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Question Number 72108 by ajfour last updated on 24/Oct/19

Answered by Tanmay chaudhury last updated on 24/Oct/19

$$Torqueduetomg\mathrm{\Gamma }=(\frac{l}{2}-a)mgsin(\frac{\pi }{2}+\alpha )$$$$(\frac{l}{2}-a)mgcos\alpha =I.\beta$$$$calculationofI=\frac{{ml}^2}{12}+m{(\frac{l}{2}-a)}^2$$$$\theta =0.t+\frac{1}{2}\beta t^2\to nowwhen\theta =\alpha t=t_0\left(say\right)$$$$t_0=\sqrt{\frac{2\alpha }{\beta }}\mathit{P}lzcheckisitcorrect...$$$$$$

Commented by Tanmay chaudhury last updated on 24/Oct/19

$$okishallgodeepinto[theproblem...$$

Commented by ajfour last updated on 24/Oct/19

$$Torque{isn}^{\prime }tjustequaltoI\left(\frac{d^2\theta }{{dt}^2}\right),$$$$TanmaySir.$$

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