Torque & Angular-Momentum Change

Take one particle and a fixed origin O. Its angular momentum is L⃗ = r⃗ × p⃗, and the product rule for a cross product keeps the factors in order: dL⃗/dt = (dr⃗/dt) × p⃗ + r⃗ × (dp⃗/dt). Because O is fixed, dr⃗/dt is the particle's velocity v⃗, and v⃗ × mv⃗ = 0. Newton's second law turns dp⃗/dt into F⃗, leaving dL⃗/dt = r⃗ × F⃗ = τ⃗. Test it on a 0.50 kg stone dropped from rest at a point 2.0 m to the right of O. Its weight, 4.905 N straight down, has a lever arm of 2.0 m, so τ_z = −9.81 N m, clockwise. Directly, p⃗ = −(0.50)(9.81)t ĵ, and only the 2.0 m horizontal part of r⃗ survives the cross product, so L_z = −9.81t kg m² s⁻¹, whose slope is exactly the torque. A stone falling in a straight line has angular momentum about O, and that angular momentum grows because gravity has a lever arm about O.

For a system, add the particle results. Each particle feels external forces and forces from the other particles. Particles 1 and 2 push on each other with F⃗₁₂ = −F⃗₂₁, and when that pair acts along the line joining them their torques sum to r⃗₁ × F⃗₁₂ + r⃗₂ × F⃗₂₁ = (r⃗₁ − r⃗₂) × F⃗₁₂ = 0, because r⃗₁ − r⃗₂ is parallel to F⃗₁₂. Every internal pair cancels, leaving τ⃗ₑₓₜ = dL⃗/dt for the whole system. So the working move is to list external forces only and find each one's torque about the chosen point. A force through that point contributes nothing, so a hinge or axle force has zero torque about the hinge, and gravity on a rigid body acts as if at its centre of mass. A uniform 2.0 kg rod 1.2 m long, hinged at one end and released horizontal, has dL/dt = (2.0)(9.81)(0.60) = 11.8 N m about the hinge at release, and the hinge force, however large, never enters.

Integrate τ⃗ = dL⃗/dt over an interval and ΔL⃗ = ∫τ⃗ dt. Along a fixed axis this is one number: the signed area under the torque–time graph, in N m s, the same unit as kg m² s⁻¹. A motor that ramps its torque on a flywheel from 0 to 6.0 N m over 2.0 s and then holds 6.0 N m for 3.0 s delivers ½(6.0)(2.0) + (6.0)(3.0) = 24 N m s. A flywheel with I = 1.5 kg m² starting from rest ends at ω = 24/1.5 = 16 rad s⁻¹, and the shape of the ramp matters only through its area. Torque against the spin counts as negative area. This is also where τ = Iα sits: when I is constant, dL_z/dt = I dω/dt = Iα. When I changes during the push, say a rider on a spinning stool pulls in her arms while someone pushes the stool, Iα gives the wrong answer and the impulse statement is still right: I₂ω₂ − I₁ω₁ = ∫τ_z dt.

τ⃗ = dL⃗/dt is three equations, one per component, and they do not interfere. Along any fixed direction τ_z = dL_z/dt, so a torque with no z-component cannot change L_z however the other components behave. That is how fixed-axis problems work. Push on a door whose hinge line is the z-axis, at a handle 0.80 m from that line, with F⃗ = (30 ĵ − 20 k̂) N, partly sideways and partly downward. Then τ⃗ = (0.80 î) × (30 ĵ − 20 k̂) = (16 ĵ + 24 k̂) N m. Only τ_z = 24 N m changes how fast the door swings: with I = 12 kg m² about the hinge line, its angular acceleration is 24/12 = 2.0 rad s⁻². The 16 N m about the horizontal y-axis tries to twist the door out of its hinges, and the hinges push back. Hinge forces act at points on the z-axis, so they can supply torque about y but none about z. The converse matters just as much: when τ_z = 0, L_z stays fixed even while the rest of L⃗ swings round, as the Hard example below shows.

Components can also be taken relative to L⃗ itself. In a short time dt the torque adds dL⃗ = τ⃗ dt. The part of τ⃗ along L⃗ changes its size at the rate τ∥; the part across L⃗ swings it sideways through an angle τ⊥ dt/L. A wheel with L = 4.0 kg m² s⁻¹ given an angular impulse of 1.0 N m s along its axis ends with 5.0 kg m² s⁻¹, pointing the same way. Give it the same impulse at right angles, with the torque's direction held fixed, and L⃗₁ = L⃗₀ + τ⃗Δt is the hypotenuse of a 4.0-by-1.0 triangle: its size is √17 = 4.12 kg m² s⁻¹, only 3% larger, and it has turned through atan(1.0/4.0) = 14.0°. Double the spin to 8.0 kg m² s⁻¹ and the same sideways impulse turns it by only 7.1°. A large L⃗ is hard to redirect, and a torque that keeps itself perpendicular to L⃗ as L⃗ turns changes its direction without changing its size at all, the idea gyroscopes are built on.

Torque and angular momentum both depend on the point they are measured about, so the law compares like with like: torque about O equals the rate of change of angular momentum about O. Any point fixed in an inertial frame works, and so does the centre of mass even while it accelerates. Choosing the point is a modelling decision, and a good choice removes unknowns. A tennis ball thrown with topspin shows that two answers can both be right. About the launch point, gravity's lever arm is the horizontal distance travelled: a 0.057 kg ball thrown horizontally at 20 m s⁻¹ is 8.0 m out after 0.40 s, so the torque about that point is (0.057)(9.81)(8.0) = 4.5 N m and growing, and L⃗ about the launch point keeps changing. About the ball's own centre of mass, gravity has no lever arm, so the spin is untouched in flight until air drag or a bounce applies a torque about the centre. Mixing points, torque about a hinge with angular momentum about a centre, gives equations that look right and are not.