A uniform solid cylinder of mass 2.00 kg rolls without slipping down a 25.0° incline through a vertical drop of 1.20 m. Its moment of inertia is I = ½mR². Determine its translational acceleration, the static-friction magnitude and direction, and its speed after the drop. Explain why static friction does no mechanical work in the ideal rolling model and compare the speed with that of a frictionless sliding block.

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a = 2.76 m s⁻² down the slope; static friction = 2.76 N up the slope; final speed = 3.96 m s⁻¹. A sliding block would be faster at 4.85 m s⁻¹. Translation along the slope gives mg sinθ − f = ma. Rotation about the centre gives fR = Iα, and no slip gives α = a/R. With I = ½mR², the torque equation gives f = ½ma. Substitution into translation yields mg sinθ = 3ma/2, so a = (2/3)g sin25° = 2.76 m s⁻². The friction is f = ½ma = 2.76 N and points uphill. It supplies the uphill torque needed to increase the cylinder's angular speed while the centre accelerates downhill. Energy gives mgh = ½mv² + ½I(v/R)² = ¾mv². Therefore v = √(4gh/3) = 3.96 m s⁻¹. The instantaneous contact point is at rest relative to the fixed slope, so ideal static friction transfers no energy at that point. A frictionless slider has v = √(2gh) = 4.85 m s⁻¹ because none of its energy is rotational.

Rolling dynamics down an incline

Choose your AP course · School year 2026–27 · May 2027 exam

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