Question-225604

Question Number 225604 by mr W last updated on 04/Nov/25

Commented by mr W last updated on 04/Nov/25

An uniform cylinder with radius R is spinned about its axis to the angular velocity ω_0 and then placed on an inclined plane as shown. The coefficient of friction between the cylinder and the incline is μ. How many turns and in what time will the cylinder accomplish before it stops momentarily?

$${An}\:{uniform}\:{cylinder}\:{with}\:{radius}\:{R} \\ $$$${is}\:{spinned}\:{about}\:{its}\:{axis}\:{to}\:{the} \\ $$$${angular}\:{velocity}\:\omega_{\mathrm{0}} \:{and}\:{then}\:{placed} \\ $$$${on}\:{an}\:{inclined}\:{plane}\:{as}\:{shown}.\:{The} \\ $$$${coefficient}\:{of}\:{friction}\:{between}\:{the} \\ $$$${cylinder}\:{and}\:{the}\:{incline}\:{is}\:\mu.\:{How} \\ $$$${many}\:{turns}\:{and}\:{in}\:{what}\:{time}\:{will} \\ $$$${the}\:{cylinder}\:{accomplish}\:{before}\:{it}\: \\ $$$${stops}\:{momentarily}? \\ $$

Commented by ajfour last updated on 04/Nov/25

https://youtu.be/Xov4wK7obqg?si=3mTg8aBbBs1iBE0B https://youtu.be/7Vpoao4rYxQ?si=dexPW13XLkYO-vh4

Answered by fantastic last updated on 04/Nov/25

Commented by fantastic last updated on 04/Nov/25

N=mgcos θ friction acting backwards. ∴ backward force: Mgsin θ+μMgcos θ Mg(sin θ+μcos θ) τ=RMg(sin θ+μcos θ) ((MR^2 )/2)α=RMg(sin θ+μcos θ) α=((2g(sin θ+μcos θ))/R) ⇒θ=(ω_0 ^2 /(2α))=((ω_0 ^2 R)/(4g(sin θ+μcos θ))) n=(θ/(2π))=((ω_0 ^2 R)/(8πg(sin θ+μcos θ)))

$${N}={mg}\mathrm{cos}\:\theta \\ $$$${friction}\:{acting}\:{backwards}. \\ $$$$\therefore\:{backward}\:{force}: \\ $$$${Mg}\mathrm{sin}\:\theta+\mu{Mg}\mathrm{cos}\:\theta \\ $$$${Mg}\left(\mathrm{sin}\:\theta+\mu\mathrm{cos}\:\theta\right) \\ $$$$\tau={RMg}\left(\mathrm{sin}\:\theta+\mu\mathrm{cos}\:\theta\right) \\ $$$$\frac{{MR}^{\mathrm{2}} }{\mathrm{2}}\alpha={RMg}\left(\mathrm{sin}\:\theta+\mu\mathrm{cos}\:\theta\right) \\ $$$$\alpha=\frac{\mathrm{2}{g}\left(\mathrm{sin}\:\theta+\mu\mathrm{cos}\:\theta\right)}{{R}} \\ $$$$\Rightarrow\theta=\frac{\omega_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}\alpha}=\frac{\omega_{\mathrm{0}} ^{\mathrm{2}} {R}}{\mathrm{4}{g}\left(\mathrm{sin}\:\theta+\mu\mathrm{cos}\:\theta\right)} \\ $$$${n}=\frac{\theta}{\mathrm{2}\pi}=\frac{\omega_{\mathrm{0}} ^{\mathrm{2}} {R}}{\mathrm{8}\pi{g}\left(\mathrm{sin}\:\theta+\mu\mathrm{cos}\:\theta\right)} \\ $$

Commented by ajfour last updated on 04/Nov/25

acceleration until it slips since start a. ma=μmgcos α−mgsin α a=(μcos α−sin α)g (μmgcos α)R=((mR^2 )/2)((d^2 ω/dt^2 )) ((d^2 ω/dt^2 ))T=ω_0 −ω_(roll) =(((2μgcos α)/R))T T=(((ω_0 −ω_(roll) )R)/(2μgcos α)) ω_(roll) R=aT (2μgcos α+a)T=ω_0 R T=((𝛚_0 R/g)/((3𝛍)cos 𝛂−sin 𝛂))

$${acceleration}\:{until}\:{it}\:{slips}\:{since} \\ $$$${start}\:\boldsymbol{{a}}. \\ $$$${m}\boldsymbol{{a}}=\mu{mg}\mathrm{cos}\:\alpha−{mg}\mathrm{sin}\:\alpha \\ $$$$\boldsymbol{{a}}=\left(\mu\mathrm{cos}\:\alpha−\mathrm{sin}\:\alpha\right)\boldsymbol{{g}} \\ $$$$\left(\mu{mg}\mathrm{cos}\:\alpha\right){R}=\frac{{mR}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{{d}^{\mathrm{2}} \omega}{{dt}^{\mathrm{2}} }\right) \\ $$$$\left(\frac{{d}^{\mathrm{2}} \omega}{{dt}^{\mathrm{2}} }\right){T}=\omega_{\mathrm{0}} −\omega_{{roll}} =\left(\frac{\mathrm{2}\mu{g}\mathrm{cos}\:\alpha}{{R}}\right){T} \\ $$$${T}=\frac{\left(\omega_{\mathrm{0}} −\omega_{{roll}} \right){R}}{\mathrm{2}\mu{g}\mathrm{cos}\:\alpha} \\ $$$$\omega_{{roll}} {R}=\boldsymbol{{a}}{T} \\ $$$$\left(\mathrm{2}\mu{g}\mathrm{cos}\:\alpha+\boldsymbol{{a}}\right)\boldsymbol{{T}}=\omega_{\mathrm{0}} {R} \\ $$$$\boldsymbol{{T}}=\frac{\boldsymbol{\omega}_{\mathrm{0}} \boldsymbol{{R}}/\boldsymbol{{g}}}{\left(\mathrm{3}\boldsymbol{\mu}\right)\boldsymbol{{cos}}\:\boldsymbol{\alpha}−\boldsymbol{{sin}}\:\boldsymbol{\alpha}} \\ $$

Answered by mahdipoor last updated on 04/Nov/25

by energy method : mgΔh=(1/2)Iω^2 +(1/2)m(v_(cg) )^2 no slip ⇒ v_(cg) =Rω , I=(1/2)mR^2 Δh=((0.5ω^2 (I+mR^2 ))/(mg))=0.75((ω^2 R^2 )/g) d=n(2πR)=Δh/sinθ ⇒n=0.375((ω^2 R)/(πgsinθ)) (true) mg.sin(θ)=f_k (fals) ⇒f_k R=Rmgsin(θ)=Iα α=((Rmg.sinθ)/(0.5mR^2 ))=((ω−0)/(Δt)) ⇒ Δt=((0.5ωR)/(gsinθ)) (fals)

$$\mathrm{by}\:\mathrm{energy}\:\mathrm{method}\:: \\ $$$$\mathrm{mg}\Delta\mathrm{h}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{I}\omega^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{m}\left(\mathrm{v}_{\mathrm{cg}} \right)^{\mathrm{2}} \\ $$$$\mathrm{no}\:\mathrm{slip}\:\Rightarrow\:\mathrm{v}_{\mathrm{cg}} =\mathrm{R}\omega\:\:\:,\:\mathrm{I}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{mR}^{\mathrm{2}} \\ $$$$\Delta\mathrm{h}=\frac{\mathrm{0}.\mathrm{5}\omega^{\mathrm{2}} \left(\mathrm{I}+\mathrm{mR}^{\mathrm{2}} \right)}{\mathrm{mg}}=\mathrm{0}.\mathrm{75}\frac{\omega^{\mathrm{2}} \mathrm{R}^{\mathrm{2}} }{\mathrm{g}} \\ $$$$\mathrm{d}=\mathrm{n}\left(\mathrm{2}\pi\mathrm{R}\right)=\Delta\mathrm{h}/\mathrm{sin}\theta \\ $$$$\Rightarrow\mathrm{n}=\mathrm{0}.\mathrm{375}\frac{\omega^{\mathrm{2}} \mathrm{R}}{\pi\mathrm{gsin}\theta}\:\:\:\:\left(\mathrm{true}\right) \\ $$$$\mathrm{mg}.\mathrm{sin}\left(\theta\right)=\mathrm{f}_{\mathrm{k}} \:\:\:\:\:\left(\mathrm{fals}\right) \\ $$$$\Rightarrow\mathrm{f}_{\mathrm{k}} \mathrm{R}=\mathrm{Rmgsin}\left(\theta\right)=\mathrm{I}\alpha \\ $$$$\alpha=\frac{\mathrm{Rmg}.\mathrm{sin}\theta}{\mathrm{0}.\mathrm{5mR}^{\mathrm{2}} }=\frac{\omega−\mathrm{0}}{\Delta\mathrm{t}}\:\Rightarrow\:\Delta\mathrm{t}=\frac{\mathrm{0}.\mathrm{5}\omega\mathrm{R}}{\mathrm{gsin}\theta}\:\:\:\left(\mathrm{fals}\right) \\ $$

Commented by fantastic last updated on 04/Nov/25

$${am}\:{i}\:{right}\:{sir}? \\ $$$${pls}\:{tell} \\ $$

Commented by mahdipoor last updated on 04/Nov/25

W_t =mgsinθ ✓ W_n =mgcosθ ✓ f_k =? (no slip : just sure abuot ∣f_k ∣≤Nμ) we have 3 general eq : (CM : center of mass) F=ma_(CM) ⇒ { ((W_t +f_k =ma_t )),((−W_n +N=ma_n )) :} ΣM_(CM) =I_(CM) α (↺+)⇒ { ((f_k R=Iα)) :} no slip : −v_(t,CM) =ωR ⇒ a_t =−αR=((−f_k R^2 )/I) and : v_(n,CM) =0 ⇒ a_n =0 so → N = W_n =mgcosθ W_t =−f_k (((mR^2 )/I)+1)=−3f_k f_k =−((mgsinθ)/3) and not =μmgcosθ or =mgsinθ so α=((f_k R)/I)=((−2)/3)(((gsinθ)/R)) t_(stop) =((0−ω_0 )/α)=1.5((Rω_0 )/(gsinθ)) ∴ θ_(stop) =(ω_0 /2)t_(stop) =0.75((Rω_0 ^2 )/(gsinθ))=n_(stop) (2π) ⇒n_(stop) =((0.375)/π).((Rω_0 ^2 )/(gsinθ)) ∴ (same answer by energy method)

$$ \\ $$$$\mathrm{W}_{\mathrm{t}} =\mathrm{mgsin}\theta\:\:\checkmark \\ $$$$\mathrm{W}_{\mathrm{n}} =\mathrm{mgcos}\theta\:\:\checkmark \\ $$$$\mathrm{f}_{\mathrm{k}} =?\:\:\left(\mathrm{no}\:\mathrm{slip}\::\:\mathrm{just}\:\mathrm{sure}\:\mathrm{abuot}\:\mid\mathrm{f}_{\mathrm{k}} \mid\leqslant\mathrm{N}\mu\right) \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{3}\:\mathrm{general}\:\mathrm{eq}\::\:\left(\mathrm{CM}\::\:\mathrm{center}\:\mathrm{of}\:\mathrm{mass}\right) \\ $$$$\boldsymbol{\mathrm{F}}=\mathrm{m}\boldsymbol{\mathrm{a}}_{\mathrm{CM}} \Rightarrow\begin{cases}{\mathrm{W}_{\mathrm{t}} +\mathrm{f}_{\mathrm{k}} =\mathrm{ma}_{\mathrm{t}} }\\{−\mathrm{W}_{\mathrm{n}} +\mathrm{N}=\mathrm{ma}_{\mathrm{n}} }\end{cases} \\ $$$$\Sigma\mathrm{M}_{\mathrm{CM}} =\mathrm{I}_{\mathrm{CM}} \alpha\:\:\left(\circlearrowleft+\right)\Rightarrow\begin{cases}{\mathrm{f}_{\mathrm{k}} \mathrm{R}=\mathrm{I}\alpha}\end{cases} \\ $$$$\mathrm{no}\:\mathrm{slip}\:: \\ $$$$−\mathrm{v}_{\mathrm{t},\mathrm{CM}} =\omega\mathrm{R}\:\:\Rightarrow\:\mathrm{a}_{\mathrm{t}} =−\alpha\mathrm{R}=\frac{−\mathrm{f}_{\mathrm{k}} \mathrm{R}^{\mathrm{2}} }{\mathrm{I}} \\ $$$$\mathrm{and}\:: \\ $$$$\mathrm{v}_{\mathrm{n},\mathrm{CM}} =\mathrm{0}\:\:\Rightarrow\:\mathrm{a}_{\mathrm{n}} =\mathrm{0} \\ $$$$\:\mathrm{so}\:\rightarrow \\ $$$$\mathrm{N}\:=\:\mathrm{W}_{\mathrm{n}} =\mathrm{mgcos}\theta \\ $$$$\mathrm{W}_{\mathrm{t}} =−\mathrm{f}_{\mathrm{k}} \left(\frac{\mathrm{mR}^{\mathrm{2}} }{\mathrm{I}}+\mathrm{1}\right)=−\mathrm{3f}_{\mathrm{k}} \\ $$$$\mathrm{f}_{\mathrm{k}} =−\frac{\mathrm{mgsin}\theta}{\mathrm{3}}\:\mathrm{and}\:\mathrm{not}\:=\mu\mathrm{mgcos}\theta\:\:\mathrm{or}\:=\mathrm{mgsin}\theta \\ $$$$\mathrm{so}\:\:\:\alpha=\frac{\mathrm{f}_{\mathrm{k}} \mathrm{R}}{\mathrm{I}}=\frac{−\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{gsin}\theta}{\mathrm{R}}\right) \\ $$$$\mathrm{t}_{\mathrm{stop}} =\frac{\mathrm{0}−\omega_{\mathrm{0}} }{\alpha}=\mathrm{1}.\mathrm{5}\frac{\mathrm{R}\omega_{\mathrm{0}} }{\mathrm{gsin}\theta}\:\:\therefore \\ $$$$\theta_{\mathrm{stop}} =\frac{\omega_{\mathrm{0}} }{\mathrm{2}}\mathrm{t}_{\mathrm{stop}} =\mathrm{0}.\mathrm{75}\frac{\mathrm{R}\omega_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{gsin}\theta}=\mathrm{n}_{\mathrm{stop}} \left(\mathrm{2}\pi\right) \\ $$$$\Rightarrow\mathrm{n}_{\mathrm{stop}} =\frac{\mathrm{0}.\mathrm{375}}{\pi}.\frac{\mathrm{R}\omega_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{gsin}\theta}\:\therefore\:\:\left(\mathrm{same}\:\mathrm{answer}\:\mathrm{by}\:\mathrm{energy}\:\mathrm{method}\right) \\ $$

Commented by mahdipoor last updated on 04/Nov/25

$$\mathrm{I}\:\mathrm{assumed}\:\mathrm{we}\:\mathrm{didnt}\:\mathrm{slip}\:… \\ $$

Commented by mr W last updated on 05/Nov/25

$${thanks}\:{for}\:{trying}\:{sirs}! \\ $$$${i}\:{think}\:{the}\:{question}\:{is}\:{a}\:{little}\:{bit} \\ $$$${more}\:{complex}\:{than}\:{it}\:{looks}\:{like}. \\ $$$${i}'{ll}\:{post}\:{my}\:{attempt}\:{later}. \\ $$

Commented by fantastic last updated on 04/Nov/25

$${can}\:{you}\:{give}\:{us}\:{any}\:{hint} \\ $$$${what}\:{we}\:{missed}\:{sir}? \\ $$$${so}\:{that}\:{we}\:{can}\:{consider} \\ $$$${that} \\ $$

Answered by mr W last updated on 05/Nov/25

at first we assume that the incline is rough enough. otherwise, if μ is very small, the grip between the cylinder and the incline is weak and the cylinder will just “slide” down along the incline, in spite of its initial rotation upwards. in this case the cylinder woulde never stop, not even momentarily. when μ is large enough, the cylinder will make following motion:

$${at}\:{first}\:{we}\:{assume}\:{that}\:{the}\:{incline} \\ $$$${is}\:{rough}\:{enough}.\:{otherwise},\:\:{if}\:\mu\:{is}\: \\ $$$${very}\:{small},\:{the}\:{grip}\:{between}\:{the} \\ $$$${cylinder}\:{and}\:{the}\:{incline}\:{is}\:{weak} \\ $$$${and}\:{the}\:{cylinder}\:{will}\:{just}\:“{slide}''\: \\ $$$${down}\:{along}\:{the}\:{incline},\:{in}\:{spite}\:{of} \\ $$$${its}\:{initial}\:{rotation}\:{upwards}. \\ $$$${in}\:{this}\:{case}\:{the}\:{cylinder}\:{woulde} \\ $$$${never}\:{stop},\:{not}\:{even}\:{momentarily}. \\ $$$${when}\:\mu\:{is}\:{large}\:{enough},\:{the}\:{cylinder} \\ $$$${will}\:{make}\:{following}\:{motion}: \\ $$

Commented by mr W last updated on 06/Nov/25

phase 1: t=0 to t_1 at t=0 the cylinder has angular velocity, but no translational velocity, therefore it will roll and slip simultaneously, till its translational velocity matches its angular velocity, i.e. v_1 =ω_1 R. after that it will roll purely. phase 2: t=t_1 to t_2 the cylinder rolls without slipping till it reaches its highest position and stops momentarily. then the cyclinder begins to roll down the incline backwards. phase 1: f=μN=μMg cos θ Ma=f−Mg sin θ=Mg(μ cos θ−sin θ) ⇒a=g(μ cos θ−sin θ) a=(v_1 /t_1 ) ⇒t_1 =(v_1 /(g(μ cos θ−sin θ))) Iα=fR ((MR^2 )/2)α=μMg cos θR ⇒α=((2μg cos θ)/R) α=((w_0 −ω_1 )/t_1 ) ⇒t_1 =(((ω_0 −ω_1 )R)/(2μg cos θ)) (((ω_0 −ω_1 )R)/(2μg cos θ))=(v_1 /(g(μ cos θ−sin θ))) with v_1 =ω_1 R (((ω_0 −ω_1 )R)/(2μg cos θ))=((ω_1 R)/(g(μ cos θ−sin θ))) ((ω_0 −ω_1 )/(2μ cos θ))=(ω_1 /(μ cos θ−sin θ)) ⇒ω_1 =(((μ−tan θ)ω_0 )/(3μ−tan θ)) rotation angle in phase 1: φ_1 =((ω_0 ^2 −ω_1 ^2 )/(2α)) =((ω_0 ^2 R)/(4μg cos θ))(1+((μ−tan θ)/(3μ−tan θ)))(1−((μ−tan θ)/(3μ−tan θ))) =(((2μ−tan θ)ω_0 ^2 R)/(g(3μ−tan θ)^2 cos θ)) t_1 =(((ω_0 −ω_1 )R)/(2μg cos θ)) =((ω_0 R)/(2μg cos θ))(1−((μ−tan θ)/(3μ−tan θ))) =((ω_0 R)/(g(3μ−tan θ) cos θ)) phase 2: energy conservation: (1/2)(((MR^2 ω_1 ^2 )/2)+Mv_1 ^2 )=Mgφ_2 R sin θ φ_2 =((3ω_1 ^2 R)/(4g sin θ)) =((3(μ−tan θ)^2 ω_0 ^2 R)/(4g(3μ−tan θ)^2 sin θ)) or: ((MR^2 )/2)α=−fR Ma=f−Mg sin θ since a=αR, MαR^2 =fR−MgR sin θ ((MR^2 )/2)α+MαR^2 =MgR sin θ ⇒α=((2g sin θ)/(3R)) φ_2 =(ω_1 ^2 /(2α))=((3R)/(4g sin θ)) ×(((μ−tan θ)^2 ω_0 ^2 )/((3μ−tan θ)^2 )) =((3(μ−tan θ)^2 ω_0 ^2 R)/(4g(3μ−tan θ)^2 sin θ)) t_2 =(ω_1 /α)=(((μ−tan θ)ω_0 )/(3μ−tan θ))×((3R)/(2g sin θ)) =((3(μ−tan θ)ω_0 R)/(2g (3μ−tan θ) sin θ)) total rotation angle: φ=φ_1 +φ_2 =(((2μ−tan θ)ω_0 ^2 R)/(g(3μ−tan θ)^2 cos θ))+((3(μ−tan θ)^2 ω_0 ^2 R)/(4g(3μ−tan θ)^2 sin θ)) =((ω_0 ^2 R)/(g(3μ−tan θ)^2 ))[((2μ−tan θ)/(cos θ))+((3(μ−tan θ)^2 )/(4 sin θ))] =(((μ+tan θ)ω_0 ^2 R)/(4g(3μ−tan θ) sin θ)) total number of turns: n=(φ/(2π))=(((μ+tan θ)ω_0 ^2 R)/(8πg(3μ−tan θ) sin θ)) total time: T= t_1 + t_2 =((ω_0 R)/(g(3μ−tan θ) cos θ))+((3(μ−tan θ)ω_0 R)/(2g(3μ−tan θ) sin θ)) =((ω_0 R)/(g(3μ−tan θ)))[(1/(cos θ))+((3(μ−tan θ))/(2 sin θ))] =((ω_0 R)/(2g sin θ)) we see such that this motion is possible, i.e. ω_1 ≥0, we must have μ−tan θ≥0, i.e. μ≥tan θ.

$$\underline{{phase}\:\mathrm{1}:\:{t}=\mathrm{0}\:{to}\:{t}_{\mathrm{1}} } \\ $$$${at}\:{t}=\mathrm{0}\:{the}\:{cylinder}\:{has}\:{angular}\: \\ $$$${velocity},\:{but}\:{no}\:{translational} \\ $$$${velocity},\:{therefore}\:{it}\:{will}\:{roll}\:{and} \\ $$$${slip}\:{simultaneously},\:{till}\:{its}\: \\ $$$${translational}\:{velocity}\:{matches} \\ $$$${its}\:{angular}\:{velocity},\:{i}.{e}.\:{v}_{\mathrm{1}} =\omega_{\mathrm{1}} {R}.\: \\ $$$${after}\:{that}\:{it}\:{will}\:{roll}\:{purely}. \\ $$$$ \\ $$$$\underline{{phase}\:\mathrm{2}:\:{t}={t}_{\mathrm{1}} \:{to}\:{t}_{\mathrm{2}} } \\ $$$${the}\:{cylinder}\:{rolls}\:{without}\:{slipping} \\ $$$${till}\:{it}\:{reaches}\:{its}\:{highest}\:{position}\: \\ $$$${and}\:{stops}\:{momentarily}.\:{then}\: \\ $$$${the}\:{cyclinder}\:{begins}\:{to}\:{roll}\:{down} \\ $$$${the}\:{incline}\:{backwards}. \\ $$$$ \\ $$$$\boldsymbol{{phase}}\:\mathrm{1}: \\ $$$${f}=\mu{N}=\mu{Mg}\:\mathrm{cos}\:\theta \\ $$$${Ma}={f}−{Mg}\:\mathrm{sin}\:\theta={Mg}\left(\mu\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow{a}={g}\left(\mu\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right) \\ $$$${a}=\frac{{v}_{\mathrm{1}} }{{t}_{\mathrm{1}} }\:\Rightarrow{t}_{\mathrm{1}} =\frac{{v}_{\mathrm{1}} }{{g}\left(\mu\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right)} \\ $$$${I}\alpha={fR} \\ $$$$\frac{{MR}^{\mathrm{2}} }{\mathrm{2}}\alpha=\mu{Mg}\:\mathrm{cos}\:\theta{R} \\ $$$$\Rightarrow\alpha=\frac{\mathrm{2}\mu{g}\:\mathrm{cos}\:\theta}{{R}} \\ $$$$\alpha=\frac{{w}_{\mathrm{0}} −\omega_{\mathrm{1}} }{{t}_{\mathrm{1}} }\:\Rightarrow{t}_{\mathrm{1}} =\frac{\left(\omega_{\mathrm{0}} −\omega_{\mathrm{1}} \right){R}}{\mathrm{2}\mu{g}\:\mathrm{cos}\:\theta} \\ $$$$\frac{\left(\omega_{\mathrm{0}} −\omega_{\mathrm{1}} \right){R}}{\mathrm{2}\mu{g}\:\mathrm{cos}\:\theta}=\frac{{v}_{\mathrm{1}} }{{g}\left(\mu\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right)} \\ $$$${with}\:{v}_{\mathrm{1}} =\omega_{\mathrm{1}} {R} \\ $$$$\frac{\left(\omega_{\mathrm{0}} −\omega_{\mathrm{1}} \right){R}}{\mathrm{2}\mu{g}\:\mathrm{cos}\:\theta}=\frac{\omega_{\mathrm{1}} {R}}{{g}\left(\mu\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right)} \\ $$$$\frac{\omega_{\mathrm{0}} −\omega_{\mathrm{1}} }{\mathrm{2}\mu\:\mathrm{cos}\:\theta}=\frac{\omega_{\mathrm{1}} }{\mu\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta} \\ $$$$\Rightarrow\omega_{\mathrm{1}} =\frac{\left(\mu−\mathrm{tan}\:\theta\right)\omega_{\mathrm{0}} }{\mathrm{3}\mu−\mathrm{tan}\:\theta} \\ $$$${rotation}\:{angle}\:{in}\:{phase}\:\mathrm{1}: \\ $$$$\phi_{\mathrm{1}} =\frac{\omega_{\mathrm{0}} ^{\mathrm{2}} −\omega_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}\alpha} \\ $$$$\:\:\:\:\:=\frac{\omega_{\mathrm{0}} ^{\mathrm{2}} {R}}{\mathrm{4}\mu{g}\:\mathrm{cos}\:\theta}\left(\mathrm{1}+\frac{\mu−\mathrm{tan}\:\theta}{\mathrm{3}\mu−\mathrm{tan}\:\theta}\right)\left(\mathrm{1}−\frac{\mu−\mathrm{tan}\:\theta}{\mathrm{3}\mu−\mathrm{tan}\:\theta}\right) \\ $$$$\:\:\:\:\:=\frac{\left(\mathrm{2}\mu−\mathrm{tan}\:\theta\right)\omega_{\mathrm{0}} ^{\mathrm{2}} {R}}{{g}\left(\mathrm{3}\mu−\mathrm{tan}\:\theta\right)^{\mathrm{2}} \:\mathrm{cos}\:\theta} \\ $$$$\:{t}_{\mathrm{1}} =\frac{\left(\omega_{\mathrm{0}} −\omega_{\mathrm{1}} \right){R}}{\mathrm{2}\mu{g}\:\mathrm{cos}\:\theta} \\ $$$$\:\:\:\:=\frac{\omega_{\mathrm{0}} {R}}{\mathrm{2}\mu{g}\:\mathrm{cos}\:\theta}\left(\mathrm{1}−\frac{\mu−\mathrm{tan}\:\theta}{\mathrm{3}\mu−\mathrm{tan}\:\theta}\right) \\ $$$$\:\:\:\:=\frac{\omega_{\mathrm{0}} {R}}{{g}\left(\mathrm{3}\mu−\mathrm{tan}\:\theta\right)\:\mathrm{cos}\:\theta} \\ $$$$ \\ $$$$\boldsymbol{{phase}}\:\mathrm{2}: \\ $$$${energy}\:{conservation}: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{MR}^{\mathrm{2}} \omega_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}}+{Mv}_{\mathrm{1}} ^{\mathrm{2}} \right)={Mg}\phi_{\mathrm{2}} {R}\:\mathrm{sin}\:\theta \\ $$$$\phi_{\mathrm{2}} =\frac{\mathrm{3}\omega_{\mathrm{1}} ^{\mathrm{2}} {R}}{\mathrm{4}{g}\:\mathrm{sin}\:\theta} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{3}\left(\mu−\mathrm{tan}\:\theta\right)^{\mathrm{2}} \omega_{\mathrm{0}} ^{\mathrm{2}} {R}}{\mathrm{4}{g}\left(\mathrm{3}\mu−\mathrm{tan}\:\theta\right)^{\mathrm{2}} \:\mathrm{sin}\:\theta} \\ $$$${or}: \\ $$$$\frac{{MR}^{\mathrm{2}} }{\mathrm{2}}\alpha=−{fR} \\ $$$${Ma}={f}−{Mg}\:\mathrm{sin}\:\theta \\ $$$${since}\:{a}=\alpha{R}, \\ $$$${M}\alpha{R}^{\mathrm{2}} ={fR}−{MgR}\:\mathrm{sin}\:\theta \\ $$$$\frac{{MR}^{\mathrm{2}} }{\mathrm{2}}\alpha+{M}\alpha{R}^{\mathrm{2}} ={MgR}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\alpha=\frac{\mathrm{2}{g}\:\mathrm{sin}\:\theta}{\mathrm{3}{R}} \\ $$$$\phi_{\mathrm{2}} =\frac{\omega_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}\alpha}=\frac{\mathrm{3}{R}}{\mathrm{4}{g}\:\mathrm{sin}\:\theta}\:×\frac{\left(\mu−\mathrm{tan}\:\theta\right)^{\mathrm{2}} \omega_{\mathrm{0}} ^{\mathrm{2}} }{\left(\mathrm{3}\mu−\mathrm{tan}\:\theta\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:=\frac{\mathrm{3}\left(\mu−\mathrm{tan}\:\theta\right)^{\mathrm{2}} \omega_{\mathrm{0}} ^{\mathrm{2}} {R}}{\mathrm{4}{g}\left(\mathrm{3}\mu−\mathrm{tan}\:\theta\right)^{\mathrm{2}} \:\mathrm{sin}\:\theta} \\ $$$$\:{t}_{\mathrm{2}} =\frac{\omega_{\mathrm{1}} }{\alpha}=\frac{\left(\mu−\mathrm{tan}\:\theta\right)\omega_{\mathrm{0}} }{\mathrm{3}\mu−\mathrm{tan}\:\theta}×\frac{\mathrm{3}{R}}{\mathrm{2}{g}\:\mathrm{sin}\:\theta} \\ $$$$\:\:\:=\frac{\mathrm{3}\left(\mu−\mathrm{tan}\:\theta\right)\omega_{\mathrm{0}} {R}}{\mathrm{2}{g}\:\left(\mathrm{3}\mu−\mathrm{tan}\:\theta\right)\:\mathrm{sin}\:\theta} \\ $$$$ \\ $$$${total}\:{rotation}\:{angle}: \\ $$$$\phi=\phi_{\mathrm{1}} +\phi_{\mathrm{2}} \\ $$$$\:\:\:=\frac{\left(\mathrm{2}\mu−\mathrm{tan}\:\theta\right)\omega_{\mathrm{0}} ^{\mathrm{2}} {R}}{{g}\left(\mathrm{3}\mu−\mathrm{tan}\:\theta\right)^{\mathrm{2}} \:\mathrm{cos}\:\theta}+\frac{\mathrm{3}\left(\mu−\mathrm{tan}\:\theta\right)^{\mathrm{2}} \omega_{\mathrm{0}} ^{\mathrm{2}} {R}}{\mathrm{4}{g}\left(\mathrm{3}\mu−\mathrm{tan}\:\theta\right)^{\mathrm{2}} \:\mathrm{sin}\:\theta} \\ $$$$\:\:\:=\frac{\omega_{\mathrm{0}} ^{\mathrm{2}} {R}}{{g}\left(\mathrm{3}\mu−\mathrm{tan}\:\theta\right)^{\mathrm{2}} }\left[\frac{\mathrm{2}\mu−\mathrm{tan}\:\theta}{\mathrm{cos}\:\theta}+\frac{\mathrm{3}\left(\mu−\mathrm{tan}\:\theta\right)^{\mathrm{2}} }{\mathrm{4}\:\mathrm{sin}\:\theta}\right] \\ $$$$\:\:\:=\frac{\left(\mu+\mathrm{tan}\:\theta\right)\omega_{\mathrm{0}} ^{\mathrm{2}} {R}}{\mathrm{4}{g}\left(\mathrm{3}\mu−\mathrm{tan}\:\theta\right)\:\mathrm{sin}\:\theta} \\ $$$$ \\ $$$${total}\:{number}\:{of}\:{turns}: \\ $$$${n}=\frac{\phi}{\mathrm{2}\pi}=\frac{\left(\mu+\mathrm{tan}\:\theta\right)\omega_{\mathrm{0}} ^{\mathrm{2}} {R}}{\mathrm{8}\pi{g}\left(\mathrm{3}\mu−\mathrm{tan}\:\theta\right)\:\mathrm{sin}\:\theta} \\ $$$$ \\ $$$${total}\:{time}: \\ $$$${T}=\:{t}_{\mathrm{1}} +\:{t}_{\mathrm{2}} \\ $$$$\:\:\:=\frac{\omega_{\mathrm{0}} {R}}{{g}\left(\mathrm{3}\mu−\mathrm{tan}\:\theta\right)\:\mathrm{cos}\:\theta}+\frac{\mathrm{3}\left(\mu−\mathrm{tan}\:\theta\right)\omega_{\mathrm{0}} {R}}{\mathrm{2}{g}\left(\mathrm{3}\mu−\mathrm{tan}\:\theta\right)\:\mathrm{sin}\:\theta} \\ $$$$\:\:\:=\frac{\omega_{\mathrm{0}} {R}}{{g}\left(\mathrm{3}\mu−\mathrm{tan}\:\theta\right)}\left[\frac{\mathrm{1}}{\mathrm{cos}\:\theta}+\frac{\mathrm{3}\left(\mu−\mathrm{tan}\:\theta\right)}{\mathrm{2}\:\mathrm{sin}\:\theta}\right] \\ $$$$\:\:\:=\frac{\omega_{\mathrm{0}} {R}}{\mathrm{2}{g}\:\mathrm{sin}\:\theta} \\ $$$$ \\ $$$${we}\:{see}\:{such}\:{that}\:{this}\:{motion}\:{is}\: \\ $$$${possible},\:{i}.{e}.\:\omega_{\mathrm{1}} \geqslant\mathrm{0},\:{we}\:{must}\:{have} \\ $$$$\mu−\mathrm{tan}\:\theta\geqslant\mathrm{0},\:{i}.{e}.\:\mu\geqslant\mathrm{tan}\:\theta. \\ $$

Commented by mr W last updated on 05/Nov/25

Commented by mr W last updated on 05/Nov/25

it′s interesting to note that T=((ω_0 R)/(2g sin θ)), which is independent from μ and is equal to the time which the same cylinder takes when it rolls upwards the incline with an initial angular velocity ω_0 and without slipping.

$${it}'{s}\:{interesting}\:{to}\:{note}\:{that} \\ $$$${T}=\frac{\omega_{\mathrm{0}} {R}}{\mathrm{2}{g}\:\mathrm{sin}\:\theta},\:{which}\:{is}\:{independent} \\ $$$${from}\:\mu\:{and}\:{is}\:{equal}\:{to}\:{the}\:{time} \\ $$$${which}\:{the}\:{same}\:{cylinder}\:{takes}\: \\ $$$${when}\:{it}\:{rolls}\:{upwards}\:{the}\:{incline} \\ $$$${with}\:{an}\:{initial}\:{angular}\:{velocity}\:\omega_{\mathrm{0}} \\ $$$${and}\:{without}\:{slipping}. \\ $$

Commented by ajfour last updated on 05/Nov/25