Question Number 52825 by ajfour last updated on 13/Jan/19 | ||
Commented by ajfour last updated on 13/Jan/19 | ||
$${Find}\:\theta\:.\:\:\:{Assume}\:{no}\:{friction}\:{and} \\ $$$${a}\:{static}\:{situation}. \\ $$ | ||
Commented by tanmay.chaudhury50@gmail.com last updated on 13/Jan/19 | ||
Commented by ajfour last updated on 14/Jan/19 | ||
Commented by mr W last updated on 14/Jan/19 | ||
$${perfectly}\:{illustrated}!\:{thanks}\:{a}\:{lot}\:{sir}! \\ $$ | ||
Answered by mr W last updated on 14/Jan/19 | ||
$${let}\:\eta=\frac{{m}}{{M}},\:\lambda=\frac{{L}}{{h}} \\ $$$$\alpha={angle}\:{between}\:{string}\:{and}\:{vertical} \\ $$$${a}={distance}\:{from}\:{ring}\:{to}\:{lower}\:{end}\:{of}\:{rod} \\ $$$${N}={contact}\:{force}\:{between}\:{ring}\:{and}\:{rod} \\ $$$$\frac{{MgL}\:\mathrm{cos}\:\theta}{\mathrm{2}}={Na} \\ $$$$\frac{{Mg}}{\mathrm{sin}\:\left(\pi−\theta\right)}=\frac{{mg}}{\mathrm{sin}\:\left(\theta−\alpha\right)}=\frac{{N}}{\mathrm{sin}\:\alpha} \\ $$$$\mathrm{sin}\:\left(\theta−\alpha\right)=\frac{{m}}{{M}}\:\mathrm{sin}\:\theta=\eta\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\theta−\alpha=\mathrm{sin}^{−\mathrm{1}} \left(\eta\:\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\alpha=\theta−\mathrm{sin}^{−\mathrm{1}} \left(\eta\:\mathrm{sin}\:\theta\right) \\ $$$${N}=\frac{{Mg}\:\mathrm{sin}\:\alpha}{\mathrm{sin}\:\theta} \\ $$$${a}=\frac{{h}\:\mathrm{sin}\:\alpha}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}+\theta−\alpha\right)}=\frac{{h}\:\mathrm{sin}\:\alpha}{\mathrm{cos}\:\left(\theta−\alpha\right)}=\frac{{h}\:\mathrm{sin}\:\alpha}{\sqrt{\mathrm{1}−\eta^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}} \\ $$$$\Rightarrow\frac{{MgL}\:\mathrm{cos}\:\theta}{\mathrm{2}}=\frac{{Mg}\:\mathrm{sin}\:\alpha}{\mathrm{sin}\:\theta}×\frac{{h}\:\mathrm{sin}\:\alpha}{\sqrt{\mathrm{1}−\eta^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}} \\ $$$$\Rightarrow\frac{\mathrm{cos}\:\theta}{\mathrm{2}}×\frac{{L}}{{h}}=\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\mathrm{sin}\:\theta\:\sqrt{\mathrm{1}−\eta^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}} \\ $$$$\Rightarrow\frac{\lambda\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\:\sqrt{\mathrm{1}−\eta^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}}{\mathrm{2}}=\mathrm{sin}^{\mathrm{2}} \:\alpha \\ $$$$\Rightarrow\lambda\:\mathrm{sin}\:\mathrm{2}\theta\:\sqrt{\mathrm{1}−\eta^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}=\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\alpha\right) \\ $$$$\Rightarrow\lambda\:\mathrm{sin}\:\mathrm{2}\theta\:\sqrt{\mathrm{1}−\eta^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}=\mathrm{2}\left[\mathrm{1}−\mathrm{cos}\:\:\mathrm{2}\left\{\theta−\mathrm{sin}^{−\mathrm{1}} \left(\eta\:\mathrm{sin}\:\theta\right)\right\}\right] \\ $$$$ \\ $$$${examples}: \\ $$$$\lambda=\frac{{L}}{{h}}=\mathrm{2},\:\eta=\frac{{m}}{{M}}=\mathrm{0}.\mathrm{2}\Rightarrow\theta=\mathrm{52}.\mathrm{7639}° \\ $$$$\lambda=\frac{{L}}{{h}}=\mathrm{2},\:\eta=\frac{{m}}{{M}}=\mathrm{0}\Rightarrow\theta=\mathrm{45}° \\ $$ | ||
Commented by mr W last updated on 13/Jan/19 | ||
Commented by ajfour last updated on 14/Jan/19 | ||
$${Alright}\:{all}-{along}\:{Sir}.\:{Not}\:{so}\:{easy}. \\ $$$${God}\:{bless}\:{you}! \\ $$ | ||