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

Two masses $$m$$ and $$2m$$ are connected by a light string over a pulley (disc) of mass $$30m$$ with radius $$r = 0.1$$ m, taking $$g = 10$$ m/s$$^2$$. When the $$2m$$ mass is released from rest and descends a height $$h = 3.6$$ m, its speed is found by applying conservation of energy.

Since the string does not slip on the pulley, energy is conserved. Let $$v$$ be the speed of the masses after the $$2m$$ mass has descended by $$h = 3.6$$ m. The $$2m$$ mass descends by $$h$$ while the $$m$$ mass rises by the same amount, so the net loss in potential energy is $$2mgh - mgh = mgh$$.

Both masses move with speed $$v$$, and the pulley rotates with angular velocity $$\omega = v/r$$. For a disc, the moment of inertia about its axis is $$I = \frac{1}{2}(30m)r^2 = 15mr^2$$.

Accordingly, the total kinetic energy is $$\frac{1}{2}(m)v^2 + \frac{1}{2}(2m)v^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2 + mv^2 + \frac{1}{2}(15mr^2)\frac{v^2}{r^2} = \frac{3}{2}mv^2 + \frac{15}{2}mv^2 = 9mv^2.$$

Equating the net potential energy loss to the total kinetic energy gives $$mgh = 9mv^2,$$ so $$v^2 = \frac{gh}{9} = \frac{10 \times 3.6}{9} = \frac{36}{9} = 4,$$ and hence $$v = 2\text{ m/s}.$$

The speed of the $$2m$$ mass after descending 3.6 m is 2 m/s.