Question Number 224556 by mr W last updated on 18/Sep/25
Commented by mr W last updated on 18/Sep/25
$${if}\:{the}\:{friction}\:{coefficient}\:{between} \\ $$$${a}\:{small}\:{ball}\:{with}\:{mass}\:\boldsymbol{{m}}\:{and}\:{a} \\ $$$${thin}\:{tube}\:{ring}\:{with}\:{radius}\:\boldsymbol{{R}}\:{and} \\ $$$${mass}\:\boldsymbol{{M}}\:{is}\:\boldsymbol{\mu},\:{find}\:{the}\:{angle}\:\boldsymbol{\theta}\:{which} \\ $$$${indicates}\:{the}\:{position}\:{of}\:{the}\:{small} \\ $$$${ball}\:{when}\:{the}\:{ring}\:{rolls}\:{down}\:{the} \\ $$$${inclined}\:{plank}. \\ $$
Commented by mahdipoor last updated on 19/Sep/25
$$\mathrm{The}\:\mathrm{direction}\:\mathrm{of}\:\mathrm{the}\:\mathrm{frictional}\:\mathrm{force}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{ball}\:\mathrm{chang}\:\mathrm{when}\:\mathrm{directin}\:\mathrm{of}\:\mathrm{velocity}\:\mathrm{chang} \\ $$$$\mathrm{this}\:\mathrm{making}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{dificult}\:\mathrm{to}\:\mathrm{solve}\:… \\ $$
Commented by fantastic last updated on 19/Sep/25
$${nice}\:{photo}\left({your}\:{profile}\:{picture}\right) \\ $$
Commented by mr W last updated on 19/Sep/25
$${i}\:{think}\:{the}\:{answer}\:{is}\:{not}\:{a}\:{single} \\ $$$${value}\:{of}\:\theta,\:{but}\:{two}\:{ranges}\:{like}\:{this}: \\ $$
Commented by mr W last updated on 19/Sep/25
Answered by mr W last updated on 20/Sep/25
Commented by mr W last updated on 20/Sep/25
![let λ=(m/M) f=kN with −μ≤k≤μ mg cos φ=f sin θ+N cos θ mg cos φ=N(k sin θ+cos θ) ⇒N=((mg cos φ)/(k sin θ+cos θ)) ma=mg sin φ−f cos θ+N sin θ ⇒(a/g)=sin φ+(((sin θ−k cos θ)cos φ)/(k sin θ+cos θ)) MR^2 ×(a/R)=f_1 R−fR ⇒f_1 =Ma+f Ma=Mg sin φ−f_1 +f cos θ−N sin θ Ma=Mg sin φ−Ma−f+f cos θ−N sin θ 2Ma=Mg sin φ−(k−k cos θ+sin θ)×((mg cos φ)/(k sin θ+cos θ)) ((2a)/g)=sin φ−((λ(k−k cos θ+sin θ)cos φ)/(k sin θ+cos θ)) 2 sin φ+((2(sin θ−k cos θ)cos φ)/(k sin θ+cos θ))=sin φ−((λ(k−k cos θ+sin θ)cos φ)/(k sin θ+cos θ)) sin φ+((2(sin θ−k cos θ)cos φ)/(k sin θ+cos θ))+((λ(k−k cos θ+sin θ)cos φ)/(k sin θ+cos θ))=0 λk+(λ+2+k tan φ)sin θ+[tan φ−(λ+2)k]cos θ=0 ⇒θ=−tan^(−1) ((tan φ−(λ+2)k)/(λ+2+k tan φ))−sin^(−1) ((λk)/( (√((λ+2+k tan φ)^2 +[tan φ−(λ+2)k]^2 ))))](https://www.tinkutara.com/question/Q224574.png)
$${let}\:\lambda=\frac{{m}}{{M}} \\ $$$${f}={kN}\:{with}\:−\mu\leqslant{k}\leqslant\mu \\ $$$${mg}\:\mathrm{cos}\:\phi={f}\:\mathrm{sin}\:\theta+{N}\:\mathrm{cos}\:\theta \\ $$$${mg}\:\mathrm{cos}\:\phi={N}\left({k}\:\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\right) \\ $$$$\Rightarrow{N}=\frac{{mg}\:\mathrm{cos}\:\phi}{{k}\:\mathrm{sin}\:\theta+\mathrm{cos}\:\theta} \\ $$$${ma}={mg}\:\mathrm{sin}\:\phi−{f}\:\mathrm{cos}\:\theta+{N}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\frac{{a}}{{g}}=\mathrm{sin}\:\phi+\frac{\left(\mathrm{sin}\:\theta−{k}\:\mathrm{cos}\:\theta\right)\mathrm{cos}\:\phi}{{k}\:\mathrm{sin}\:\theta+\mathrm{cos}\:\theta} \\ $$$${MR}^{\mathrm{2}} ×\frac{{a}}{{R}}={f}_{\mathrm{1}} {R}−{fR} \\ $$$$\Rightarrow{f}_{\mathrm{1}} ={Ma}+{f} \\ $$$${Ma}={Mg}\:\mathrm{sin}\:\phi−{f}_{\mathrm{1}} +{f}\:\mathrm{cos}\:\theta−{N}\:\mathrm{sin}\:\theta \\ $$$${Ma}={Mg}\:\mathrm{sin}\:\phi−{Ma}−{f}+{f}\:\mathrm{cos}\:\theta−{N}\:\mathrm{sin}\:\theta \\ $$$$\mathrm{2}{Ma}={Mg}\:\mathrm{sin}\:\phi−\left({k}−{k}\:\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)×\frac{{mg}\:\mathrm{cos}\:\phi}{{k}\:\mathrm{sin}\:\theta+\mathrm{cos}\:\theta} \\ $$$$\frac{\mathrm{2}{a}}{{g}}=\mathrm{sin}\:\phi−\frac{\lambda\left({k}−{k}\:\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)\mathrm{cos}\:\phi}{{k}\:\mathrm{sin}\:\theta+\mathrm{cos}\:\theta} \\ $$$$\mathrm{2}\:\mathrm{sin}\:\phi+\frac{\mathrm{2}\left(\mathrm{sin}\:\theta−{k}\:\mathrm{cos}\:\theta\right)\mathrm{cos}\:\phi}{{k}\:\mathrm{sin}\:\theta+\mathrm{cos}\:\theta}=\mathrm{sin}\:\phi−\frac{\lambda\left({k}−{k}\:\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)\mathrm{cos}\:\phi}{{k}\:\mathrm{sin}\:\theta+\mathrm{cos}\:\theta} \\ $$$$\mathrm{sin}\:\phi+\frac{\mathrm{2}\left(\mathrm{sin}\:\theta−{k}\:\mathrm{cos}\:\theta\right)\mathrm{cos}\:\phi}{{k}\:\mathrm{sin}\:\theta+\mathrm{cos}\:\theta}+\frac{\lambda\left({k}−{k}\:\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)\mathrm{cos}\:\phi}{{k}\:\mathrm{sin}\:\theta+\mathrm{cos}\:\theta}=\mathrm{0} \\ $$$$\lambda{k}+\left(\lambda+\mathrm{2}+{k}\:\mathrm{tan}\:\phi\right)\mathrm{sin}\:\theta+\left[\mathrm{tan}\:\phi−\left(\lambda+\mathrm{2}\right){k}\right]\mathrm{cos}\:\theta=\mathrm{0} \\ $$$$\Rightarrow\theta=−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:\phi−\left(\lambda+\mathrm{2}\right){k}}{\lambda+\mathrm{2}+{k}\:\mathrm{tan}\:\phi}−\mathrm{sin}^{−\mathrm{1}} \frac{\lambda{k}}{\:\sqrt{\left(\lambda+\mathrm{2}+{k}\:\mathrm{tan}\:\phi\right)^{\mathrm{2}} +\left[\mathrm{tan}\:\phi−\left(\lambda+\mathrm{2}\right){k}\right]^{\mathrm{2}} }} \\ $$
Commented by fantastic last updated on 20/Sep/25
$${you}\:{are}\:{very}\:{good}\:{at}\:{physics} \\ $$
Commented by mr W last updated on 20/Sep/25
$${example}:\:\phi=\mathrm{30}°,\:\lambda=\frac{{m}}{{M}}=\mathrm{0}.\mathrm{2},\:\mu=\mathrm{0}.\mathrm{6} \\ $$
Commented by mr W last updated on 20/Sep/25
Commented by mr W last updated on 20/Sep/25
Commented by mr W last updated on 20/Sep/25
$${if}\:{no}\:{friction}: \\ $$$${k}=\mathrm{0} \\ $$$$\left(\lambda+\mathrm{2}\right)\:\mathrm{sin}\:\theta+\mathrm{tan}\:\phi\:\mathrm{cos}\:\theta=\mathrm{0} \\ $$$$\Rightarrow\theta=−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{tan}\:\phi}{\lambda+\mathrm{2}}\right) \\ $$