Apply Rigid Body Mechanics — Apply A.4 Rigid body mechanics...

SECTION 1

Historical Context & Motivation

For thousands of years, builders and engineers have wrestled with a deceptively simple question: why do some structures stand while others topple? The ancient Egyptians balanced enormous stone blocks to construct pyramids, and Greek engineers designed lever systems that could move loads far heavier than a person could lift. Yet none of these civilizations had a formal mathematical framework to explain why these systems worked. The study of rigid body mechanics — the physics of objects that can rotate and translate without deforming — filled that gap over several centuries of scientific progress.

~250 BCE

Archimedes & the Lever

Archimedes formalised the law of the lever, showing that two masses balance when their distances from the pivot are inversely proportional to their weights. This was the first quantitative treatment of rotational equilibrium.

1687

Newton's Laws of Motion

Isaac Newton published the Principia, establishing the three laws that govern translational motion. These laws became the foundation upon which rotational analogues — torque, angular acceleration, and moment of inertia — were later built.

1736

Euler's Rigid Body Dynamics

Leonhard Euler extended Newton's framework to rotating bodies, introducing the concept of moment of inertia and deriving the rotational equations of motion that we still use today in IB Physics.

1800s–Today

Engineering Applications

Rigid body mechanics became essential in designing bridges, engines, gyroscopes, and spacecraft attitude-control systems. Modern robotics and biomechanics continue to rely on these same principles.

In your IB Physics course, Topic A.4 asks you to move beyond point-particle models and treat objects as extended bodies that can spin, tip, and roll. The central question is: How do forces applied at different points on an object determine whether it accelerates, rotates, or stays in equilibrium? Answering that question requires torque, rotational inertia, and angular momentum — the tools you will master in this lesson.

SECTION 2

Core Principles & Definitions

A rigid body is an idealised object whose shape does not change when forces act on it. Unlike a point particle, a rigid body has size and shape, so the location at which a force is applied matters enormously. The same force pushing through the centre of mass produces pure translation, but applied off-centre it also causes rotation. Understanding this distinction is the heart of rigid body mechanics.

Torque (τ)

The rotational equivalent of force. Torque equals the force multiplied by the perpendicular distance from the pivot (the moment arm). A larger moment arm means a greater turning effect, even with the same force.

Moment of Inertia (I)

The rotational equivalent of mass. It measures how difficult it is to change an object's rotational motion. It depends on both the mass and how that mass is distributed relative to the axis of rotation.

Angular Momentum (L)

The rotational equivalent of linear momentum, defined as L = Iω. Like linear momentum, angular momentum is conserved when no external net torque acts on the system.

Rotational Equilibrium

An object is in rotational equilibrium when the sum of all torques about any point equals zero (Στ = 0). Combined with translational equilibrium (ΣF = 0), this defines full static equilibrium.

Centre of Mass

The single point where the entire weight of an object can be considered to act. For symmetric, uniform objects it lies at the geometric centre. The centre of mass is the natural pivot for analysing free rotation.

✦ KEY TAKEAWAY

Think of opening a heavy door. Pushing near the hinges barely moves it, but pushing at the handle (far from the pivot) swings it easily. You are applying the same force in both cases — what changes is the moment arm, and therefore the torque. Rigid body mechanics is all about recognising that where and how a force is applied matters just as much as how large the force is.

SECTION 3

Visualising Torque & Equilibrium

A uniform beam balanced on a fulcrum (pivot). The blue upward force F₁ at distance r₁ creates a counter-clockwise torque, while the red force F₂ at distance r₂ creates a clockwise torque. For rotational equilibrium, these torques must be equal.

In the diagram above, notice that the shorter moment arm (r₂ = 0.40 m) requires a larger force (60 N) to balance the torque produced by the smaller force (40 N) acting at the longer moment arm (r₁ = 0.60 m). This is precisely the principle behind levers, seesaws, and even the way you use a wrench to loosen a bolt — extending the handle gives you a bigger moment arm and therefore a bigger torque for the same effort.

When solving IB problems, always start by choosing a pivot point. A smart choice of pivot eliminates unknown forces that act at that point (their moment arm is zero, so their torque is zero). Then set up the condition Στ = 0 by summing clockwise and counter-clockwise torques.

SECTION 4

Mathematical Framework

Rigid body mechanics extends Newton's laws into the rotational domain. For every translational quantity, there is a rotational analogue. The equations below are the core toolkit for IB A.4 problems.

TORQUE

τ = F × r × sin θ

τ = torque (N·m), F = applied force (N), r = distance from the pivot to the point of application (m), θ = angle between F and r. When the force is perpendicular to r, sin θ = 1 and τ = Fr.

NEWTON'S SECOND LAW FOR ROTATION

Στ = Iα

Στ = net torque (N·m), I = moment of inertia (kg·m²), α = angular acceleration (rad/s²). This is the rotational version of ΣF = ma.

ANGULAR MOMENTUM

L = Iω

L = angular momentum (kg·m²/s), I = moment of inertia (kg·m²), ω = angular velocity (rad/s). Angular momentum is the rotational analogue of p = mv.

CONSERVATION OF ANGULAR MOMENTUM

I₁ω₁ = I₂ω₂ (when Στ_ext = 0)

When no external net torque acts, the product Iω remains constant. If I decreases (mass moves closer to the axis), ω increases — this is why a figure skater spins faster when pulling their arms in.

📐 Rotational Kinematic Equations

Just as translational kinematics has v = u + at and s = ut + ½at², rotational kinematics has ω = ω₀ + αt and θ = ω₀t + ½αt². Use these when angular acceleration is constant, for example a wheel speeding up uniformly.

Translational vs. rotational analogues

Translational Quantity	Symbol	Rotational Analogue	Symbol
Displacement	s	Angular displacement	θ
Velocity	v	Angular velocity	ω
Acceleration	a	Angular acceleration	α
Force	F	Torque	τ
Mass	m	Moment of inertia	I
Momentum (p = mv)	p	Angular momentum (L = Iω)	L

SECTION 5

Moment of Inertia — Shape Matters

The moment of inertia depends on how mass is distributed relative to the rotation axis. A solid sphere of a given mass is easier to spin than a hollow sphere of the same mass and radius, because more of the hollow sphere's mass sits at a greater distance from the centre. The IB formula booklet provides expressions for common shapes; you don't need to derive them, but you must understand why they differ.

Common moments of inertia required for IB Physics A.4. The dashed vertical lines represent the axis of rotation. Note that the hollow shapes always have a larger moment of inertia than their solid counterparts of the same mass and radius, because more mass is far from the axis.

The numerical coefficient in front of mr² tells you how 'spread out' the mass is. A point mass has a coefficient of 1 (all mass at distance r). A solid cylinder has ½ because some mass is close to the axis, bringing the average down. A solid sphere has ⅖ because mass is distributed in three dimensions, placing even more mass near the axis. When an IB question asks you to predict which object reaches the bottom of a ramp first, the one with the smallest coefficient accelerates the fastest, because less torque is 'used up' fighting rotational inertia.

SECTION 6

Worked Example — Beam in Equilibrium

A uniform plank of mass 12 kg and length 4.0 m is supported at its left end by a pivot and at a point 3.0 m from the left end by a vertical cable. A 5.0 kg box is placed 1.0 m from the left end. Find the tension in the cable and the reaction force at the pivot. Take g = 9.8 m/s².

Uniform Plank with Cable Support

Step 1 — Draw a Free-Body Diagram and Identify ForcesThere are four forces acting on the plank: (1) the reaction force R at the pivot (upward, at x = 0); (2) the weight of the box W_box = 5.0 × 9.8 = 49 N downward at x = 1.0 m; (3) the weight of the plank W_plank = 12 × 9.8 = 117.6 N downward at the centre of mass (x = 2.0 m); (4) the cable tension T upward at x = 3.0 m.

Step 2 — Take Torques About the PivotChoosing the pivot as the rotation point eliminates R from the torque equation (its moment arm is zero). Clockwise torques (weights): τ_box = 49 × 1.0 = 49.0 N·m, τ_plank = 117.6 × 2.0 = 235.2 N·m. Counter-clockwise torque: τ_cable = T × 3.0.

Step 3 — Apply Rotational Equilibrium (Στ = 0)Setting counter-clockwise torques equal to clockwise torques: T × 3.0 = 49.0 + 235.2 = 284.2 N·m. Solving: T = 284.2 / 3.0 = 94.7 N.

T ≈ 95 N (2 s.f.)

Step 4 — Apply Translational Equilibrium (ΣF = 0) to Find RUpward forces = downward forces: R + T = W_box + W_plank. R + 94.7 = 49.0 + 117.6. R = 166.6 − 94.7 = 71.9 N.

R ≈ 72 N (2 s.f.)

Step 5 — VerifyCheck: total upward = 72 + 95 = 167 N. Total downward = 49 + 117.6 ≈ 167 N. ✓ Forces balance. Torques also balance by construction. The solution is self-consistent.

SECTION 7

Strengths & Limitations of the Rigid Body Model

When the rigid body model works well and when it breaks down

Strengths	Limitations
Simplifies complex objects to a single shape with a defined moment of inertia, making problems tractable.	Ignores internal deformations — real materials bend, flex, and vibrate under load.
Accurately predicts equilibrium and motion for stiff objects like steel beams, wheels, and spinning discs.	Fails for soft or elastic bodies (e.g., a bouncing rubber ball deforms significantly on impact).
Conservation of angular momentum applies universally — from ice skaters to collapsing stars.	Does not account for friction-generated heat or energy losses inside the body itself.
Extends Newton's laws seamlessly into rotational problems with direct analogies.	For very large or very fast objects, relativistic corrections may be needed (beyond IB scope).

✦ KEY TAKEAWAY

The rigid body model is like a perfectly stiff LEGO brick — it keeps its shape no matter what you do to it. Real objects are more like modelling clay, flexing and deforming. For most IB-level problems, the rigid body approximation is excellent, but always ask yourself: does the object significantly change shape during the process? If yes, you may need a more advanced model.

SECTION 8

Connection to Advanced Rotational Dynamics

At the IB level, you work primarily with objects rotating about a single fixed axis. In university-level mechanics, the picture becomes richer. Objects can rotate about multiple axes simultaneously, leading to phenomena like precession (a spinning top slowly tracing a cone) and nutation (a wobble superimposed on precession). The moment of inertia becomes a tensor — a 3×3 matrix rather than a single number — capturing how resistance to rotation varies with direction.

IB-level vs. university-level rigid body mechanics

Feature	IB A.4 (This Lesson)	University Mechanics
Rotation axes	Fixed, single axis	Any axis, possibly changing
Moment of inertia	Scalar (I)	Inertia tensor (3×3 matrix)
Equations of motion	Στ = Iα	Euler's equations (coupled differential equations)
Calculus required?	No	Yes — integral and vector calculus
Real-world examples	Seesaws, wheels, doors	Gyroscopes, spacecraft, spinning molecules

Don't worry about tensors or Euler's equations for now. The key insight to carry forward is that the principles you learn in A.4 are not simplified toys — they are the genuine building blocks of advanced rotational dynamics. Mastering torque, moment of inertia, and angular momentum here gives you a solid launchpad for any future study in engineering, astrophysics, or biomechanics.

SECTION 9

Practice Problems

PROBLEM 1 — CONCEPTUAL

A uniform metre rule is balanced on a pivot at the 50 cm mark. A student places a 2.0 N weight at the 80 cm mark. Without calculating, explain on which side of the pivot the student should place a 4.0 N weight to restore balance, and whether it should be closer to or further from the pivot than the 2.0 N weight.

PROBLEM 2 — BASIC CALCULATION

A force of 25 N is applied perpendicularly at the end of a 0.40 m wrench to tighten a bolt. Calculate the torque applied to the bolt.

PROBLEM 3 — INTERMEDIATE

A solid disc of mass 4.0 kg and radius 0.30 m is free to rotate about its central axis. A constant tangential force of 6.0 N is applied at the rim. Calculate (a) the moment of inertia, (b) the angular acceleration, and (c) the angular velocity after 5.0 s starting from rest.

PROBLEM 4 — APPLIED

A figure skater spins with her arms extended, giving her a moment of inertia of 3.6 kg·m² and an angular velocity of 2.0 rad/s. She then pulls her arms in, reducing her moment of inertia to 1.2 kg·m². (a) Calculate her new angular velocity. (b) Compare her rotational kinetic energies before and after pulling in her arms, and explain where the extra energy comes from.

PROBLEM 5 — CRITICAL THINKING

A solid sphere and a hollow sphere of the same mass and radius are released from rest at the top of the same inclined ramp. Both roll without slipping. Without performing a full calculation, use the concept of moment of inertia to predict which sphere reaches the bottom first, and explain your reasoning. Then discuss whether the result would change if the masses were different.

SUMMARY

Lesson Summary

Rigid body mechanics extends Newton's laws to objects that can rotate as well as translate. The key rotational quantity is torque (τ = Fr sin θ), which measures a force's turning effect about a pivot. An object's resistance to angular acceleration is captured by its moment of inertia (I), which depends on both mass and mass distribution. Newton's second law for rotation, Στ = Iα, relates net torque to angular acceleration in the same way ΣF = ma relates net force to linear acceleration.

For equilibrium problems, apply two conditions simultaneously: ΣF = 0 (translational equilibrium) and Στ = 0 (rotational equilibrium). Choosing the pivot wisely — at the point where an unknown force acts — simplifies the algebra considerably. For dynamic problems, conservation of angular momentum (L = Iω = constant when Στ_ext = 0) is a powerful tool, explaining phenomena from spinning skaters to rolling objects on ramps. Always remember that for rolling without slipping, energy splits between translational and rotational kinetic energy, and the shape factor in I determines which objects accelerate fastest.