System of Particles and Rotational Motion · Lesson 5 of 9
Torque and Angular Momentum
“Calculate moments of force and momentum and derive the external-torque law for particles and systems.”
• Calculate torque using perpendicular lever arms and cross products. • Find angular-momentum components and their direction. • Derive dℓ/dt = τ and the system law with central internal forces. • Apply angular-momentum conservation and interpret the bicycle-rim observation.
Torque and its reference point
Force alone does not specify turning effect
Push a door near its hinge and then at its outer edge using the same force. The outer push usually turns it more effectively. The turning effect depends on the point of application, the force direction and the reference point or axis. Torque, also called moment of force, combines these features. A force can accelerate a body’s CM while also changing its rotation.
The force’s line of action is the entire straight line along which it acts. The lever arm is perpendicular to that line, not necessarily the length of a rod. Resolve the force into a radial component parallel to r and a perpendicular component. The radial part has no moment; only the perpendicular part contributes. Torque is zero if force vanishes, its application point is the origin, or its line of action passes through the origin.
Direction, components and reference point
Use the right-hand rule for r × F. In a plane, choose an out-of-plane positive axis; anticlockwise moments are then positive and clockwise moments negative. Add torques algebraically only when referring to the same axis. Reversing F alone reverses torque. Reversing both r and F preserves it. A different origin usually changes r and therefore changes the torque.
Problem
Let r = (1, −1, 1) m and F = (7, 3, −5) N.
- 1.The x component is (−1)(−5) − (1)(3) = 2 N m.
- 2.The y component is (1)(7) − (1)(−5) = 12 N m.
- 3.The z component is (1)(3) − (−1)(7) = 10 N m. Thus τ = (2,12,10) N m.
Torque has dimensions ML²T⁻², the same as energy, and SI unit newton metre. Nevertheless, torque is an axial vector describing turning effect, while work is a scalar describing energy transfer. We normally retain N m for torque rather than labelling it joules. Matching dimensions do not make physical quantities interchangeable.
Angular momentum and external torque
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.
For motion confined to the x–y plane through the chosen origin, z = 0 and pz = 0. Substituting these into the component expressions makes ℓx = ℓy = 0, leaving ℓz = xpy − ypx. Angular momentum is therefore perpendicular to that plane; its sign selects +z or −z. Angular momentum has SI unit kg m²/s and dimensions M L² T⁻¹. Torque has unit N m and dimensions M L² T⁻², consistent with torque being its time rate of change.
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.
Check your understanding
Quiz
The moment arm is which distance?
The force times this shortest distance gives the torque magnitude.
Torque and energy share what?
Their dimensions match but torque is a vector and energy a scalar.
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.
Key Takeaways
• Torque and angular momentum require a specified reference origin. • Torque is r × F; angular momentum is r × p. • Internal torques cancel under the central-force assumptions used in the PDF. • Zero total external torque conserves total angular momentum about that reference.