Consider the arrangement shown in figure. The string is wrapped around a uniform cylinder which rolls without slipping. The other end of the string is passed over a masslessm frictionless pulley to a falling weight, determine the acceleration of the falling mass `m` in terms of only the mass of the cylinder `M`, the mass `m` and `g`

Explanation

let `T` be the tension in the string and `f` the force of (static) friction, between the cylinder and the surface.
`a_(1)=` acceleration of centre of mass of cylinder
towards right
`a_(2)=` downward acceleration of block m
`alpha=` angular acceleration of cylinder (clockwise)
Acceleration equations
for block `mg-T=ma_(2)`
for cylinder `T+f=Ma_(1)` ..(ii)
`alpha=((T-f)R)/((1)/(2)MR^(2))` ..(iii)
Contact equations
The string is attached to the mass `m` at the highest point of the cylinder, hence
`v_(m)=v_(COM)+Romega` ..
differentiating we get `a_(2)=a_(1)+Ralpha` ..(iv)
We also have (for rolling without slipping)
`a_(1)=Ralpha` ..(v)
Solving these equations, we get
`a_(2)=(8mg)/(3M+8m)`
alternate solution (energy methode)
since, there is not slipping at all contact mechanical energy of the system will remain conserved.
`therefore` decrease in gravitational potential energy of block m in time `t=` increase in translational kinetic energy of block `+` increase in rotational as well as translational kinetic energy of cylinder
`thereforemgh=(1)/(2)mv_(2)^(2)+(1)/(2)Iomega^(2)+(1)/(2)Mv_(1)^(2)`
or `mg((1)/(2)a_(2)t^(2))=(1)/(2)m(a_(2)t)^(2)+(1)/(2)((1)/(2)MR^(2))(alphat)^(2)+(1)/(2)M(a_(1)t)^(2)` ..(vi)
Solving Eqs. (iv) (V) and (vi) we get the same result.