Question Number 129660 by ajfour last updated on 17/Jan/21
Commented by ajfour last updated on 17/Jan/21
$${If}\:{a}\:{plank}\:{is}\:{released}\:{in}\:{contact} \\ $$$${with}\:{a}\:{smooth}\:{cylinder}\:{of}\:{radius} \\ $$$${R},\:{at}\:{an}\:{angle}\:\alpha.\:{Find}\:{the}\:{angle} \\ $$$$\theta\:{when}\:{plank}\:{is}\:{about}\:{to}\:{leave} \\ $$$${contact}\:{with}\:{cylinder}. \\ $$
Answered by mr W last updated on 18/Jan/21
Commented by mr W last updated on 19/Jan/21
$${assumed}\:{that}\:{the}\:{contact}\:{point}\:{A}\:{is} \\ $$$${the}\:{end}\:{of}\:{the}\:{rod}\:{when}\:{the}\:{contact} \\ $$$${is}\:{about}\:{to}\:{get}\:{lost},\:{i}.{e}.\:\theta\leqslant\theta_{{m}} =\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{{R}}{{L}}. \\ $$$${let}\:\lambda=\frac{{L}}{{R}} \\ $$$${y}_{{A}} ={R}+{R}\:\mathrm{cos}\:\varphi={L}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\mathrm{cos}\:\varphi=\lambda\:\mathrm{sin}\:\theta−\mathrm{1} \\ $$$$\Rightarrow\mathrm{sin}\:\varphi=\sqrt{\mathrm{1}−\left(\lambda\:\mathrm{sin}\:\theta−\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{\lambda\mathrm{sin}\:\theta\left(\mathrm{2}−\lambda\mathrm{sin}\:\theta\right)} \\ $$$${y}_{{C}} =\frac{{L}}{\mathrm{2}}\:\mathrm{sin}\:\theta={R}×\frac{\lambda}{\mathrm{2}}\:\mathrm{sin}\:\theta \\ $$$${x}_{{C}} ={R}\:\mathrm{sin}\:\varphi+\frac{{L}}{\mathrm{2}}\:\mathrm{cos}\:\theta={R}\left[\sqrt{\lambda\:\mathrm{sin}\:\theta\left(\mathrm{2}−\lambda\:\mathrm{sin}\:\theta\right)}+\frac{\lambda}{\mathrm{2}}\:\mathrm{cos}\:\theta\right] \\ $$$$\omega=\frac{{d}\theta}{{dt}} \\ $$$${v}_{{C},{x}} =\frac{{dx}_{{C}} }{{dt}}=\omega\frac{{dx}_{{C}} }{{d}\theta}=\frac{\omega\lambda{R}}{\mathrm{2}}\left[\frac{\mathrm{2cos}\:\theta\left(\mathrm{1}−\lambda\mathrm{sin}\:\theta\right)}{\:\sqrt{\lambda\:\mathrm{sin}\:\theta\left(\mathrm{2}−\lambda\:\mathrm{sin}\:\theta\right)}}−\mathrm{sin}\:\theta\right] \\ $$$${let}\:\eta=\frac{\mathrm{cos}\:\theta\left(\mathrm{1}−\lambda\mathrm{sin}\:\theta\right)}{\:\sqrt{\lambda\:\mathrm{sin}\:\theta\left(\mathrm{2}−\lambda\:\mathrm{sin}\:\theta\right)}} \\ $$$$\Rightarrow{v}_{{C},{x}} =\frac{\omega\lambda{R}}{\mathrm{2}}\left(\mathrm{2}\eta−\mathrm{sin}\:\theta\right) \\ $$$${v}_{{C},{y}} =\frac{{dy}_{{C}} }{{dt}}=\omega\frac{{dy}_{{C}} }{{d}\theta}=\frac{\omega\lambda{R}}{\mathrm{2}}\mathrm{cos}\:\theta \\ $$$${v}_{{C}} ^{\mathrm{2}} ={v}_{{C},{x}} ^{\mathrm{2}} +{v}_{{C},{y}} ^{\mathrm{2}} =\left(\frac{\omega\lambda{R}}{\mathrm{2}}\right)^{\mathrm{2}} \left\{\left(\mathrm{2}\eta−\mathrm{sin}\:\theta\right)^{\mathrm{2}} +\mathrm{cos}^{\mathrm{2}} \:\theta\right\} \\ $$$$\Rightarrow{v}_{{C}} ^{\mathrm{2}} =\left(\frac{\omega{L}}{\mathrm{2}}\right)^{\mathrm{2}} \left[\mathrm{4}\eta\left(\eta−\mathrm{sin}\:\theta\right)+\mathrm{1}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{mv}_{{C}} ^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{mL}^{\mathrm{2}} }{\mathrm{12}}\right)\omega^{\mathrm{2}} ={mg}\frac{{L}}{\mathrm{2}}\left(\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta\right) \\ $$$${v}_{{C}} ^{\mathrm{2}} +\frac{{L}^{\mathrm{2}} \omega^{\mathrm{2}} }{\mathrm{12}}={gL}\left(\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta\right) \\ $$$$\frac{\omega^{\mathrm{2}} {L}^{\mathrm{2}} }{\mathrm{4}}\left[\mathrm{4}\eta\left(\eta−\mathrm{sin}\:\theta\right)+\mathrm{1}\right]+\frac{{L}^{\mathrm{2}} \omega^{\mathrm{2}} }{\mathrm{12}}={gL}\left(\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta\right) \\ $$$$\omega^{\mathrm{2}} \left[\mathrm{3}\eta\left(\eta−\mathrm{sin}\:\theta\right)+\mathrm{1}\right]=\frac{\mathrm{3}{g}}{{L}}\left(\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\omega=\sqrt{\frac{\mathrm{3}{g}}{{L}}}×\sqrt{\frac{\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta}{\mathrm{3}\eta\left(\eta−\mathrm{sin}\:\theta\right)+\mathrm{1}}} \\ $$$${v}_{{C},{x}} =\frac{\omega{L}}{\mathrm{2}}\left(\mathrm{2}\eta−\mathrm{sin}\:\theta\right) \\ $$$${a}_{{C},{x}} =\frac{{dv}_{{C},{x}} }{{dt}}=\omega\frac{{dv}_{{C},{x}} }{{d}\theta}=\omega\frac{{d}}{{d}\theta}\left\{\frac{\omega{L}}{\mathrm{2}}\left(\mathrm{2}\eta−\mathrm{sin}\:\theta\right)\right\} \\ $$$${N}\:\mathrm{sin}\:\varphi={ma}_{{C},{x}} \\ $$$${N}=\mathrm{0}\:\Rightarrow\:{a}_{{C},{x}} =\mathrm{0} \\ $$$$\Rightarrow\frac{{d}}{{d}\theta}\left\{\sqrt{\frac{\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta}{\mathrm{3}\eta\left(\eta−\mathrm{sin}\:\theta\right)+\mathrm{1}}}\left(\eta−\frac{\mathrm{sin}\:\theta}{\mathrm{2}}\right)\right\}=\mathrm{0} \\ $$$${that}\:{means} \\ $$$$\sqrt{\frac{\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta}{\mathrm{3}\eta\left(\eta−\mathrm{sin}\:\theta\right)+\mathrm{1}}}\left(\eta−\frac{\mathrm{sin}\:\theta}{\mathrm{2}}\right)={extremal} \\ $$$$ \\ $$$${example}: \\ $$$$\lambda=\frac{{L}}{{R}}=\mathrm{3},\:\alpha=\mathrm{60}° \\ $$$$\Rightarrow\theta\approx\mathrm{32}.\mathrm{1111}°\:<\:\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}\approx\mathrm{36}.\mathrm{87}°\checkmark \\ $$
Commented by ajfour last updated on 18/Jan/21
$${Thank}\:{you},\:{Sir},\:{for}\:{the}\:{divine} \\ $$$${light},\:{i}\:{was}\:{in}\:{great}\:{need}\:{of}\:{it}.. \\ $$
Commented by ajfour last updated on 17/Jan/21
$${Thanks}\:{Sir},\:{lets}\:{continue}\: \\ $$$${the}\:{discussion}\:{tomorrow}.. \\ $$
Commented by mr W last updated on 18/Jan/21
Commented by mr W last updated on 18/Jan/21
$${phase}\:{I}:\:\:\theta_{{m}} <\theta<\alpha \\ $$$${phase}\:{II}:\:\:\theta\:\leqslant\:\theta_{{m}} \\ $$
Commented by mr W last updated on 18/Jan/21
Answered by mr W last updated on 18/Jan/21
Commented by mr W last updated on 22/Jan/21
$${phase}\:{I} \\ $$$$\lambda=\frac{{L}}{{R}} \\ $$$$\theta_{{m}} =\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\lambda} \\ $$$$\theta_{{m}} <\theta<\alpha \\ $$$${x}_{{C}} =\frac{{R}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}−\frac{{L}}{\mathrm{2}}\mathrm{cos}\:\theta={R}\left(\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}−\frac{\lambda\mathrm{cos}\:\theta}{\mathrm{2}}\right) \\ $$$${y}_{{C}} =\frac{{L}}{\mathrm{2}}\mathrm{sin}\:\theta={R}\frac{\lambda\:\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$$\omega=\frac{{d}\theta}{{dt}} \\ $$$${v}_{{C},} {x}=\frac{{dx}_{{C}} }{{dt}}=\omega\frac{{dx}_{{C}} }{{d}\theta}=\frac{\omega{R}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}+\lambda\:\mathrm{sin}\:\theta\right) \\ $$$${v}_{{C},{y}} =\omega\frac{{dy}_{{C}} }{{d}\theta}=\frac{\omega{R}}{\mathrm{2}}\lambda\:\mathrm{cos}\:\theta \\ $$$${v}_{{C}} ^{\mathrm{2}} =\frac{\omega^{\mathrm{2}} {R}^{\mathrm{2}} }{\mathrm{4}}\left[\left(−\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}+\lambda\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta\right] \\ $$$${v}_{{C}} ^{\mathrm{2}} =\frac{\omega^{\mathrm{2}} {R}^{\mathrm{2}} }{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}}}−\frac{\mathrm{4}\lambda}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}+\lambda^{\mathrm{2}} \right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{mv}_{{C}} ^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{mL}^{\mathrm{2}} }{\mathrm{12}}\right)\omega^{\mathrm{2}} ={mg}\frac{{L}}{\mathrm{2}}\left(\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta\right) \\ $$$$\frac{\omega^{\mathrm{2}} {R}^{\mathrm{2}} }{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}}}−\frac{\mathrm{4}\lambda}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}+\lambda^{\mathrm{2}} \right)+\left(\frac{{L}^{\mathrm{2}} }{\mathrm{12}}\right)\omega^{\mathrm{2}} ={gL}\left(\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta\right) \\ $$$$\left[\frac{\mathrm{3}}{\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}}}−\frac{\mathrm{12}\lambda}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}+\mathrm{4}\lambda^{\mathrm{2}} \right]\omega^{\mathrm{2}} =\frac{\mathrm{12}{g}}{{L}}\lambda^{\mathrm{2}} \left(\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\omega=\sqrt{\frac{\mathrm{12}{g}}{{L}}}×\sqrt{\frac{\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta}{\frac{\mathrm{3}}{\lambda^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}}}−\frac{\mathrm{12}}{\lambda\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}+\mathrm{4}}} \\ $$$${v}_{{C},{x}} =\frac{\omega{R}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}+\lambda\:\mathrm{sin}\:\theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:={R}\sqrt{\frac{\mathrm{3}{g}}{{L}}}×\sqrt{\frac{\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta}{\frac{\mathrm{3}}{\lambda^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}}}−\frac{\mathrm{12}}{\lambda\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}+\mathrm{4}}}\left(−\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}+\lambda\:\mathrm{sin}\:\theta\right) \\ $$$${a}_{{C},{x}} =\omega\frac{{dv}_{{C},{x}} }{{d}\theta} \\ $$$${a}_{{C},{x}} =\omega{R}\sqrt{\frac{\mathrm{3}{g}}{{L}}}×\:\frac{{d}}{{d}\theta}\left[\sqrt{\frac{\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta}{\frac{\mathrm{3}}{\lambda^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}}}−\frac{\mathrm{12}}{\lambda\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}+\mathrm{4}}}\left(−\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}+\lambda\:\mathrm{sin}\:\theta\right)\right] \\ $$$${N}\:\mathrm{sin}\:\theta={ma}_{{C},{x}} \\ $$$${N}=\mathrm{0}\:\Rightarrow\:{a}_{{C},{x}} =\mathrm{0} \\ $$$$\Rightarrow\frac{{d}}{{d}\theta}\left[\sqrt{\frac{\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta}{\frac{\mathrm{3}}{\lambda^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}}}−\frac{\mathrm{12}}{\lambda\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}+\mathrm{4}}}\left(−\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}+\lambda\:\mathrm{sin}\:\theta\right)\right]=\mathrm{0} \\ $$$${i}.{e}. \\ $$$$\sqrt{\frac{\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta}{\frac{\mathrm{3}}{\lambda^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}}}−\frac{\mathrm{12}}{\lambda\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}+\mathrm{4}}}\left(−\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}+\lambda\:\mathrm{sin}\:\theta\right)={extermal} \\ $$$${with}\:\theta_{{m}} <\theta<\alpha \\ $$$${this}\:{has}\:{normally}\:{no}\:{solution},\:{i}.{e}. \\ $$$${in}\:{phase}\:{I}\:{we}\:{can}'{t}\:{get}\:{the}\:{case}\:{N}=\mathrm{0}. \\ $$
Commented by mr W last updated on 19/Jan/21
Answered by mr W last updated on 19/Jan/21
Commented by mr W last updated on 20/Jan/21
$$\boldsymbol{{Phase}}\:\boldsymbol{{II}}\:−\:\boldsymbol{{an}}\:\boldsymbol{{other}}\:\boldsymbol{{way}} \\ $$$$ \\ $$$${R}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)={L}\:\mathrm{sin}\:\theta \\ $$$$\mathrm{cos}\:\varphi=\frac{{L}\:\mathrm{sin}\:\theta}{{R}}−\mathrm{1}=\lambda\:\mathrm{sin}\:\theta−\mathrm{1} \\ $$$$\mathrm{tan}\:\varphi=\frac{\sqrt{\lambda\:\mathrm{sin}\:\theta\:\left(\mathrm{2}−\lambda\:\mathrm{sin}\:\theta\right)}}{\lambda\:\mathrm{sin}\:\theta−\mathrm{1}} \\ $$$${SB}={R}+\frac{{OB}}{\mathrm{tan}\:\varphi}={R}+\frac{{L}\:\mathrm{cos}\:\theta+{R}\:\mathrm{sin}\:\varphi}{\mathrm{tan}\:\varphi} \\ $$$${SB}=\left(\mathrm{1}+\mathrm{cos}\:\varphi+\frac{\lambda\:\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}\right){R} \\ $$$$\Rightarrow{SB}={L}\left(\mathrm{sin}\:\theta+\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}\right) \\ $$$${SC}^{\mathrm{2}} =\left(\frac{{L}}{\mathrm{2}}\right)^{\mathrm{2}} +{L}^{\mathrm{2}} \left(\mathrm{sin}\:\theta+\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}\right)^{\mathrm{2}} −\mathrm{2}×\frac{{L}}{\mathrm{2}}×{L}\left(\mathrm{sin}\:\theta+\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}\right)×\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−\theta\right) \\ $$$${SC}^{\mathrm{2}} ={L}^{\mathrm{2}} \left[\frac{\mathrm{1}}{\mathrm{4}}+\left(\mathrm{sin}\:\theta+\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}\right)^{\mathrm{2}} −\mathrm{sin}\:\theta\:\left(\mathrm{sin}\:\theta+\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}\right)\right] \\ $$$$\Rightarrow{SC}^{\mathrm{2}} =\left[\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}\left(\mathrm{sin}\:\theta+\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}\right)\right]{L}^{\mathrm{2}} \\ $$$${I}_{{S}} ={I}+{m}×{SC}^{\mathrm{2}} =\frac{{mL}^{\mathrm{2}} }{\mathrm{12}}+\left[\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}\left(\mathrm{sin}\:\theta+\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}\right)\right]{mL}^{\mathrm{2}} \\ $$$$\Rightarrow{I}_{{S}} =\left[\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}\left(\mathrm{sin}\:\theta+\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}\right)\right]{mL}^{\mathrm{2}} \\ $$$$\omega=\frac{{d}\theta}{{dt}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{I}_{{S}} \omega^{\mathrm{2}} ={mg}\frac{{L}}{\mathrm{2}}\left(\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta\right) \\ $$$$\left[\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}\left(\mathrm{sin}\:\theta+\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}\right)\right]{mL}^{\mathrm{2}} \omega^{\mathrm{2}} ={mgL}\left(\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\omega^{\mathrm{2}} =\frac{\mathrm{3}{g}}{{L}}×\frac{\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta}{\mathrm{1}+\frac{\mathrm{3}\:\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}\left(\mathrm{sin}\:\theta+\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}\right)} \\ $$$${let}\:\xi=\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}=\frac{\mathrm{cos}\:\theta\:\left(\lambda\:\mathrm{sin}\:\theta−\mathrm{1}\right)}{\:\sqrt{\lambda\:\mathrm{sin}\:\theta\:\left(\mathrm{2}−\lambda\:\mathrm{sin}\:\theta\right)}}\: \\ $$$$\Rightarrow\omega=\sqrt{\frac{\mathrm{3}{g}}{{L}}}×\sqrt{\frac{\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta}{\mathrm{1}+\mathrm{3}\xi\left(\mathrm{sin}\:\theta+\xi\right)}} \\ $$$${v}_{{C},{x}} =\omega\left({SB}−\frac{{L}}{\mathrm{2}}\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow{v}_{{C},{x}} =\omega\left(\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}+\frac{\mathrm{sin}\:\theta}{\mathrm{2}}\right){L}=\omega\left(\xi+\frac{\mathrm{sin}\:\theta}{\mathrm{2}}\right){L} \\ $$$${a}_{{C},{x}} =\omega\frac{{dv}_{{C},{x}} }{{d}\theta}=\omega\sqrt{\mathrm{3}{gL}}×\frac{{d}}{{d}\theta}\left\{\left(\xi+\frac{\mathrm{sin}\:\theta}{\mathrm{2}}\right)\sqrt{\frac{\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta}{\mathrm{1}+\mathrm{3}\xi\left(\mathrm{sin}\:\theta+\xi\right)}}\right\} \\ $$$${N}\:\mathrm{sin}\:\varphi={ma}_{{C},{x}} ={m}\omega\sqrt{\mathrm{3}{gL}}×\frac{{d}}{{d}\theta}\left\{\left(\xi+\frac{\mathrm{sin}\:\theta}{\mathrm{2}}\right)\sqrt{\frac{\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta}{\mathrm{1}+\mathrm{3}\xi\left(\mathrm{sin}\:\theta+\xi\right)}}\right\} \\ $$$$\frac{{N}}{{mg}}=\frac{\mathrm{3}}{\mathrm{sin}\:\varphi}\sqrt{\frac{\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta}{\mathrm{1}+\mathrm{3}\xi\left(\mathrm{sin}\:\theta+\xi\right)}}×\frac{{d}}{{d}\theta}\left\{\left(\xi+\frac{\mathrm{sin}\:\theta}{\mathrm{2}}\right)\sqrt{\frac{\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta}{\mathrm{1}+\mathrm{3}\xi\left(\mathrm{sin}\:\theta+\xi\right)}}\right\} \\ $$$$\Rightarrow\frac{{N}}{{mg}}=\mathrm{3}\sqrt{\frac{\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta}{\lambda\:\mathrm{sin}\:\theta\:\left(\mathrm{2}−\lambda\:\mathrm{sin}\:\theta\right)\left[\mathrm{1}+\mathrm{3}\xi\left(\mathrm{sin}\:\theta+\xi\right)\right]}}×\frac{{d}}{{d}\theta}\left\{\left(\xi+\frac{\mathrm{sin}\:\theta}{\mathrm{2}}\right)\sqrt{\frac{\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta}{\mathrm{1}+\mathrm{3}\xi\left(\mathrm{sin}\:\theta+\xi\right)}}\right\} \\ $$$$ \\ $$$${for}\:{N}=\mathrm{0}\:\Rightarrow\:{a}_{{C},{x}} =\mathrm{0} \\ $$$$\Rightarrow\frac{{d}}{{d}\theta}\left\{\left(\xi+\frac{\mathrm{sin}\:\theta}{\mathrm{2}}\right)\sqrt{\frac{\mathrm{sin}\:\alpha−\mathrm{sin}\:\theta}{\mathrm{1}+\mathrm{3}\xi\left(\mathrm{sin}\:\theta+\xi\right)}}\right\}=\mathrm{0} \\ $$
Commented by ajfour last updated on 23/Jan/21
$${Reviewing}\:{these}\:{solutions},\:{Sir}. \\ $$$${Thank}\:{you}\:{plentifully}! \\ $$
Commented by mr W last updated on 21/Jan/21
$${best}\:{wish}\:{for}\:{your}\:{health}\:{sir}! \\ $$