Let the speed of the masses be v and the angular velocity of the pulley be ω=v/r since the string does not slip.
The pulley is a disc of mass M=30m and radius r. Its moment of inertia is I=21Mr2=21(30m)r2=15mr2.
Using the principle of conservation of mechanical energy, the loss in potential energy of the system equals the gain in kinetic energy.
When the mass 2m descends by h=3.6 m, the mass m ascends by h=3.6 m.
Loss in potential energy ΔU=(2m)gh−(m)gh=mgh.
Gain in kinetic energy ΔK=21(2m)v2+21(m)v2+21Iω2.
Substituting I and ω: ΔK=23mv2+21(15mr2)(rv)2=23mv2+215mv2=9mv2.
Equating ΔU=ΔK: mgh=9mv2.
v2=9gh.
Given g=10 m/s2 and h=3.6 m:
v2=910×3.6=936=4.
v=4=2 m/s.