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Question Number 130350 by ajfour last updated on 24/Jan/21
Commented by ajfour last updated on 24/Jan/21
A hollow cylinder (radius R, mass M) rolling down a long incline, has a small ball at an angular position 𝛗. Determine 𝛗 in terms of α, g, m, M, and R.
Ahollowcylinder(radiusR,massM)rollingdownalongincline,hasasmallballatanangularpositionϕ.Determineϕintermsofα,g,m,M,andR.
Answered by mr W last updated on 25/Jan/21
Commented by mr W last updated on 25/Jan/21
N cos φ=mg cos β ⇒N=((cos β)/(cos φ))mg N sin φ+mg sin β=ma ⇒a=(tan φ cos β+sin β)g I_1 =I_0 +MR^2 =MR^2 +MR^2 =2MR^2 I_1 α=Mg R sin β−NR sin φ 2MR^2 α=Mg R sin β−((cos β)/(cos φ))mgRsin φ ⇒α=(sin β−(m/M) cos β tan φ)(g/(2R)) a=αR(1−cos φ)=(g/2)(sin β−(m/M) cos β tan φ)(1−cos φ) (g/2)(sin β−(m/M) cos β tan φ)(1−cos φ)=(tan φ cos β+sin β)g (sin β−(m/M) cos β tan φ)(1−cos φ)=2(tan φ cos β+sin β) (tan β−(m/M)tan φ)(1−cos φ)=2(tan φ+tan β) ⇒(m/M) sin φ=(2+(m/M))tan φ+(1+cos φ)tan β
Ncosϕ=mgcosβ⇒N=cosβcosϕmgNsinϕ+mgsinβ=ma⇒a=(tanϕcosβ+sinβ)gI1=I0+MR2=MR2+MR2=2MR2I1α=MgRsinβ−NRsinϕ2MR2α=MgRsinβ−cosβcosϕmgRsinϕ⇒α=(sinβ−mMcosβtanϕ)g2Ra=αR(1−cosϕ)=g2(sinβ−mMcosβtanϕ)(1−cosϕ)g2(sinβ−mMcosβtanϕ)(1−cosϕ)=(tanϕcosβ+sinβ)g(sinβ−mMcosβtanϕ)(1−cosϕ)=2(tanϕcosβ+sinβ)(tanβ−mMtanϕ)(1−cosϕ)=2(tanϕ+tanβ)⇒mMsinϕ=(2+mM)tanϕ+(1+cosϕ)tanβ
Commented by ajfour last updated on 26/Jan/21
Sir i think, a=αR
Sirithink,a=αR
Commented by mr W last updated on 26/Jan/21
you are right sir!
youarerightsir!
Commented by ajfour last updated on 26/Jan/21
MgRsin β−NRsin φ=2MR^2 α Ncos φ=mgcos β Nsin φ+mgsin β=ma a=αR −−−−−−−−−−−−−−− MgRsin β−mgRcos βtan φ =2MR(gsin β+gcos βtan φ) tan φ=((−Msin β)/((2M+m)cos β)) ⇒ tan φ=−((tan β)/((2+(m/M))))
MgRsinβ−NRsinϕ=2MR2αNcosϕ=mgcosβNsinϕ+mgsinβ=maa=αR−−−−−−−−−−−−−−−MgRsinβ−mgRcosβtanϕ=2MR(gsinβ+gcosβtanϕ)tanϕ=−Msinβ(2M+m)cosβ⇒tanϕ=−tanβ(2+mM)
Commented by ajfour last updated on 26/Jan/21
but then either, φ<0 or φ>(π/2) I wonder which one? what if m→0 ?
buttheneither,ϕ<0orϕ>π2Iwonderwhichone?whatifm→0?
Commented by mr W last updated on 26/Jan/21
since N must be >0, so φ<0 is correct.
sinceNmustbe>0,soϕ<0iscorrect.
Commented by ajfour last updated on 26/Jan/21
(N/m)=((gcos β)/(cos φ))=((a−gsin β)/(sin φ)) a=((Mgsin β)/((M+(I_(cm) /R^2 ))))=((gsin β)/2) ⇒ tan φ=−((tan β)/2) But (N/m)>0 ⇒ ((gcos β)/(cos φ))>0 and even (N/m)>0 ⇒ ((−gsin β)/(sin φ))>0 ⇒ cos φ>0 , sin φ<0 ⇒ −(π/2) <−tan^(−1) (((tan β)/2))= φ < 0
Nm=gcosβcosϕ=a−gsinβsinϕa=Mgsinβ(M+IcmR2)=gsinβ2⇒tanϕ=−tanβ2ButNm>0⇒gcosβcosϕ>0andevenNm>0⇒−gsinβsinϕ>0⇒cosϕ>0,sinϕ<0⇒−π2<−tan−1(tanβ2)=ϕ<0
Answered by mr W last updated on 26/Jan/21
Commented by mr W last updated on 26/Jan/21
the question is the same as if a small mass is suspended with a string on the center of the cylinder. say the length of the string is r. N cos φ=mg cos β ⇒N=((cos β)/(cos φ))mg mg sin β−N sin φ=ma ⇒a=(sin β−cos β tan φ)g I_1 =I_0 +MR^2 =2MR^2 I_1 α=Mg R sin β+NRsin φ ⇒α=(sin β+(m/M)×cos β tan φ)(g/(2R)) a=αR (sin β+(m/M)×cos β tan φ)(g/2)=(sin β−cos β tan φ)g (2+(m/M)) tan φ=tan β ⇒tan φ=((tan β)/(2+(m/M))) ⇒φ=tan^(−1) (((tan β)/(2+(m/M)))) it is to see that φ<β and this is independent from the length of the string r! in our case r=R.
thequestionisthesameasifasmallmassissuspendedwithastringonthecenterofthecylinder.saythelengthofthestringisr.Ncosϕ=mgcosβ⇒N=cosβcosϕmgmgsinβ−Nsinϕ=ma⇒a=(sinβ−cosβtanϕ)gI1=I0+MR2=2MR2I1α=MgRsinβ+NRsinϕ⇒α=(sinβ+mM×cosβtanϕ)g2Ra=αR(sinβ+mM×cosβtanϕ)g2=(sinβ−cosβtanϕ)g(2+mM)tanϕ=tanβ⇒tanϕ=tanβ2+mM⇒ϕ=tan−1(tanβ2+mM)itistoseethatϕ<βandthisisindependentfromthelengthofthestringr!inourcaser=R.
Answered by mr W last updated on 27/Jan/21
Commented by mr W last updated on 28/Jan/21
friction coefficient between ball and cylinder =μ N cos φ+μN sin φ=mg cos β ⇒N=((cos β)/(cos φ+μ sin φ))mg ma=N sin φ−μN cos φ+mg sin β ⇒a=[((cos β (tan φ−μ))/(1+μ tan φ))+sin β]g 2MR^2 α=MgR sin β−NR sin φ−μNR(1−cos φ) 2Rα={sin β−(m/M)×((cos β(μ+sin φ−μ cos φ))/(cos φ+μ sin φ))}g 2a={sin β−(m/M)×((cos β(μ+sin φ−μ cos φ))/(cos φ+μ sin φ))}g ((2(tan φ−μ))/(1+μ tan φ))+2tan β=tan β−(m/M)×((μ+sin φ−μ cos φ)/(cos φ+μ sin φ)) ⇒((((mμ)/(Mcos φ))+(2+(m/M))(tan φ−μ))/(1+μ tan φ))=−tan β let λ=(m/M), η=tan β ((λμ+(2+λ)(sin φ−μ cos φ))/(cos φ+μ sin φ))=−η (2+λ+μη)sin φ−(2μ+λμ−η) cos φ=−λμ let δ=tan^(−1) ((2μ+λμ−η)/(2+λ+μη)) sin (φ−δ)=−((λμ)/( (√((2+λ+μη)^2 +(2μ+λμ−η)^2 )))) φ=tan^(−1) ((2μ+λμ−η)/(2+λ+μη))−sin^(−1) ((λμ)/( (√((2+λ+μη)^2 +(2μ+λμ−η)^2 )))) ⇒φ=tan^(−1) (((2+(m/M))μ−tan β)/(2+(m/M)+μ tan β))−sin^(−1) (((μm)/M)/( (√((1+μ^2 ) [tan^2 β+(2+(m/M))^2 ])))) examples: β=30°, (m/M)=0.2, μ=1 ⇒φ=26.7306° β=30°, (m/M)=0.2, μ=0.5 ⇒φ=9.6067° β=30°, (m/M)=0.2, μ=((√3)/6)=0.289 ⇒φ=0 β=30°, (m/M)=0.2, μ=0 ⇒φ=−14.7047°
frictioncoefficientbetweenballandcylinder=μNcosϕ+μNsinϕ=mgcosβ⇒N=cosβcosϕ+μsinϕmgma=Nsinϕ−μNcosϕ+mgsinβ⇒a=[cosβ(tanϕ−μ)1+μtanϕ+sinβ]g2MR2α=MgRsinβ−NRsinϕ−μNR(1−cosϕ)2Rα={sinβ−mM×cosβ(μ+sinϕ−μcosϕ)cosϕ+μsinϕ}g2a={sinβ−mM×cosβ(μ+sinϕ−μcosϕ)cosϕ+μsinϕ}g2(tanϕ−μ)1+μtanϕ+2tanβ=tanβ−mM×μ+sinϕ−μcosϕcosϕ+μsinϕ⇒mμMcosϕ+(2+mM)(tanϕ−μ)1+μtanϕ=−tanβletλ=mM,η=tanβλμ+(2+λ)(sinϕ−μcosϕ)cosϕ+μsinϕ=−η(2+λ+μη)sinϕ−(2μ+λμ−η)cosϕ=−λμletδ=tan−12μ+λμ−η2+λ+μηsin(ϕ−δ)=−λμ(2+λ+μη)2+(2μ+λμ−η)2ϕ=tan−12μ+λμ−η2+λ+μη−sin−1λμ(2+λ+μη)2+(2μ+λμ−η)2⇒ϕ=tan−1(2+mM)μ−tanβ2+mM+μtanβ−sin−1μmM(1+μ2)[tan2β+(2+mM)2]examples:β=30°,mM=0.2,μ=1⇒ϕ=26.7306°β=30°,mM=0.2,μ=0.5⇒ϕ=9.6067°β=30°,mM=0.2,μ=36=0.289⇒ϕ=0β=30°,mM=0.2,μ=0⇒ϕ=−14.7047°
Commented by ajfour last updated on 28/Jan/21
Great generalisation, Sir. Thanks for the analysis.
Greatgeneralisation,Sir.Thanksfortheanalysis.

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