Mechanics Of A Rotating Cylinder

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$Equating\quad horizontal\quad forces\quad as\\ acceleration\quad is\quad zero.\\ k(mg-kN)\quad =\quad N\quad \\ N\quad =\quad \frac { kmg }{ 1+{ k }^{ 2 } } \quad .....(1)\\ Torque\quad about\quad C.O.M\quad :\\ kRN\quad +\quad kR(mg-kN)\quad =\quad \frac { m{ R }^{ 2 } }{ 2 } \alpha \quad .....(2)\\ from\quad (1)\quad and\quad (2)\quad :\\ \alpha \quad =\quad \frac { 2kg(1+k) }{ R(1+{ k }^{ 2 }) } \quad \\ Cylinder\quad will\quad stop\quad rotating\quad when\quad its\quad angular\\ velocity\quad will\quad be\quad zero.\\ { \omega }^{ 2 }\quad =\quad 2\alpha (2\pi .n)\\ n\quad =\quad \frac { { \omega }^{ 2 }R(1+{ k }^{ 2 }) }{ 8\pi k(1+k)g }$ $\boxed{a+b+c+d+e=14}$