A solid cylinder is released from rest from the top of an inclined plane of inclination $$30°$$ and length 60 cm. If the cylinder rolls without slipping, its speed upon reaching the bottom of the inclined plane is ______ $$\text{m s}^{-1}$$. (Given $$g = 10 \text{ m s}^{-2}$$)

Vertical height through which the cylinder descends is $$h = L \sin \theta = 0.6 \text{ m} \times \sin(30^{\circ})$$

$$h = 0.6 \times 0.5 = 0.3 \text{ m}$$

For a body starting from rest and rolling without slipping, the total energy at the bottom is $$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

For a solid cylinder, $$I = \frac{1}{2}mr^2$$ and for pure rolling, $$\omega = \frac{v}{r}$$:
$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{v}{r}\right)^2$$

$$mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$$

$$v^2 = \frac{4}{3}gh$$

$$v = \sqrt{\frac{4}{3} \times 10 \times 0.3}$$

$$v = \sqrt{\frac{4}{3} \times 3} = \sqrt{4} = 2 \text{ m s}^{-1}$$