Consider the situation as shown in the diagram. The pulley has radius `R` and moment of inertia `I`. The block is release when the spring (spring constant `k`) is unstretched. Find the speed of block when the vlock has fallen a distance `h`. Assume no slipping.

Explanation

Decrease in the gravitation potential energy of block `=` gain in the kinetic energy of (block `+` pulley) `+` spring energy
`mgh=(1)/(2)mv^(2)+(1)/(2)Iomega^(2)+(1)/(2)kh^(2)`
`omega=(v)/(R )`
`mgh-(1)/(2)kh^(2)=(1)/(2)mv^(2)+(1)/(2)I((v)/(R))^(2)`
`=(1)/(2)mv^(2)(1+(I)/(mR^(2)))`
`v=sqrt((2mgh-kh^(2))/(m(1+(I)/(mR^(2)))))`