Two masses m and 2m are connected over a pulley of mass 30m and radius 0.1 m. Find final speed of 2m after descending 3.6 m.

Solution:

Using conservation of energy:

Loss in potential energy:

ΔPE = (2m - m)gh

ΔPE = mgh

h = 3.6 m

Kinetic energy:

KE = (1/2)mv² + (1/2)(2m)v² + (1/2)Iω²

For pulley:

I = (1/2)MR²

M = 30m

I = (1/2)(30m)(0.1)²

Using:

ω = v/R

Rotational KE:

= (1/2) × (1/2)(30m)R² × (v²/R²)

= 7.5mv²

Total KE:

= 0.5mv² + mv² + 7.5mv²

= 9mv²

Equating:

mgh = 9mv²

10 × 3.6 = 9v²

v² = 4

v = 2 m/s

Answer: 2 m/s