Question Number 225054 by fantastic last updated on 16/Oct/25
Commented by fantastic last updated on 17/Oct/25
$${how} \\ $$
Commented by mr W last updated on 16/Oct/25
$${T}=\frac{{l}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}{Rv}_{\mathrm{0}} } \\ $$
Answered by mr W last updated on 19/Oct/25
![solution for the case if friction≠0 v=Rω(λ−θ) (see above) a=(dv/dt)=ω(dv/dθ)=ωR[(λ−θ)(dω/dθ)−ω]=−μg ω(λ−θ)(dω/dθ)=ω^2 −((μg)/R) (λ−θ)((d(ω^2 −((μg)/R)))/dθ)=2(ω^2 −((μg)/R)) let Φ=ω^2 −((μg)/R) (((λ−θ)dΦ)/dθ)=2Φ (dΦ/Φ)=((2dθ)/(λ−θ)) ⇒ln Φ=−2 ln (λ−θ)+C ⇒Φ=(k/((λ−θ)^2 )) with k=constant ⇒ω^2 =(k/((λ−θ)^2 ))+((μg)/R) at θ=0: ω_0 =(v_0 /l_0 ) ((v_0 /l_0 ))^2 =(k/λ^2 )+((μg)/R) ⇒k=λ^2 ((v_0 ^2 /l_0 ^2 )−((μg)/R)) ω^2 =((v_0 ^2 /l_0 ^2 )−((μg)/R))(λ^2 /((λ−θ)^2 ))+((μg)/R) ω=(dθ/dt)=(√(((v_0 ^2 /l_0 ^2 )−((μg)/R))(λ^2 /((λ−θ)^2 ))+((μg)/R))) dt=(dθ/( (√(((v_0 ^2 /l_0 ^2 )−((μg)/R))(λ^2 /((λ−θ)^2 ))+((μg)/R))))) dt=(((λ−θ)dθ)/( (√(((v_0 ^2 /l_0 ^2 )−((μg)/R))λ^2 +((μg)/R)(λ−θ)^2 )))) T=∫_0 ^λ (((λ−θ)dθ)/( (√(((v_0 ^2 /l_0 ^2 )−((μg)/R))λ^2 +((μg)/R)(λ−θ)^2 )))) =−(R/(μg))[(√(((v_0 ^2 /l_0 ^2 )−((μg)/R))λ^2 +((μg)/R)(λ−θ)^2 ))]_0 ^λ =(R/(μg))[(√(((v_0 ^2 /l_0 ^2 )−((μg)/R))λ^2 +((μgλ^2 )/R)))−(√(((v_0 ^2 /l_0 ^2 )−((μg)/R))λ^2 ))] =((Rλ)/(μg))[(v_0 /l_0 )−(√((v_0 ^2 /l_0 ^2 )−((μg)/R)))] =(λ/((v_0 /l_0 )+(√((v_0 ^2 /l_0 ^2 )−((μg)/R))))) ⇒T=(l_0 ^2 /(R(v_0 +(√(v_0 ^2 −((μgl_0 ^2 )/R)))))) ✓ such that it′s possible, v_0 ^2 −((μgl_0 ^2 )/R)≥0 ⇒v_0 ≥l_0 (√((μg)/R)) otherwise, motion only till θ=θ_(max) . at θ_(max) : v=0, w=0 ω=(√(((v_0 ^2 /l_0 ^2 )−((μg)/R))(λ^2 /((λ−θ_(max) )^2 ))+((μg)/R)))=0 (v_0 ^2 /l_0 ^2 )=((μg)/R)[1+(1/((1−(θ_(max) /λ))^2 ))] ((Rv_0 ^2 )/(μgl_0 ^2 ))−1=(1/((1−(θ_(max) /λ))^2 )) ⇒θ_(max) =(l_0 /R)(√(1−(1/(((Rv_0 ^2 )/(μgl_0 ^2 ))−1))))](https://www.tinkutara.com/question/Q225143.png)
$${solution}\:{for}\:{the}\:{case}\:{if}\:{friction}\neq\mathrm{0} \\ $$$${v}={R}\omega\left(\lambda−\theta\right)\:\:\left({see}\:{above}\right) \\ $$$${a}=\frac{{dv}}{{dt}}=\omega\frac{{dv}}{{d}\theta}=\omega{R}\left[\left(\lambda−\theta\right)\frac{{d}\omega}{{d}\theta}−\omega\right]=−\mu{g} \\ $$$$\omega\left(\lambda−\theta\right)\frac{{d}\omega}{{d}\theta}=\omega^{\mathrm{2}} −\frac{\mu{g}}{{R}} \\ $$$$\left(\lambda−\theta\right)\frac{{d}\left(\omega^{\mathrm{2}} −\frac{\mu{g}}{{R}}\right)}{{d}\theta}=\mathrm{2}\left(\omega^{\mathrm{2}} −\frac{\mu{g}}{{R}}\right) \\ $$$${let}\:\Phi=\omega^{\mathrm{2}} −\frac{\mu{g}}{{R}} \\ $$$$\frac{\left(\lambda−\theta\right){d}\Phi}{{d}\theta}=\mathrm{2}\Phi \\ $$$$\frac{{d}\Phi}{\Phi}=\frac{\mathrm{2}{d}\theta}{\lambda−\theta} \\ $$$$\Rightarrow\mathrm{ln}\:\Phi=−\mathrm{2}\:\mathrm{ln}\:\left(\lambda−\theta\right)+{C} \\ $$$$\Rightarrow\Phi=\frac{{k}}{\left(\lambda−\theta\right)^{\mathrm{2}} }\:{with}\:{k}={constant} \\ $$$$\Rightarrow\omega^{\mathrm{2}} =\frac{{k}}{\left(\lambda−\theta\right)^{\mathrm{2}} }+\frac{\mu{g}}{{R}} \\ $$$${at}\:\theta=\mathrm{0}:\:\omega_{\mathrm{0}} =\frac{{v}_{\mathrm{0}} }{{l}_{\mathrm{0}} } \\ $$$$\left(\frac{{v}_{\mathrm{0}} }{{l}_{\mathrm{0}} }\right)^{\mathrm{2}} =\frac{{k}}{\lambda^{\mathrm{2}} }+\frac{\mu{g}}{{R}} \\ $$$$\Rightarrow{k}=\lambda^{\mathrm{2}} \left(\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{l}_{\mathrm{0}} ^{\mathrm{2}} }−\frac{\mu{g}}{{R}}\right) \\ $$$$\omega^{\mathrm{2}} =\left(\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{l}_{\mathrm{0}} ^{\mathrm{2}} }−\frac{\mu{g}}{{R}}\right)\frac{\lambda^{\mathrm{2}} }{\left(\lambda−\theta\right)^{\mathrm{2}} }+\frac{\mu{g}}{{R}} \\ $$$$\omega=\frac{{d}\theta}{{dt}}=\sqrt{\left(\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{l}_{\mathrm{0}} ^{\mathrm{2}} }−\frac{\mu{g}}{{R}}\right)\frac{\lambda^{\mathrm{2}} }{\left(\lambda−\theta\right)^{\mathrm{2}} }+\frac{\mu{g}}{{R}}} \\ $$$${dt}=\frac{{d}\theta}{\:\sqrt{\left(\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{l}_{\mathrm{0}} ^{\mathrm{2}} }−\frac{\mu{g}}{{R}}\right)\frac{\lambda^{\mathrm{2}} }{\left(\lambda−\theta\right)^{\mathrm{2}} }+\frac{\mu{g}}{{R}}}} \\ $$$${dt}=\frac{\left(\lambda−\theta\right){d}\theta}{\:\sqrt{\left(\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{l}_{\mathrm{0}} ^{\mathrm{2}} }−\frac{\mu{g}}{{R}}\right)\lambda^{\mathrm{2}} +\frac{\mu{g}}{{R}}\left(\lambda−\theta\right)^{\mathrm{2}} }} \\ $$$${T}=\int_{\mathrm{0}} ^{\lambda} \frac{\left(\lambda−\theta\right){d}\theta}{\:\sqrt{\left(\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{l}_{\mathrm{0}} ^{\mathrm{2}} }−\frac{\mu{g}}{{R}}\right)\lambda^{\mathrm{2}} +\frac{\mu{g}}{{R}}\left(\lambda−\theta\right)^{\mathrm{2}} }} \\ $$$$\:\:=−\frac{{R}}{\mu{g}}\left[\sqrt{\left(\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{l}_{\mathrm{0}} ^{\mathrm{2}} }−\frac{\mu{g}}{{R}}\right)\lambda^{\mathrm{2}} +\frac{\mu{g}}{{R}}\left(\lambda−\theta\right)^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\lambda} \\ $$$$\:\:=\frac{{R}}{\mu{g}}\left[\sqrt{\left(\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{l}_{\mathrm{0}} ^{\mathrm{2}} }−\frac{\mu{g}}{{R}}\right)\lambda^{\mathrm{2}} +\frac{\mu{g}\lambda^{\mathrm{2}} }{{R}}}−\sqrt{\left(\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{l}_{\mathrm{0}} ^{\mathrm{2}} }−\frac{\mu{g}}{{R}}\right)\lambda^{\mathrm{2}} }\right] \\ $$$$\:\:=\frac{{R}\lambda}{\mu{g}}\left[\frac{{v}_{\mathrm{0}} }{{l}_{\mathrm{0}} }−\sqrt{\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{l}_{\mathrm{0}} ^{\mathrm{2}} }−\frac{\mu{g}}{{R}}}\right] \\ $$$$\:\:=\frac{\lambda}{\frac{{v}_{\mathrm{0}} }{{l}_{\mathrm{0}} }+\sqrt{\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{l}_{\mathrm{0}} ^{\mathrm{2}} }−\frac{\mu{g}}{{R}}}} \\ $$$$\Rightarrow{T}=\frac{{l}_{\mathrm{0}} ^{\mathrm{2}} }{{R}\left({v}_{\mathrm{0}} +\sqrt{{v}_{\mathrm{0}} ^{\mathrm{2}} −\frac{\mu{gl}_{\mathrm{0}} ^{\mathrm{2}} }{{R}}}\right)}\:\:\checkmark \\ $$$${such}\:{that}\:{it}'{s}\:{possible}, \\ $$$${v}_{\mathrm{0}} ^{\mathrm{2}} −\frac{\mu{gl}_{\mathrm{0}} ^{\mathrm{2}} }{{R}}\geqslant\mathrm{0} \\ $$$$\Rightarrow{v}_{\mathrm{0}} \geqslant{l}_{\mathrm{0}} \sqrt{\frac{\mu{g}}{{R}}} \\ $$$${otherwise},\:{motion}\:{only}\:{till}\:\theta=\theta_{{max}} . \\ $$$${at}\:\theta_{{max}} :\:{v}=\mathrm{0},\:{w}=\mathrm{0} \\ $$$$\omega=\sqrt{\left(\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{l}_{\mathrm{0}} ^{\mathrm{2}} }−\frac{\mu{g}}{{R}}\right)\frac{\lambda^{\mathrm{2}} }{\left(\lambda−\theta_{{max}} \right)^{\mathrm{2}} }+\frac{\mu{g}}{{R}}}=\mathrm{0} \\ $$$$\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{l}_{\mathrm{0}} ^{\mathrm{2}} }=\frac{\mu{g}}{{R}}\left[\mathrm{1}+\frac{\mathrm{1}}{\left(\mathrm{1}−\frac{\theta_{{max}} }{\lambda}\right)^{\mathrm{2}} }\right] \\ $$$$\frac{{Rv}_{\mathrm{0}} ^{\mathrm{2}} }{\mu{gl}_{\mathrm{0}} ^{\mathrm{2}} }−\mathrm{1}=\frac{\mathrm{1}}{\left(\mathrm{1}−\frac{\theta_{{max}} }{\lambda}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\theta_{{max}} =\frac{{l}_{\mathrm{0}} }{{R}}\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\frac{{Rv}_{\mathrm{0}} ^{\mathrm{2}} }{\mu{gl}_{\mathrm{0}} ^{\mathrm{2}} }−\mathrm{1}}} \\ $$
Commented by fantastic last updated on 17/Oct/25
$${wow}\:{sir}. \\ $$$$ \\ $$
Commented by mr W last updated on 19/Oct/25
$${not}\:{a}\:{single}\:{but}\:{more}\:{equations}\:{are} \\ $$$${needed}\:{for}\:{the}\:{animation}. \\ $$
Commented by mr W last updated on 19/Oct/25
Commented by mr W last updated on 19/Oct/25
Commented by fantastic last updated on 19/Oct/25
$${what}\:{equation}\:{did}\:{you} \\ $$$${write}??{sir} \\ $$
Answered by mr W last updated on 17/Oct/25
Commented by mr W last updated on 17/Oct/25
$${no}\:{sir}.\:{but}\:{you}\:{are}\:{just}\:{talking}\:{too} \\ $$$${much}\:{about}\:{nothing}\:{these}\:{days}. \\ $$
Commented by mr W last updated on 17/Oct/25
![say λ=(l_0 /R) let ω=(dθ/dt) b=l_0 −Rθ=R(λ−θ) at position A′: x=R sin θ+b cos θ =R sin θ+R(λ−θ)cos θ y=R−R cos θ+b sin θ =R−R cos θ+R(λ−θ)sin θ v_x =(dx/dt)=ω(dx/dθ)=−ωR(λ−θ)sin θ v_y =(dy/dt)=ω(dy/dθ)=ωR(λ−θ)cos θ v=(√(v_x ^2 +v_y ^2 ))=v_0 ωR(λ−θ)=v_0 [or we can get this from: v=bω=v_0 ] ⇒ω=(dθ/dt)=(v_0 /(R(λ−θ))) dt=((R(λ−θ)dθ)/v_0 ) ∫_0 ^T dt=∫_0 ^(l_0 /R) ((R(λ−θ)dθ)/v_0 ) ⇒T=(R/v_0 )∫_0 ^λ (λ−θ)dθ=((Rλ^2 )/(2v_0 ))=(l_0 ^2 /(2Rv_0 )) ✓](https://www.tinkutara.com/question/Q225064.png)
$${say}\:\lambda=\frac{{l}_{\mathrm{0}} }{{R}} \\ $$$${let}\:\omega=\frac{{d}\theta}{{dt}} \\ $$$${b}={l}_{\mathrm{0}} −{R}\theta={R}\left(\lambda−\theta\right) \\ $$$${at}\:{position}\:{A}': \\ $$$${x}={R}\:\mathrm{sin}\:\theta+{b}\:\mathrm{cos}\:\theta \\ $$$$\:\:={R}\:\mathrm{sin}\:\theta+{R}\left(\lambda−\theta\right)\mathrm{cos}\:\theta \\ $$$${y}={R}−{R}\:\mathrm{cos}\:\theta+{b}\:\mathrm{sin}\:\theta \\ $$$$\:\:={R}−{R}\:\mathrm{cos}\:\theta+{R}\left(\lambda−\theta\right)\mathrm{sin}\:\theta \\ $$$${v}_{{x}} =\frac{{dx}}{{dt}}=\omega\frac{{dx}}{{d}\theta}=−\omega{R}\left(\lambda−\theta\right)\mathrm{sin}\:\theta \\ $$$${v}_{{y}} =\frac{{dy}}{{dt}}=\omega\frac{{dy}}{{d}\theta}=\omega{R}\left(\lambda−\theta\right)\mathrm{cos}\:\theta \\ $$$${v}=\sqrt{{v}_{{x}} ^{\mathrm{2}} +{v}_{{y}} ^{\mathrm{2}} }={v}_{\mathrm{0}} \\ $$$$\omega{R}\left(\lambda−\theta\right)={v}_{\mathrm{0}} \\ $$$$\left[{or}\:{we}\:{can}\:{get}\:{this}\:{from}:\:{v}={b}\omega={v}_{\mathrm{0}} \right] \\ $$$$\Rightarrow\omega=\frac{{d}\theta}{{dt}}=\frac{{v}_{\mathrm{0}} }{{R}\left(\lambda−\theta\right)} \\ $$$${dt}=\frac{{R}\left(\lambda−\theta\right){d}\theta}{{v}_{\mathrm{0}} } \\ $$$$\int_{\mathrm{0}} ^{{T}} {dt}=\int_{\mathrm{0}} ^{\frac{{l}_{\mathrm{0}} }{{R}}} \frac{{R}\left(\lambda−\theta\right){d}\theta}{{v}_{\mathrm{0}} } \\ $$$$\Rightarrow{T}=\frac{{R}}{{v}_{\mathrm{0}} }\int_{\mathrm{0}} ^{\lambda} \left(\lambda−\theta\right){d}\theta=\frac{{R}\lambda^{\mathrm{2}} }{\mathrm{2}{v}_{\mathrm{0}} }=\frac{{l}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}{Rv}_{\mathrm{0}} }\:\checkmark \\ $$
Commented by fantastic last updated on 17/Oct/25
$${thank}\:{you} \\ $$
Commented by mr W last updated on 17/Oct/25
$${if}\:{friction}\:{is}\:{present},\:{the}\:{question} \\ $$$${will}\:{become}\:{very}\:{hard}. \\ $$
Commented by fantastic last updated on 17/Oct/25
$${I}\:{give}\:{you}\:{a}\:{challenge}\: \\ $$$${solve}\:{the}\:{q}\:{but}\:{f}\:\:{is}\:{present} \\ $$
Commented by mr W last updated on 17/Oct/25
$${you}\:{don}'{t}\:{need}\:{to}\:{give}\:{me}\:{any}\: \\ $$$${challenge}!\: \\ $$
Commented by fantastic last updated on 17/Oct/25
$${sir}\:{are}\:{you}\:{angry}\:{whith}\:{me}\: \\ $$
Commented by fantastic last updated on 17/Oct/25
$${I}\:\:{have}\:{some}\:{friends} \\ $$$${but}\:{they}\:{are}\:{not}\:{talking} \\ $$$${to}\:{me}\:{for}\:{some}\:{weeks}\left({online}\right) \\ $$$${thats}\:{why}… \\ $$
Commented by fantastic last updated on 17/Oct/25
$${if}\:{F}.\:{coff}.\:\:\:\:{is}\:\mu_{{k}} \\ $$$${then}\:{i}\:{think}\: \\ $$$${v}_{\mathrm{0}_{{min}} } \:{to}\:{hit}\:{that}\:{cylender}\:={l}_{\mathrm{0}} \sqrt{\frac{{g}\mu_{{k}} }{{R}}} \\ $$
Commented by mr W last updated on 17/Oct/25
$${can}\:{you}\:{provide}\:{a}\:{working}\:{for}\:{this}? \\ $$
Commented by fantastic last updated on 17/Oct/25
$${wait} \\ $$
Answered by fantastic last updated on 17/Oct/25
Commented by fantastic last updated on 17/Oct/25
$${tell}\:{me}\::\left(\right. \\ $$
Commented by mr W last updated on 17/Oct/25
$${v}_{\mathrm{0}} \geqslant{l}_{\mathrm{0}} \sqrt{\frac{\mu{g}}{{R}}}\:{is}\:{correct}. \\ $$
Commented by fantastic last updated on 03/Nov/25
![At first we are going ti find the distant s ℓ_0 −Rθ−Rdθ=ℓ_0 −Rθ[θ is tooooo small] so ds=(ℓ_0 −Rθ)dθ ∴s=∫_0 ^ψ (ℓ_0 −Rθ)dθ now Rψ=ℓ_0 so ψ=(ℓ_0 /R) so s=∫_0 ^(ℓ_0 /R) (ℓ_0 −Rθ)dθ =[ℓ_0 θ−R(θ^2 /2)]_0 ^(ℓ_0 /R) =(ℓ_0 ^2 /R)−(ℓ_0 ^2 /(2R)) =(ℓ_0 ^2 /(2R)) So s=(ℓ_0 ^2 /(2R)) now only kinetic friction will work here v_(final) =0 s=(ℓ_0 ^2 /(2R)) u=v_0 a_r =((mgμ_k )/m)=gμ_k (v_(final) )^2 =v_0 ^2 −2as ⇒v_0 =ℓ_0 (√((gμ_k )/R))](https://www.tinkutara.com/question/Q225119.png)
$${At}\:{first}\:{we}\:{are}\:{going}\:{ti}\:{find}\:{the}\:{distant}\:{s} \\ $$$$\ell_{\mathrm{0}} −{R}\theta−{Rd}\theta=\ell_{\mathrm{0}} −{R}\theta\left[\theta\:{is}\:{tooooo}\:{small}\right] \\ $$$${so}\:{ds}=\left(\ell_{\mathrm{0}} −{R}\theta\right){d}\theta \\ $$$$\therefore{s}=\underset{\mathrm{0}} {\overset{\psi} {\int}}\left(\ell_{\mathrm{0}} −{R}\theta\right){d}\theta \\ $$$${now}\:{R}\psi=\ell_{\mathrm{0}} \\ $$$${so}\:\psi=\frac{\ell_{\mathrm{0}} }{{R}} \\ $$$${so} \\ $$$${s}=\overset{\ell_{\mathrm{0}} /{R}} {\int}_{\mathrm{0}} \left(\ell_{\mathrm{0}} −{R}\theta\right){d}\theta \\ $$$$=\left[\ell_{\mathrm{0}} \theta−{R}\frac{\theta^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\frac{\ell_{\mathrm{0}} }{{R}}} \\ $$$$=\frac{\ell_{\mathrm{0}} ^{\mathrm{2}} }{{R}}−\frac{\ell_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}{R}} \\ $$$$=\frac{\ell_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}{R}} \\ $$$${So}\:{s}=\frac{\ell_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}{R}} \\ $$$${now}\:{only}\:{kinetic}\:{friction}\:{will}\:{work}\:{here} \\ $$$${v}_{{final}} =\mathrm{0} \\ $$$${s}=\frac{\ell_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}{R}} \\ $$$${u}={v}_{\mathrm{0}} \\ $$$${a}_{{r}} =\frac{{mg}\mu_{{k}} }{{m}}={g}\mu_{{k}} \\ $$$$\left({v}_{{final}} \right)^{\mathrm{2}} ={v}_{\mathrm{0}} ^{\mathrm{2}} −\mathrm{2}{as} \\ $$$$\Rightarrow{v}_{\mathrm{0}} =\ell_{\mathrm{0}} \sqrt{\frac{{g}\mu_{{k}} }{{R}}} \\ $$
Commented by fantastic last updated on 17/Oct/25
$${sir}\:{am}\:{i}\:{correct} \\ $$$${i}\:{think}\:{i}\:{am}\:{not}… \\ $$
Commented by fantastic last updated on 17/Oct/25
$${thanks}\:{for}\:{your}\:{confirmation}\:{sir} \\ $$