Let the mass of the flywheel be M=3 kg and its radius be R=5 m.
Let the hanging mass be m=3 kg and the distance descended be h=3 m.
The flywheel is a disk, so its moment of inertia is I=21MR2.
By conservation of energy, the loss in potential energy of the hanging mass equals the gain in kinetic energy of the system.
mgh=21mv2+21Iω2
Since the string does not slip, v=Rω, so ω=Rv.
mgh=21mv2+21(21MR2)(Rv)2
mgh=21mv2+41Mv2
Substituting the values: 3×10×3=21(3)v2+41(3)v2
90=23v2+43v2=49v2
v2=990×4=40 m2/s2.
The kinetic energy of the wheel is Kwheel=21Iω2=41Mv2.
Kwheel=41×3×40=30 J.