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Question Number 134064 by mr W last updated on 27/Feb/21

Commented by mr W last updated on 27/Feb/21

$$findtheangle\theta atwhichthefalling$$$$rodbeginstosliponthegroundif$$$$thefrictioncoefficientbetweenrod$$$$andgroundis\mu .$$

Answered by mr W last updated on 27/Feb/21

Commented by mr W last updated on 28/Feb/21

$$C(x_C,y_C)$$$$x_C=\frac{L\sin \theta }{2}$$$$y_C=\frac{L\cos \theta }{2}$$$$v_{C,x}=\frac{L\cos \theta \omega }{2}$$$$a_{C,x}=\frac{L}{2}(\alpha \cos \theta -{\omega }^2\sin \theta )$$$$v_{C,y}=-\frac{{dy}_C}{dt}=\frac{L\sin \theta }{2}\omega$$$$a_{C,y}=\frac{L}{2}(\alpha \sin \theta +{\omega }^2\cos \theta )$$$$\frac{{mL}^2}{3}\alpha =mg\frac{L\sin \theta }{2}$$$$\alpha =\frac{3g}{2L}\sin \theta$$$$\omega \frac{d\omega }{d\theta }=\frac{3g}{2L}\sin \theta$$$${\int }_0^{\omega }\omega d\omega =\frac{3g}{2L}{\int }_0^{\theta }\sin \theta d\theta$$$$\frac{{\omega }^2}{2}=\frac{3g}{2L}(1-\cos \theta )$$$${\omega }^2=\frac{3g}{L}(1-\cos \theta )$$$$mg-N={ma}_{C,y}=m\frac{L}{2}(\alpha \sin \theta +{\omega }^2\cos \theta )$$$$\Rightarrow N=\frac{9mg}{4}{(\cos \theta -\frac{1}{3})}^2⩾0$$$$F={ma}_{C,x}=m\frac{L}{2}(\alpha \cos \theta -{\omega }^2\sin \theta )$$$$\Rightarrow F=\frac{9mg}{4}(\cos \theta -\frac{2}{3})\sin \theta$$$$$$$$slippingoccurswhen\mid F\mid ⩾\mu N$$$$\pm \frac{9mg}{4}(\cos \theta -\frac{2}{3})\sin \theta =\mu \frac{9mg}{4}{(\cos \theta -\frac{1}{3})}^2$$$$\Rightarrow \pm (\cos \theta -\frac{2}{3})\sin \theta =\mu {(\cos \theta -\frac{1}{3})}^2$$

Commented by mr W last updated on 01/Mar/21

Commented by mr W last updated on 01/Mar/21

Commented by mr W last updated on 01/Mar/21

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