Question Number 73297 by mr W last updated on 10/Nov/19
Commented by mr W last updated on 10/Nov/19
$${a}\:{block}\:{of}\:{mass}\:{m}\:{is}\:{released}\:{at}\:{the} \\ $$$${top}\:{of}\:{a}\:{hemisphere}. \\ $$$${find}\:{the}\:{time}\:{for}\:{the}\:{block}\:{to}\:{hit}\:{the} \\ $$$${ground}\:{and}\:{the}\:{position}\:{of}\:{the}\:{hit}. \\ $$$${the}\:{hemisphere}\:{is}\:{fixed}\:{on}\:{the}\:{ground} \\ $$$${and}\:{there}\:{is}\:{no}\:{friction}. \\ $$
Answered by ajfour last updated on 10/Nov/19
$${let}\:{angle}\:{descended}\:{until}\:{in}\: \\ $$$${contact}\:{be}\:\theta. \\ $$$${mg}\mathrm{cos}\:\theta=\frac{{mv}^{\mathrm{2}} }{{R}} \\ $$$$\frac{{mv}^{\mathrm{2}} }{\mathrm{2}}={mgR}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$$\Rightarrow\:\:{mgR}\mathrm{cos}\:\theta=\mathrm{2}{mgR}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$$\Rightarrow\:\:\theta=\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{2}/\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:{v}=\sqrt{\frac{\mathrm{2}{gR}}{\mathrm{3}}} \\ $$$$\:\:\left({v}\mathrm{cos}\:\theta\right){t}={d}−{R}\mathrm{sin}\:\theta \\ $$$$\Rightarrow\:\:{d}=\frac{\sqrt{\mathrm{5}}{R}}{\mathrm{3}}+\frac{\mathrm{2}{vt}}{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:…..\left({i}\right) \\ $$$$\:\:{R}\mathrm{cos}\:\theta=\left({v}\mathrm{sin}\:\theta\right){t}+\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\:\:\frac{\mathrm{2}{R}}{\mathrm{3}}=\frac{\sqrt{\mathrm{5}}{v}}{\mathrm{3}}{t}+\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{3}{gt}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{5}}{vt}−\mathrm{4}{R}=\mathrm{0} \\ $$$$\:\:{t}=\frac{−\mathrm{2}\sqrt{\mathrm{5}}{v}+\sqrt{\mathrm{20}{v}^{\mathrm{2}} +\mathrm{48}{Rg}}}{\mathrm{6}{g}} \\ $$$$\:\:\:\:=\frac{−\mathrm{2}\sqrt{\mathrm{5}}\sqrt{\frac{\mathrm{2}{gR}}{\mathrm{3}}}+\sqrt{\left(\frac{\mathrm{40}}{\mathrm{3}}+\mathrm{48}\right){Rg}}}{\mathrm{6}{g}} \\ $$$$\:\:\:{t}\:=\sqrt{\frac{{R}}{{g}}}×\frac{\left(\sqrt{\mathrm{46}}−\sqrt{\mathrm{10}}\right)}{\mathrm{3}\sqrt{\mathrm{3}}}\:\:\:\:\:\:….\left({ii}\right) \\ $$$${Now}\:{from}\:\left({i}\right) \\ $$$${d}=\frac{\sqrt{\mathrm{5}}{R}}{\mathrm{3}}+\frac{\mathrm{2}{vt}}{\mathrm{3}} \\ $$$$\underset{−} {{time}\:{for}\:{circular}\:{arc}\:{path}} \\ $$$$\:\:\:{mg}\mathrm{sin}\:\phi=\frac{{mdv}}{{dt}} \\ $$$$\:\:\:{v}=\sqrt{\mathrm{2}{gR}\left(\mathrm{1}−\mathrm{cos}\:\phi\right)}=\mathrm{2}\sqrt{{gR}}\mathrm{sin}\:\frac{\phi}{\mathrm{2}} \\ $$$$\:\:{dv}=\sqrt{{gR}}\mathrm{cos}\:\frac{\phi}{\mathrm{2}}{d}\phi \\ $$$$\Rightarrow\:\:\int_{\mathrm{0}} ^{\:{T}} {dt}=\sqrt{\frac{{R}}{{g}}}\int_{\mathrm{0}} ^{\:\theta} \frac{\mathrm{cos}\:\left(\phi/\mathrm{2}\right)}{\mathrm{2sin}\:\frac{\phi}{\mathrm{2}}\mathrm{cos}\:\frac{\phi}{\mathrm{2}}}{d}\phi \\ $$$$\:\:\:{T}\:=\:\sqrt{\frac{{R}}{{g}}}\mathrm{ln}\:\mid\frac{\mathrm{1}}{\mathrm{sin}\:\frac{\phi}{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\phi}{\mathrm{2}}}\mid_{\mathrm{0}} ^{\theta} \\ $$$$\:\:\:\:{T}\:=\sqrt{\frac{{R}}{{g}}}\left\{\mathrm{ln}\:\left(\frac{\mathrm{1}}{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}\right)\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:−\underset{\phi\rightarrow\mathrm{0}} {\mathrm{lim}ln}\:\left(\frac{\mathrm{2sin}\:^{\mathrm{2}} \frac{\phi}{\mathrm{4}}}{\mathrm{2sin}\:\frac{\phi}{\mathrm{4}}\mathrm{cos}\:\frac{\phi}{\mathrm{4}}}\right)\right\} \\ $$$${second}\:{term}\rightarrow−\infty \\ $$$${T}\:{not}\:{defined}. \\ $$$$\:\:\mathrm{tan}\:\theta=\frac{\mathrm{2tan}\:\frac{\theta}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}\:\:\:\:{let}\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}={z} \\ $$$$\Rightarrow\:\:\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}=\frac{\mathrm{2}{z}}{\mathrm{1}−{z}^{\mathrm{2}} }\:\:\:\Rightarrow\:\:{z}^{\mathrm{2}} +\frac{\mathrm{4}}{\:\sqrt{\mathrm{5}}}{z}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:{z}=\frac{−\frac{\mathrm{4}}{\:\sqrt{\mathrm{5}}}+\sqrt{\frac{\mathrm{16}}{\mathrm{5}}+\mathrm{4}}}{\mathrm{2}} \\ $$$$\:\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$${d}=\frac{\sqrt{\mathrm{5}}{R}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\frac{\mathrm{2}{gR}}{\mathrm{3}}}\left[\sqrt{\frac{{R}}{{g}}}×\frac{\left(\sqrt{\mathrm{46}}−\sqrt{\mathrm{10}}\right)}{\mathrm{3}\sqrt{\mathrm{3}}}\right] \\ $$$$\Rightarrow \\ $$$$\:{d}=\frac{{R}}{\mathrm{3}}\left\{\sqrt{\mathrm{5}}+\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{9}}\left(\sqrt{\mathrm{46}}−\sqrt{\mathrm{10}}\right)\right\} \\ $$$$\:\:=\left(\frac{\mathrm{4}\sqrt{\mathrm{23}}+\mathrm{5}\sqrt{\mathrm{5}}}{\mathrm{27}}\right){R}\:. \\ $$$${t}_{{total}} ={T}+\sqrt{\frac{{R}}{{g}}}\left\{\frac{\left(\sqrt{\mathrm{46}}−\sqrt{\mathrm{10}}\right)}{\mathrm{3}\sqrt{\mathrm{3}}}\right\} \\ $$$${T}\:{is}\:{not}\:{defined}. \\ $$
Commented by mr W last updated on 10/Nov/19
$${thanks}\:{sir}! \\ $$$${time}\:{along}\:{hemisphere}\:{is}\:{not}\:{to} \\ $$$${determine},\:{it}'{s}\:+\infty. \\ $$
Answered by mr W last updated on 10/Nov/19
Commented by mr W last updated on 10/Nov/19
$${at}\:\theta: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} ={mgR}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$${v}=\sqrt{\mathrm{2}{gR}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}=\mathrm{2}\sqrt{{gR}}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}} \\ $$$${v}={R}\frac{{d}\theta}{{dt}} \\ $$$$\frac{{d}\theta}{{dt}}=\mathrm{2}\sqrt{\frac{{g}}{{R}}}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\theta} \frac{{d}\left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}=\sqrt{\frac{{g}}{{R}}}\int_{\mathrm{0}} ^{{t}_{\mathrm{1}} } {dt} \\ $$$$−\int_{\mathrm{0}} ^{\theta} \frac{{d}\left(\mathrm{cos}\:\frac{\theta}{\mathrm{2}}\right)}{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}={t}_{\mathrm{1}} \sqrt{\frac{{g}}{{R}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\:\frac{\mathrm{1}−\mathrm{cos}\:\frac{\theta}{\mathrm{2}}}{\mathrm{1}+\mathrm{cos}\:\frac{\theta}{\mathrm{2}}}\right]_{\mathrm{0}} ^{\theta} ={t}_{\mathrm{1}} \sqrt{\frac{{g}}{{R}}} \\ $$$${let}\:{F}\left(\theta\right)=\mathrm{ln}\:\frac{\mathrm{1}−\mathrm{cos}\:\frac{\theta}{\mathrm{2}}}{\mathrm{1}+\mathrm{cos}\:\frac{\theta}{\mathrm{2}}} \\ $$$$\Rightarrow{t}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{{R}}{{g}}}\:\left[{F}\left(\theta\right)−\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}{F}\left(\theta\right)\right] \\ $$$${since}\:\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}{F}\left(\theta\right)=−\infty,\:{t}_{\mathrm{1}} \:{can}\:{not}\:{be} \\ $$$${determined},\:{t}_{\mathrm{1}} \rightarrow+\infty. \\ $$$$ \\ $$$${at}\:\theta=\varphi: \\ $$$${N}=\mathrm{0} \\ $$$${mg}\:\mathrm{cos}\:\varphi={m}\frac{{v}^{\mathrm{2}} }{{R}}=\mathrm{2}{mg}\left(\mathrm{1}−\mathrm{cos}\:\varphi\right) \\ $$$$\mathrm{cos}\:\varphi=\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\varphi\right) \\ $$$$\Rightarrow\mathrm{cos}\:\varphi=\frac{\mathrm{2}}{\mathrm{3}}=\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\varphi}{\mathrm{2}}−\mathrm{1} \\ $$$$\Rightarrow\mathrm{cos}\:\frac{\varphi}{\mathrm{2}}=\sqrt{\frac{\mathrm{5}}{\mathrm{6}}},\:\mathrm{sin}\:\frac{\varphi}{\mathrm{2}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}} \\ $$$$\Rightarrow\mathrm{sin}\:\varphi=\frac{\sqrt{\mathrm{5}}}{\mathrm{3}} \\ $$$${v}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{6}}}\sqrt{{gR}} \\ $$$${R}\:\mathrm{cos}\:\varphi={v}\:\mathrm{sin}\:\varphi\:{t}_{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{gt}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\mathrm{0}=−\frac{\mathrm{4}}{\mathrm{3}}+\frac{\mathrm{2}\sqrt{\mathrm{30}}}{\mathrm{9}}\:\left(\sqrt{\frac{{g}}{{R}}}{t}_{\mathrm{2}} \right)+\left(\sqrt{\frac{{g}}{{R}}}{t}_{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{\frac{{g}}{{R}}}{t}_{\mathrm{2}} =\frac{\sqrt{\mathrm{138}}−\sqrt{\mathrm{30}}}{\mathrm{9}} \\ $$$$\Rightarrow{t}_{\mathrm{2}} =\frac{\sqrt{\mathrm{138}}−\sqrt{\mathrm{30}}}{\mathrm{9}}\sqrt{\frac{{R}}{{g}}} \\ $$$$ \\ $$$${d}={R}\:\mathrm{sin}\:\varphi+{v}\:\mathrm{cos}\:\varphi\:{t}_{\mathrm{2}} \\ $$$${d}={R}\:\frac{\sqrt{\mathrm{5}}}{\mathrm{3}}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{6}}}×\frac{\mathrm{2}}{\mathrm{3}}×\frac{\left(\sqrt{\mathrm{138}}−\sqrt{\mathrm{30}}\right){R}}{\mathrm{9}} \\ $$$${d}=\frac{\left(\mathrm{4}\sqrt{\mathrm{23}}+\mathrm{5}\sqrt{\mathrm{5}}\right){R}}{\mathrm{27}}\approx\mathrm{1}.\mathrm{124}{R} \\ $$
Commented by ajfour last updated on 11/Nov/19
$${thanks}\:{for}\:{solving}\:{and}\:{confirming}, \\ $$$${mrW}\:{Sir}. \\ $$