# Question-131823

Question Number 131823 by mr W last updated on 09/Feb/21
Commented by mr W last updated on 10/Feb/21
$${a}\:{small}\:{sphere}\:{of}\:{mass}\:{m}\:{is}\:{released} \\$$$${from}\:{rest}\:{at}\:{position}\:{as}\:{shown}. \\$$$${both}\:{the}\:{sphere}\:{and}\:{the}\:{semi}\:{cylinder} \\$$$${move}\:{without}\:{friction}. \\$$$${find}\:{the}\:{time}\:{the}\:{sphere}\:{needs}\:{to} \\$$$${hit}\:{the}\:{ground}. \\$$
Answered by mr W last updated on 09/Feb/21
Commented by mr W last updated on 10/Feb/21
Commented by mr W last updated on 10/Feb/21
Commented by mr W last updated on 09/Feb/21
$$\theta_{\mathrm{0}} =\mathrm{cos}^{−\mathrm{1}} \frac{{R}−{r}}{{R}+{r}} \\$$$$\theta_{\mathrm{1}} =\mathrm{sin}^{−\mathrm{1}} \frac{{r}}{{R}+{r}} \\$$
Commented by mr W last updated on 11/Feb/21
$${let}\:\omega=\frac{{d}\theta}{{dt}},\:\alpha=\frac{{d}\omega}{{dt}},\:\mu=\frac{{M}}{{m}} \\$$$${x}_{{Q}} ={r}+\left({R}+{r}\right)\mathrm{cos}\:\theta \\$$$${V}=\frac{{dx}_{{Q}} }{{dt}}=\omega\frac{{dx}_{{Q}} }{{d}\theta}=−\left({R}+{r}\right)\omega\:\mathrm{sin}\:\theta \\$$$${A}=\frac{{dV}}{{dt}}=−\left({R}+{r}\right)\left(\alpha\:\mathrm{sin}\:\theta+\omega^{\mathrm{2}} \:\mathrm{cos}\:\theta\right) \\$$$$\\$$$${y}_{{P}} =\left({R}+{r}\right)\mathrm{sin}\:\theta \\$$$${v}=−\frac{{dy}_{{P}} }{{dt}}=−\omega\frac{{dy}_{{P}} }{{d}\theta}=−\left({R}+{r}\right)\omega\:\mathrm{cos}\:\theta \\$$$${a}=\frac{{dv}}{{dt}}=−\left({R}+{r}\right)\left(\alpha\:\mathrm{cos}\:\theta−\omega^{\mathrm{2}} \:\mathrm{sin}\:\theta\right) \\$$$$\\$$$$\frac{{MV}^{\mathrm{2}} +{mv}^{\mathrm{2}} }{\mathrm{2}}={mg}\left({R}+{r}\right)\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right) \\$$$$\frac{{M}}{{m}}{V}^{\mathrm{2}} +{v}^{\mathrm{2}} =\mathrm{2}{g}\left({R}+{r}\right)\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right) \\$$$$\\$$$$\left[\frac{{M}}{{m}}\:\mathrm{sin}^{\mathrm{2}} \:\theta+\:\mathrm{cos}^{\mathrm{2}} \:\theta\right]\omega^{\mathrm{2}} \left({R}+{r}\right)^{\mathrm{2}} =\mathrm{2}{g}\left({R}+{r}\right)\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right) \\$$$$\Rightarrow\omega^{\mathrm{2}} =\frac{\mathrm{2}{g}}{{R}+{r}}×\frac{\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta}{\mu\:\mathrm{sin}^{\mathrm{2}} \:\theta+\:\mathrm{cos}^{\mathrm{2}} \:\theta} \\$$$$\\$$$${N}\:\mathrm{cos}\:\theta={MA} \\$$$$\Rightarrow{N}=−\frac{{M}\left({R}+{r}\right)}{\mathrm{cos}\:\theta}\left(\alpha\:\mathrm{sin}\:\theta+\omega^{\mathrm{2}} \:\mathrm{cos}\:\theta\right) \\$$$${mg}−{N}\:\mathrm{sin}\:\theta={ma} \\$$$${N}=\frac{{m}\left({g}−{a}\right)}{\mathrm{sin}\:\theta} \\$$$$\Rightarrow{N}=\frac{{m}\left({R}+{r}\right)}{\mathrm{sin}\:\theta}\left(\frac{{g}}{{R}+{r}}+\alpha\:\mathrm{cos}\:\theta−\omega^{\mathrm{2}} \:\mathrm{sin}\:\theta\right) \\$$$$\\$$$$\frac{{m}\left({R}+{r}\right)}{\mathrm{sin}\:\theta}\left(\frac{{g}}{{R}+{r}}+\alpha\:\mathrm{cos}\:\theta−\omega^{\mathrm{2}} \:\mathrm{sin}\:\theta\right)=−\frac{{M}\left({R}+{r}\right)}{\mathrm{cos}\:\theta}\left(\alpha\:\mathrm{sin}\:\theta+\omega^{\mathrm{2}} \:\mathrm{cos}\:\theta\right) \\$$$$\frac{{g}}{{R}+{r}}+\alpha\:\mathrm{cos}\:\theta−\omega^{\mathrm{2}} \:\mathrm{sin}\:\theta=−\frac{\mu\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\left(\alpha\:\mathrm{sin}\:\theta+\omega^{\mathrm{2}} \:\mathrm{cos}\:\theta\right) \\$$$$\alpha=−\frac{\mathrm{cos}\:\theta}{\mu\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{cos}^{\mathrm{2}} \:\theta}\left[\frac{{g}}{{R}+{r}}+\left(\mu−\mathrm{1}\right)\mathrm{sin}\:\theta\:\omega^{\mathrm{2}} \right] \\$$$$\Rightarrow\alpha=−\frac{{g}}{{R}+{r}}×\frac{\mathrm{cos}\:\theta}{\mu\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{cos}^{\mathrm{2}} \:\theta}\left[\mathrm{1}+\frac{\mathrm{2}\left(\mu−\mathrm{1}\right)\mathrm{sin}\:\theta\:\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right)}{\mu\:\mathrm{sin}^{\mathrm{2}} \:\theta+\:\mathrm{cos}^{\mathrm{2}} \:\theta}\right] \\$$$$\\$$$$\Rightarrow{N}=−{M}\left({R}+{r}\right)\left(\frac{\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\alpha+\omega^{\mathrm{2}} \right) \\$$$$\Rightarrow{N}={Mg}\left\{\frac{\mathrm{sin}\:\theta}{\mu\:\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{cos}^{\mathrm{2}} \:\theta}\left[\mathrm{1}+\frac{\mathrm{2}\left(\mu−\mathrm{1}\right)\mathrm{sin}\:\theta\:\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right)}{\mu\:\mathrm{sin}^{\mathrm{2}} \:\theta+\:\mathrm{cos}^{\mathrm{2}} \:\theta}\right]−\frac{\mathrm{2}\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right)}{\mu\:\mathrm{sin}^{\mathrm{2}} \:\theta+\:\mathrm{cos}^{\mathrm{2}} \:\theta}\right\} \\$$$$\Rightarrow\frac{{N}}{{mg}}=\frac{\mu}{\mu\:\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{cos}^{\mathrm{2}} \:\theta}\left[\mathrm{sin}\:\theta−\frac{\mathrm{2}\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right)}{\mu\:\mathrm{sin}^{\mathrm{2}} \:\theta+\:\mathrm{cos}^{\mathrm{2}} \:\theta}\right] \\$$$$\\$$$${say}\:{N}=\mathrm{0}\:{at}\:\theta=\theta_{\mathrm{2}} . \\$$$${N}=\mathrm{0}\:\Rightarrow\:\mathrm{sin}\:\theta−\frac{\mathrm{2}\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right)}{\mu\:\mathrm{sin}^{\mathrm{2}} \:\theta+\:\mathrm{cos}^{\mathrm{2}} \:\theta}=\mathrm{0} \\$$$$\mathrm{sin}\:\theta=\frac{\mathrm{2}\left(\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta\right)}{\left(\mu−\mathrm{1}\right)\:\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{1}} \\$$$$\mathrm{sin}^{\mathrm{3}} \:\theta+\frac{\mathrm{3}}{\mu−\mathrm{1}}\:\mathrm{sin}\:\theta−\frac{\mathrm{2}}{\mu−\mathrm{1}}\:\mathrm{sin}\:\theta_{\mathrm{0}} =\mathrm{0} \\$$$${if}\:\mu=\mathrm{1}:\: \\$$$$\mathrm{3}\:\mathrm{sin}\:\theta−\mathrm{2}\:\mathrm{sin}\:\theta_{\mathrm{0}} =\mathrm{0} \\$$$$\Rightarrow\mathrm{sin}\:\theta=\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{sin}\:\theta_{\mathrm{0}} \\$$$$\Rightarrow\theta_{\mathrm{2}} =\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}\:\mathrm{sin}\:\theta_{\mathrm{0}} }{\mathrm{3}}\right) \\$$$${for}\:\mu\neq\mathrm{1}: \\$$$${let}\:\lambda=\frac{\mathrm{1}}{\mu−\mathrm{1}}=\frac{{m}}{{M}−{m}} \\$$$$\mathrm{sin}^{\mathrm{3}} \:\theta+\mathrm{3}\lambda\:\mathrm{sin}\:\theta−\mathrm{2}\lambda\:\mathrm{sin}\:\theta_{\mathrm{0}} =\mathrm{0} \\$$$$\Delta=\lambda^{\mathrm{3}} +\lambda^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta_{\mathrm{0}} =\lambda^{\mathrm{2}} \left(\lambda+\mathrm{sin}^{\mathrm{2}} \:\theta_{\mathrm{0}} \right) \\$$$${for}\:\lambda+\mathrm{sin}^{\mathrm{2}} \:\theta_{\mathrm{0}} =\frac{\mathrm{1}}{\mu−\mathrm{1}}+\mathrm{sin}^{\mathrm{2}} \:\theta_{\mathrm{0}} >\mathrm{0},\:{i}.{e}. \\$$$${for}\:\mu>\mathrm{1}: \\$$$$\mathrm{sin}\:\theta=\sqrt[{\mathrm{3}}]{\lambda\left(\sqrt{\lambda+\mathrm{sin}^{\mathrm{2}} \:\theta_{\mathrm{0}} }+\mathrm{sin}\:\theta_{\mathrm{0}} \right)}−\sqrt[{\mathrm{3}}]{\lambda\left(\sqrt{\lambda+\mathrm{sin}^{\mathrm{2}} \:\theta_{\mathrm{0}} }−\mathrm{sin}\:\theta_{\mathrm{0}} \right)} \\$$$$\Rightarrow\theta_{\mathrm{2}} =\mathrm{sin}^{−\mathrm{1}} \left[\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mu−\mathrm{1}}\left(\sqrt{\frac{\mathrm{1}}{\mu−\mathrm{1}}+\mathrm{sin}^{\mathrm{2}} \:\theta_{\mathrm{0}} }+\mathrm{sin}\:\theta_{\mathrm{0}} \right)}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mu−\mathrm{1}}\left(\sqrt{\frac{\mathrm{1}}{\mu−\mathrm{1}}+\mathrm{sin}^{\mathrm{2}} \:\theta_{\mathrm{0}} }−\mathrm{sin}\:\theta_{\mathrm{0}} \right)}\right] \\$$$${for}\:\mu<\mathrm{1}: \\$$$$\mathrm{sin}\:\theta=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\mu}}\:\mathrm{sin}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\mathrm{1}−\mu}\:\mathrm{sin}\:\theta_{\mathrm{0}} \right)+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right] \\$$$$\Rightarrow\theta_{\mathrm{2}} =\mathrm{sin}^{−\mathrm{1}} \left\{\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\mu}}\:\mathrm{sin}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\mathrm{1}−\mu}\:\mathrm{sin}\:\theta_{\mathrm{0}} \right)\right]\right\} \\$$$${we}\:{get}\:\theta_{\mathrm{2}} >\theta_{\mathrm{1}} ,{see}\:{diagram}. \\$$$${that}\:{means}\:{the}\:{sphere}\:{losses}\:{contact} \\$$$${before}\:{it}\:{hits}\:{the}\:{ground}.\:{after}\:{this} \\$$$${point}\:{the}\:{sphere}\:{has}\:{free}\:{fall}. \\$$$$\\$$$${T}_{\mathrm{1}} =\:{time}\:{for}\:\theta\:{from}\:\theta_{\mathrm{0}} \:{to}\:\theta_{\mathrm{2}} \\$$$${T}_{\mathrm{2}} =\:{time}\:{for}\:{free}\:{fall}\:{after}\:\theta_{\mathrm{2}} \\$$$$\Rightarrow\omega=\frac{{d}\theta}{{dt}}=−\sqrt{\frac{\mathrm{2}{g}}{{R}+{r}}}×\sqrt{\frac{\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta}{\mu\:\mathrm{sin}^{\mathrm{2}} \:\theta+\:\mathrm{cos}^{\mathrm{2}} \:\theta}} \\$$$$\Rightarrow{T}_{\mathrm{1}} =\sqrt{\frac{{R}+{r}}{\mathrm{2}{g}}}\int_{\theta_{\mathrm{2}} } ^{\theta_{\mathrm{0}} } \sqrt{\frac{\mathrm{1}+\left(\mu−\mathrm{1}\right)\:\mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta}}{d}\theta \\$$$$\\$$$${v}=\sqrt{\mathrm{2}{g}\left({R}+{r}\right)}×\mathrm{cos}\:\theta\sqrt{\frac{\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta}{\mu\:\mathrm{sin}^{\mathrm{2}} \:\theta+\:\mathrm{cos}^{\mathrm{2}} \:\theta}} \\$$$${v}_{\mathrm{2}} =\sqrt{\mathrm{2}{g}\left({R}+{r}\right)}×\mathrm{cos}\:\theta_{\mathrm{2}} \sqrt{\frac{\mathrm{sin}\:\theta_{\mathrm{0}} −\mathrm{sin}\:\theta_{\mathrm{2}} }{\mu\:\mathrm{sin}^{\mathrm{2}} \:\theta_{\mathrm{2}} +\:\mathrm{cos}^{\mathrm{2}} \:\theta_{\mathrm{2}} }} \\$$$$\Delta{h}=\left({R}+{r}\right)\mathrm{sin}\:\theta_{\mathrm{2}} −{r} \\$$$$\Delta{h}={v}_{\mathrm{2}} {T}_{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{gT}_{\mathrm{2}} ^{\mathrm{2}} \\$$$${T}_{\mathrm{2}} =\frac{\mathrm{1}}{{g}}\left(\sqrt{{v}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{2}{g}\Delta{h}}−{v}_{\mathrm{2}} \right) \\$$$$\\$$$${total}\:{time}: \\$$$${T}={T}_{\mathrm{1}} +{T}_{\mathrm{2}} \\$$
Commented by mr W last updated on 10/Feb/21
Commented by mr W last updated on 10/Feb/21
$${God}\:{bless}\:{you}\:{too}! \\$$
Commented by otchereabdullai@gmail.com last updated on 10/Feb/21
$$\mathrm{Amen} \\$$
Commented by otchereabdullai@gmail.com last updated on 10/Feb/21
$$\mathrm{You}\:\mathrm{are}\:\mathrm{too}\:\mathrm{much}\:\mathrm{prof}\:\mathrm{W}\:\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\$$
Commented by mr W last updated on 10/Feb/21
Commented by ajfour last updated on 10/Feb/21
$${N}\mathrm{cos}\:\theta={MA} \\$$$${mg}−{N}\mathrm{sin}\:\theta={ma} \\$$$${a}\mathrm{sin}\:\theta={A}\mathrm{cos}\:\theta \\$$$$\Rightarrow\:\:{mg}={ma}+\frac{{MA}\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta} \\$$$$\Rightarrow\:\:{mg}={ma}+\frac{{Ma}\mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{cos}\:^{\mathrm{2}} \theta} \\$$$${a}=\frac{{g}}{\mathrm{1}+\lambda\mathrm{tan}\:^{\mathrm{2}} \theta}\:\:\:\:\:\:\:{where}\:\lambda=\frac{{M}}{{m}} \\$$$${A}=\frac{{g}\mathrm{tan}\:\theta}{\mathrm{1}+\lambda\mathrm{tan}\:^{\mathrm{2}} \theta} \\$$$$\Rightarrow\:\:{Normal}\:{reaction}\:{N}\:{is}\:{not} \\$$$${zero}\:\:{before}\:{ball}\:{touches}\:{ground}; \\$$$${so}\:{it}\:{appears}\:{from}\:{the} \\$$$${acceleration}\:{expressions}… \\$$$${What}\:{d}'{ya}\:{think}\:{Sir}? \\$$$$\frac{{dv}}{{dt}}=\frac{{g}}{\mathrm{1}+\lambda\mathrm{tan}\:^{\mathrm{2}} \theta} \\$$$${y}=\left({R}+{r}\right)\mathrm{sin}\:\theta \\$$$${v}=\left(−\frac{{d}\theta}{{dt}}\right)\left({R}+{r}\right)\mathrm{cos}\:\theta \\$$$$\Rightarrow\int{vdv}=−{g}\left({R}+{r}\right)\int\frac{\mathrm{cos}\:\theta{d}\theta}{\mathrm{1}+\lambda\mathrm{tan}\:^{\mathrm{2}} \theta} \\$$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}}={g}\left({R}+{r}\right)\int_{\theta_{\mathrm{0}} } ^{\:\theta} \frac{\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \theta\right){d}\left(\mathrm{sin}\:\theta\right)}{\mathrm{1}+\left(\lambda−\mathrm{1}\right)\mathrm{sin}\:^{\mathrm{2}} \theta} \\$$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}}=\frac{{g}\left({R}+{r}\right)}{\lambda−\mathrm{1}}\int\:\frac{\left(\mathrm{1}−{s}^{\mathrm{2}} \right){ds}}{{s}^{\mathrm{2}} +\frac{\mathrm{1}}{\lambda−\mathrm{1}}} \\$$$$\frac{{v}^{\mathrm{2}} \left(\lambda−\mathrm{1}\right)}{\mathrm{2}{g}\left({R}+{r}\right)}=−\int{ds}+\frac{\mathrm{1}}{\lambda−\mathrm{1}}\int\frac{{ds}}{{s}^{\mathrm{2}} +\frac{\mathrm{1}}{\lambda−\mathrm{1}}} \\$$$$\left.\frac{{v}^{\mathrm{2}} \left(\lambda−\mathrm{1}\right)}{\mathrm{2}{g}\left({R}+{r}\right)}=−{s}+\frac{\mathrm{1}}{\:\sqrt{\lambda−\mathrm{1}}}\mathrm{tan}^{−\mathrm{1}} \left({s}\sqrt{\lambda−\mathrm{1}}\right)\right) \\$$$$\\$$
Commented by mr W last updated on 10/Feb/21
$${thanks}\:{so}\:{far}\:{sir}! \\$$$${i}\:{agree}\:{with}\:{v}\:\mathrm{sin}\:\theta={V}\:\mathrm{cos}\:\theta,\:{but}\:{it} \\$$$${doesn}'{t}\:{follow}\:{that}\:{a}\mathrm{sin}\:\theta={A}\mathrm{cos}\:\theta. \\$$$${please}\:{recheck}! \\$$
Commented by mr W last updated on 11/Feb/21