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Question-226334

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Question Number 226334 by fantastic2 last updated on 25/Nov/25
Commented by mr W last updated on 25/Nov/25
is the ground smooth or the roller rolls on the ground without slipping either?
$${is}\:{the}\:{ground}\:{smooth}\:{or}\:{the}\:{roller} \\ $$$${rolls}\:{on}\:{the}\:{ground}\:{without}\:{slipping} \\ $$$${either}? \\ $$
Commented by fantastic2 last updated on 26/Nov/25
the roller doesnot slip
$${the}\:{roller}\:{doesnot}\:{slip} \\ $$
Answered by mr W last updated on 26/Nov/25
Commented by mr W last updated on 26/Nov/25
case 1 no slipping between plank and roller no slipping between roller and ground A=acceleration of plank a=acceleration of center of cylinder α=angular acceleration of cylinder α=(a/R) A=a+αR=2a m_2 A=F−f_2 ⇒f_2 =F−2m_2 a ...(i) m_1 a=f_2 +f_1 ...(ii) ((m_1 R^2 )/2)α=R(f_2 −f_1 ) ((m_1 a)/2)=f_2 −f_1 ...(iii) (ii)+(iii): ((3m_1 a)/2)=2f_2 =2(F−2m_2 a) (3m_1 +8m_2 )a=4F ⇒a=((4F)/(3m_1 +8m_2 )) ⇒α=((4F)/(R(3m_1 +8m_2 ))) ⇒f_2 =((3m_1 F)/(3m_1 +8m_2 ))
$${case}\:\mathrm{1} \\ $$$${no}\:{slipping}\:{between}\:{plank}\:{and}\:{roller} \\ $$$${no}\:{slipping}\:{between}\:{roller}\:{and}\:{ground} \\ $$$$ \\ $$$${A}={acceleration}\:{of}\:{plank} \\ $$$${a}={acceleration}\:{of}\:{center}\:{of}\:{cylinder} \\ $$$$\alpha={angular}\:{acceleration}\:{of}\:{cylinder} \\ $$$$\alpha=\frac{{a}}{{R}} \\ $$$${A}={a}+\alpha{R}=\mathrm{2}{a} \\ $$$${m}_{\mathrm{2}} {A}={F}−{f}_{\mathrm{2}} \\ $$$$\Rightarrow{f}_{\mathrm{2}} ={F}−\mathrm{2}{m}_{\mathrm{2}} {a}\:\:\:…\left({i}\right) \\ $$$${m}_{\mathrm{1}} {a}={f}_{\mathrm{2}} +{f}_{\mathrm{1}} \:\:\:…\left({ii}\right) \\ $$$$\frac{{m}_{\mathrm{1}} {R}^{\mathrm{2}} }{\mathrm{2}}\alpha={R}\left({f}_{\mathrm{2}} −{f}_{\mathrm{1}} \right) \\ $$$$\frac{{m}_{\mathrm{1}} {a}}{\mathrm{2}}={f}_{\mathrm{2}} −{f}_{\mathrm{1}} \:\:\:…\left({iii}\right) \\ $$$$\left({ii}\right)+\left({iii}\right): \\ $$$$\frac{\mathrm{3}{m}_{\mathrm{1}} {a}}{\mathrm{2}}=\mathrm{2}{f}_{\mathrm{2}} =\mathrm{2}\left({F}−\mathrm{2}{m}_{\mathrm{2}} {a}\right) \\ $$$$\left(\mathrm{3}{m}_{\mathrm{1}} +\mathrm{8}{m}_{\mathrm{2}} \right){a}=\mathrm{4}{F} \\ $$$$\Rightarrow{a}=\frac{\mathrm{4}{F}}{\mathrm{3}{m}_{\mathrm{1}} +\mathrm{8}{m}_{\mathrm{2}} } \\ $$$$\Rightarrow\alpha=\frac{\mathrm{4}{F}}{{R}\left(\mathrm{3}{m}_{\mathrm{1}} +\mathrm{8}{m}_{\mathrm{2}} \right)} \\ $$$$\Rightarrow{f}_{\mathrm{2}} =\frac{\mathrm{3}{m}_{\mathrm{1}} {F}}{\mathrm{3}{m}_{\mathrm{1}} +\mathrm{8}{m}_{\mathrm{2}} } \\ $$
Commented by fantastic2 last updated on 26/Nov/25
great sir.all correct
$${great}\:{sir}.{all}\:{correct} \\ $$
Answered by mr W last updated on 26/Nov/25
Commented by mr W last updated on 26/Nov/25
case 2 no slipping between plank and roller no friction between roller and ground A=acceleration of plank a=acceleration of center of cylinder α=angular acceleration of cylinder A=a+αR m_2 A=F−f_2 ⇒f_2 =F−m_2 (a+αR) m_1 a=f_2 =F−m_2 (a+αR) m_1 a=f_2 =F−m_2 a−m_2 αR) (m_1 +m_2 )a=f_2 =F−m_2 αR (m_1 +m_2 )a+m_2 Rα=F ...(i) ((m_1 R^2 )/2)α=Rf_2 ((m_1 R)/2)α=f_2 =F−m_2 (a+αR) ((m_1 /2)+m_2 )Rα+m_2 a=F m_2 a+((m_1 /2)+m_2 )Rα=F ...(ii) (i)−(ii): m_1 a−((m_1 Rα)/2)=0 ⇒Rα=2a into (i): (m_1 +m_2 )a+2m_2 a=F ⇒a=(F/(m_1 +3m_2 )) ⇒α=((2F)/(R(m_1 +3m_2 ))) A=a+αR=3a ⇒A=((3F)/(m_1 +3m_2 ))
$${case}\:\mathrm{2} \\ $$$${no}\:{slipping}\:{between}\:{plank}\:{and}\:{roller} \\ $$$${no}\:{friction}\:{between}\:{roller}\:{and}\:{ground} \\ $$$$ \\ $$$${A}={acceleration}\:{of}\:{plank} \\ $$$${a}={acceleration}\:{of}\:{center}\:{of}\:{cylinder} \\ $$$$\alpha={angular}\:{acceleration}\:{of}\:{cylinder} \\ $$$$ \\ $$$${A}={a}+\alpha{R} \\ $$$${m}_{\mathrm{2}} {A}={F}−{f}_{\mathrm{2}} \\ $$$$\Rightarrow{f}_{\mathrm{2}} ={F}−{m}_{\mathrm{2}} \left({a}+\alpha{R}\right) \\ $$$${m}_{\mathrm{1}} {a}={f}_{\mathrm{2}} ={F}−{m}_{\mathrm{2}} \left({a}+\alpha{R}\right)\: \\ $$$$\left.{m}_{\mathrm{1}} {a}={f}_{\mathrm{2}} ={F}−{m}_{\mathrm{2}} {a}−{m}_{\mathrm{2}} \alpha{R}\right) \\ $$$$\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right){a}={f}_{\mathrm{2}} ={F}−{m}_{\mathrm{2}} \alpha{R} \\ $$$$\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right){a}+{m}_{\mathrm{2}} {R}\alpha={F}\:\:\:…\left({i}\right) \\ $$$$\frac{{m}_{\mathrm{1}} {R}^{\mathrm{2}} }{\mathrm{2}}\alpha={Rf}_{\mathrm{2}} \\ $$$$\frac{{m}_{\mathrm{1}} {R}}{\mathrm{2}}\alpha={f}_{\mathrm{2}} ={F}−{m}_{\mathrm{2}} \left({a}+\alpha{R}\right) \\ $$$$\left(\frac{{m}_{\mathrm{1}} }{\mathrm{2}}+{m}_{\mathrm{2}} \right){R}\alpha+{m}_{\mathrm{2}} {a}={F} \\ $$$${m}_{\mathrm{2}} {a}+\left(\frac{{m}_{\mathrm{1}} }{\mathrm{2}}+{m}_{\mathrm{2}} \right){R}\alpha={F}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$${m}_{\mathrm{1}} {a}−\frac{{m}_{\mathrm{1}} {R}\alpha}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow{R}\alpha=\mathrm{2}{a} \\ $$$${into}\:\left({i}\right): \\ $$$$\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right){a}+\mathrm{2}{m}_{\mathrm{2}} {a}={F}\: \\ $$$$\Rightarrow{a}=\frac{{F}}{{m}_{\mathrm{1}} +\mathrm{3}{m}_{\mathrm{2}} } \\ $$$$\Rightarrow\alpha=\frac{\mathrm{2}{F}}{{R}\left({m}_{\mathrm{1}} +\mathrm{3}{m}_{\mathrm{2}} \right)} \\ $$$${A}={a}+\alpha{R}=\mathrm{3}{a} \\ $$$$\Rightarrow{A}=\frac{\mathrm{3}{F}}{{m}_{\mathrm{1}} +\mathrm{3}{m}_{\mathrm{2}} }\: \\ $$
Commented by fantastic2 last updated on 26/Nov/25
you cooked IIT Q like nothing
$${you}\:{cooked}\:{IIT}\:{Q}\:{like}\:{nothing} \\ $$

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