The Rolling Constraint & Pulleys With Mass
- Apply the rolling-without-slipping constraint v = ωR and a = αR
- Explain why static rather than kinetic friction acts on a rolling object
- Solve a system with a massive pulley, where the two string tensions differ
Rolling links rotation to translation
An object that rolls without slipping has its rotation and its translation locked together: for every radian it turns, it advances one radius along the ground. That gives v_cm = ωR and a_cm = αR. This constraint is what makes rolling problems solvable — it supplies the extra equation that lets you connect the rotational equation Στ = Iα to the translational one ΣF = ma.
The contact point is instantaneously at rest
This is the idea that makes rolling counterintuitive. The bottom of a rolling wheel has translational velocity v forward and rotational velocity ωR backward relative to the center — and since v = ωR, those cancel exactly. The contact point is momentarily stationary. That is why rolling involves static friction rather than kinetic: the surfaces are not sliding past each other. And static friction does no work, which is why a ball can roll down a hill without losing mechanical energy while a block sliding down loses it steadily.
The top of the wheel moves at 2v
The same superposition run at the top gives translational v forward plus rotational ωR forward, so the top of a rolling wheel moves at 2v — twice the speed of the car it belongs to. This is real and observable: it is why the top of a spinning bicycle wheel blurs while the bottom looks sharp in a photograph.
A pulley with mass breaks the equal-tension rule
An ideal pulley is massless, so no net torque is needed to change its rotation and the tension is the same on both sides. Give the pulley rotational inertia and that stops being true: the pulley must be angularly accelerated, and the only thing that can do it is a difference in the two tensions. So T₁ ≠ T₂, and (T₁ − T₂)R = Iα. The system also accelerates more slowly than the ideal case, because some of the available energy now goes into spinning the pulley.
A 2.0 kg block hangs from a string over a pulley modeled as a solid disk of mass 1.0 kg and radius 0.10 m. Find the block's acceleration. (g = 10 m/s²)
- 1.Block: mg − T = ma → 20 − T = 2.0a.
- 2.Pulley: TR = Iα, with I = ½MR² = ½(1.0)(0.10)² = 0.0050 kg·m².
- 3.Rolling constraint at the rim: α = a/R, so TR = I(a/R) → T = Ia/R² = (0.0050)a ÷ 0.010 = 0.50a.
- 4.Substitute: 20 − 0.50a = 2.0a → 20 = 2.5a.
- 5.Solve: a = 8.0 m/s², less than the 10 m/s² a massless pulley would allow.
A rolling object needs friction to roll, yet loses no mechanical energy to it. There is no contradiction: static friction acts at a point that is instantaneously at rest, and a force acting through zero displacement does zero work. Kinetic friction, acting on a sliding object, does dissipate energy.
A wheel of radius 0.25 m rolls without slipping at 4.0 m/s. What is its angular speed?
For a wheel rolling without slipping, the speed of the point in contact with the ground, relative to the ground, is:
A block hangs from a string over a pulley that has significant rotational inertia. Compared with an ideal massless pulley, the block's acceleration is:
Whenever a problem says "rolls without slipping", write v = ωR and a = αR immediately. That constraint is the equation that closes the system, and rubrics award it as a separate point from the dynamics equations it connects.
Answer the 3 checkpoints as you read.
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