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Question Number 28349 by ajfour last updated on 24/Jan/18

Commented by ajfour last updated on 24/Jan/18

$$Q.28339\left(answer\right)$$

Answered by ajfour last updated on 24/Jan/18

$$\mathit{\omega }=\frac{\mathit{d}\mathit{\theta }}{\mathit{d}\mathit{t}},$$$${\mathit{y}}_1=\mathit{a}\tan \mathit{\theta },{\mathit{y}}_2=\mathit{a}\sec \mathit{\theta }-\mathit{a}$$$$\frac{{dy}_1}{dt}=\omega a\sec {}^2\theta ,\frac{{dy}_2}{dt}=\omega a\sec \theta \tan \theta$$$${\mathit{m}\mathit{g}\mathit{y}}_1-{\mathit{m}\mathit{g}\mathit{y}}_2=\frac{1}{2}{\mathit{m}}_1{\left(\frac{{\mathit{d}\mathit{y}}_1}{\mathit{d}\mathit{t}}\right)}^2+$$$$\frac{1}{2}{\mathit{m}}_2{\left(\frac{{\mathit{d}\mathit{y}}_2}{\mathit{d}\mathit{t}}\right)}^2$$$$\Rightarrow 2m_{1}ga\tan \theta -2m_{2}ga(\sec \theta -1)$$$$=m_1{\omega }^2{a}^2\sec {}^4\theta +m_2{\omega }^2{a}^2\sec {}^2\theta \tan {}^2\theta$$$$or$$$$2ga(m_1\tan \theta -m_2\sec \theta +m_2)=$$$${\omega }^2{a}^2\sec {}^2\theta (m_1\sec {}^2\theta +m_2\tan {}^2\theta )$$$$\Rightarrow {\mathit{\omega }}^2=\frac{2\mathit{g}}{\mathit{a}}\frac{({\mathit{m}}_1\tan \mathit{\theta }-{\mathit{m}}_2\sec \mathit{\theta }+{\mathit{m}}_2)}{({\mathit{m}}_1\sec {}^2\mathit{\theta }+{\mathit{m}}_2\tan {}^2\mathit{\theta })}\cos {}^2\mathit{\theta }.$$$$\omega varieswith\theta ,$$$$at\theta =0$$$${\omega }^2=0\Rightarrow motionispossible$$$$T\to \mathrm{\infty }\Rightarrow tendencyofirreversible$$$$motionatthestart.$$

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

$$forring:$$$$m_{1}g-T\sin \theta =m_1{a}_1$$$$forblock$$$$T-m_{2}g=m_2{a}_2$$$$forequilibrium{a}_1=a_2=0$$$$\Rightarrow m_{1}g-T\sin \theta =T-m_{2}g=0$$$$atequilibriumT=m_{2}g$$$$\sin \theta =\frac{m_1}{m_2}$$$$equilibriumcannotbepossible$$$$for{m}_1⩾m_2$$$$butinthecase{m}_1=\frac{m_2}{2}$$$$equilibriumat\mathit{\theta }={\sin }^{-1}\left(\frac{m_1}{m_2}\right)=\frac{\pi }{6}.$$

Commented by Tinkutara last updated on 25/Jan/18

Thank you very much Sir! I got the answer.

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