With variable masses\Large{\blue{With\space variable \space masses}}

In the above pulley mass system mass mm decreases with a constant rate of ww.

At time=0s,m=Mm=Mwttime=0s,m=M\Rightarrow m=M-wt

If x,yx,y are the horizontal distances of m,Mm,M from the pulley respectively and TT is the tension in the string y+x=constant\Rightarrow y+x=constant y˙=x˙\Rightarrow\dot{y}=-\dot{x} y¨=x¨\Rightarrow \ddot{y}=-\ddot{x} Applying newtons second law on each massApplying \space newton's \space second \space law \space on \space each \space mass My¨=MgT..........[1]M\ddot{y}=Mg-T..........[1] dmx˙dt=mgT\dfrac{dm\dot{x}}{dt}=mg-T dmy˙dt=Tmg..........[2]\Rightarrow \dfrac{dm\dot{y}}{dt}=T-mg..........[2] [1]+[2]dmy˙dt+My¨=(Mm)g[1]+[2]\Rightarrow\dfrac{dm\dot{y}}{dt}+M\ddot{y}=(M-m)g d[(Mwt)y˙]dt+My¨=wtg\Rightarrow \dfrac{d[(M-wt)\dot{y}]}{dt}+M\ddot{y}=wtg t=0t=t1d[(Mwt)y˙]+t=0t=t1My¨dt=t=0t=t1wtgdt\Rightarrow \int_{t=0}^{t=t_1}{d[(M-wt)\dot{y}]}+\int_{t=0}^{t=t_1}M\ddot{y}{dt}=\int_{t=0}^{t=t_1}wtg{dt} My˙+(Mwt1)y˙=wt12g2\Rightarrow M\dot{y}+(M-wt_1)\dot{y}=\dfrac{wt_1^2g}{2} Since we can replace t1 with tSince \space we \space can \space replace \space t_1 \space with\space t y˙=wt2g2(2Mwt)\Rightarrow\dot{y}=\dfrac{wt^2g}{2(2M-wt)} y=wg20t1t22Mwtdt\Rightarrow \triangle y=\dfrac{wg}{2}\int_{0}^{t_1}\dfrac{t^2}{2M-wt}dt let I=t2abtdtlet \space I=\int\dfrac{t^2}{\red{a-bt}}\red{dt} =t2bdlnabt   dtabt=dlnabtb=-\int\dfrac{t^2}{b}\red{d\ln|a-bt|}\space\space\space\blue{\because\dfrac{dt}{a-bt}=-\dfrac{-d\ln|a-bt|}{b}} =1b(t2lna2b2tlnabtdt)=-\dfrac{1}{b}\left({t^2\ln|a-2b|} -2\int t\ln|a-bt|dt\right) =1b(2tlnabtdtt2lna2b)= \dfrac{1}{b}\left(\red{2\int t\ln|a-bt|dt}-{t^2\ln|a-2b|} \right) =1b(Jt2lna2b)=\dfrac{1}{b}\left(\red{J}-{t^2\ln|a-2b|} \right) J2=tlnabtdt\dfrac{J}{2}=\int t\ln|a-bt|dt d[xlnabt]abdlnabtdt=lnabtdt\blue{\because d[x\ln|a-bt|]-\dfrac{a}{b}d\ln|a-bt|-dt=\ln|a-bt|dt} J2=td[xlnabt]abtdlnabttdt\therefore \dfrac{J}{2}=\int td[x\ln|a-bt|] -\dfrac{a}{b}\int td\ln|a-bt|-\int tdt =(t2lnabttlnabtdt)ab(tlnabtlnabtdt)t22+C=\left(t^2\ln|a-bt|-\red{\int t\ln|a-bt|dt}\right)-\dfrac{a}{b}\left(t\ln|a-bt|-\int \ln|a-bt|dt\right)-\dfrac{t^2}{2}+\blue{C} J2=t2lnabtJ2ab(tlnabt(tab)lnabt+t)t22+C\Rightarrow\red{\dfrac{J}{2}}=t^2\ln|a-bt|-\red{\dfrac{J}{2}}-\dfrac{a}{b}\left(\cancel{t\ln|a-bt|}-\left(\cancel{t}-\dfrac{a}{b}\right)\ln|a-bt|+t\right)-\dfrac{t^2}{2}+\blue{C} J=t2lnabta2b2lnabtatbt22+C\Rightarrow J=t^2\ln|a-bt|-\dfrac{a^2}{b^2}\ln|a-bt|-\dfrac{at}{b}-\dfrac{t^2}{2}+\blue{C} J=(t2a2b2)lnabt(atb+t22)+C\Rightarrow J=\left(t^2-\dfrac{a^2}{b^2}\right)\ln|a-bt|-\left(\dfrac{at}{b}+\dfrac{t^2}{2}\right)+\blue{C} I=1b((t2a2b2)lnabt(atb+t22)+Ct2lna2b)\Rightarrow I=\dfrac{1}{b}\left(\red{\left(\cancel{t^2}-\dfrac{a^2}{b^2}\right)\ln|a-bt|-\left(\dfrac{at}{b}+\dfrac{t^2}{2}\right)+{C}}-\cancel{t^2\ln|a-2b|} \right) =1b(a2b2lnabt+atb+t22+C)=\dfrac{-1}{b}\left(\dfrac{ a^2}{b^2}\ln| a-bt|+\dfrac{ at}{b}+\dfrac{t^2}{2}+{C} \right) =1w(4M2w2ln2Mwt+2Mtw+t22+C)=\dfrac{-1}{w}\left(\dfrac{4M^2}{w^2}\ln|2M-wt|+\dfrac{2Mt}{w}+\dfrac{t^2}{2}+{C} \right) y=wg2[I]0t\Rightarrow \triangle y=\dfrac{wg}{2}\red{[I]_{0}^{t}} =wg2w(4M2w2ln(2M2Mwt)2Mtwt22)=\dfrac{\cancel{w}g}{2\cancel{w}}\left(\dfrac{4M^2}{w^2}\ln\left(\left|\dfrac{2M}{2M-wt}\right|\right) -\dfrac{2Mt}{w}-\dfrac{t^2}{2}\right) y=g(2M2w2ln(2M2Mwt)Mtwt24)\Rightarrow \boxed{\triangle y={g}\left(\dfrac{2M^2}{w^2}\ln\left(\left|\dfrac{2M}{2M-wt}\right|\right) -\dfrac{Mt}{w}-\dfrac{t^2}{4}\right)}

Note by Zakir Husain
3 months, 1 week ago

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Bonus Problem\Large{\boxed{Bonus\space Problem}} What if w is not constant but instead is a function of time?\boxed{\red{What \space if\space w \space is\space not\space constant\space but\space instead \space is \space a\space function\space of\space time ?}} (Answer is in the reply)

Zakir Husain - 3 months, 1 week ago

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y=gt=0t=t1fdt2Mf\triangle y = g\int_{t=0}^{t=t_1}\dfrac{fdt}{2M-f'} where f=wdtdtwhere\space f=\iint wdt dt

Zakir Husain - 3 months, 1 week ago

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This is good mathematical physics. I miss doing proper physics and maths 😭 school takes a lot of my time

Krishna Karthik - 3 months ago

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Bonus Problem\boxed{\Large{\purple{Bonus \space Problem}}} What if the pulley have a considerable moment of inertia = I?\boxed{What\space if\space the \space pulley \space have \space a \space considerable \space moment \space of \space inertia\space = \space I ?}

Zakir Husain - 2 months, 3 weeks ago

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