Menu Close

Question-224531

Tinku Tara 0 Comments
Pin




Question Number 224531 by ajfour last updated on 17/Sep/25
Commented by ajfour last updated on 17/Sep/25
A plank inclined at angle phi to the level ground has coefficient of friction such that it just permits enough friction for a ring of mass M and radius R to purely roll down the fixed incline. Now our plan is to weld a bead at the periphery of the ring so that this new ring plus bead combination when olaced upright on the same incline only slides down not rotating at all. we hsve to find the minimum bead mass. assume coefficient of static and kinetic friction to be same.
Commented by mahdipoor last updated on 17/Sep/25
i get m=(M/(((2sin(θ−φ))/(tanφ))+cos(θ−φ)−1))
$${i}\:{get}\: \\ $$$${m}=\frac{{M}}{\frac{\mathrm{2}{sin}\left(\theta−\phi\right)}{{tan}\phi}+{cos}\left(\theta−\phi\right)−\mathrm{1}} \\ $$
Commented by ajfour last updated on 17/Sep/25
looks genuine. i shall attempt too, its due. Google Gemini says m=((Msin φ)/(2−sin φ))
$${looks}\:{genuine}.\:{i}\:{shall}\:{attempt}\:{too}, \\ $$$${its}\:{due}.\:{Google}\:{Gemini}\:{says} \\ $$$${m}=\frac{{M}\mathrm{sin}\:\phi}{\mathrm{2}−\mathrm{sin}\:\phi} \\ $$
Commented by mr W last updated on 17/Sep/25
i think mahdipoor sir is right.
$${i}\:{think}\:{mahdipoor}\:{sir}\:{is}\:{right}. \\ $$
Answered by mahdipoor last updated on 17/Sep/25
Consider an oblique coordinate system cenered at the ring such that the x−axis is aigned with the slope direction. center of mass : CM=−S(cosψ,sinψ) S=((mR)/(M+m)) ψ=θ−90−φ for only slip → α=0⇒ΣM_(CM) =I_(CM) α=0 1)for Weight (mg,Mg) → Σ_1 M_(CM) =0 2)for N & f=μN → min μ=((tanφ)/2) (when m=0, ring no slip) N=g(m+M)cosφ Σ_2 M_(CM) =d×ΣF=0 (S.cosψ , −S.sinψ−R)×(−μN,N)= N(Scosψ−((tanφ)/2)(Ssinψ+R))=0 ⇒mRcosψ=((tanφ)/2)(mRsinψ+(M+m)R) ⇒m=(M/(((2cosψ)/(tanφ))−sinψ−1))=(M/(((2sin(θ−φ))/(tanφ))+cos(θ−ψ)−1))
$$\mathrm{Consider}\:\mathrm{an}\:\mathrm{oblique}\:\mathrm{coordinate}\:\mathrm{system} \\ $$$$\mathrm{cenered}\:\mathrm{at}\:\mathrm{the}\:\mathrm{ring}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{x}−\mathrm{axis} \\ $$$$\mathrm{is}\:\mathrm{aigned}\:\mathrm{with}\:\mathrm{the}\:\mathrm{slope}\:\mathrm{direction}. \\ $$$${center}\:{of}\:{mass}\:: \\ $$$${CM}=−{S}\left({cos}\psi,{sin}\psi\right) \\ $$$${S}=\frac{{mR}}{{M}+{m}}\:\:\:\:\psi=\theta−\mathrm{90}−\phi \\ $$$${for}\:{only}\:{slip}\:\rightarrow\:\alpha=\mathrm{0}\Rightarrow\Sigma{M}_{{CM}} ={I}_{{CM}} \alpha=\mathrm{0} \\ $$$$\left.\mathrm{1}\right){for}\:{Weight}\:\left({mg},{Mg}\right)\:\rightarrow\:\sum_{\mathrm{1}} {M}_{{CM}} =\mathrm{0} \\ $$$$\left.\mathrm{2}\right){for}\:{N}\:\&\:{f}=\mu{N}\:\rightarrow\: \\ $$$${min}\:\mu=\frac{{tan}\phi}{\mathrm{2}}\:\left({when}\:{m}=\mathrm{0},\:{ring}\:{no}\:{slip}\right) \\ $$$${N}={g}\left({m}+{M}\right){cos}\phi \\ $$$$\sum_{\mathrm{2}} {M}_{{CM}} =\boldsymbol{\mathrm{d}}×\Sigma\boldsymbol{\mathrm{F}}=\mathrm{0} \\ $$$$\left({S}.{cos}\psi\:,\:−{S}.{sin}\psi−{R}\right)×\left(−\mu{N},{N}\right)= \\ $$$${N}\left({Scos}\psi−\frac{{tan}\phi}{\mathrm{2}}\left({Ssin}\psi+{R}\right)\right)=\mathrm{0}\: \\ $$$$\Rightarrow{mRcos}\psi=\frac{{tan}\phi}{\mathrm{2}}\left({mRsin}\psi+\left({M}+{m}\right){R}\right) \\ $$$$\Rightarrow{m}=\frac{{M}}{\frac{\mathrm{2}{cos}\psi}{{tan}\phi}−{sin}\psi−\mathrm{1}}=\frac{{M}}{\frac{\mathrm{2}{sin}\left(\theta−\phi\right)}{{tan}\phi}+{cos}\left(\theta−\psi\right)−\mathrm{1}} \\ $$
Answered by mr W last updated on 17/Sep/25
Commented by mr W last updated on 17/Sep/25
Commented by mr W last updated on 17/Sep/25
case 1: ring without bead mass I=Mr^2 pure rolling ⇒α=(a/r) f=μN=μMg cos φ Ma=Mg sin φ−f ⇒a=g(sin φ−μ cos φ) Iα=rf Mr^2 ×((g(sin φ−μ cos φ))/r)=rμMg cos φ sin φ−μ cos φ=μ cos φ ⇒μ=((tan φ)/2) case 2: ring with bead mass e=((mr)/(m+M)) tan ϕ=μ=((tan φ)/2) such that no rolling, (e/(sin ϕ))=(r/(sin (θ+ϕ))) ((mr)/((m+M) sin ϕ))=(r/(sin θ cos ϕ+cos θ sin ϕ)) (m/(m+M))=(1/(((sin θ)/(tan ϕ))+cos θ)) (M/m)+1=((sin θ)/(tan ϕ))+cos θ (M/m)=((2 sin θ)/(tan φ))+cos θ−1=Φ, say m_(min) ⇔ Φ_(max) (dΦ/dθ)=((2 cos θ)/(tan φ))−sin θ=0 ⇒tan θ=(2/(tan φ))=(1/(tan ϕ)) ⇒θ+ϕ=(π/2) ! Φ_(max) =(2/(tan φ))×(2/( (√(4+tan^2 φ))))+((tan φ)/( (√(4+tan^2 φ))))−1 =((4+tan^2 φ)/( tan φ(√(4+tan^2 φ))))−1 =(((√(4+tan^2 φ))−tan φ)/(tan φ)) ⇒(m_(min) /M)=((tan φ)/( (√(4+tan^2 φ))−tan φ)) =((tan φ ((√(4+tan^2 φ))+tan φ))/( 4))
$$\boldsymbol{{case}}\:\mathrm{1}:\:\boldsymbol{{ring}}\:\boldsymbol{{without}}\:\boldsymbol{{bead}}\:\boldsymbol{{mass}} \\ $$$${I}={Mr}^{\mathrm{2}} \\ $$$${pure}\:{rolling}\:\Rightarrow\alpha=\frac{{a}}{{r}} \\ $$$${f}=\mu{N}=\mu{Mg}\:\mathrm{cos}\:\phi \\ $$$${Ma}={Mg}\:\mathrm{sin}\:\phi−{f} \\ $$$$\Rightarrow{a}={g}\left(\mathrm{sin}\:\phi−\mu\:\mathrm{cos}\:\phi\right) \\ $$$${I}\alpha={rf} \\ $$$${Mr}^{\mathrm{2}} ×\frac{{g}\left(\mathrm{sin}\:\phi−\mu\:\mathrm{cos}\:\phi\right)}{{r}}={r}\mu{Mg}\:\mathrm{cos}\:\phi \\ $$$$\mathrm{sin}\:\phi−\mu\:\mathrm{cos}\:\phi=\mu\:\mathrm{cos}\:\phi \\ $$$$\Rightarrow\mu=\frac{\mathrm{tan}\:\phi}{\mathrm{2}} \\ $$$$ \\ $$$$\boldsymbol{{case}}\:\mathrm{2}:\:\boldsymbol{{ring}}\:\boldsymbol{{with}}\:\boldsymbol{{bead}}\:\boldsymbol{{mass}} \\ $$$${e}=\frac{{mr}}{{m}+{M}} \\ $$$$\mathrm{tan}\:\varphi=\mu=\frac{\mathrm{tan}\:\phi}{\mathrm{2}} \\ $$$${such}\:{that}\:{no}\:{rolling}, \\ $$$$\frac{{e}}{\mathrm{sin}\:\varphi}=\frac{{r}}{\mathrm{sin}\:\left(\theta+\varphi\right)} \\ $$$$\frac{{mr}}{\left({m}+{M}\right)\:\mathrm{sin}\:\varphi}=\frac{{r}}{\mathrm{sin}\:\theta\:\mathrm{cos}\:\varphi+\mathrm{cos}\:\theta\:\mathrm{sin}\:\varphi} \\ $$$$\frac{{m}}{{m}+{M}}=\frac{\mathrm{1}}{\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\varphi}+\mathrm{cos}\:\theta} \\ $$$$\frac{{M}}{{m}}+\mathrm{1}=\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\varphi}+\mathrm{cos}\:\theta \\ $$$$\frac{{M}}{{m}}=\frac{\mathrm{2}\:\mathrm{sin}\:\theta}{\mathrm{tan}\:\phi}+\mathrm{cos}\:\theta−\mathrm{1}=\Phi,\:{say} \\ $$$${m}_{{min}} \:\Leftrightarrow\:\Phi_{{max}} \\ $$$$\frac{{d}\Phi}{{d}\theta}=\frac{\mathrm{2}\:\mathrm{cos}\:\theta}{\mathrm{tan}\:\phi}−\mathrm{sin}\:\theta=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{\mathrm{2}}{\mathrm{tan}\:\phi}=\frac{\mathrm{1}}{\mathrm{tan}\:\varphi}\:\Rightarrow\theta+\varphi=\frac{\pi}{\mathrm{2}}\:! \\ $$$$\Phi_{{max}} =\frac{\mathrm{2}}{\mathrm{tan}\:\phi}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}+\mathrm{tan}^{\mathrm{2}} \:\phi}}+\frac{\mathrm{tan}\:\phi}{\:\sqrt{\mathrm{4}+\mathrm{tan}^{\mathrm{2}} \:\phi}}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}+\mathrm{tan}^{\mathrm{2}} \:\phi}{\:\mathrm{tan}\:\phi\sqrt{\mathrm{4}+\mathrm{tan}^{\mathrm{2}} \:\phi}}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{4}+\mathrm{tan}^{\mathrm{2}} \:\phi}−\mathrm{tan}\:\phi}{\mathrm{tan}\:\phi} \\ $$$$\Rightarrow\frac{{m}_{{min}} }{{M}}=\frac{\mathrm{tan}\:\phi}{\:\sqrt{\mathrm{4}+\mathrm{tan}^{\mathrm{2}} \:\phi}−\mathrm{tan}\:\phi} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{tan}\:\phi\:\left(\sqrt{\mathrm{4}+\mathrm{tan}^{\mathrm{2}} \:\phi}+\mathrm{tan}\:\phi\right)}{\:\mathrm{4}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *