Question Number 146303 by ajfour last updated on 12/Jul/21
Commented by ajfour last updated on 12/Jul/21
$${If}\:{released}\:{as}\:{shown},\:{and} \\ $$$${assuming}\:{no}\:{friction},\:{find} \\ $$$${after}\:{what}\:{time}\:{does}\:{the}\: \\ $$$${upper}\:{ball}\:{hit}\:{the}\:{ground}? \\ $$
Commented by ajfour last updated on 13/Jul/21
$${just}\:{saw}\:{a}\:{notification}\:{to}\:{this} \\ $$$${question},\:{but}\:{nothing}\:{in}\:{here}.. \\ $$
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
$${i}\:{added}\:{an}\:{image}\:{comment}\:{and}\:{then}\: \\ $$$${deleted}. \\ $$
Answered by mr W last updated on 14/Jul/21
Commented by mr W last updated on 15/Jul/21
$$\phi=\mathrm{60}° \\ $$$${s}\:\mathrm{sin}\:\phi+{r}\:\mathrm{cos}\:\phi={r}+\mathrm{2}{r}\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow{s}=\frac{{r}\left(\mathrm{1}−\mathrm{cos}\:\phi\right)}{\mathrm{sin}\:\phi}+\frac{\mathrm{2}{r}\:\mathrm{cos}\:\theta}{\mathrm{sin}\:\phi} \\ $$$${h}+{s}\:\mathrm{cos}\:\phi={r}\:\mathrm{sin}\:\phi+\mathrm{2}{r}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow{h}={r}\left(\mathrm{sin}\:\phi−\frac{\mathrm{1}−\mathrm{cos}\:\phi}{\mathrm{tan}\:\phi}\right)+\mathrm{2}{r}\left(\mathrm{sin}\:\theta−\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\phi}\right) \\ $$$${let}\:\omega=\frac{{d}\theta}{{dt}} \\ $$$${u}=\frac{{ds}}{{dt}}=−\frac{\mathrm{2}{r}\:\mathrm{sin}\:\theta}{\mathrm{sin}\:\phi}\omega \\ $$$${U}=\frac{{dh}}{{dt}}=\mathrm{2}{r}\left(\mathrm{cos}\:\theta+\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\phi}\right)\omega \\ $$$${y}_{{B}} ={r}+\mathrm{2}{r}\:\mathrm{cos}\:\theta \\ $$$$\Delta\left({y}_{{B}\mathrm{0}} −{y}_{{B}} \right)=\mathrm{2}{r}\left(\mathrm{sin}\:\phi−\mathrm{cos}\:\theta\right) \\ $$$$\frac{{m}}{\mathrm{2}}\left({u}^{\mathrm{2}} +{U}^{\mathrm{2}} \right)={mg}\mathrm{2}{r}\left(\mathrm{sin}\:\phi−\mathrm{cos}\:\theta\right) \\ $$$${u}^{\mathrm{2}} +{U}^{\mathrm{2}} =\mathrm{4}{gr}\left(\mathrm{sin}\:\phi−\mathrm{cos}\:\theta\right) \\ $$$$\frac{\mathrm{4}{r}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{sin}^{\mathrm{2}} \:\phi}\omega^{\mathrm{2}} +\mathrm{4}{r}^{\mathrm{2}} \left(\mathrm{cos}\:\theta+\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\phi}\right)^{\mathrm{2}} \omega^{\mathrm{2}} =\mathrm{4}{gr}\left(\mathrm{sin}\:\phi−\mathrm{cos}\:\theta\right) \\ $$$$\left[\frac{\mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{sin}^{\mathrm{2}} \:\phi}+\left(\mathrm{cos}\:\theta+\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\phi}\right)^{\mathrm{2}} \right]\omega^{\mathrm{2}} =\frac{{g}}{{r}}\left(\mathrm{sin}\:\phi−\mathrm{cos}\:\theta\right) \\ $$$$\omega=\sqrt{\frac{{g}}{{r}}}×\sqrt{\frac{\mathrm{sin}\:\phi−\mathrm{cos}\:\theta}{\frac{\mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{sin}^{\mathrm{2}} \:\phi}+\left(\mathrm{cos}\:\theta+\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\phi}\right)^{\mathrm{2}} }} \\ $$$${let}\:{f}\left(\theta\right)=\sqrt{\frac{\mathrm{sin}\:\phi−\mathrm{cos}\:\theta}{\frac{\mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{sin}^{\mathrm{2}} \:\phi}+\left(\mathrm{cos}\:\theta+\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\phi}\right)^{\mathrm{2}} }} \\ $$$${f}\left(\theta\right)=\mathrm{sin}\:\phi\sqrt{\frac{\mathrm{sin}\:\phi−\mathrm{cos}\:\theta}{\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{sin}^{\mathrm{2}} \:\left(\theta+\phi\right)}} \\ $$$$\omega={f}\left(\theta\right)\sqrt{\frac{{g}}{{r}}} \\ $$$${A}=\frac{{dU}}{{dt}}=\omega\frac{{dU}}{{d}\theta}=\mathrm{2}{r}\left[\omega^{\mathrm{2}} \left(−\mathrm{sin}\:\theta+\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\phi}\right)+\omega\left(\mathrm{cos}\:\theta+\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\phi}\right)\frac{{d}\omega}{{d}\theta}\right] \\ $$$$ \\ $$$${N}\mathrm{sin}\:\theta={mA} \\ $$$${when}\:{contact}\:{gets}\:{lost}, \\ $$$${N}=\mathrm{0}\:\Rightarrow{A}=\mathrm{0} \\ $$$$\omega\left(−\mathrm{sin}\:\theta+\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\phi}\right)+\left(\mathrm{cos}\:\theta+\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\phi}\right)\frac{{d}\omega}{{d}\theta}=\mathrm{0} \\ $$$${f}\left(\theta\right)\left(−\mathrm{sin}\:\theta+\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\phi}\right)+\left(\mathrm{cos}\:\theta+\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\phi}\right)\frac{{df}\left(\theta\right)}{{d}\theta}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\left(\theta+\phi\right)\frac{{df}\left(\theta\right)}{{d}\theta}+{f}\left(\theta\right)=\mathrm{0} \\ $$$$\Rightarrow\theta=… \\ $$$$ \\ $$$${examples}:\: \\ $$$$\phi=\mathrm{60}°,\:\theta_{\mathrm{1}} \approx\mathrm{35}.\mathrm{3946}° \\ $$$$\phi=\mathrm{45}°,\:\theta_{\mathrm{1}} \approx\mathrm{50}.\mathrm{3307}° \\ $$$$ \\ $$$$\omega=\frac{{d}\theta}{{dt}}=\sqrt{\frac{{g}}{{r}}}\:{f}\left(\theta\right) \\ $$$${dt}=\sqrt{\frac{{r}}{{g}}}×\frac{{d}\theta}{{f}\left(\theta\right)} \\ $$$${t}_{\mathrm{1}} =\sqrt{\frac{{r}}{{g}}}×\int_{\frac{\pi}{\mathrm{2}}−\phi} ^{\theta_{\mathrm{1}} } \frac{{d}\theta}{{f}\left(\theta\right)} \\ $$$${t}_{\mathrm{1}} =\sqrt{\frac{{r}}{{g}}}×\frac{\mathrm{1}}{\mathrm{sin}\:\phi}\int_{\frac{\pi}{\mathrm{2}}−\phi} ^{\theta_{\mathrm{1}} } \sqrt{\frac{\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{sin}^{\mathrm{2}} \:\left(\theta+\phi\right)}{\mathrm{sin}\:\phi−\mathrm{cos}\:\theta}}\:{d}\theta \\ $$
Commented by ajfour last updated on 15/Jul/21
$${yes}\:{sir},\:{i}\:{will}\:{match}\:{with}\:{yours}, \\ $$$${this}\:{time},\:{thanks},\:{very}\:{nice} \\ $$$${solution}! \\ $$
Commented by mr W last updated on 15/Jul/21
$${when}\:{there}\:{is}\:{no}\:{friction},\:{i}\:{think}\:{the} \\ $$$${balls}\:{don}'{t}\:{rotate}. \\ $$
Commented by otchereabdullai@gmail.com last updated on 16/Jul/21
$$\mathrm{wow}! \\ $$
Answered by ajfour last updated on 14/Jul/21