Q.(a) Discuss rolling on inclined plane and arrive at the expression for the acceleration. OR

A body rolling down an incline both slides down (translates) AND spins (rotates); part of the gravitational potential energy/force goes into each motion, so the linear acceleration down the slope is less than gsinθ (the value for pure sliding without friction) — the exact reduction depends on the body's radius of gyration.

Setup: A rigid body (sphere, cylinder, etc.) of mass m, radius R, and moment of inertia I = mk² about its own axis through the centre of mass (k = radius of gyration) rolls WITHOUT SLIPPING down an incline of angle θ.

Forces along the incline: component of gravity mg sinθ (down the slope, driving the motion), and a static friction force f (up the slope, at the point of contact) which provides the torque needed to make the body rotate as it rolls.

Translational equation of motion (Newton's second law along the incline):

mg sinθ − f = ma ... (1)

Rotational equation of motion (torque about the centre of mass, produced only by friction f, since gravity and the normal force pass through/near the axis):

f R = I α

Since rolling without slipping means a = Rα, i.e., α = a/R:

f R = I (a/R)

f = I a / R² = (mk²) a / R² ... (2)

Substitute (2) into (1):

mg sinθ − (mk²a/R²) = ma

mg sinθ = ma + (mk²a/R²)

mg sinθ = ma [1 + k²/R²]

Solving for acceleration:

a = g sinθ / [1 + k²/R²]

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