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Rolling Down An Incline (numerical)

Hardphysics

A solid cylinder is released from rest and rolls without slipping down an incline making an angle whose sine equals 0.30, with gravitational acceleration taken as 10 m/s². What is its linear acceleration?

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About This Question

Subject
physics
Chapter
rotational motion
Topic
rolling down an incline (numerical)
Difficulty
Hard
Year
2025
Tags
rolling accelerationsolid cylinderinclined planeshape factorrolling without slipping

Solution

Correct Answer:

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.

This hard difficulty physics question is from the chapter rotational motion, covering the topic of rolling down an incline (numerical). It appeared in the 2025 exam.

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