System of Particles and Rotational Motion · Lesson 8 of 15
Angular Momentum and External Torque
“Derive how torque changes angular momentum for a particle and a system.”
• Calculate r × p and its components. • Derive dL/dt = external torque. • State the internal-force assumptions. • Explain conservation and precession qualitatively.
Momentum with a moment arm
A moving particle can have angular momentum about an origin even while travelling in a straight line. Its perpendicular distance from the origin to the motion line supplies the moment arm. Angular momentum vanishes when momentum is zero or the motion line passes through the origin. Its direction is perpendicular to the r–p plane. It is generally origin dependent.
Deriving the torque law
For a system, differentiate and sum the particle angular momenta. Internal action–reaction pairs cancel in force, but torque cancellation additionally requires their forces to lie along the line joining the pair. A pair contributes (rᵢ − rⱼ) × Fᵢⱼ, which vanishes for a central internal force. Under these assumptions, only external torque changes total angular momentum.
Problem
Why is angular momentum about any fixed origin constant for a force-free particle?
- 1.Its speed and straight motion line remain unchanged. The perpendicular distance from a fixed origin to that line is constant.
- 2.Therefore mvd⊥ is constant. The plane containing the line and origin is also fixed, giving a constant angular-momentum direction.
- 3.Equivalently, zero resultant force gives zero torque about every fixed origin, so the torque law gives dℓ/dt = 0. The value can differ between origins even though it is constant in time about each.
Problem
Two particles of mass m move with opposite velocities of magnitude v along lines separated by d. Show why total L is origin independent.
- 1.Their total linear momentum is zero. If the origin shifts by vector a, angular momentum changes by −a × P = 0.
- 2.Choose an origin on the first motion line. That particle contributes zero and the other contributes magnitude mvd.
- 3.Hence total angular momentum has magnitude mvd about every origin, with direction fixed by the circulation sense.
The bicycle-rim observation
A fast-spinning bicycle rim has angular momentum directed along its axle. If supported at one axle end, gravity acting at its CM produces torque about the support. The torque can change the direction of L rather than simply its magnitude, giving precession. A small change ΔL points along torque times Δt. The wheel’s plane turns while it keeps spinning. A nonspinning rim tends to tilt downward instead. This qualitative demonstration connects direction change with vector dynamics; deriving the full precession rate is beyond this chapter.
Quiz
A particle on a straight path can have nonzero angular momentum about an off-path origin?
An offset motion line gives a nonzero perpendicular moment arm.
Which extra condition makes internal pair torques cancel?
Equal and opposite noncentral forces alone need not cancel torque.
Zero external torque implies which conserved quantity?
The torque law directly gives constant L.
For motion in the x–y plane, angular momentum points along which axis?
The cross product is perpendicular to the motion plane.
Gravity-induced precession chiefly illustrates what?
A torque perpendicular to L turns its direction.
Key Takeaways
• Angular momentum is the moment of linear momentum. • A particle need not move in a circle to have angular momentum. • Central internal-force pairs leave only external torque in the system law. • Conservation concerns L, not automatically speed or kinetic energy.