Angular Impulse-Momentum — Apply angular impulse–momentum about a point (intro-to-standard) - Statics and Dynamics

SECTION 1

Historical Context & Motivation

The concept of angular momentum and its relationship to applied torques developed over several centuries, rooted in the study of planetary motion and rotating bodies. Newton's laws, originally formulated for translational motion, were soon extended to rotational problems by mathematicians and mechanicians who recognized that the moment of a force about a point plays the same role for rotation that force itself plays for translation. The principle of angular impulse–momentum emerged as the integral form of this rotational Newton's second law, providing a powerful tool for analyzing systems where torques act over finite time intervals—collisions, impacts, and thrust maneuvers being among the most prominent applications in engineering dynamics.

1687 • Newton's Principia

Isaac Newton published the laws of motion, establishing that the rate of change of (linear) momentum equals the applied force. The rotational analogue—moment equals rate of change of angular momentum—was implicit in his treatment of planetary orbits and the equal-area law.

1736 • Euler's Mechanica

Leonhard Euler formalized the equations of rotational motion for rigid bodies and introduced the concept of moment of inertia, laying the groundwork for systematic angular momentum analysis about fixed and moving points.

1788 • Lagrange's Analytical Mechanics

Joseph-Louis Lagrange generalized mechanics using energy methods, within which conservation of angular momentum appeared naturally whenever the Lagrangian was invariant under rotation—connecting symmetry to conservation laws.

1918 • Noether's Theorem

Emmy Noether proved that every continuous symmetry of a physical system corresponds to a conserved quantity; rotational symmetry yields conservation of angular momentum, providing the deepest justification for the impulse–momentum relation in the absence of net external moments.

In an introductory dynamics course, the central question this principle addresses is: How does a time-varying moment about a point alter the angular momentum of a particle or system? Rather than resolving instantaneous accelerations, the angular impulse–momentum theorem lets the engineer integrate the effect of moments over a time interval, directly linking initial and final angular momentum states. This is especially valuable in impact and impulsive-force problems where forces are large but act over very short durations, making direct force analysis impractical.

SECTION 2

Core Principles & Definitions

Before applying the angular impulse–momentum theorem, several foundational quantities must be clearly defined. Each builds upon Newton's second law extended to rotation, and together they form the conceptual scaffold for every problem in this topic.

Angular Momentum about a Point

For a particle of mass m with position vector r measured from point O and velocity v, the angular momentum is H_O = r × (mv). It is a vector perpendicular to the plane of r and v, with magnitude |r||mv|sin θ.

Moment of a Force about a Point

The moment (torque) of force F about point O is M_O = r × F. Only the component of F perpendicular to r contributes to this moment; the radial component produces zero torque.

Angular Impulse

The angular impulse about O over [t₁, t₂] is the time integral ∫M_O dt. It represents the cumulative rotational effect of all external moments during that interval.

The Angular Impulse–Momentum Theorem

The theorem states (H_O)₁ + ∫M_O dt = (H_O)₂. The initial angular momentum plus the angular impulse equals the final angular momentum.

Conservation of Angular Momentum

When the net external moment about O is zero over the interval, the angular impulse vanishes and (H_O)₁ = (H_O)₂. This is the rotational analogue of conservation of linear momentum.

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KEY TAKEAWAY

Think of angular impulse–momentum as the rotational equivalent of the linear impulse–momentum theorem (F·Δt = Δ(mv)). Just as pushing a shopping cart over time changes its linear momentum, applying a torque over time changes angular momentum. The angular impulse is like the 'rotational push' accumulated over a time interval: a small torque acting for a long time can produce the same angular impulse as a large torque acting briefly. When no external torques act, angular momentum is conserved—analogous to how a spacecraft coasts at constant velocity in the absence of external forces.

SECTION 3

Visual Explanation — Angular Momentum About a Point

A particle of mass m moves with velocity v at position r from fixed point O. The angular momentum vector H_O = r × mv is perpendicular to the plane formed by r and v, directed out of that plane by the right-hand rule. The magnitude is |H_O| = m|r||v|sin θ.

In the diagram above, point O is the fixed reference about which we compute angular momentum. The position vector r extends from O to the particle, while the velocity v defines the particle's instantaneous motion. The cross product r × (mv) yields the angular momentum vector $H_O$ , which points perpendicular to the plane containing r and v. For planar problems—the vast majority of introductory applications—H_O reduces to a scalar whose sign indicates the sense of rotation (positive counterclockwise by convention). The choice of point O is crucial: selecting a point through which unknown forces pass eliminates those forces from the moment equation, greatly simplifying the analysis.

SECTION 4

Mathematical Framework

The angular impulse–momentum theorem is derived directly from Newton's second law applied in moment form. Starting from the equation of motion for a particle, we take the cross product of the position vector with both sides, then integrate over time. The derivation proceeds as follows.

Derivation from Newton's Second Law

For a particle of mass m subject to resultant force F, Newton's second law gives F = m(dv/dt). Taking the cross product of the position vector r (from fixed point O) with both sides: r × F = r × m(dv/dt). The left side is the moment M_O of the resultant force about O. For the right side, note that d(r × mv)/dt = (dr/dt) × mv + r × m(dv/dt). Since dr/dt = v and v × mv = 0, the first term vanishes, giving d(r × mv)/dt = r × m(dv/dt). Therefore M_O = dH_O/dt, confirming that the net moment about a fixed point equals the time rate of change of angular momentum about that point.

ANGULAR MOMENTUM (PARTICLE)

H_O = r × (mv)

where H_O = angular momentum about O (kg·m²/s), r = position vector from O to particle (m), m = mass (kg), v = velocity (m/s).

MOMENT–ANGULAR MOMENTUM RELATION

M_O = dH_O / dt

The net external moment about O equals the time derivative of the angular momentum about O. This is the rotational analogue of F = dp/dt.

ANGULAR IMPULSE–MOMENTUM THEOREM

(H_O)₁ + ∫₁² M_O dt = (H_O)₂

Integrating M_O = dH_O/dt from t₁ to t₂: the initial angular momentum plus the angular impulse (the integral of the moment) equals the final angular momentum.

SCALAR FORM (PLANAR PROBLEMS)

(H_O)₁ + ∫₁² M_O dt = (H_O)₂ where H_O = m·v·d

For 2-D planar motion, angular momentum simplifies to H_O = m·v·d, where d is the perpendicular (moment arm) distance from O to the line of action of mv. Counterclockwise is taken as positive.

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Important Sign Convention

In planar problems, counterclockwise angular momentum and moments are typically taken as positive. Be consistent with your sign convention throughout a problem. If a velocity component creates a clockwise tendency about O, its contribution to H_O is negative.

SECTION 5

Angular Impulse — Detailed Breakdown

The angular impulse ∫M_O dt is the rotational analogue of linear impulse ∫F dt. Understanding how to evaluate this integral under different loading conditions is essential for applying the theorem correctly. In many introductory problems, the moment is constant, varies linearly, or is impulsive—each case requiring a different treatment.

The angular impulse–momentum diagram mirrors the linear impulse–momentum diagram. The particle's initial angular momentum (H_O)₁ is shown at left. The central panel illustrates the angular impulse as the area under the M_O vs. t curve. The final angular momentum (H_O)₂ appears at right. This three-panel layout is a powerful problem-solving framework.

The three-panel layout shown above is a systematic approach to every angular impulse–momentum problem. On the left, sketch the system at time t₁ and compute the initial angular momentum about O. In the center, identify every external moment about O and integrate over the time interval. On the right, express the final angular momentum in terms of the unknowns. Setting the equation (H_O)₁ + ∫M_O dt = (H_O)₂ then yields a single scalar equation (in 2-D) that relates the unknowns to the given information.

Loading Type	$M_O$ (t)	Angular Impulse ∫ $M_O$ dt
Constant moment	M₀ (constant)	M₀ × Δt
Linear ramp	M₀(t/T)	M₀T/2 (triangle area)
Impulsive moment	Very large, brief	Given as finite value (impulse)
Zero net moment	$M_O$ = 0	0 → conservation of $H_O$

Common loading scenarios and the resulting angular impulse expressions.

SECTION 6

Worked Example

A 2 kg ball is attached to a light inextensible string and moves in a horizontal circle of radius 1.5 m about a fixed pivot O at an initial speed of 4 m/s. A constant tangential braking force of 3 N is applied to the ball for 1.2 s. Determine the ball's speed at the end of the braking period, assuming the string remains taut and the motion stays circular.

Constant Tangential Brake on a Circular-Path Particle

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Step 1 — Identify the Reference Point and Given Values

Choose the fixed pivot O as the reference point for angular momentum. Given: m = 2 kg, r = 1.5 m (constant since the string is inextensible), v₁ = 4 m/s, F_brake = 3 N (tangential, opposing motion), Δt = 1.2 s. The string tension and the weight/normal force all pass through O or are balanced, so they produce zero moment about O.

Only the braking force contributes a moment about O.
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Step 2 — Compute the Initial Angular Momentum

Since the velocity is tangential (perpendicular to r), the moment arm d = r = 1.5 m. Therefore (H_O)₁ = m × v₁ × r = 2 × 4 × 1.5 = 12 kg·m²/s. Taking the initial direction of motion as positive (counterclockwise).

(H_O)₁ = +12 kg·m²/s
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Step 3 — Compute the Angular Impulse

The braking force is tangential and opposes the direction of motion, so its moment about O is M_O = −F_brake × r = −3 × 1.5 = −4.5 N·m. Since this moment is constant, the angular impulse is ∫M_O dt = M_O × Δt = (−4.5)(1.2) = −5.4 kg·m²/s.

Angular impulse = −5.4 kg·m²/s
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Step 4 — Apply the Angular Impulse–Momentum Theorem

(H_O)₁ + ∫M_O dt = (H_O)₂. Substituting: 12 + (−5.4) = (H_O)₂ = 6.6 kg·m²/s.

(H_O)₂ = 6.6 kg·m²/s
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Step 5 — Solve for the Final Speed

Since the string remains taut and the radius is constant, (H_O)₂ = m × v₂ × r. Therefore v₂ = (H_O)₂ / (m × r) = 6.6 / (2 × 1.5) = 6.6 / 3 = 2.2 m/s.

v₂ = 2.2 m/s (counterclockwise)

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Verification

We can verify using linear impulse–momentum: the tangential impulse is F·Δt = 3 × 1.2 = 3.6 N·s. Therefore m(v₂ − v₁) = −3.6, giving v₂ = 4 − 3.6/2 = 4 − 1.8 = 2.2 m/s. This matches the angular impulse–momentum result, confirming internal consistency. Both methods are valid; the angular approach is preferred when moments about a point naturally eliminate unknown reaction forces.

SECTION 7

Strengths, Limitations & Comparison to Other Methods

The angular impulse–momentum theorem is one of several tools in the engineer's dynamics toolkit. Understanding when to deploy it—versus Newton's second law in moment form, the work–energy theorem, or conservation of angular momentum—is critical for efficient problem solving. The table below contrasts these approaches.

Method	Best For	Limitations
Angular Impulse–Momentum	Problems involving moments acting over time intervals; impulsive torques; eliminating unknown forces passing through the reference point	Cannot directly determine positions or displacements; requires time data or moment–time relationships
ΣM = dH/dt (instantaneous)	Finding instantaneous angular acceleration or rate of change of angular momentum at a specific instant	Requires integration or ODE solving for finite-interval problems; not directly finite-time
Work–Energy (T₁ + U₁₋₂ = T₂)	Finding speeds when displacements are known; no time information needed	Scalar equation—no directional velocity info; friction work can be complex; cannot determine time
Conservation of Angular Momentum	Central-force problems; collisions where net moment about a point is zero; satellite orbit changes	Only applicable when Σ $M_O$ = 0; a special case of the impulse–momentum theorem

Comparison of angular impulse–momentum with related methods in dynamics.

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WHEN TO USE ANGULAR IMPULSE–MOMENTUM

Use angular impulse–momentum when the problem involves a time interval and asks for velocity or angular momentum changes. The method shines in collision, impact, and impulsive-force scenarios. It is also the tool of choice when taking moments about a strategically chosen point eliminates unknown constraint forces—much like choosing a pivot in statics to eliminate a pin reaction from the moment equation. If the problem gives displacements rather than times, consider work–energy instead.

SECTION 8

Connection to Advanced Theory — Rigid Bodies & 3-D Systems

The particle-based angular impulse–momentum theorem introduced here is the foundation for more advanced formulations encountered in intermediate and graduate dynamics courses. As you progress, the same principle extends to rigid bodies (where H_O = I_Oω for planar motion), 3-D systems (requiring the inertia tensor), and systems of particles (where internal forces cancel by Newton's third law). The table below maps the introductory concepts to their advanced counterparts.

Introductory Concept	Advanced Extension
$H_O$ = r × mv (particle)	$H_O$ = $I_O$ · ω (rigid body, planar); $H_O$ = [I] · ω (rigid body, 3-D tensor)
Fixed point O	Moving reference point (mass center G), yielding $M_G$ = dH_G/dt for rigid bodies
Scalar (2-D) angular impulse	Vector (3-D) angular impulse; Euler's equations for spinning tops and gyroscopes
Conservation when Σ $M_O$ = 0	Noether's theorem: rotational symmetry ↔ angular momentum conservation; Lagrangian/Hamiltonian mechanics

From introductory particle theory to advanced rigid-body and Lagrangian formulations.

Mastering the particle formulation now builds the conceptual and computational fluency needed for these extensions. The selection of a convenient reference point, the sign convention for moments and angular momenta, and the systematic impulse–momentum diagram approach all carry directly into rigid-body dynamics and beyond.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A particle moves in a straight line that does not pass through point O. Is the angular momentum of the particle about O constant, increasing, or decreasing? Explain your reasoning without performing any calculations.

Show answer

If no external force acts on the particle, then the net moment about any point is zero, so angular momentum about O is conserved. Even though the perpendicular distance d from O to the velocity line changes as the particle moves, the product m·v·d remains constant because v is constant and d decreases while the component of position along the velocity direction increases—they compensate exactly. This is essentially Kepler's equal-area law applied to free motion: the particle sweeps out equal areas per unit time relative to O. If a net force does act, one must check whether that force's line of action passes through O; only forces with nonzero moment arms about O will change

H_O

Question 2: BASIC CALCULATION

A 0.5 kg puck slides on frictionless ice. At time t₁, its position relative to point O is r₁ = (3î + 0ĵ) m and its velocity is v₁ = (0î + 6ĵ) m/s. Compute the angular momentum

H_O

at this instant.

Show answer

The angular momentum is

H_O

= r × (mv) = (3î × 0.5 × 6ĵ) = 3 × 3 (î × ĵ) = 9k̂ kg·m²/s. The cross product î × ĵ = k̂ by the right-hand rule, confirming

H_O

is directed out of the x-y plane. Numerically,

H_O

= +9 kg·m²/s in the k̂ direction, indicating counterclockwise angular momentum when viewed from above. Since the ice is frictionless and no other horizontal forces act (gravity and normal cancel), this angular momentum is conserved for all future times.

Question 3: INTERMEDIATE

A 3 kg particle moves in a circular path of radius 2 m about fixed point O with an initial speed of 8 m/s. A tangential force whose moment about O varies as

M_O

(t) = 12 − 4t N·m (counterclockwise positive) is applied from t = 0 to t = 3 s. Find the angular momentum

H_O

at t = 3 s and the particle's speed at that instant.

Show answer

First, compute the initial angular momentum: (

H_O

)₁ = m·v₁·r = 3 × 8 × 2 = 48 kg·m²/s. Next, compute the angular impulse: ∫₀³ (12 − 4t) dt = [12t − 2t²]₀³ = (36 − 18) − 0 = 18 kg·m²/s. Apply the theorem: (

H_O

)₂ = 48 + 18 = 66 kg·m²/s. Finally, since r = 2 m remains constant on the circular path, v₂ = (

H_O

)₂ / (m·r) = 66 / (3 × 2) = 11 m/s. The positive result confirms the particle is still moving counterclockwise, and the increasing angular momentum is consistent with the net angular impulse being positive over the interval.

Question 4: APPLIED

A satellite modeled as a 500 kg point mass orbits a planet in a circular orbit of radius 7000 km at speed 7.5 km/s. The satellite fires a short thruster burst that produces a tangential force of 2500 N for 4 s. Taking the planet's center as point O, determine the satellite's angular momentum before and after the burn, and estimate the new orbital speed assuming the radius has not yet changed appreciably during the brief burn.

Show answer

Before the burn: (

H_O

)₁ = m·v₁·r = 500 × 7500 × 7.0 × 10⁶ = 2.625 × 10¹³ kg·m²/s. The thruster force is tangential, so its moment arm about the planet center is r = 7.0 × 10⁶ m. The angular impulse is

M_O

·Δt = (2500 × 7.0 × 10⁶) × 4 = 7.0 × 10¹⁰ kg·m²/s. After the burn: (

H_O

)₂ = 2.625 × 10¹³ + 7.0 × 10¹⁰ = 2.6320 × 10¹³ kg·m²/s. The new speed (assuming constant radius) is v₂ = (

H_O

)₂ / (m·r) = 2.6320 × 10¹³ / (500 × 7.0 × 10⁶) ≈ 7520 m/s ≈ 7.52 km/s. The speed increase of about 20 m/s is consistent with the linear impulse check: Δv = F·Δt/m = 2500 × 4 / 500 = 20 m/s.

Question 5: CRITICAL THINKING

Consider a particle that spirals inward along a smooth frictionless track toward a fixed center O under no external tangential force, but the track exerts a normal (radial) force that constrains the motion. Prove, using the angular impulse–momentum theorem, that the particle's angular momentum about O is conserved throughout the spiral, and then explain why the particle speeds up as it moves inward despite no tangential force being applied.

Show answer

Since the track is frictionless, the only contact force is normal to the velocity, directed radially toward O. A radial force passes through O, so its moment about O is zero:

M_O

= r × F = 0 (since r and F are parallel). By the angular impulse–momentum theorem, ∫

M_O

dt = 0, which gives (

H_O

)₁ = (

H_O

)₂ for any two instants. Since

H_O

= m·v·d where d is the perpendicular distance from O to the velocity vector, and for the radially-constrained spiral d = r·sin α (α being the angle between r and v), we have m·v·r·sin α = constant. As r decreases, v must increase to keep the product constant. For a tightly spiraling path where v is nearly tangential (sin α ≈ 1), this reduces approximately to m·v·r = constant, directly showing v ∝ 1/r. The particle speeds up because the normal constraint force, while doing no work in the tangential direction, redirects momentum inward; conservation of angular momentum requires the tangential speed to increase as the radius shrinks. This is the same principle that causes an ice skater to spin faster when pulling in their arms.

SUMMARY

Summary — Angular Impulse–Momentum About a Point

Summary

The angular impulse–momentum theorem states that the initial angular momentum of a particle about a fixed point O plus the angular impulse (time integral of the net moment about O) equals the final angular momentum. Mathematically: (H_O)₁ + ∫M_O dt = (H_O)₂. For a particle, $H_O$ = r × mv, which in planar problems reduces to H_O = m·v·d, where d is the moment arm from O to the line of action of the momentum vector.

This theorem is the rotational analogue of linear impulse–momentum and is especially powerful for problems involving impulsive forces, collisions, and orbital mechanics. When the net external moment about O is zero, angular momentum is conserved. The strategic choice of reference point O—such that unknown reaction forces pass through it—is the single most important problem-solving skill in applying this theorem. The three-panel impulse–momentum diagram (initial state + impulse = final state) provides a systematic framework for solving any problem of this type.