Chapter 08 · Rotational Mechanics · Full book

Rotational Mechanics: When Particles Are Not Enough

Core intuition: Chapter Roadmap

Newton's Laws tell you how a single particle answers a force. But a flywheel, a rolling cylinder, a spinning top, a diver curling into a tuck—these are extended bodies, and a force applied off-centre does something a point mass can never do: it makes the body turn. This chapter builds the mechanics of that turning, one rotational analogue at a time.

Angular kinematics — describing rigid rotation with angular displacement, velocity, and acceleration \(\theta,\ \omega,\ \alpha\), linked to their linear cousins by the radius.
Torque — the turning ability of a force, \(\vec{\tau} = \vec{r}\times\vec{F}\), set not by how hard you push but by where and in which direction.
Angular momentum — the rotational analogue of \(\vec{p}\), both for a single particle (\(\vec{L} = \vec{r}\times\vec{p}\)) and for a rigid body about a fixed axis (\(L = I\omega\)).
Moment of inertia — rotational mass, \(I = \sum m_i r_i^2\), the chapter's one genuinely new idea: response depends not just on how much mass but on how it is distributed about the axis. The parallel-axis theorem \(I = I_{\text{cm}} + Md^2\) moves that figure from one axis to another.
Rolling motion — translation and rotation locked by the constraint \(v = R\omega\), with kinetic energy split in a fixed shape-dependent ratio between the two.
Conservation of angular momentum — when no external torque acts, \(L\) stays fixed; pull the mass inward and the spin must speed up.
Gyroscopic effects — why a fast spin resists toppling and instead precesses, \(\Omega = \tau/L\).
Choosing your tool — a decision algorithm for picking torque-dynamics, energy, the rolling constraint, or conservation of angular momentum as the shortest path through a problem.

Master these analogues and rotation stops being a separate subject: it becomes translational mechanics, read off a new axis, with geometry now doing visible work.

Perplexing Questions

Perplexing questions

Before we begin, let these questions sit with you. Do not reach for formulas. You will return to each one—with full answers—before this chapter ends.

The Door Puzzle: You push a heavy door at its handle and it swings open easily. Now push with the same force right next to the hinge. The door barely moves. The force is identical—what has changed?
The Skater's Trick: A figure skater spins slowly with arms outstretched, then pulls her arms inward and suddenly spins much faster. No external torque acts on her. Where does the extra rotational speed come from?
The Rolling Race: A solid sphere, a hollow sphere, and a solid cylinder—all of the same mass and radius—are released together from the top of the same incline. They all experience the same gravitational pull. Yet they reach the bottom at different times. What decides the winner?
The Falling Chimney: A tall chimney topples over. Near the end of its fall the tip is moving faster than a freely falling object dropped from the same height. How can a rigid body beat free fall?
The Bicycle Paradox: A bicycle wheel rolls along a flat road without slipping. The topmost point moves at twice the speed of the centre, while the contact point is instantaneously at rest. Every particle of the wheel has a different velocity. How do you describe such motion?

Each question targets a trap that catches students every year.
By the end of this chapter, every one of them will be transparent.

Why Rotational Mechanics Deserves Its Own Chapter

Until now, every object in this book has been a particle: a point mass with no size, no shape, no internal structure. That abstraction was spectacularly successful. We described projectiles, blocks on inclines, collisions, and even systems of particles using just three ideas—Newton's Laws, energy, and momentum.

But watch a wheel roll, a gymnast somersault, or a planet spin on its axis, and the particle model breaks. Different parts of the wheel move at different speeds. Different parts of the gymnast accelerate in different directions. No single velocity vector describes the whole object.

The moment an object has spatial extent and turns about some axis, we need a richer language.

What Makes Rotation Different?

Consider a rigid disc spinning about a fixed axle. Every point on the disc traces a circle, but these circles differ in radius. A point near the rim moves fast; a point near the axle barely moves. Their speeds are different, their accelerations are different, and yet the disc is a single object executing a single motion.

The key observation: although the linear velocities differ from point to point, there is one quantity that is the same for every point—the angular velocity. Every point sweeps through the same angle in the same time.

This is the conceptual seed of the chapter: rotational motion is best described not by \(v\), \(a\), and \(F\), but by their angular counterparts—\(\omega\), \(\alpha\), and \(\tau\).

New Quantities, Same Physics

The good news is that you do not need an entirely new theory. Rotational mechanics is built from Newton's Laws applied to extended bodies. The mathematics generates a set of rotational analogues:

Translation	Rotation
Displacement \(x\)	Angular displacement \(\theta\)
Velocity \(v\)	Angular velocity \(\omega\)
Acceleration \(a\)	Angular acceleration \(\alpha\)
Force \(F\)	Torque \(\tau\)
Mass \(m\)	Moment of inertia \(I\)
Momentum \(p = mv\)	Angular momentum \(L = I\omega\)
Kinetic energy \(\tfrac{1}{2}mv^2\)	Kinetic energy \(\tfrac{1}{2}I\omega^2\)

Every equation of translational mechanics has a rotational twin. \(F = ma\) becomes \(\tau = I\alpha\). The impulse–momentum theorem becomes the angular impulse–angular momentum theorem. Conservation of linear momentum has a rotational sibling: conservation of angular momentum.

The new ingredient is moment of inertia, \(I\). Mass tells you how hard it is to accelerate a body in a straight line. Moment of inertia tells you how hard it is to spin it about an axis. Crucially, \(I\) depends not only on how much mass the body has but on how that mass is distributed relative to the axis. This is why the rolling race has a winner: the three objects have different mass distributions and therefore different moments of inertia.

Core intuition: The Central Insight

Rotational motion obeys laws that are structurally identical to translational motion. But whereas translational inertia depends only on total mass, rotational inertia depends on the distribution of mass about the axis of rotation. Geometry enters physics in a way it never did for a point particle. This single fact—that where the mass sits matters as much as how much mass there is—drives every new result in this chapter.

Connections to Earlier Chapters

This chapter does not start from scratch. It builds on everything you already know:

Newton's Laws provide the foundation. Torque and angular momentum are derived from \(\vec{F} = m\vec{a}\) applied to every particle of a rigid body.
Work–Energy–Power extends directly. The work done by a torque is \(W = \int \tau\,d\theta\); the rotational kinetic energy is \(\tfrac{1}{2}I\omega^2\). Rolling problems require you to track both translational and rotational kinetic energy.
Centre of Mass and Linear Momentum remain essential. In rolling motion, the centre of mass translates while the body rotates about it. The total kinetic energy splits cleanly into a translational part and a rotational part—a fact that simplifies every rolling problem.
Circular Motion already introduced radial acceleration and the idea of centripetal force. Rotational mechanics extends this from a single particle on a circle to an entire rigid body spinning about an axis.

The moral: you are not learning a new subject. You are extending familiar laws to a richer class of motions.

Misconception alert

“Rotation is a separate theory, independent of Newton's Laws.” Wrong. Every rotational equation—\(\tau = I\alpha\), \(L = I\omega\), conservation of angular momentum—is derived from \(\vec{F} = m\vec{a}\) applied to each particle of the body. Rotational mechanics is Newton's Laws, rewritten for extended objects.
“A rigid body either translates or rotates, never both.” Wrong. A rolling wheel translates and rotates simultaneously. A thrown baton tumbles through the air while its centre of mass follows a parabolic trajectory. The general motion of a rigid body is always translation of the centre of mass plus rotation about the centre of mass.

✓ Checkpoint · Before We Begin

Can an object rotate without translating? Can it translate without rotating? Give one physical example of each.
Yes to both. A spinning top on a fixed pivot rotates without translating. A block sliding on a frictionless surface translates without rotating.
A solid disc and a ring of the same mass and radius roll down an incline from the same height. Without calculation, predict which reaches the bottom first and explain your reasoning in one sentence.
The solid disc wins. Its mass is concentrated closer to the axis, giving it a smaller moment of inertia; less kinetic energy is “diverted” into rotation, leaving more for translation.
You push a door with force \(F\) at a distance \(d\) from the hinge. Your friend pushes with the same force at distance \(d/2\) from the hinge. Who produces the larger torque, and by what factor?
You produce a torque \(Fd\); your friend produces \(Fd/2\). Your torque is larger by a factor of \(2\).
State the rotational analogue of each: (a) \(F = ma\), (b) \(p = mv\), (c) \(\text{KE} = \tfrac{1}{2}mv^2\).
(a) \(\tau = I\alpha\). (b) \(L = I\omega\). (c) \(\text{KE}_{\text{rot}} = \tfrac{1}{2}I\omega^2\).

Angular Kinematics of Rigid Bodies

From Point Particles to Extended Bodies

In the circular-motion chapter we developed the full angular kinematic language—angular displacement \(\theta\), angular velocity \(\omega = d\theta/dt\), angular acceleration \(\alpha = d\omega/dt\), and the bridge relations \(v = r\omega\), \(a_t = r\alpha\)—for a single particle moving on a circle. We also derived the constant-\(\alpha\) equations of motion (\(\omega = \omega_0 + \alpha t\), etc.) by direct analogy with linear kinematics.

That entire machinery carries over to rigid bodies. But there is a crucial new idea: a rigid body is not a single point. It is an extended collection of particles, each at a different distance from the rotation axis, each tracing a different circle. The question is: how can one angular variable describe them all?

The answer lies in the defining property of a rigid body.

The Rigidity Constraint

Fig 7.1 A rigid body is a fixed lattice of particles — every inter-particle distance is locked. Because no strut can stretch, the whole body turns as one: a single angular velocity \(\omega\) about the axis describes every particle at once.

A rigid body is an idealised object in which the distance between every pair of constituent particles remains constant, no matter what forces act on the body. Real objects deform under stress, but for a wide range of problems in mechanics the deformation is negligible and the rigid-body model is excellent.

The rigidity constraint has a powerful kinematic consequence. Consider a rigid body rotating about a fixed axis. Pick any two points \(A\) and \(B\) in the body, at distances \(r_A\) and \(r_B\) from the axis. In a small time \(dt\), point \(A\) sweeps through arc \(ds_A = r_A\,d\theta_A\) and point \(B\) sweeps through arc \(ds_B = r_B\,d\theta_B\).

If \(d\theta_A \neq d\theta_B\), the line \(AB\) would stretch or compress—violating rigidity. Therefore:

\[ d\theta_A = d\theta_B = d\theta \qquad\text{for all points in the body} \]

Dividing by \(dt\):

\[ \omega_A = \omega_B = \omega \]

Differentiating again:

\[ \alpha_A = \alpha_B = \alpha \]

Core intuition: The Fundamental Property of Rigid Rotation

When a rigid body rotates about a fixed axis, every point in the body shares the same:

angular displacement \(\theta(t)\)
angular velocity \(\omega(t)\)
angular acceleration \(\alpha(t)\)

This is what makes the angular description so powerful for rigid bodies: one function \(\theta(t)\) completely specifies the rotational state of the entire object.

What differs from point to point is the linear kinematics. A point at distance \(r\) from the axis has: \[\begin{aligned} v &= r\omega \\[4pt] a_t &= r\alpha \\[4pt] a_c &= r\omega^2 = \frac{v^2}{r} \end{aligned}\]

Points farther from the axis move faster and experience larger centripetal acceleration, even though they rotate through the same angle per second. On a spinning grinding wheel, a grain at the rim moves many times faster than a grain near the hub—but both complete each revolution in exactly the same time.

Angular Displacement of a Rigid Body

When a door swings open, every point on the door—the handle, the hinges (excepting the axis itself), the centre of the door—sweeps through the same angle. That common angle is the angular displacement of the rigid body.

Sign convention. As established earlier: counterclockwise rotation (viewed from a standard orientation) is positive; clockwise is negative. This convention is consistent with the right-hand rule, which we will use extensively once angular quantities are treated as vectors.

Units. Radians, always. The relations \(v = r\omega\) and \(a_t = r\alpha\) require \(\theta\) in radians. A common exam error is to substitute an angle in degrees or revolutions into a kinematic formula. Convert first.

Angular Velocity and Angular Acceleration of a Rigid Body

For a rigid body rotating about a fixed axis:

\[ \boxed{\omega = \frac{d\theta}{dt}} \qquad\text{(same for every point)} \]

\[ \boxed{\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}} \qquad\text{(same for every point)} \]

These are not new definitions—they are identical to equations and from the circular motion chapter. What is new is the emphasis: for a rigid body, \(\omega\) and \(\alpha\) are properties of the body as a whole, not of any individual particle.

When we say “the wheel has \(\omega = 50\,\mathrm{rad/s}\),” we mean every particle in the wheel rotates at \(50\,\mathrm{rad/s}\). The rim particle at \(r = 0.3\,\mathrm{m}\) has linear speed \(v = 15\,\mathrm{m/s}\); a point at \(r = 0.1\,\mathrm{m}\) has \(v = 5\,\mathrm{m/s}\). The angular velocity is shared; the linear speed is not.

Misconception alert: “Faster Points Have Larger \(\omega\)”

Wrong. All points on a rigid body share the same \(\omega\). The rim of a wheel moves faster than the hub, but it does not have a larger angular velocity. It has a larger linear velocity: \(v = r\omega\). The difference is entirely due to the factor \(r\). This confusion is widespread and is tested directly in JEE and NEET conceptual questions.

Kinematic Equations for Constant Angular Acceleration

When a rigid body rotates with constant angular acceleration \(\alpha\), the angular analogues of the linear equations of motion apply (as derived in the circular-motion chapter):

\[ \boxed{\omega = \omega_0 + \alpha\,t} \]

\[ \boxed{\theta = \theta_0 + \omega_0\,t + \tfrac{1}{2}\alpha\,t^2} \]

\[ \boxed{\omega^2 = \omega_0^2 + 2\alpha\,(\theta - \theta_0)} \]

These are not new results. They are the same equations, applied now to extended rigid bodies rather than point particles. The single new insight is that one set of equations describes every point in the body simultaneously.

Using these equations. They require \(\theta\) in radians, \(\omega\) in \(\mathrm{rad/s}\), \(\alpha\) in \(\mathrm{rad/s^2}\). Problems often give information in revolutions or rpm—convert before substituting.

The Linear–Angular Bridge for Rigid Bodies

Fig 7.2 Every point shares the one \(\omega\), but linear speed is set by radius: \(v=\omega r\). The particle at \(2r\) sweeps the same angle in the same time, so it must travel twice as fast as the particle at \(r\).

The bridge relations derived in circular motion now acquire a richer meaning. For a rigid body rotating about a fixed axis, a particle at distance \(r\) from the axis has:

Quantity	Relation	Shared?
Arc length	\(s = r\theta\)	No (\(r\) varies)
Linear speed	\(v = r\omega\)	No
Tangential acceleration	\(a_t = r\alpha\)	No
Centripetal acceleration	\(a_c = r\omega^2\)	No
Angular displacement	\(\theta\)	Yes
Angular velocity	\(\omega\)	Yes
Angular acceleration	\(\alpha\)	Yes

The angular quantities are properties of the body. The linear quantities are properties of a point on the body and depend on where that point sits relative to the axis.

This table encodes the entire kinematics of fixed-axis rotation. Every problem in this chapter reduces to: (i) find \(\theta(t)\), \(\omega(t)\), or \(\alpha(t)\) for the body, then (ii) convert to linear quantities at the specific point of interest using the bridge relations.

Example 1 · Spinning wheel with constant angular acceleration

A grinding wheel starts from rest and accelerates uniformly at \(\alpha = 3\,\mathrm{rad/s^2}\) for \(10\,\mathrm{s}\).

Find the angular velocity at the end of \(10\,\mathrm{s}\).
Find the total angle turned through.
Find the number of revolutions completed.
If the wheel has radius \(R = 0.2\,\mathrm{m}\), find the linear speed and centripetal acceleration of a point on the rim at \(t = 10\,\mathrm{s}\).

Solution. Given: \(\omega_0 = 0\), \(\alpha = 3\,\mathrm{rad/s^2}\), \(t = 10\,\mathrm{s}\), \(R = 0.2\,\mathrm{m}\). (a) From : \[ \omega = 0 + 3 \times 10 = 30\,\mathrm{rad/s} \] (b) From (with \(\theta_0 = 0\)): \[ \theta = 0 + 0 + \tfrac{1}{2}(3)(100) = 150\,\mathrm{rad} \] (c) Number of revolutions: \[ n = \frac{\theta}{2\pi} = \frac{150}{2\pi} \approx 23.9 \;\text{revolutions} \] (d) At the rim (\(r = R = 0.2\,\mathrm{m}\)): \[ v = R\omega = 0.2 \times 30 = 6\,\mathrm{m/s} \] \[ a_c = R\omega^2 = 0.2 \times 900 = 180\,\mathrm{m/s^2} \] Sanity check. A point halfway to the rim (\(r = 0.1\,\mathrm{m}\)) would have \(v = 3\,\mathrm{m/s}\) and \(a_c = 90\,\mathrm{m/s^2}\)—half the linear speed and half the centripetal acceleration, as expected from \(v = r\omega\) and \(a_c = r\omega^2\). The angular quantities (\(\omega = 30\,\mathrm{rad/s}\), \(\theta = 150\,\mathrm{rad}\)) are the same for both points.

Example 2 · Decelerating flywheel

A flywheel spinning at \(600\,\mathrm{rpm}\) is brought to rest uniformly in \(30\,\mathrm{s}\). Find the angular deceleration and the number of revolutions completed before stopping.

Solution. Convert: \(\omega_0 = 600 \times 2\pi/60 = 20\pi\,\mathrm{rad/s}\). \(\omega = 0\). Angular deceleration: from : \[ 0 = 20\pi + \alpha(30) \quad\Longrightarrow\quad \alpha = -\frac{20\pi}{30} = -\frac{2\pi}{3}\,\mathrm{rad/s^2} \approx -2.09\,\mathrm{rad/s^2} \] Total angle: from : \[ 0 = (20\pi)^2 + 2\!\left(-\frac{2\pi}{3}\right)\theta \quad\Longrightarrow\quad \theta = \frac{400\pi^2}{4\pi/3} = 300\pi\,\mathrm{rad} \] Revolutions: \(n = 300\pi/(2\pi) = 150\). Quick check: average angular velocity \(= (20\pi + 0)/2 = 10\pi\,\mathrm{rad/s}\). In \(30\,\mathrm{s}\): \(\theta = 10\pi \times 30 = 300\pi\,\mathrm{rad}\). Confirmed.

✓ Checkpoint · Angular Kinematics of Rigid Bodies

A rigid disc rotates about its central axis. Point \(A\) is on the rim; point \(B\) is halfway between the centre and the rim.
1. Compare their angular velocities.
2. Compare their linear speeds.
3. Compare their centripetal accelerations.
(a) \(\omega_A = \omega_B\) (rigidity: all points share the same \(\omega\)). (b) \(v_A = 2v_B\) (since \(v = r\omega\) and \(r_A = 2r_B\)). (c) \(a_{c,A} = 2a_{c,B}\) (since \(a_c = r\omega^2\) and \(r_A = 2r_B\)).
A turntable accelerates uniformly from rest to \(45\,\mathrm{rpm}\) in \(15\,\mathrm{s}\). How many revolutions does it complete during this time?
\(\omega_f = 45 \times 2\pi/60 = 3\pi/2\,\mathrm{rad/s}\). \(\bar{\omega} = (0 + 3\pi/2)/2 = 3\pi/4\). \(\theta = \bar{\omega}\,t = (3\pi/4)(15) = 45\pi/4\,\mathrm{rad}\). \(n = 45\pi/(4\times 2\pi) = 45/8 = 5.625\) revolutions.
Can a point on a rotating rigid body have zero linear velocity while the body's angular velocity is nonzero? If so, which point?
Yes: any point on the rotation axis itself has \(r = 0\), so \(v = r\omega = 0\) even though \(\omega \neq 0\). The axis of rotation is the locus of points with zero linear velocity during pure rotation.

Torque: The Rotational Analogue of Force

◈ Torque — the turning bench

live bench

Push a lever at a chosen point and angle. Only the perpendicular component turns the body: \(\tau = rF\sin\theta\). Slide the application point toward the axis and the same force loses all its bite.

◈ loading the interactive bench…

↗ open in the Simulations page

Why Force Alone Is Not Enough

Push the handle of a door and it swings open effortlessly. Push the door near its hinges with the same force and it barely budges. The force is identical; the rotational effect is not.

Tighten a bolt with a short wrench and you struggle. Attach an extension pipe to the wrench handle and the same muscular effort produces a much larger turning effect. Again, the force has not changed—but something about where and how it is applied has.

Translation is governed by force. Rotation is governed by a different quantity—one that encodes not only how hard you push, but how far from the axis and in what direction. That quantity is torque.

The Lever Arm: Physical Intuition

Fig 7.3 The lever arm \(r_\perp\) is the perpendicular distance from the axis \(O\) to the line of action of \(\vec F\) — not to the point where it is applied. Torque is \(\tau=r_\perp F = rF\sin\theta\); sliding \(\vec F\) along its own line changes nothing.

Consider a rigid body free to rotate about a fixed axis. A force \(\vec{F}\) is applied at a point whose position relative to the axis is \(\vec{r}\).

Not all of \(\vec{F}\) contributes to rotation. The component of \(\vec{F}\) directed toward or away from the axis pulls the body radially—it tries to translate the axis, not to rotate the body about it. Only the component of \(\vec{F}\) perpendicular to \(\vec{r}\) produces a turning effect.

The lever arm (or moment arm) is the perpendicular distance from the axis of rotation to the line of action of the force. If \(\vec{r}\) makes angle \(\theta\) with \(\vec{F}\), the lever arm is: \[ d = r\sin\theta \]

A longer lever arm means a greater turning effect for the same force. This is why door handles are placed far from the hinges, why long wrenches are more effective than short ones, and why a force applied perpendicular to the door produces the maximum rotation.

Magnitude of Torque

The torque (or moment of force) produced by a force \(\vec{F}\) about a given axis is defined as:

\[ \boxed{\tau = rF\sin\theta} \]

where:

\(r\) is the distance from the axis to the point of application of the force,
\(F\) is the magnitude of the force,
\(\theta\) is the angle between \(\vec{r}\) and \(\vec{F}\).

This equation can be read in two equivalent ways:

\[ \tau = F \times \underbrace{(r\sin\theta)}_{\text{lever arm}} = r \times \underbrace{(F\sin\theta)}_{F_{\perp}} \]

First reading: torque equals the full force times the perpendicular distance (lever arm) from the axis to the line of action.

Second reading: torque equals the full distance times the component of force perpendicular to \(\vec{r}\).

Both give the same number. Use whichever decomposition the geometry of the problem makes easier.

Units. Torque has dimensions of \([\text{force}]\times[\text{distance}] = \mathrm{N\cdot m}\). Although this is dimensionally the same as energy (joule), torque is not energy. The units are kept as \(\mathrm{N\cdot m}\) (never written as \(\mathrm{J}\)) to maintain the distinction.

When Is Torque Maximum? When Is It Zero?

From \(\tau = rF\sin\theta\):

Maximum torque (\(\tau = rF\)): when \(\theta = 90°\), i.e. the force is applied perpendicular to the position vector. This is why you push a door at right angles to its surface.

Zero torque (\(\tau = 0\)):

\(\theta = 0°\) or \(180°\): the force is directed along \(\vec{r}\) (toward or away from the axis). A radial force has zero lever arm and produces no rotation.
\(r = 0\): the force is applied at the axis itself. No lever arm, no torque.
\(F = 0\): no force, no torque (trivially).

Pushing a door edge-on. If you push a door parallel to its surface (along the edge), the force is directed along \(\vec{r}\) from the hinge. \(\theta = 0\), so \(\tau = 0\). No matter how hard you push, the door does not rotate. This is a favourite conceptual question on competitive exams.

The Cross Product and the Vector Definition of Torque

Fig 7.4 The vector definition \(\vec\tau=\vec r\times\vec F\). With \(\vec r\) and \(\vec F\) both in the page, the right-hand rule points \(\vec\tau\) straight out of the page (\(\odot\)); its magnitude is \(rF\sin\theta\), largest when the force is applied at a right angle.

Torque has a direction: it tells you which way the rotation tends to occur. To capture this, we need the cross product (or vector product) of two vectors.

The Cross Product: A Brief Introduction

For two vectors \(\vec{A}\) and \(\vec{B}\), the cross product \(\vec{A}\times\vec{B}\) is a vector defined by:

Magnitude: \[ |\vec{A}\times\vec{B}| = AB\sin\theta \] where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\) (\(0 \le \theta \le 180°\)).

Direction: perpendicular to the plane containing \(\vec{A}\) and \(\vec{B}\), determined by the right-hand rule: curl the fingers of your right hand from \(\vec{A}\) toward \(\vec{B}\) through the smaller angle; your thumb points in the direction of \(\vec{A}\times\vec{B}\).

Key properties:

Anti-commutative: \(\vec{A}\times\vec{B} = -\vec{B}\times\vec{A}\). Order matters—swapping the vectors reverses the direction.
Parallel vectors: \(\vec{A}\times\vec{B} = \vec{0}\) when \(\theta = 0\) or \(180°\).
Perpendicular vectors: \(|\vec{A}\times\vec{B}| = AB\) (maximum) when \(\theta = 90°\).
Unit vector identities: \(\hat{\imath}\times\hat{\jmath} = \hat{k}\), \(\hat{\jmath}\times\hat{k} = \hat{\imath}\), \(\hat{k}\times\hat{\imath} = \hat{\jmath}\) (cyclic).

Torque as a Cross Product

Birth of the formula: Vector Definition of Torque

The torque produced by a force \(\vec{F}\) applied at position \(\vec{r}\) relative to the chosen axis (or point) is: \[ \boxed{\vec{\tau} = \vec{r}\times\vec{F}} \] Magnitude: \(|\vec{\tau}| = rF\sin\theta\), recovering . Direction: perpendicular to the plane of \(\vec{r}\) and \(\vec{F}\), given by the right-hand rule. For a body rotating in the \(xy\)-plane about the \(z\)-axis: a torque that tends to produce counterclockwise rotation points in the \(+z\) direction; a torque tending to produce clockwise rotation points in the \(-z\) direction.

The vector form automatically encodes magnitude, direction, and sign. It will be essential when we define angular momentum and write the rotational analogue of Newton's second law.

Core intuition: Force Causes Translation; Torque Causes Rotation

Force and torque are fundamentally different quantities. A net force \(\vec{F}_{\text{net}}\) on a body changes its translational motion: \(\vec{F}_{\text{net}} = m\vec{a}_{\text{cm}}\). A net torque \(\vec{\tau}_{\text{net}}\) about an axis changes the body's rotational motion about that axis. The turning effect of a force depends on two things: how hard you push (\(F\)) and how far from the axis you push (\(r\sin\theta\), the lever arm). A small force at a large lever arm can overpower a large force at a small lever arm. This is the principle of the lever, known since antiquity.

Net Torque

When several forces act on a rigid body, each produces its own torque about the chosen axis. The net torque is the vector sum of all individual torques:

\[ \vec{\tau}_{\text{net}} = \sum_i \vec{r}_i \times \vec{F}_i \]

For rotation about a fixed axis (say the \(z\)-axis), only the \(z\)-component of each torque matters. Torques that tend to rotate the body counterclockwise (positive \(z\)) are positive; clockwise torques are negative. The net torque is the algebraic sum:

\[ \tau_{\text{net}} = \sum_i \tau_i \qquad\text{(with appropriate signs)} \]

The rotational state of the body is determined by \(\tau_{\text{net}}\)—not by any individual torque. This is the rotational analogue of \(\vec{F}_{\text{net}}\) determining translational acceleration.

Misconception alert: Common Torque Errors

1. Confusing torque with force. A force of \(100\,\mathrm{N}\) applied at the axis (\(r = 0\)) produces zero torque. A force of \(10\,\mathrm{N}\) applied at \(r = 2\,\mathrm{m}\) perpendicular to \(\vec{r}\) produces \(20\,\mathrm{N\cdot m}\) of torque. The smaller force wins the rotation contest. 2. Using the wrong distance. The lever arm is the perpendicular distance from the axis to the line of action of the force—not the distance from the axis to the point of application. If the force is not perpendicular to \(\vec{r}\), you must use \(r\sin\theta\) (or equivalently, \(F_\perp \cdot r\)). 3. Ignoring direction. Torques have signs (or directions). Two equal torques in opposite senses cancel. Always assign a sign convention before summing torques.

Example 3 · Door handle versus hinge

A door of width \(0.9\,\mathrm{m}\) is hinged along one vertical edge. A horizontal force of \(15\,\mathrm{N}\) is applied perpendicular to the door. Compare the torque about the hinge when the force is applied: (a) at the handle, \(0.8\,\mathrm{m}\) from the hinge, and (b) at a point \(0.2\,\mathrm{m}\) from the hinge.

Solution. The force is perpendicular to \(\vec{r}\) in both cases (\(\theta = 90°\), so \(\sin\theta = 1\)). (a) \(\tau = rF = 0.8 \times 15 = 12\,\mathrm{N\cdot m}\). (b) \(\tau = rF = 0.2 \times 15 = 3\,\mathrm{N\cdot m}\). The torque at the handle is four times larger. This is why door handles are placed as far from the hinge as possible.

Example 4 · Force at an angle

A force of \(50\,\mathrm{N}\) is applied to a wrench at a point \(0.3\,\mathrm{m}\) from the bolt, at an angle of \(60°\) to the wrench handle. Find the torque about the bolt. What fraction of the applied force actually contributes to rotation?

Solution. \[ \tau = rF\sin\theta = 0.3 \times 50 \times \sin 60° = 15 \times \frac{\sqrt{3}}{2} = \frac{15\sqrt{3}}{2} \approx 13.0\,\mathrm{N\cdot m} \] The perpendicular component of the force is: \(F_\perp = F\sin 60° = 50\times\sqrt{3}/2 \approx 43.3\,\mathrm{N}\). So \(43.3/50 \approx 86.6\%\) of the applied force contributes to rotation. The remaining \(\approx 13.4\%\) (the radial component \(F\cos 60°\)) tries to push the wrench along its length and produces no torque.

Example 5 · Balancing torques

A uniform beam of length \(2\,\mathrm{m}\) and mass \(10\,\mathrm{kg}\) is supported at its centre. A \(5\,\mathrm{kg}\) mass is hung \(0.6\,\mathrm{m}\) to the left of the support. Where must a \(3\,\mathrm{kg}\) mass be hung on the right side to balance the beam? (\(g = 10\,\mathrm{m/s^2}\).)

Solution. The beam's weight acts at its centre (the support point), so it produces zero torque about the support. For rotational equilibrium: net torque about the support \(= 0\). Taking counterclockwise as positive: \[ \tau_{\text{left}} = -(5)(10)(0.6) = -30\,\mathrm{N\cdot m} \qquad\text{(clockwise)} \] \[ \tau_{\text{right}} = +(3)(10)(d) = 30d\,\mathrm{N\cdot m} \qquad\text{(counterclockwise)} \] Setting \(\tau_{\text{net}} = 0\): \[ -30 + 30d = 0 \quad\Longrightarrow\quad d = 1\,\mathrm{m} \] The \(3\,\mathrm{kg}\) mass must be hung \(1\,\mathrm{m}\) to the right of the support.

✓ Checkpoint · Torque

Why is it easier to open a door by pushing at the handle than by pushing near the hinge, even with the same force?
The lever arm is larger at the handle. Since \(\tau = rF\sin\theta\) and the force is the same, a larger \(r\) produces a larger torque. The handle is placed far from the hinge precisely to maximise the lever arm.
Can a large force produce zero torque? Give a specific example.
Yes. Any force directed along the line from the axis to the point of application (\(\theta = 0° \text{ or } 180°\)) has zero lever arm and produces zero torque, regardless of its magnitude. Example: pushing a door edge-on (parallel to its surface).
A force \(\vec{F} = (3\hat{\imath} + 4\hat{\jmath})\,\mathrm{N}\) acts at the point \(\vec{r} = (2\hat{\imath} - \hat{\jmath})\,\mathrm{m}\) relative to an axis through the origin. Find the torque about the origin.
\(\vec{\tau} = \vec{r}\times\vec{F} = (2\hat{\imath} - \hat{\jmath})\times(3\hat{\imath} + 4\hat{\jmath}) = 2(4)(\hat{\imath}\times\hat{\jmath}) + (-1)(3)(\hat{\jmath}\times\hat{\imath}) = 8\hat{k} + 3\hat{k} = 11\hat{k}\;\mathrm{N\cdot m}\). The torque is \(11\,\mathrm{N\cdot m}\) in the \(+z\) direction (counterclockwise when viewed from above).

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · Spinning up a flywheel

A flywheel starts from rest and reaches \(300\,\mathrm{rev/min}\) in \(20\,\mathrm{s}\) under constant angular acceleration. Find (a) the angular acceleration and (b) the number of revolutions turned in that time.

Solution. Given: \(\omega_0 = 0\), \(\omega_f = 300\,\mathrm{rev/min}\), \(t = 20\,\mathrm{s}\), constant \(\alpha\).
Find: \(\alpha\) and the revolutions \(N\).
Setup: Convert to SI first: \(\omega_f = 300\times\frac{2\pi}{60} = 10\pi\,\mathrm{rad/s}\). With constant \(\alpha\), use \(\alpha = (\omega_f-\omega_0)/t\) and \(\theta = \tfrac12\alpha t^2\).
Solve: \[ \alpha = \frac{10\pi - 0}{20} = \frac{\pi}{2} \approx 1.57\,\mathrm{rad/s^2}, \] \[ \theta = \tfrac12\alpha t^2 = \tfrac12\left(\tfrac{\pi}{2}\right)(20)^2 = 100\pi\,\mathrm{rad}, \qquad N = \frac{\theta}{2\pi} = 50. \] Answer: \(\boxed{\alpha = \tfrac{\pi}{2} \approx 1.57\,\mathrm{rad/s^2},\quad N = 50\ \text{rev}}\)
Check: Average-rate route: the mean angular speed is \(\tfrac12(0+10\pi) = 5\pi\,\mathrm{rad/s}\), so \(\theta = (5\pi)(20) = 100\pi\,\mathrm{rad}\), i.e. \(50\) revolutions—independent of the \(\alpha\) value, and in agreement. ✓

Example 2 · Torque from a force at an angle

A spanner of length \(r = 0.30\,\mathrm{m}\) is used to turn a bolt. A force \(F = 50\,\mathrm{N}\) is applied at the far end, directed at \(60^\circ\) to the handle. Find the torque about the bolt.

Solution. Given: \(r = 0.30\,\mathrm{m}\), \(F = 50\,\mathrm{N}\), \(\theta = 60^\circ\).
Find: the torque \(\tau\) about the bolt.
Setup: Only the component of \(F\) perpendicular to the handle produces torque, so \(\tau = rF\sin\theta\).
Solve: \[ \tau = rF\sin\theta = (0.30)(50)\sin 60^\circ = 15\times\frac{\sqrt{3}}{2} = 7.5\sqrt{3} \approx 13.0\,\mathrm{N\,m}. \] Answer: \(\boxed{\tau = 7.5\sqrt{3} \approx 13.0\,\mathrm{N\,m}}\)
Check: Moment-arm route: the perpendicular distance from the bolt to the line of action is \(r\sin\theta = 0.30(0.866) = 0.260\,\mathrm{m}\), so \(\tau = F\times 0.260 = 50(0.260) = 13.0\,\mathrm{N\,m}\)—same value by the alternative construction. ✓

Example 3 · Moment of inertia of three masses

Three point masses, each \(m = 2\,\mathrm{kg}\), sit at the vertices of an equilateral triangle of side \(a = 1\,\mathrm{m}\). Find the moment of inertia about the axis through the centroid, perpendicular to the plane.

Solution. Given: three masses \(m = 2\,\mathrm{kg}\), side \(a = 1\,\mathrm{m}\).
Find: \(I\) about the centroidal perpendicular axis.
Setup: Use \(I = \sum m_i r_i^2\). Each vertex of an equilateral triangle lies a distance \(r = a/\sqrt{3}\) from the centroid (the circumradius), so \(r^2 = a^2/3\).
Solve: \[ I = 3\,m\,r^2 = 3(2)\left(\frac{a^2}{3}\right) = 2a^2 = 2(1)^2 = 2\,\mathrm{kg\,m^2}. \] Answer: \(\boxed{I = 2\,\mathrm{kg\,m^2}}\)
Check: Dimensions: \([\mathrm{kg}][\mathrm{m^2}] = \mathrm{kg\,m^2}\) ✓. Symmetry sanity: all three masses are equidistant from the axis, so \(I = (3m)r^2 = (6)(1/3) = 2\,\mathrm{kg\,m^2}\), confirming the lumped form. ✓

Example 4 · Angular momentum of a spinning disc

A uniform disc of mass \(M = 3\,\mathrm{kg}\) and radius \(R = 0.40\,\mathrm{m}\) spins about its central axis at \(\omega = 10\,\mathrm{rad/s}\). Find its angular momentum.

Solution. Given: \(M = 3\,\mathrm{kg}\), \(R = 0.40\,\mathrm{m}\), \(\omega = 10\,\mathrm{rad/s}\).
Find: the angular momentum \(L\).
Setup: For a disc about its centre, \(I = \tfrac12 MR^2\); then \(L = I\omega\).
Solve: \[ I = \tfrac12(3)(0.40)^2 = \tfrac12(3)(0.16) = 0.24\,\mathrm{kg\,m^2}, \qquad L = I\omega = 0.24(10) = 2.4\,\mathrm{kg\,m^2/s}. \] Answer: \(\boxed{L = 2.4\,\mathrm{kg\,m^2/s}}\)
Check: Energy cross-check: the rotational kinetic energy is \(\tfrac12 I\omega^2 = \tfrac12(0.24)(100) = 12\,\mathrm{J}\); since \(K = L^2/2I\), we recover \(L = \sqrt{2IK} = \sqrt{2(0.24)(12)} = \sqrt{5.76} = 2.4\,\mathrm{kg\,m^2/s}\). ✓

Example 5 · Velocities on a rolling wheel

A wheel of radius \(R = 0.35\,\mathrm{m}\) rolls without slipping along level ground; its centre moves at \(v = 7\,\mathrm{m/s}\). Find (a) the angular speed, (b) the speed of the topmost point, and (c) the speed of the contact point.

Solution. Given: \(R = 0.35\,\mathrm{m}\), \(v = 7\,\mathrm{m/s}\), rolling without slipping.
Find: \(\omega\), \(v_{\text{top}}\), \(v_{\text{contact}}\).
Setup: Rolling without slipping gives \(\omega = v/R\). Each point's velocity is the centre's velocity plus the rotational part \(\omega\times(\text{position from centre})\), which adds at the top and cancels at the contact.
Solve: \[ \omega = \frac{v}{R} = \frac{7}{0.35} = 20\,\mathrm{rad/s}, \] \[ v_{\text{top}} = v + \omega R = 7 + (20)(0.35) = 14\,\mathrm{m/s}, \qquad v_{\text{contact}} = v - \omega R = 0. \] Answer: \(\boxed{\omega = 20\,\mathrm{rad/s},\quad v_{\text{top}} = 14\,\mathrm{m/s} = 2v,\quad v_{\text{contact}} = 0}\)
Check: Instantaneous-axis view: a rolling wheel rotates momentarily about its contact point, so speeds scale with distance from that point: the top is \(2R\) away (\(\Rightarrow 2v\)), the centre is \(R\) away (\(\Rightarrow v\)), the contact is \(0\) away (\(\Rightarrow 0\))—all consistent. ✓