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Question-28349

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Question Number 28349 by ajfour last updated on 24/Jan/18
Commented by ajfour last updated on 24/Jan/18
Q. 28339 (answer)
$${Q}.\:\mathrm{28339}\:\:\:\left({answer}\right) \\ $$
Answered by ajfour last updated on 24/Jan/18
𝛚=(d𝛉/dt) , y_1 =atan 𝛉 , y_2 =asec 𝛉−a (dy_1 /dt)=ωasec^2 θ , (dy_2 /dt)=ωasec θtan θ mgy_1 −mgy_2 =(1/2)m_1 ((dy_1 /dt))^2 + (1/2)m_2 ((dy_2 /dt))^2 ⇒ 2m_1 gatan θ−2m_2 ga(sec θ−1) =m_1 ω^2 a^2 sec^4 θ+m_2 ω^2 a^2 sec^2 θtan^2 θ or 2ga(m_1 tan θ−m_2 sec θ+m_2 )= ω^2 a^2 sec^2 θ(m_1 sec^2 θ+m_2 tan^2 θ) ⇒ 𝛚^2 =((2g)/a) (((m_1 tan 𝛉−m_2 sec 𝛉+m_2 ))/((m_1 sec^2 𝛉+m_2 tan^2 𝛉))) cos^2 𝛉 . ω varies with θ , at θ=0 ω^2 =0 ⇒ motion is possible T→∞ ⇒ tendency of irreversible motion at the start.
$$\:\:\boldsymbol{\omega}=\frac{\boldsymbol{{d}\theta}}{\boldsymbol{{dt}}}\:, \\ $$$$\:\boldsymbol{{y}}_{\mathrm{1}} =\boldsymbol{{a}}\mathrm{tan}\:\boldsymbol{\theta}\:\:,\:\:\boldsymbol{{y}}_{\mathrm{2}} =\boldsymbol{{a}}\mathrm{sec}\:\boldsymbol{\theta}−\boldsymbol{{a}} \\ $$$$\frac{{dy}_{\mathrm{1}} }{{dt}}=\omega{a}\mathrm{sec}\:^{\mathrm{2}} \theta\:\:,\:\frac{{dy}_{\mathrm{2}} }{{dt}}=\omega{a}\mathrm{sec}\:\theta\mathrm{tan}\:\theta \\ $$$$\boldsymbol{{mgy}}_{\mathrm{1}} −\boldsymbol{{mgy}}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{m}}_{\mathrm{1}} \left(\frac{\boldsymbol{{dy}}_{\mathrm{1}} }{\boldsymbol{{dt}}}\right)^{\mathrm{2}} + \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{m}}_{\mathrm{2}} \left(\frac{\boldsymbol{{dy}}_{\mathrm{2}} }{\boldsymbol{{dt}}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{2}{m}_{\mathrm{1}} {ga}\mathrm{tan}\:\theta−\mathrm{2}{m}_{\mathrm{2}} {ga}\left(\mathrm{sec}\:\theta−\mathrm{1}\right) \\ $$$$\:\:={m}_{\mathrm{1}} \omega^{\mathrm{2}} {a}^{\mathrm{2}} \mathrm{sec}\:^{\mathrm{4}} \theta+{m}_{\mathrm{2}} \omega^{\mathrm{2}} {a}^{\mathrm{2}} \mathrm{sec}\:^{\mathrm{2}} \theta\mathrm{tan}\:^{\mathrm{2}} \theta \\ $$$${or} \\ $$$$\mathrm{2}{ga}\left({m}_{\mathrm{1}} \mathrm{tan}\:\theta−{m}_{\mathrm{2}} \mathrm{sec}\:\theta+{m}_{\mathrm{2}} \right)= \\ $$$$\:\:\:\:\:\:\:\omega^{\mathrm{2}} {a}^{\mathrm{2}} \mathrm{sec}\:^{\mathrm{2}} \theta\left({m}_{\mathrm{1}} \mathrm{sec}\:^{\mathrm{2}} \theta+{m}_{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \theta\right) \\ $$$$\Rightarrow\:\:\boldsymbol{\omega}^{\mathrm{2}} =\frac{\mathrm{2}\boldsymbol{{g}}}{\boldsymbol{{a}}}\:\frac{\left(\boldsymbol{{m}}_{\mathrm{1}} \mathrm{tan}\:\boldsymbol{\theta}−\boldsymbol{{m}}_{\mathrm{2}} \mathrm{sec}\:\boldsymbol{\theta}+\boldsymbol{{m}}_{\mathrm{2}} \right)}{\left(\boldsymbol{{m}}_{\mathrm{1}} \mathrm{sec}\:^{\mathrm{2}} \boldsymbol{\theta}+\boldsymbol{{m}}_{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \boldsymbol{\theta}\right)}\:\mathrm{cos}\:^{\mathrm{2}} \boldsymbol{\theta}\:. \\ $$$$\omega\:{varies}\:{with}\:\theta\:, \\ $$$$\:{at}\:\theta=\mathrm{0} \\ $$$$\:\:\:\omega^{\mathrm{2}} =\mathrm{0}\:\:\Rightarrow\:\:\:{motion}\:{is}\:{possible} \\ $$$$\:{T}\rightarrow\infty\:\Rightarrow\:{tendency}\:{of}\:\:{irreversible} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{motion}\:{at}\:{the}\:{start}. \\ $$
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_1 g−Tsin θ=m_1 a_1 for block T−m_2 g=m_2 a_2 for equilibrium a_1 =a_2 =0 ⇒ m_1 g−Tsin θ=T−m_2 g=0 at equilibrium T= m_2 g sin θ=(m_1 /m_2 ) equilibrium cannot be possible for m_1 ≥ m_2 but in the case m_1 =(m_2 /2) equilibrium at 𝛉=sin^(−1) ((m_1 /m_2 ))=(π/6) .
$$\:{for}\:{ring}: \\ $$$${m}_{\mathrm{1}} {g}−{T}\mathrm{sin}\:\theta={m}_{\mathrm{1}} {a}_{\mathrm{1}} \\ $$$${for}\:{block} \\ $$$${T}−{m}_{\mathrm{2}} {g}={m}_{\mathrm{2}} {a}_{\mathrm{2}} \\ $$$${for}\:{equilibrium}\:{a}_{\mathrm{1}} ={a}_{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:{m}_{\mathrm{1}} {g}−{T}\mathrm{sin}\:\theta={T}−{m}_{\mathrm{2}} {g}=\mathrm{0} \\ $$$${at}\:{equilibrium}\:\:\:\:{T}=\:\:{m}_{\mathrm{2}} {g} \\ $$$$\mathrm{sin}\:\theta=\frac{{m}_{\mathrm{1}} }{{m}_{\mathrm{2}} }\:\:\: \\ $$$${equilibrium}\:{cannot}\:{be}\:{possible} \\ $$$${for}\:\:{m}_{\mathrm{1}} \geqslant\:{m}_{\mathrm{2}} \\ $$$${but}\:{in}\:{the}\:{case}\:\:{m}_{\mathrm{1}} =\frac{{m}_{\mathrm{2}} }{\mathrm{2}} \\ $$$${equilibrium}\:{at}\:\boldsymbol{\theta}=\mathrm{sin}^{−\mathrm{1}} \left(\frac{{m}_{\mathrm{1}} }{{m}_{\mathrm{2}} }\right)=\frac{\pi}{\mathrm{6}}\:. \\ $$
Commented by Tinkutara last updated on 25/Jan/18
Thank you very much Sir! I got the answer.

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