Combined Translation & Rotation

Release a uniform solid cylinder on a 30° incline and ask for its acceleration. A sliding block needs one equation; the cylinder needs two, because it does two things at once. Its centre of mass moves down the slope, and that motion obeys ΣF = m a_cm no matter where on the body the forces act. It also spins up about its centre, and that obeys Στ_cm = I_cm α, a law that survives the centre's own acceleration, which is what makes the centre the safe axis for a body that is going somewhere. With down the slope positive, the force law reads m g sinθ − f = m a_cm. For the torque law, the weight and the normal force both act along lines through the centre, so they have no torque, and friction at the rim is left alone: f R = I_cm α. Now count. Two equations, three unknowns: a_cm, α and f. No amount of redrawing the free-body diagram will supply a third equation, because the missing information is not a force. It is a statement about how the rim moves relative to the ramp.

If the cylinder rolls without slipping, the point touching the ramp is momentarily at rest. The rim's backward speed Rω relative to the centre must cancel the centre's forward speed, so v_cm = Rω, and differentiating, a_cm = Rα. Put α = a_cm/R into the torque equation: f = I_cm a_cm/R² = β m a_cm, where β = I_cm/(m R²). Put that into the force equation: m g sinθ − β m a_cm = m a_cm, so a_cm = g sinθ/(1 + β). For the cylinder β = 1/2 and a_cm = 9.81 × 0.500/1.5 = 3.27 m s⁻², against 4.91 m s⁻² for a frictionless slide. Friction follows: f = β m g sinθ/(1 + β), one third of m g sinθ; for a 4.0 kg cylinder that is 19.62/3 = 6.54 N, up the slope. Notice what never appeared: μₛ. The friction came out of the two laws and the constraint, set by the shape of the body, and the coefficient played no part. Mass and radius cancelled too, so every solid cylinder shares this acceleration, and only β separates a sphere from a hoop.

On a level floor, pull a 4.0 kg solid cylinder of radius R with a horizontal force F = 12 N applied at height h above its centre. Call friction f, positive forwards, without deciding in advance which way it points. Force: F + f = m a_cm. Torque about the centre, in the rolling sense: F h − f R = I_cm α. Constraint: a_cm = Rα. Solving, a_cm = F (1 + h/R)/(m (1 + β)) and f = F (h/R − β)/(1 + β). Pull at the axle, h = 0: a_cm = 12/6.0 = 2.0 m s⁻² and f = −4.0 N, backwards, because friction alone must supply the spin. Pull along the top, h = R: a_cm = 24/6.0 = 4.0 m s⁻² and f = +4.0 N, forwards. The pull now over-supplies torque, the rim tries to skid backwards, and friction pushes the cylinder along, so its centre accelerates faster than F/m = 3.0 m s⁻². At h = βR friction vanishes altogether. The slogan that friction opposes motion fails in two of these three cases. Friction opposes the slipping that would otherwise occur at the contact, and the sign of the solved f tells you which way that is.

Solving gives the friction rolling needs. The surface can supply at most μₛ N = μₛ m g cosθ, so the solution is legitimate only while β m g sinθ/(1 + β) ≤ μₛ m g cosθ, that is tanθ ≤ μₛ (1 + β)/β. With μₛ = 0.30 a solid cylinder rolls up to tanθ = 0.90, θ = 42.0°; a hoop on the same surface only to tanθ = 0.60, θ = 31.0°, because a larger share of its motion is spin and more friction is needed to drive it. Past the threshold the model changes character. Friction becomes kinetic, f = μₖ m g cosθ, a known force rather than an unknown, and the constraint is dropped. The force law alone gives a_cm = g (sinθ − μₖ cosθ); the torque law alone gives α = μₖ m g cosθ R/I_cm. For the cylinder at 45° with μₖ = 0.25, a_cm = 9.81 × 0.7071 × 0.75 = 5.20 m s⁻² while Rα = μₖ g cosθ/β = 3.47 m s⁻². The rim no longer keeps up with the centre, and the contact point skids down the ramp with relative acceleration 1.73 m s⁻².

The same three-line pattern runs a pulley whose mass matters. Hang m₁ = 3.0 kg and m₂ = 2.0 kg over a pulley with I = 0.040 kg m² and R = 0.10 m, the string gripping the rim. Draw three diagrams, not two. Heavier mass, down positive: m₁ g − T₁ = m₁ a. Lighter mass, up positive: T₂ − m₂ g = m₂ a. Pulley, about its axle: (T₁ − T₂) R = I α. A string that does not slip makes the rim move with it, which is the rolling constraint again: a = Rα, so T₁ − T₂ = (I/R²) a = 4.0 kg × a. Add the three equations and the tensions cancel: (m₁ − m₂) g = (m₁ + m₂ + I/R²) a, so a = 9.81/9.0 = 1.09 m s⁻². Then T₁ = m₁ (g − a) = 26.2 N and T₂ = m₂ (g + a) = 21.8 N. Their difference, 4.36 N, is exactly what the pulley needs: 4.36 N × 0.10 m = 0.436 N m, and I α = 0.040 × 10.9 = 0.436 N m. Two things generalise. I/R² enters as an effective mass, M/2 for a uniform disc. And writing one tension for the whole string is a claim that I = 0, nothing more.

Static friction at a non-slipping contact acts on a point that is momentarily at rest, so it does no work, and a body rolling on a rigid ramp keeps its mechanical energy: m g h = ½ m v_cm² + ½ I_cm ω² = ½ m v_cm² (1 + β). The cylinder of the first two sections, after 2.0 m along its 30° slope, has dropped h = 1.0 m, so v_cm = √(2 × 9.81 × 1.0/1.5) = 3.62 m s⁻¹. The force route gives the same number, v_cm = √(2 a_cm s) = √(2 × 3.27 × 2.0) = 3.62 m s⁻¹, as it must: the energy equation is the force and torque equations integrated along the path, not new information. Adding it to the three equations you already have cannot fix a fourth unknown; it only trades accelerations for speeds. Choose by the question asked. Speeds after a given drop come fastest from energy. Friction, tensions and the rolling test come only from the force method, because energy cannot see a force that does no work. And once the body slips, kinetic friction does do work, and the conservation statement fails.