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Question Number 27997 by ajfour last updated on 18/Jan/18

Commented by mrW2 last updated on 18/Jan/18

$$\omega =\sqrt{\frac{3g}{2a\cos \alpha }}$$

Commented by ajfour last updated on 18/Jan/18

$$yessir,pleaseseemysolution$$$$whenipost.Beforehandcani$$$$havesomeworkingstepstothis$$$$answer..$$

Commented by ajfour last updated on 18/Jan/18

$$Whatisthemaximumvalueof$$$$\omega forwhichtheplatecanstayin$$$$verticalplane?$$

Answered by mrW2 last updated on 18/Jan/18

Commented by mrW2 last updated on 18/Jan/18

$$\rho =\frac{M}{ab}$$$$dm=\rho dydx$$$$dF=r{\omega }^{2}dm$$$${dF}_{\mathrm{\perp }}=\cos \theta dF=\rho {\omega }^{2}r\cos \theta dydx$$$$r^2={\left(x\sin \alpha \right)}^2+y^2$$$$r\cos \theta =x\sin \alpha$$$$\Rightarrow {dF}_{\mathrm{\perp }}=\rho {\omega }^2\sin \alpha xdydx$$$$\frac{Mga\sin \alpha }{2}=\int x\cos \alpha {dF}_{\mathrm{\perp }}=\rho {\omega }^2\sin \alpha \cos \alpha {\int }_0^a{\int }_{-\frac{b}{2}}^{\frac{b}{2}}{x}^{2}dydx$$$$\frac{Mga\sin \alpha }{2}=\rho {\omega }^2\sin \alpha \cos \alpha \frac{a^{3}b}{3}$$$$\frac{Mg}{2}=\rho {\omega }^2\cos \alpha \frac{a^{2}b}{3}$$$$\frac{g}{2}={\omega }^2\cos \alpha \frac{a}{3}$$$$\Rightarrow \omega =\sqrt{\frac{3g}{2a\cos \alpha }}$$$$$$$$for\alpha =0$$$$\Rightarrow {\omega }_{max}=\sqrt{\frac{3g}{2a}}$$

Commented by ajfour last updated on 18/Jan/18

Thank you too much Sir. I enjoyed thoroughly following the solution Sir.

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