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Question-134064

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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
find the angle θ at which the falling rod begins to slip on the ground if the friction coefficient between rod and ground is μ.
$${find}\:{the}\:{angle}\:\theta\:{at}\:{which}\:{the}\:{falling} \\ $$$${rod}\:{begins}\:{to}\:{slip}\:{on}\:{the}\:{ground}\:{if} \\ $$$${the}\:{friction}\:{coefficient}\:{between}\:{rod} \\ $$$${and}\:{ground}\:{is}\:\mu. \\ $$
Answered by mr W last updated on 27/Feb/21
Commented by mr W last updated on 28/Feb/21
C(x_C ,y_C ) x_C =((Lsin θ)/2) y_C =((Lcos θ)/2) v_(C,x) =((Lcos θ ω)/2) a_(C,x) =(L/2)(αcos θ−ω^2 sin θ) v_(C,y) =−(dy_C /dt)=((Lsin θ)/2)ω a_(C,y) =(L/2)(α sin θ+ω^2 cos θ) ((mL^2 )/3)α=mg((Lsin θ)/2) α=((3g)/(2L))sin θ ω(dω/dθ)=((3g)/(2L))sin θ ∫_0 ^ω ωdω=((3g)/(2L))∫_0 ^θ sin θdθ (ω^2 /2)=((3g)/(2L))(1−cos θ) ω^2 =((3g)/L)(1−cos θ) mg−N=ma_(C,y) =m(L/2)(α sin θ+ω^2 cos θ) ⇒N=((9mg)/4)(cos θ−(1/3))^2 ≥0 F=ma_(C,x) =m(L/2)(αcos θ−ω^2 sin θ) ⇒F=((9mg)/4)(cos θ−(2/3))sin θ slipping occurs when ∣F∣≥μN ±((9mg)/4)(cos θ−(2/3))sin θ=μ((9mg)/4)(cos θ−(1/3))^2 ⇒±(cos θ−(2/3))sin θ=μ(cos θ−(1/3))^2
$${C}\left({x}_{{C}} ,{y}_{{C}} \right) \\ $$$${x}_{{C}} =\frac{{L}\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$${y}_{{C}} =\frac{{L}\mathrm{cos}\:\theta}{\mathrm{2}} \\ $$$${v}_{{C},{x}} =\frac{{L}\mathrm{cos}\:\theta\:\omega}{\mathrm{2}} \\ $$$${a}_{{C},{x}} =\frac{{L}}{\mathrm{2}}\left(\alpha\mathrm{cos}\:\theta−\omega^{\mathrm{2}} \mathrm{sin}\:\theta\right) \\ $$$${v}_{{C},{y}} =−\frac{{dy}_{{C}} }{{dt}}=\frac{{L}\mathrm{sin}\:\theta}{\mathrm{2}}\omega \\ $$$${a}_{{C},{y}} =\frac{{L}}{\mathrm{2}}\left(\alpha\:\mathrm{sin}\:\theta+\omega^{\mathrm{2}} \mathrm{cos}\:\theta\right) \\ $$$$\frac{{mL}^{\mathrm{2}} }{\mathrm{3}}\alpha={mg}\frac{{L}\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$$\alpha=\frac{\mathrm{3}{g}}{\mathrm{2}{L}}\mathrm{sin}\:\theta \\ $$$$\omega\frac{{d}\omega}{{d}\theta}=\frac{\mathrm{3}{g}}{\mathrm{2}{L}}\mathrm{sin}\:\theta \\ $$$$\int_{\mathrm{0}} ^{\omega} \omega{d}\omega=\frac{\mathrm{3}{g}}{\mathrm{2}{L}}\int_{\mathrm{0}} ^{\theta} \mathrm{sin}\:\theta{d}\theta \\ $$$$\frac{\omega^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{3}{g}}{\mathrm{2}{L}}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$$\omega^{\mathrm{2}} =\frac{\mathrm{3}{g}}{{L}}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$${mg}−{N}={ma}_{{C},{y}} ={m}\frac{{L}}{\mathrm{2}}\left(\alpha\:\mathrm{sin}\:\theta+\omega^{\mathrm{2}} \mathrm{cos}\:\theta\right) \\ $$$$\Rightarrow{N}=\frac{\mathrm{9}{mg}}{\mathrm{4}}\left(\mathrm{cos}\:\theta−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${F}={ma}_{{C},{x}} ={m}\frac{{L}}{\mathrm{2}}\left(\alpha\mathrm{cos}\:\theta−\omega^{\mathrm{2}} \mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow{F}=\frac{\mathrm{9}{mg}}{\mathrm{4}}\left(\mathrm{cos}\:\theta−\frac{\mathrm{2}}{\mathrm{3}}\right)\mathrm{sin}\:\theta \\ $$$$ \\ $$$${slipping}\:{occurs}\:{when}\:\mid{F}\mid\geqslant\mu{N} \\ $$$$\pm\frac{\mathrm{9}{mg}}{\mathrm{4}}\left(\mathrm{cos}\:\theta−\frac{\mathrm{2}}{\mathrm{3}}\right)\mathrm{sin}\:\theta=\mu\frac{\mathrm{9}{mg}}{\mathrm{4}}\left(\mathrm{cos}\:\theta−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\pm\left(\mathrm{cos}\:\theta−\frac{\mathrm{2}}{\mathrm{3}}\right)\mathrm{sin}\:\theta=\mu\left(\mathrm{cos}\:\theta−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$
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

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