittoday..$$

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