Two masses $m$ and 2 m are connected by a light string going over a pulley (disc) of mass 30 m with radius $r=0.1 \mathrm{~m}$. The pulley is mounted in a vertical plane and it is free to rotate about its axis. The 2 m mass is released from rest and its speed when it has descended through a height of 3.6 m is $\_\_\_\_$ $\mathrm{m} / \mathrm{s}$. (Assume string does not slip and $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$)

Let the speed of the masses be $v$ and the angular velocity of the pulley be $\omega = v/r$ since the string does not slip.
The pulley is a disc of mass $M = 30m$ and radius $r$. Its moment of inertia is $I = \frac{1}{2}Mr^2 = \frac{1}{2}(30m)r^2 = 15mr^2$.
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 $\Delta U = (2m)gh - (m)gh = mgh$.
Gain in kinetic energy $\Delta K = \frac{1}{2}(2m)v^2 + \frac{1}{2}(m)v^2 + \frac{1}{2}I\omega^2$.
Substituting $I$ and $\omega$: $\Delta K = \frac{3}{2}mv^2 + \frac{1}{2}(15mr^2)(\frac{v}{r})^2 = \frac{3}{2}mv^2 + \frac{15}{2}mv^2 = 9mv^2$.
Equating $\Delta U = \Delta K$: $mgh = 9mv^2$.
$v^2 = \frac{gh}{9}$.
Given $g = 10$ m/s$^2$ and $h = 3.6$ m:
$v^2 = \frac{10 \times 3.6}{9} = \frac{36}{9} = 4$.
$v = \sqrt{4} = 2$ m/s.