Rotational motion & torque
The idea
Rotation gets its own copy of Newton's second law: torque replaces force, angular acceleration replaces linear acceleration, and moment of inertia replaces mass, giving τ = Iα. Torque is force times lever arm — the perpendicular distance from the rotation axis to the line of the force — so where and in what direction you push matters as much as how hard. You already know F = ma and circular motion; this is the same logic applied to spinning.
Moment of inertia is the rotational stubbornness of a body, and unlike mass it depends on how the mass is distributed: I = Σmr², so material far from the axis counts quadratically more. A hoop (I = MR²) resists spin-up twice as hard as a uniform disk of the same mass and radius (I = MR²/2). When a rope unwinds from a rotating object, the constraint a = αR ties the linear and angular worlds together.
The classic mistake in falling-mass-and-pulley problems is to set the torque equal to mgR. The string tension, not the hanging weight, acts on the pulley rim — and tension drops below mg as soon as the mass accelerates. Solve the linear and rotational equations together rather than guessing the rim force.
Worked example
A light string is wound around a uniform solid disk of mass 2.0 kg and radius 0.20 m that spins on a frictionless fixed axle. A 1.0 kg mass hangs from the free end. Using g = 9.8 m/s², find the downward acceleration of the mass, the string tension, and the angular acceleration of the disk.
- Write Newton's second law for the hanging mass, taking down as positive: mg − T = ma, with the tension unknown because the mass accelerates.
- Write the rotational law for the disk: the tension acts at the rim, so TR = Iα with I = MR²/2 = 0.5 × 2.0 × 0.20² = 0.040 kg·m².
- Use the string constraint a = αR to eliminate α: TR = (MR²/2)(a/R) gives T = (M/2)a = 1.0a.
- Substitute into the linear equation: 1.0 × 9.8 − 1.0a = 1.0a, so 9.8 = 2.0a and a = 4.9 m/s²; then T = 1.0 × 4.9 = 4.9 N.
- Finish with α = a/R = 4.9/0.20 = 24.5 rad/s², and check the torque route independently: τ = TR = 4.9 × 0.20 = 0.98 N·m, and τ/I = 0.98/0.040 = 24.5 rad/s² — consistent.
- Interpret: the mass falls at half of g because the disk effectively adds M/2 = 1.0 kg of inertia, and the tension (4.9 N) is well below the 9.8 N weight, as it must be.
Answer. The mass accelerates down at 4.9 m/s², the tension is 4.9 N, and the disk has angular acceleration 24.5 rad/s².
Check your understanding
- Why does a hoop accelerate more slowly than a uniform disk of equal mass and radius under the same torque?
- What exactly does the constraint a = αR encode, and when does a string slip enough to break it?
- How would you convince a friend that pushing on a door near its hinge is wasted effort, using the torque definition?
- Where does the energy go in this pulley system — how would an energy method reproduce the 4.9 m/s² answer?
Build the foundations first
Rotational motion & torque builds on these concepts. If any feel shaky, start there.