# Question-134064

Question Number 134064 by mr W last updated on 27/Feb/21
Commented by mr W last updated on 27/Feb/21
$${find}\:{the}\:{angle}\:\theta\:{at}\:{which}\:{the}\:{falling} \\$$$${rod}\:{begins}\:{to}\:{slip}\:{on}\:{the}\:{ground}\:{if} \\$$$${the}\:{friction}\:{coefficient}\:{between}\:{rod} \\$$$${and}\:{ground}\:{is}\:\mu. \\$$
Answered by mr W last updated on 27/Feb/21
Commented by mr W last updated on 28/Feb/21
$${C}\left({x}_{{C}} ,{y}_{{C}} \right) \\$$$${x}_{{C}} =\frac{{L}\mathrm{sin}\:\theta}{\mathrm{2}} \\$$$${y}_{{C}} =\frac{{L}\mathrm{cos}\:\theta}{\mathrm{2}} \\$$$${v}_{{C},{x}} =\frac{{L}\mathrm{cos}\:\theta\:\omega}{\mathrm{2}} \\$$$${a}_{{C},{x}} =\frac{{L}}{\mathrm{2}}\left(\alpha\mathrm{cos}\:\theta−\omega^{\mathrm{2}} \mathrm{sin}\:\theta\right) \\$$$${v}_{{C},{y}} =−\frac{{dy}_{{C}} }{{dt}}=\frac{{L}\mathrm{sin}\:\theta}{\mathrm{2}}\omega \\$$$${a}_{{C},{y}} =\frac{{L}}{\mathrm{2}}\left(\alpha\:\mathrm{sin}\:\theta+\omega^{\mathrm{2}} \mathrm{cos}\:\theta\right) \\$$$$\frac{{mL}^{\mathrm{2}} }{\mathrm{3}}\alpha={mg}\frac{{L}\mathrm{sin}\:\theta}{\mathrm{2}} \\$$$$\alpha=\frac{\mathrm{3}{g}}{\mathrm{2}{L}}\mathrm{sin}\:\theta \\$$$$\omega\frac{{d}\omega}{{d}\theta}=\frac{\mathrm{3}{g}}{\mathrm{2}{L}}\mathrm{sin}\:\theta \\$$$$\int_{\mathrm{0}} ^{\omega} \omega{d}\omega=\frac{\mathrm{3}{g}}{\mathrm{2}{L}}\int_{\mathrm{0}} ^{\theta} \mathrm{sin}\:\theta{d}\theta \\$$$$\frac{\omega^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{3}{g}}{\mathrm{2}{L}}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\$$$$\omega^{\mathrm{2}} =\frac{\mathrm{3}{g}}{{L}}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\$$$${mg}−{N}={ma}_{{C},{y}} ={m}\frac{{L}}{\mathrm{2}}\left(\alpha\:\mathrm{sin}\:\theta+\omega^{\mathrm{2}} \mathrm{cos}\:\theta\right) \\$$$$\Rightarrow{N}=\frac{\mathrm{9}{mg}}{\mathrm{4}}\left(\mathrm{cos}\:\theta−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\$$$${F}={ma}_{{C},{x}} ={m}\frac{{L}}{\mathrm{2}}\left(\alpha\mathrm{cos}\:\theta−\omega^{\mathrm{2}} \mathrm{sin}\:\theta\right) \\$$$$\Rightarrow{F}=\frac{\mathrm{9}{mg}}{\mathrm{4}}\left(\mathrm{cos}\:\theta−\frac{\mathrm{2}}{\mathrm{3}}\right)\mathrm{sin}\:\theta \\$$$$\\$$$${slipping}\:{occurs}\:{when}\:\mid{F}\mid\geqslant\mu{N} \\$$$$\pm\frac{\mathrm{9}{mg}}{\mathrm{4}}\left(\mathrm{cos}\:\theta−\frac{\mathrm{2}}{\mathrm{3}}\right)\mathrm{sin}\:\theta=\mu\frac{\mathrm{9}{mg}}{\mathrm{4}}\left(\mathrm{cos}\:\theta−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \\$$$$\Rightarrow\pm\left(\mathrm{cos}\:\theta−\frac{\mathrm{2}}{\mathrm{3}}\right)\mathrm{sin}\:\theta=\mu\left(\mathrm{cos}\:\theta−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \\$$
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