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Question Number 146303 by ajfour last updated on 12/Jul/21

Commented by ajfour last updated on 12/Jul/21

$$Ifreleasedasshown,and$$$$assumingnofriction,find$$$$afterwhattimedoesthe$$$$upperballhittheground?$$

Commented by ajfour last updated on 13/Jul/21

$$justsawanotificationtothis$$$$question,butnothinginhere..$$

Commented by Tinku Tara last updated on 13/Jul/21

$$\mathrm{Maybe}\mathrm{somebody}\mathrm{added}\mathrm{something}$$$$\mathrm{and}\mathrm{then}\mathrm{deleted}$$

Commented by mr W last updated on 13/Jul/21

$$iaddedanimagecommentandthen$$$$deleted.$$

Answered by mr W last updated on 14/Jul/21

Commented by mr W last updated on 15/Jul/21

$$\varphi =60°$$$$s\sin \varphi +r\cos \varphi =r+2r\cos \theta$$$$\Rightarrow s=\frac{r(1-\cos \varphi )}{\sin \varphi }+\frac{2r\cos \theta }{\sin \varphi }$$$$h+s\cos \varphi =r\sin \varphi +2r\sin \theta$$$$\Rightarrow h=r(\sin \varphi -\frac{1-\cos \varphi }{\tan \varphi })+2r(\sin \theta -\frac{\cos \theta }{\tan \varphi })$$$$let\omega =\frac{d\theta }{dt}$$$$u=\frac{ds}{dt}=-\frac{2r\sin \theta }{\sin \varphi }\omega$$$$U=\frac{dh}{dt}=2r(\cos \theta +\frac{\sin \theta }{\tan \varphi })\omega$$$$y_B=r+2r\cos \theta$$$$\mathrm{\Delta }(y_{B0}-y_B)=2r(\sin \varphi -\cos \theta )$$$$\frac{m}{2}(u^2+U^2)=mg2r(\sin \varphi -\cos \theta )$$$$u^2+U^2=4gr(\sin \varphi -\cos \theta )$$$$\frac{4r^2{\sin }^2\theta }{{\sin }^2\varphi }{\omega }^2+4r^2{(\cos \theta +\frac{\sin \theta }{\tan \varphi })}^2{\omega }^2=4gr(\sin \varphi -\cos \theta )$$$$[\frac{{\sin }^2\theta }{{\sin }^2\varphi }+{(\cos \theta +\frac{\sin \theta }{\tan \varphi })}^2]{\omega }^2=\frac{g}{r}(\sin \varphi -\cos \theta )$$$$\omega =\sqrt{\frac{g}{r}}\times \sqrt{\frac{\sin \varphi -\cos \theta }{\frac{{\sin }^2\theta }{{\sin }^2\varphi }+{(\cos \theta +\frac{\sin \theta }{\tan \varphi })}^2}}$$$$letf\left(\theta \right)=\sqrt{\frac{\sin \varphi -\cos \theta }{\frac{{\sin }^2\theta }{{\sin }^2\varphi }+{(\cos \theta +\frac{\sin \theta }{\tan \varphi })}^2}}$$$$f\left(\theta \right)=\sin \varphi \sqrt{\frac{\sin \varphi -\cos \theta }{{\sin }^2\theta +{\sin }^2(\theta +\varphi )}}$$$$\omega =f\left(\theta \right)\sqrt{\frac{g}{r}}$$$$A=\frac{dU}{dt}=\omega \frac{dU}{d\theta }=2r[{\omega }^2(-\sin \theta +\frac{\cos \theta }{\tan \varphi })+\omega (\cos \theta +\frac{\sin \theta }{\tan \varphi })\frac{d\omega }{d\theta }]$$$$$$$$N\sin \theta =mA$$$$whencontactgetslost,$$$$N=0\Rightarrow A=0$$$$\omega (-\sin \theta +\frac{\cos \theta }{\tan \varphi })+(\cos \theta +\frac{\sin \theta }{\tan \varphi })\frac{d\omega }{d\theta }=0$$$$f\left(\theta \right)(-\sin \theta +\frac{\cos \theta }{\tan \varphi })+(\cos \theta +\frac{\sin \theta }{\tan \varphi })\frac{df\left(\theta \right)}{d\theta }=0$$$$\Rightarrow \tan (\theta +\varphi )\frac{df\left(\theta \right)}{d\theta }+f\left(\theta \right)=0$$$$\Rightarrow \theta =...$$$$$$$$examples:$$$$\varphi =60°,{\theta }_1\approx 35.3946°$$$$\varphi =45°,{\theta }_1\approx 50.3307°$$$$$$$$\omega =\frac{d\theta }{dt}=\sqrt{\frac{g}{r}}f\left(\theta \right)$$$$dt=\sqrt{\frac{r}{g}}\times \frac{d\theta }{f\left(\theta \right)}$$$$t_1=\sqrt{\frac{r}{g}}\times {\int }_{\frac{\pi }{2}-\varphi }^{{\theta }_1}\frac{d\theta }{f\left(\theta \right)}$$$$t_1=\sqrt{\frac{r}{g}}\times \frac{1}{\sin \varphi }{\int }_{\frac{\pi }{2}-\varphi }^{{\theta }_1}\sqrt{\frac{{\sin }^2\theta +{\sin }^2(\theta +\varphi )}{\sin \varphi -\cos \theta }}d\theta$$

Commented by ajfour last updated on 15/Jul/21

$$yessir,iwillmatchwithyours,$$$$thistime,thanks,verynice$$$$solution!$$

Commented by mr W last updated on 15/Jul/21

$$whenthereisnofriction,ithinkthe$$$$balls{don}^{\prime }trotate.$$

Commented by otchereabdullai@gmail.com last updated on 16/Jul/21

$$\mathrm{wow}!$$

Answered by ajfour last updated on 14/Jul/21

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