Will you roll with me?

$Lets\quad say\quad acceleration\quad of\quad bigger\quad cylinder\quad is\quad { a }_{ 1 }\\ and\quad acceleration\quad of\quad smaller\quad cylinder\quad is\quad { a }_{ 2 }\\ PURE\quad ROLLING:\\ { a }_{ 1 }=(2R)\alpha \quad .....(1)\\ { a }_{ 2 }=(R)\alpha \quad ......(2)\\ { a }_{ 1 }=2{ a }_{ 2 }\quad ....(3)\\ for\quad c-1:\\ mgsin\theta -T-\eta mgcos\theta \quad =\quad m{ a }_{ 1 }\\ for\quad c-2:\\ T+mgsin\theta -\eta mgcos\theta \quad =\quad m{ a }_{ 2 }\\ eliminating\quad T:\\ 2mgsin\theta -2\eta mgcos\theta \quad =\quad 3m{ a }_{ 2 }\quad ......(4)\\ Torque\quad about\quad C.O.M\quad of\quad bigger\quad cylinder:\\ \eta mgcos\theta (2R)\quad =\quad \frac { m{ (2R) }^{ 2 } }{ 2 } .\alpha \\ \eta mgcos\theta \quad =\quad m{ a }_{ 2 }\quad .......(5)\\ using\quad (4)\quad and\quad (5)\quad :\\ 2mgsin\theta \quad =\quad 5m{ a }_{ 2 }\\ { a }_{ 2 }\quad =\quad 1.962\quad \frac { m }{ { s }^{ 2 } }$

Smaller cylinder can be taken as winding upwards. Writing torque equation about centre of mass for small cylinder we get - Let the tension in the string be T Therefore , RT + Rmgsin30 = mr^2/2 * (a/r) ........ (1) Writing torque equation about inclined on bigger sphere - 2RT = 3mr^2/2 * (a/2R) 2T = 3ma putting in (1) a = 1.96m/s^2