In this solution, I am going to assume that the cylinder is fixed so its center of mass can not move, even though this is not explicitly stated in the problem. This means the cylinder will not be moving as a whole. It will, however, rotate due to the string unwinding when the weight is released.
From the consideration of the conservation of energy, the change of the gravitational potential energy of the weight will equal the rotational kinetic energy of the cylinder and the kinetic energy of the weight:
`Delta U_(gr) = DeltaK_(rot) + DeltaK `
The gravitational potential energy of the weight, once the string is unwound, will change by
`DeltaU_(gr) = mgL` .
The change of kinetic energy of the weight at that time, since it starts moving from rest, will be
`DeltaK = 1/2mv^2` , where v is the velocity we are looking for.
The change of the rotational kinetic energy of the cylinder, which also starts moving from rest, is
`DeltaK_(rot) = 1/2Iomega^2` , where I is the cylinder's moment of inertia around its axis
`I = 1/2MR^2`
is the angular velocity, which can be found considering that the point on the rim of the cylinder has the same linear velocity as that of the weight, v:
`omega = v/R` .
Plugging this into the equation of energy equation, we get
`mgL = 1/2*1/2MR^2*(v/R)^2 + 1/2mv^2`
`mgL = 1/4Mv^2+1/2mv^2 =(M+2m)/4 v^2 `
`v = sqrt((4m)/(M+2m)gL)` .
This is the velocity of the weight when the string is unwound, in terms of m, M, L and g.