Rolling Down An Incline (numerical)

Building on NCERT Class 11, Chapter 7 (System of Particles and Rotational Motion), a body rolling without slipping down an incline has linear acceleration a = g sinθ / (1 + k²/R²), where the denominator accounts for the share of gravitational energy that must be diverted into rotational motion rather than pure translation. The term k²/R² is the dimensionless shape factor, equal to the square of the radius of gyration divided by the square of the radius. For a uniform solid cylinder this factor is k²/R² = ½, so the denominator becomes 1 + ½ = 3/2. Substituting the given quantities yields a = (10 × 0.30) / (3/2) = 3 / 1.5 = 2 m/s². The option 3 m/s² is the value g sinθ alone would give if the cylinder slid frictionlessly without rotating, which ignores rotational inertia entirely. The option 1 m/s² over-divides the result and assumes an incorrect shape factor. The option 5 m/s² has no valid derivation from the given data. A plausibility check confirms that the rolling acceleration of 2 m/s² is correctly smaller than the frictionless sliding value of 3 m/s², which is exactly what we expect because some of the available energy must go into spinning the cylinder up rather than translating it.