Rotational Form of Newton's Second Law

Fix an axis and pick one particle of the body, mass m at perpendicular distance r. Its path is a circle, so the net force component along that circle obeys Fₜ = m aₜ = m r α. Multiply by r: the left side, r Fₜ, is the particle's torque about the axis, and the right side is (m r²) α. One factor of r turned α into a linear acceleration; the other turned a force into a torque. Now add over every particle. Each internal force belongs to a third-law pair, equal, opposite and along the line joining the two particles, so each pair's torques cancel: nothing a body does to itself can spin it up. The external torques remain, and α is shared by the whole rigid body, so it factors out: Στₑₓₜ = (Σ m r²) α = Iα. Check it on two 0.25 kg beads clamped to a light spoke at 0.15 m and 0.45 m, driven by a 0.30 N m torque about the hub: I = 0.25 (0.15² + 0.45²) = 0.05625 kg m², α = 5.33 rad s⁻², and the beads' tangential accelerations are 0.80 and 2.40 m s⁻², three to one like their radii. The spoke pushes them tangentially with 0.20 N and 0.60 N, whose torques, 0.030 and 0.270 N m, add back to 0.30 N m.

The rotational law is only as good as the list of torques it is fed, so build the list before the algebra. Take the axis through the axle: the bearing's supporting force then passes through the axis and drops out; its friction torque does not. Choose which sense is positive and keep it. For each remaining force, find its lever arm about the axle and write a signed torque; bearing friction acts against the current rotation ω, not against α, so its sign follows the spin, not the speeding up. A flat belt runs on a pulley of radius 0.15 m keyed to a flywheel, and the whole rotor has I = 0.60 kg m². The tight side pulls with 90 N and the slack side with 30 N, tangent to the rim on opposite sides, and the bearings exert 1.2 N m of friction. Taking the belt's sense as positive, Στ = (90 − 30)(0.15) − 1.2 = 9.0 − 1.2 = 7.8 N m, so α = 7.8/0.60 = 13 rad s⁻². Rotational kinematics finishes it: from rest, ω = 13t. Throw the belt off at ω = 40 rad s⁻¹ and only friction remains, α = −1.2/0.60 = −2.0 rad s⁻², so the rotor coasts for 20 s and turns through ω²/(2|α|) = 400 rad, about 64 revolutions.

Στ = Iα says nothing about the force at the axle, because that force has no lever arm about it. To find it, apply the translational law to the whole body, ΣF = M a for the centre of mass. A rotor whose axle passes through its centre of mass has a centre that does not move, so ΣF = 0 and the axle balances every other force. In the belt drive, both runs of the belt leave the pulley horizontally to the left, so together they pull 90 + 30 = 120 N leftward, and the 13 kg rotor weighs 128 N. The bearings must push 120 N to the right and 128 N up, a force of √(120² + 128²) = 175 N at 47° above the horizontal. Notice what did not enter: the 60 N difference between the belt tensions set the torque, but their 120 N sum loads the bearings, so an over-tightened belt delivers the same torque and wears the bearings for nothing. When the axle is off centre, the centre of mass moves on a circle and ΣF is no longer zero: the axle must also supply M ω² d toward itself and M α d along the motion, with d the offset, which is why an unbalanced wheel shakes harder the faster it spins.

Wind a light string on a drum of moment of inertia I and radius R and hang a block of mass m from it. Two bodies, two laws, and a constraint. The block obeys mg − T = ma with down positive; the drum obeys TR = Iα about its frictionless axle, since the string is its only torque; and a string that does not slip must unwind as fast as the block falls, a = Rα. Eliminate α and T: mg − (I/R²) a = ma, so a = g/(1 + I/(mR²)) and T = mg − ma. The combination I/R² is the drum's inertia expressed as an equivalent mass in kg. Take m = 1.2 kg on a drum with I = 0.030 kg m² and R = 0.10 m, the starting state of the interactive figure: I/R² = 3.0 kg, so a = 9.81/(1 + 2.5) = 2.80 m s⁻², T = 1.2 (9.81 − 2.80) = 8.41 N and α = a/R = 28.0 rad s⁻². Test the limits before trusting the number. A massless drum gives a = g and T = 0, free fall on a string that pulls nothing; an enormous drum gives a → 0 and T → mg, the static value. The tension always lies between those, so any answer with T above mg, or a above g, has a sign error in the first line.

Hang m₁ = 3.0 kg and m₂ = 2.0 kg from a string over a pulley that is a uniform disc of mass 2.0 kg, so I/R² = ½M = 1.0 kg, on a frictionless axle. Three bodies, three laws, one constraint. Heavy side, down positive: m₁g − T₁ = m₁a. Light side, up positive: T₂ − m₂g = m₂a. Pulley: (T₁ − T₂) R = Iα, with α = a/R because the string does not slip. Add the first two and use the third: (m₁ − m₂) g = (m₁ + m₂ + I/R²) a, so a = 9.81/6.0 = 1.64 m s⁻², against 1.96 m s⁻² for a massless pulley. Back-substitute: T₁ = 3.0 (9.81 − 1.64) = 24.5 N and T₂ = 2.0 (9.81 + 1.64) = 22.9 N. The difference, 1.6 N, is exactly (I/R²) a, the net rim force the disc needs to speed up. That is the physical content: a pulley with mass cannot have one tension, because equal tensions would give it zero torque, and Στ = Iα would then hold it at constant ω while the string slid over it. The system shortcut, treating the pulley as an extra mass I/R² riding on the string, is legitimate for a but skips the tensions, and it is the tensions that say whether the string or the axle will hold.

A uniform rod of mass 1.5 kg and length 1.2 m hangs from a frictionless pin at one end, is raised to horizontal, and released. About the pin, I = ⅓ML² = 0.72 kg m², and the weight acts at the centre with lever arm L/2, so at the instant of release Στ = Mg (L/2) = 8.83 N m and α = 8.83/0.72 = 12.3 rad s⁻², which is 3g/(2L) for any uniform rod. The tip's tangential acceleration is αL = 14.7 m s⁻², half again as large as g: a coin resting on the tip is left behind, and the point that falls at exactly g is 2L/3 = 0.80 m from the pin. The pin force at release comes from ΣF: the centre accelerates down at αL/2 = 7.36 m s⁻², so the pin supplies Mg − M (7.36) = 3.7 N upward, a quarter of the weight. Now the caution. As the rod swings through angle θ below horizontal the lever arm shrinks to (L/2) cos θ, so α = (3g/(2L)) cos θ is not constant and ω = αt is wrong past the first instant. Calculus still works: ω dω = α dθ integrates to ω² = (3g/L) sin θ, which is what the energy method of the next topic returns directly.