Question Number 28349 by ajfour last updated on 24/Jan/18
Commented by ajfour last updated on 24/Jan/18
$${Q}.\:\mathrm{28339}\:\:\:\left({answer}\right) \\ $$
Answered by ajfour last updated on 24/Jan/18
$$\:\:\boldsymbol{\omega}=\frac{\boldsymbol{{d}\theta}}{\boldsymbol{{dt}}}\:, \\ $$$$\:\boldsymbol{{y}}_{\mathrm{1}} =\boldsymbol{{a}}\mathrm{tan}\:\boldsymbol{\theta}\:\:,\:\:\boldsymbol{{y}}_{\mathrm{2}} =\boldsymbol{{a}}\mathrm{sec}\:\boldsymbol{\theta}−\boldsymbol{{a}} \\ $$$$\frac{{dy}_{\mathrm{1}} }{{dt}}=\omega{a}\mathrm{sec}\:^{\mathrm{2}} \theta\:\:,\:\frac{{dy}_{\mathrm{2}} }{{dt}}=\omega{a}\mathrm{sec}\:\theta\mathrm{tan}\:\theta \\ $$$$\boldsymbol{{mgy}}_{\mathrm{1}} −\boldsymbol{{mgy}}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{m}}_{\mathrm{1}} \left(\frac{\boldsymbol{{dy}}_{\mathrm{1}} }{\boldsymbol{{dt}}}\right)^{\mathrm{2}} + \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{m}}_{\mathrm{2}} \left(\frac{\boldsymbol{{dy}}_{\mathrm{2}} }{\boldsymbol{{dt}}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{2}{m}_{\mathrm{1}} {ga}\mathrm{tan}\:\theta−\mathrm{2}{m}_{\mathrm{2}} {ga}\left(\mathrm{sec}\:\theta−\mathrm{1}\right) \\ $$$$\:\:={m}_{\mathrm{1}} \omega^{\mathrm{2}} {a}^{\mathrm{2}} \mathrm{sec}\:^{\mathrm{4}} \theta+{m}_{\mathrm{2}} \omega^{\mathrm{2}} {a}^{\mathrm{2}} \mathrm{sec}\:^{\mathrm{2}} \theta\mathrm{tan}\:^{\mathrm{2}} \theta \\ $$$${or} \\ $$$$\mathrm{2}{ga}\left({m}_{\mathrm{1}} \mathrm{tan}\:\theta−{m}_{\mathrm{2}} \mathrm{sec}\:\theta+{m}_{\mathrm{2}} \right)= \\ $$$$\:\:\:\:\:\:\:\omega^{\mathrm{2}} {a}^{\mathrm{2}} \mathrm{sec}\:^{\mathrm{2}} \theta\left({m}_{\mathrm{1}} \mathrm{sec}\:^{\mathrm{2}} \theta+{m}_{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \theta\right) \\ $$$$\Rightarrow\:\:\boldsymbol{\omega}^{\mathrm{2}} =\frac{\mathrm{2}\boldsymbol{{g}}}{\boldsymbol{{a}}}\:\frac{\left(\boldsymbol{{m}}_{\mathrm{1}} \mathrm{tan}\:\boldsymbol{\theta}−\boldsymbol{{m}}_{\mathrm{2}} \mathrm{sec}\:\boldsymbol{\theta}+\boldsymbol{{m}}_{\mathrm{2}} \right)}{\left(\boldsymbol{{m}}_{\mathrm{1}} \mathrm{sec}\:^{\mathrm{2}} \boldsymbol{\theta}+\boldsymbol{{m}}_{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \boldsymbol{\theta}\right)}\:\mathrm{cos}\:^{\mathrm{2}} \boldsymbol{\theta}\:. \\ $$$$\omega\:{varies}\:{with}\:\theta\:, \\ $$$$\:{at}\:\theta=\mathrm{0} \\ $$$$\:\:\:\omega^{\mathrm{2}} =\mathrm{0}\:\:\Rightarrow\:\:\:{motion}\:{is}\:{possible} \\ $$$$\:{T}\rightarrow\infty\:\Rightarrow\:{tendency}\:{of}\:\:{irreversible} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{motion}\:{at}\:{the}\:{start}. \\ $$
Commented by Tinkutara last updated on 24/Jan/18
This �� is answer in book. Please explain this.
Commented by Tinkutara last updated on 24/Jan/18
Commented by ajfour last updated on 24/Jan/18
$$\:{for}\:{ring}: \\ $$$${m}_{\mathrm{1}} {g}−{T}\mathrm{sin}\:\theta={m}_{\mathrm{1}} {a}_{\mathrm{1}} \\ $$$${for}\:{block} \\ $$$${T}−{m}_{\mathrm{2}} {g}={m}_{\mathrm{2}} {a}_{\mathrm{2}} \\ $$$${for}\:{equilibrium}\:{a}_{\mathrm{1}} ={a}_{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:{m}_{\mathrm{1}} {g}−{T}\mathrm{sin}\:\theta={T}−{m}_{\mathrm{2}} {g}=\mathrm{0} \\ $$$${at}\:{equilibrium}\:\:\:\:{T}=\:\:{m}_{\mathrm{2}} {g} \\ $$$$\mathrm{sin}\:\theta=\frac{{m}_{\mathrm{1}} }{{m}_{\mathrm{2}} }\:\:\: \\ $$$${equilibrium}\:{cannot}\:{be}\:{possible} \\ $$$${for}\:\:{m}_{\mathrm{1}} \geqslant\:{m}_{\mathrm{2}} \\ $$$${but}\:{in}\:{the}\:{case}\:\:{m}_{\mathrm{1}} =\frac{{m}_{\mathrm{2}} }{\mathrm{2}} \\ $$$${equilibrium}\:{at}\:\boldsymbol{\theta}=\mathrm{sin}^{−\mathrm{1}} \left(\frac{{m}_{\mathrm{1}} }{{m}_{\mathrm{2}} }\right)=\frac{\pi}{\mathrm{6}}\:. \\ $$
Commented by Tinkutara last updated on 25/Jan/18
Thank you very much Sir! I got the answer.