Given: Mass of the cylinder, $m = 500 \text{ g} = 0.5 \text{ kg}$ Radius, $r = 10 \text{ cm} = 0.1 \text{ m}$ Assuming the cylinder to be a uniform solid cylinder, the moment of inertia about its natural (longitudinal) axis is given by: $I = \frac{1}{2}mr^2$ Substituting the given values: $I = \frac{1}{2} \times 0.5 \times (0.1)^2$ $I = 0.25 \times 0.01 = 0.0025 \text{ kg m}^2 = 2.5 \times 10^{-3} \text{ kg m}^2$.