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Question Number 71611 by ajfour last updated on 17/Oct/19

Commented by ajfour last updated on 17/Oct/19

Find time it takes the cylinder  to unwind and slide and hit the  ground.

Findtimeittakesthecylindertounwindandslideandhittheground.

Commented by ajfour last updated on 18/Oct/19

Answered by Tanmay chaudhury last updated on 18/Oct/19

mg−Tcosα=ma  Tsinα=N  T.r=I((a/r))→T=((Ia)/r^2 )  mg−((Iacosα)/r^2 )=ma  a(((Icosα)/r^2 )+m)=mg  a=((mg)/(((Icosα)/r^2 )+m))→h=(1/2)at^2 →t=(√((2h)/a))  t=(√(((2h)/(mg))×(m+((Icosα)/r^2 )) ))  t=(√(((2h)/(mg))×(m+((mr^2 cosα)/(2r^2 ))))) [I=((mr^2 )/2)]  t=(√(((2h)/(mg))×m(1+((r^2 cosα)/(2r^2 )))))   t=(√(((2h)/g)×(1+((cosα)/2))))     pls check is it correct...

mgTcosα=maTsinα=NT.r=I(ar)T=Iar2mgIacosαr2=maa(Icosαr2+m)=mga=mgIcosαr2+mh=12at2t=2hat=2hmg×(m+Icosαr2)t=2hmg×(m+mr2cosα2r2)[I=mr22]t=2hmg×m(1+r2cosα2r2)t=2hg×(1+cosα2)plscheckisitcorrect...

Commented by Tanmay chaudhury last updated on 18/Oct/19

do friction present...pls clarify...because i have  ignored friction

dofrictionpresent...plsclarify...becauseihaveignoredfriction

Commented by ajfour last updated on 19/Oct/19

frictionless, yes, but α goes  changing sir.

frictionless,yes,butαgoeschangingsir.

Answered by mr W last updated on 19/Oct/19

Commented by mr W last updated on 19/Oct/19

i use β to replace α in the question  AC=h+(r/(tan (β/2)))=s+(r/(tan (θ/2)))=k  tan (θ/2)=(r/(k−s))  cos θ=((1−((r/(k−s)))^2 )/(1+((r/(k−s)))^2 ))=1−((2r^2 )/((k−s)^2 +r^2 ))  mg−T cos θ=ma  Tr=Iα  I=((mr^2 )/2)  α=(a/r)  Tr=((mr^2 )/2)×(a/r)  T=((ma)/2)  mg−((ma cos θ)/2)=ma  ⇒a=((2g)/(2+cos θ))=((2g)/(3−((2r^2 )/((k−s)^2 +r^2 ))))=((2g[(k−s)^2 +r^2 ])/(3(k−s)^2 +r^2 ))  ⇒−v(dv/ds)=2g×(((k−s)^2 +r^2 )/(3(k−s)^2 +r^2 ))  ⇒∫_0 ^v vdv=−2g∫_h ^s (((k−s)^2 +r^2 )/(3(k−s)^2 +r^2 ))ds  ⇒(v^2 /2)=2g∫_h ^s (((k−s)^2 +r^2 )/(3(k−s)^2 +r^2 ))d(k−s)  ⇒(v^2 /(4g))=∫_(k−h) ^(k−s) ((u^2 +r^2 )/(3u^2 +r^2 ))du  ⇒(v^2 /(4g))=(1/3)[((2r)/(√3)) tan^(−1) (((√3)u)/r)+u]_(k−h) ^(k−s)   ⇒(v^2 /(4g))=(1/3)[((2r)/(√3))(tan^(−1) ((√3)/(tan (θ/2)))−tan^(−1) ((√3)/(tan (β/2))))+(r/(tan (θ/2)))−(r/(tan (β/2)))]  ⇒((3v^2 )/(4gr))=(2/(√3))(tan^(−1) ((√3)/(tan (θ/2)))−tan^(−1) ((√3)/(tan (β/2))))+(1/(tan (θ/2)))−(1/(tan (β/2)))  ⇒v=((2(√(gr)))/(√3))(√((2/(√3))(tan^(−1) ((√3)/(tan (θ/2)))−tan^(−1) ((√3)/(tan (β/2))))+(1/(tan (θ/2)))−(1/(tan (β/2)))))  ⇒(ds/dt)=((2(√(gr)))/(√3))(√((2/(√3))(tan^(−1) (((√3)(k−s))/r)−tan^(−1) ((√3)/(tan (β/2))))+((k−s)/r)−(1/(tan (β/2)))))  ⇒((2(√(gr)))/(√3))∫_0 ^t dt=∫_h ^r (ds/(√((2/(√3))(tan^(−1) (((√3)(k−s))/r)−tan^(−1) ((√3)/(tan (β/2))))+((k−s)/r)−(1/(tan (β/2))))))  ⇒t=((√3)/(2(√(gr))))∫_h ^r (ds/(√((2/(√3))(tan^(−1) (((√3)(k−s))/r)−tan^(−1) ((√3)/(tan (β/2))))+((k−s)/r)−(1/(tan (β/2))))))  ......

iuseβtoreplaceαinthequestionAC=h+rtanβ2=s+rtanθ2=ktanθ2=rkscosθ=1(rks)21+(rks)2=12r2(ks)2+r2mgTcosθ=maTr=IαI=mr22α=arTr=mr22×arT=ma2mgmacosθ2=maa=2g2+cosθ=2g32r2(ks)2+r2=2g[(ks)2+r2]3(ks)2+r2vdvds=2g×(ks)2+r23(ks)2+r20vvdv=2ghs(ks)2+r23(ks)2+r2dsv22=2ghs(ks)2+r23(ks)2+r2d(ks)v24g=khksu2+r23u2+r2duv24g=13[2r3tan13ur+u]khksv24g=13[2r3(tan13tanθ2tan13tanβ2)+rtanθ2rtanβ2]3v24gr=23(tan13tanθ2tan13tanβ2)+1tanθ21tanβ2v=2gr323(tan13tanθ2tan13tanβ2)+1tanθ21tanβ2dsdt=2gr323(tan13(ks)rtan13tanβ2)+ksr1tanβ22gr30tdt=hrds23(tan13(ks)rtan13tanβ2)+ksr1tanβ2t=32grhrds23(tan13(ks)rtan13tanβ2)+ksr1tanβ2......

Commented by ajfour last updated on 19/Oct/19

Commented by ajfour last updated on 19/Oct/19

Sir, it seems to me that      vcos θ=ωr   , please comment...

Sir,itseemstomethatvcosθ=ωr,pleasecomment...

Commented by mr W last updated on 19/Oct/19

i thought exactly so at the begining,  then i changed my mind. let′s say  the length of wire is AD=l,  l+s=k=constant  the speed of unwinding ωr=(dl/dt)=−(ds/dt)=v.  actually i am still confused what is  really true.

ithoughtexactlysoatthebegining,thenichangedmymind.letssaythelengthofwireisAD=l,l+s=k=constantthespeedofunwindingωr=dldt=dsdt=v.actuallyiamstillconfusedwhatisreallytrue.

Commented by ajfour last updated on 19/Oct/19

same here Sir, confused since  last night..

samehereSir,confusedsincelastnight..

Commented by mr W last updated on 20/Oct/19

Commented by mr W last updated on 20/Oct/19

i have clarified:  let AB=s  v=(ds/dt)  rotation of cylinder dϕ=(θ+dθ)+φ−θ=(ds/r)+dθ  with φ=(ds/r)=rotation due to unwinding  ω=(dϕ/dt)=(v/r)+(dθ/dt)=(v/r)+v(dθ/ds)  tan (θ/2)=(r/s)  (dθ/(2 cos^2  (θ/2)))=−((rds)/s^2 )=−tan^2  (θ/2)(ds/r)  ⇒(dθ/ds)=−(1/r)2 sin^2  (θ/2)  ⇒ω=(v/r)+v(dθ/ds)=(v/r)(1−2 sin^2  (θ/2))=((v cos θ)/r)  ⇒rω=v cos θ

ihaveclarified:letAB=sv=dsdtrotationofcylinderdφ=(θ+dθ)+ϕθ=dsr+dθwithϕ=dsr=rotationduetounwindingω=dφdt=vr+dθdt=vr+vdθdstanθ2=rsdθ2cos2θ2=rdss2=tan2θ2dsrdθds=1r2sin2θ2ω=vr+vdθds=vr(12sin2θ2)=vcosθrrω=vcosθ

Commented by mr W last updated on 20/Oct/19

new attempt:  s_0 =(r/(tan (β/2)))  s_1 =(r/(tan (β/2)))+h−r  mg(s−s_0 )=(1/2)mv^2 +(1/2)Iω^2   g(s−s_0 )=(1/2)v^2 +(r^2 /4)ω^2   g(s−s_0 )=(1/4)v^2 (2+cos^2  θ)  tan (θ/2)=(r/s)  cos θ=((1−((r/s))^2 )/(1+((r/s))^2 ))=((s^2 −r^2 )/(s^2 +r^2 ))  g(s−s_0 )=(1/4)v^2 [2+(((s^2 −r^2 )/(s^2 +r^2 )))^2 ]  ⇒v=2(√((g(s−s_0 ))/(2+(((s^2 −r^2 )/(s^2 +r^2 )))^2 )))  ⇒(ds/dt)=2(√g)(√((s−s_0 )/(2+(((s^2 −r^2 )/(s^2 +r^2 )))^2 )))  ⇒t=(1/(2(√g)))∫_s_0  ^s_1  (√((2+(((s^2 −r^2 )/(s^2 +r^2 )))^2 )/(s−s_0 )))ds

newattempt:s0=rtanβ2s1=rtanβ2+hrmg(ss0)=12mv2+12Iω2g(ss0)=12v2+r24ω2g(ss0)=14v2(2+cos2θ)tanθ2=rscosθ=1(rs)21+(rs)2=s2r2s2+r2g(ss0)=14v2[2+(s2r2s2+r2)2]v=2g(ss0)2+(s2r2s2+r2)2dsdt=2gss02+(s2r2s2+r2)2t=12gs0s12+(s2r2s2+r2)2ss0ds

Commented by ajfour last updated on 21/Oct/19

Thanks sir, i cant find much  time to attempt myself sir, i  shall however try it today..

Thankssir,icantfindmuchtimetoattemptmyselfsir,ishallhowevertryittoday..

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