Question Number 23073 by ajfour last updated on 25/Oct/17
Commented by ajfour last updated on 25/Oct/17
$${Q}.\mathrm{23059}\:\left({solution}\right) \\ $$
Answered by ajfour last updated on 25/Oct/17
![Let acc. of spool down the incline be a and let A be acc. of plank down the incline. Then as the rope is not in speed, the point of spool where rope touches it finally is zero. ⇒ a−𝛂R=0 ....(i) Further there is no slipping b/w plank and spool, So A=a−2𝛂R ...(ii) Torque eq. TR−f(2R)=I𝛂 ....(iii) Force eqns. Mgsin 𝛉−f =MA ...(iv) mgsin 𝛉+f−T = ma ...(v) Now from (i) αR=a using this in (ii): A=−a from (iii) T=((Iα)/R)+2f =((Ia)/R^2 )+2f using in (v): mgsin θ+f−2f−((Ia)/R^2 ) =ma ⇒ mgsin θ−f= ((Ia)/R^2 )+ma subtracting eq. (iv) mgsin θ−Mgsin θ=((Ia)/R^2 )+ma−MA as A=−a=−αR we get (m−M)gsin θ=a(M+m+(I/R^2 )) 𝛂=(a/R) =[(((m−M)gsin 𝛉)/((M+m)R^2 +I))]R . A=−a =(((M−m)gR^2 sin 𝛉)/((M+m)R^2 +I)) .](https://www.tinkutara.com/question/Q23076.png)
$${Let}\:{acc}.\:{of}\:{spool}\:{down}\:{the}\:{incline} \\ $$$${be}\:\boldsymbol{{a}}\:{and}\:{let}\:\boldsymbol{{A}}\:{be}\:{acc}.\:{of}\:{plank} \\ $$$${down}\:{the}\:{incline}. \\ $$$${Then}\:\:{as}\:{the}\:{rope}\:{is}\:{not}\:{in}\:{speed}, \\ $$$${the}\:{point}\:{of}\:{spool}\:{where}\:{rope} \\ $$$${touches}\:{it}\:{finally}\:{is}\:{zero}. \\ $$$$\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\:\:\:\:\boldsymbol{{a}}−\boldsymbol{\alpha{R}}=\mathrm{0}\:\:\:\:….\left(\boldsymbol{{i}}\right) \\ $$$${Further}\:{there}\:{is}\:{no}\:{slipping}\:{b}/{w} \\ $$$${plank}\:{and}\:{spool},\:{So} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{A}}=\boldsymbol{{a}}−\mathrm{2}\boldsymbol{\alpha{R}}\:\:\:\:\:…\left(\boldsymbol{{ii}}\right) \\ $$$${Torque}\:{eq}.\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{TR}}−\boldsymbol{{f}}\left(\mathrm{2}\boldsymbol{{R}}\right)=\boldsymbol{{I}\alpha}\:\:\:\:\:….\left(\boldsymbol{{iii}}\right) \\ $$$$\:{Force}\:{eqns}. \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{Mg}}\mathrm{sin}\:\boldsymbol{\theta}−\boldsymbol{{f}}\:=\boldsymbol{{MA}}\:\:\:\:\:…\left(\boldsymbol{{iv}}\right) \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{mg}}\mathrm{sin}\:\boldsymbol{\theta}+\boldsymbol{{f}}−\boldsymbol{{T}}\:=\:\boldsymbol{{ma}}\:\:\:…\left(\boldsymbol{{v}}\right) \\ $$$${Now}\:{from}\:\left({i}\right)\:\:\:\:\alpha{R}={a} \\ $$$${using}\:{this}\:{in}\:\left({ii}\right): \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{A}=−{a}\: \\ $$$${from}\:\left({iii}\right)\:\:\:\:{T}=\frac{{I}\alpha}{{R}}+\mathrm{2}{f}\:=\frac{{Ia}}{{R}^{\mathrm{2}} }+\mathrm{2}{f} \\ $$$$\:\:{using}\:{in}\:\left({v}\right): \\ $$$${mg}\mathrm{sin}\:\theta+{f}−\mathrm{2}{f}−\frac{{Ia}}{{R}^{\mathrm{2}} }\:={ma} \\ $$$$\Rightarrow\:\:\:\:\:{mg}\mathrm{sin}\:\theta−{f}=\:\frac{{Ia}}{{R}^{\mathrm{2}} }+{ma}\:\: \\ $$$${subtracting}\:{eq}.\:\left({iv}\right)\: \\ $$$${mg}\mathrm{sin}\:\theta−{Mg}\mathrm{sin}\:\theta=\frac{{Ia}}{{R}^{\mathrm{2}} }+{ma}−{MA} \\ $$$${as}\:{A}=−{a}=−\alpha{R}\:\:{we}\:{get} \\ $$$$\left({m}−{M}\right){g}\mathrm{sin}\:\theta={a}\left({M}+{m}+\frac{{I}}{{R}^{\mathrm{2}} }\right) \\ $$$$\:\:\boldsymbol{\alpha}=\frac{\boldsymbol{{a}}}{\boldsymbol{{R}}}\:=\left[\frac{\left(\boldsymbol{{m}}−\boldsymbol{{M}}\right)\boldsymbol{{g}}\mathrm{sin}\:\boldsymbol{\theta}}{\left(\boldsymbol{{M}}+\boldsymbol{{m}}\right)\boldsymbol{{R}}^{\mathrm{2}} +\boldsymbol{{I}}}\right]\boldsymbol{{R}}\:. \\ $$$$\:\boldsymbol{{A}}=−\boldsymbol{{a}}\:=\frac{\left(\boldsymbol{{M}}−\boldsymbol{{m}}\right)\boldsymbol{{gR}}^{\mathrm{2}} \mathrm{sin}\:\boldsymbol{\theta}}{\left(\boldsymbol{{M}}+\boldsymbol{{m}}\right)\boldsymbol{{R}}^{\mathrm{2}} +\boldsymbol{{I}}}\:. \\ $$
Commented by Physics lover last updated on 25/Oct/17
$${OMG},\:{Sir},\:{thats}\:{too}\:{good}. \\ $$$${thank}\:{you}\:{so}\:{much}.\:{I}\:{really}\: \\ $$$${appreciate}\:{your}\:{work}\:{on}\:{this}\: \\ $$$${forum}\:. \\ $$