Given data

• Inclination of plane: $\theta = 30^\circ$
• Length of incline: $L = 60\text{ cm} = 0.6\text{ m}$
• Acceleration due to gravity: $g = 10\text{ m s}^{-2}$
• Body: solid cylinder rolling without slipping

Find the vertical height descended

The cylinder moves along the incline of length $L$, so the vertical drop is $h = L\sin\theta = 0.6\times \sin 30^\circ = 0.6\times \frac{1}{2} = 0.3\text{ m}.$

Use conservation of mechanical energy

Since the cylinder rolls without slipping, $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2.$

For a solid cylinder, $I = \frac{1}{2}mR^2.$

Also, rolling without slipping gives $\omega = \frac{v}{R}.$

Substitute these into the energy equation: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mR^2\right)\left(\frac{v}{R}\right)^2.$

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

Solve for $v$ $gh = \frac{3}{4}v^2$ $v^2 = \frac{4}{3}gh = \frac{4}{3}\times 10\times 0.3 = 4.$

Therefore, $v = 2\text{ m s}^{-1}.$

Final integer answer $\boxed{2}$

Comparison with stored correct answer

• Derived answer: $2$
• Stored correct answer: $2$
• They match.