Newton’s Second Law for Rotation
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 6 practice questions.
Net torque causes angular acceleration: Στ = Iα, the rotational version of ΣF = ma. For systems with pulleys, write ΣF = ma for each block, Στ = Iα for each pulley, and connect them with a = αR.
1. A pulley with mass
2. Things to watch
If a pulley has mass, the tensions on its two sides must differ: their difference times R is the net torque that spins it up.
For a massless pulley, the tensions are equal (I = 0 means zero net torque).
Rod released from horizontal (pivot at one end): τ = Mg(L/2), I = ML²/3, so α = 3g/(2L). The free end starts with acceleration 3g/2, faster than free fall.
3. Example, then try it
A wheel with I = 0.80 kg·m² spins at 40 rad/s. A constant friction torque of 4.0 N·m acts on it. How long does it take to stop, and through what angle does it turn?
α = τ/I = −4.0 ÷ 0.80 = −5.0 rad/s², so t = 40 ÷ 5.0 = 8.0 s.
θ = (average ω)(t) = (20)(8.0) = 160 rad (about 25 rev).
A 3.0 kg block hangs from a light string wrapped around a uniform 2.0 kg disk pulley of radius 0.10 m that turns on a frictionless axle (the figure below). Find the block’s acceleration, the tension, and the pulley’s angular acceleration. (g = 10 m/s²)
Pulley: TR = (½MR²)(a/R), so T = ½Ma = (1.0)a.
Block: mg − T = ma → 30 − 1.0a = 3.0a.
Solve: a = 30 ÷ 4.0 = 7.5 m/s², so T = 7.5 N.
Pulley: α = a/R = 7.5 ÷ 0.10 = 75 rad/s². Without the pulley’s mass, the block would fall at the full 10 m/s².
Answer: 7.5 m/s², 7.5 N, 75 rad/s².
4. Common mistakes
In the figure below, the pulley is replaced by one with twice the mass (same radius and shape). What happens to the block’s acceleration?
A heavier pulley has more rotational inertia, so more of the block’s weight goes into spinning it up: a = mg/(m + ½M) gets smaller.
Ready to practice?
6 questions on Newton’s Second Law for Rotation, each with a hint and an explanation for every wrong answer.