Angular momentum problem studio

Before writing a conservation law, answer three questions on paper. Which bodies are in the system? Forces between them act in pairs along one line, so their torques cancel and cannot change the total L. About which point? Στ = dL/dt holds about a point fixed in an inertial frame or about the centre of mass, and nowhere else without a correction. Over which event? A collision lasts milliseconds; the swing after it lasts a second. Then list every external force with its lever arm about that point during that event. A 30 kg child running at 3.0 m s⁻¹ along the rim line jumps onto a roundabout 1.5 m from the axle, bringing 30 × 3.0 × 1.5 = 135 kg m² s⁻¹. The axle force passes through the axis, gravity and the normal force are parallel to it, and bearing friction of about 2 N m acting for 0.1 s changes L by 0.2 kg m² s⁻¹, roughly 0.15%. L about the axle is conserved for the landing. Nothing else is.

Hang a uniform 2.0 kg rod, 1.00 m long, from a frictionless pivot and fire a 0.50 kg lump of putty into it at 6.0 m s⁻¹, 0.90 m below the pivot, where it sticks. During the impact the pivot exerts whatever horizontal force keeps the top of the rod in place, so linear momentum is not a safe statement. About the pivot that force has no lever arm, and the rod's weight acts along the vertical through it, so L about O is conserved: 0.50 × 6.0 × 0.90 = 2.70 kg m² s⁻¹ = (ML²/3 + md²)ω = 1.072ω, giving ω = 2.52 rad s⁻¹. Now test the law you rejected. Just after, the momentum is (ML/2 + md)ω = 1.45 × 2.52 = 3.65 N s against 3.00 N s before, so the pivot pushed forwards with J = +0.65 N s. Strike at 2L/3 and J is zero; strike nearer the pivot and it pushes backwards.

Now take moments about the rod's centre C, 0.50 m below the pivot. Before the impact the putty passes C with lever arm 0.40 m, so L_C = 0.50 × 6.0 × 0.40 = 1.20 kg m² s⁻¹. Just after, the rod contributes (ML²/12)ω about its own centre and the putty m d(d − L/2)ω, a total of 0.347 × 2.52 = 0.87 kg m² s⁻¹. L about C fell by 0.33 kg m² s⁻¹, exactly the pivot impulse times its lever arm about C: (L/2)J = 0.50 × 0.65. The physics is fine; C is simply a point about which an unknown impulse has a moment. Assume L_C conserved anyway and 1.20 = 0.347ω gives ω = 3.46 rad s⁻¹, 37% too fast. In general, ΔL about a fixed point A differs from ΔL about O by r⃗ × ΔP⃗, with r⃗ running from A to O, so two origins agree only when the linear impulse vanishes or lies along the line joining them. Choose the point the unknown impulse passes through.

A walker and a turntable push on each other through friction at the feet, which is internal, and a frictionless axle exerts no torque about its own axis, so L about the axle keeps its starting value. The trap is the walker's speed. A 60 kg person stands on the rim of a 120 kg m² platform of radius 2.0 m, both at rest, and starts walking at u = 1.5 m s⁻¹ relative to the rim. Angular momentum needs her velocity relative to the ground, u + ωR, so 0 = 120ω + 60 × 2.0 × (1.5 + 2.0ω), which gives ω = −180/360 = −0.50 rad s⁻¹ and a ground speed of only 0.50 m s⁻¹. Treating 1.5 m s⁻¹ as the ground speed gives −1.5 rad s⁻¹, three times too fast. Kinetic energy has gone from zero to 22.5 J. Her legs supplied it, and nothing about conserved L forbids that.

A solid ball of mass m and radius R is set down sliding at v₀ with no spin. Kinetic friction slows its centre and spins it up until v = ωR, and the size of that friction, μₖ m g, is exactly what you would rather not need. About a fixed point P on the floor line, friction acts along the floor and has no lever arm. The weight and the normal force lie on the same vertical line through the contact point and cancel in torque as they do in force. So L_P is conserved through the whole skid: m v₀ R = m v R + (2/5)m R² ω, and with v = ωR at the end, v = 5v₀/7. A ball launched at 7.0 m s⁻¹ settles into rolling at 5.0 m s⁻¹ on rubber or on ice; μₖ fixes only how long the skid lasts and where it ends. About the ball's centre, friction exerts torque μₖ m g R, and you would need μₖ to get anywhere: the same physics, a worse origin.

Real problems join several events, and each law belongs to its own interval. Finish the putty and rod. Across the impact, L about the pivot was the only safe statement, and kinetic energy fell from 9.00 J to L²/(2I) = 2.70²/(2 × 1.072) = 3.40 J. After the impact the pivot still does no work, but gravity now exerts a torque, so L is no longer conserved and energy is: 3.40 J = (ML/2 + md) g (1 − cos θ) = 1.45 × 9.81 × (1 − cos θ), so the rod swings up to θ = 40.5°. Carrying energy straight across the collision would put 9.00 J into the swing, cos θ = 0.37, a 68° swing that never happens. A free body needs a different count. Putty sticking to a rod on an air table leaves two unknowns, the centre-of-mass velocity and ω, and linear momentum together with L about the combined centre of mass supplies exactly two equations.