Good Ol' Rotation - from the Dark Knight

net force acting on the cylinder parallel to incline
for motion of cylinder
f_r=frictional force between cylinder and support
ma'=mgsin@-f_r

COM of cylinder is stationary that means a'=d²xdt²=0(assuming x axis along incline)

so,

f_r=mgsin@

torque on cylinder abt point G=f_rR=(mR²/2)α
f_r=mRα/2=ma/2
ma/2=mgsin@
a=2gsin@

Excellent Eragon, ridden the dragon well.

But i've one little doubt here.

Could we have done it by a procedure where we first take the block's acceleration as a...and a pseudo acceleration a for the Cylinder ??

And why are we considering that the cylinder and the block move together such dat der's a common acceleration a ???? If it's so then there should be no force of friction between the cylinder and the block !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

friction is actually zero !
rath sir concept, again !