To calculate torque, first choose the axis or pivot. Then draw the position vector r from the pivot to the point where the force acts, and identify the force component perpendicular to r. The torque magnitude is τ = Fr sinθ. You can also use τ = Fd, where d is the perpendicular distance from the pivot to the line of action of the force. Forces acting through the pivot have zero torque. In rotational equilibrium or dynamics, signs matter: choose clockwise or counterclockwise as positive and keep that convention.

Torque is the turning effect of a force about an axis or pivot.

Magnitude is τ = Fr sinθ, where θ is the angle between the radius/lever arm and the force.

Equivalently, torque = force × perpendicular distance from the pivot to the force line of action.

Torque direction or sign is clockwise/counterclockwise according to the chosen convention.

Torque has unit N m; do not treat it as energy just because the unit dimensions match joules.

Build the torque relation for a force about a chosen pivot.

Formula

Target formula τ = Fr sinθ = Fd_perpendicular

torque about the chosen axis

N m

force magnitude

distance from pivot to point of force application

angle between r and F

d_perpendicular

perpendicular distance to force line of action

1Choose the pivot or axis of rotation.axis fixed

2Draw r from pivot to point where force acts.r

3Use the perpendicular force component or perpendicular distance.F_perp = F sinθ

4Calculate torque magnitude.τ = Fr sinθ = Fd_perpendicular

5Assign clockwise/counterclockwise sign.chosen rotational sign convention

Build the formula

Calculate the torque of a force about a specified axis and state its clockwise or counterclockwise sense.

choose the axis/pivot

use perpendicular component or perpendicular distance

write τ = Fr sinθ or Fd

give unit N m

assign clockwise/counterclockwise sign if required

IB HL questions often give an angled force or ask for the perpendicular distance to the line of action.

Torque about the pivot is τ = Fr sinθ, where θ is between r and F. Equivalently τ = Fd_perpendicular. Its sign depends on the chosen clockwise/counterclockwise convention.

Using the distance along the rod instead of perpendicular distance, or forgetting that a force through the pivot has zero torque.

Calculate the torque of a force about a specified axis and state its clockwise or counterclockwise sense.

Choose

Choose one pivot or axis, assign clockwise and anticlockwise torque signs, then add all torques about that same point. In rotational equilibrium the signed sum is zero, so clockwise and anticlockwise turning effects cancel. Picking a pivot through an unknown force often simplifies the equation because that force has no perpendicular lever arm. If the resultant torque is not zero, the rigid body has angular acceleration rather than rotational equilibrium.

Rotational equilibrium means the resultant torque about the chosen axis is zero: Στ = 0.

With a consistent sign convention, clockwise torques balance anticlockwise torques.

Each torque must be calculated about the same pivot or axis using the perpendicular lever arm.

A force whose line of action passes through the pivot has zero torque about that pivot.

For complete static equilibrium, resultant force must also be zero; ΣF = 0 alone is not enough for rotational equilibrium.

Label the quantities needed to test rotational equilibrium about one pivot.

Label

Labels

A rigid bar is in rotational equilibrium about a pivot. State the torque condition and explain why a force whose line of action passes through the pivot can be ignored in the torque equation.

Common mark losses are writing only ΣF = 0, mixing clockwise and anticlockwise signs, or using distances from different pivots in the same equation.

A rigid bar is in rotational equilibrium about a pivot. State the torque condition and explain why a force whose line of action passes through the pivot can be ignored in the torque equation.

Choose

Calculate torques about the same axis, include their clockwise or anticlockwise signs, and add them. If the resultant torque is non-zero, the rigid body’s angular velocity changes, so it has angular acceleration. This does not require a non-zero resultant force: a couple can have zero resultant force but still produce a resultant torque and therefore angular acceleration. If the signed torques cancel, there is no angular acceleration.

An unbalanced torque means the signed resultant torque is not zero: Στ ≠ 0.

A non-zero resultant torque causes angular acceleration, the rotational equivalent of F_resultant causing linear acceleration.

The angular acceleration has the same rotational sense as the resultant torque under the chosen sign convention.

If Στ = 0, the body has no angular acceleration; it stays at rest or keeps a constant angular velocity.

For a rigid body about a fixed axis, the link is Στ = Iα; this card focuses on the cause, not extended calculations.

Build the cause-and-effect relation between resultant torque and angular acceleration.

Formula

Target formula Στ = Iα; if Στ ≠ 0 then α ≠ 0

Στ

signed resultant torque about the chosen axis

N m

moment of inertia about that axis

kg m^2

angular acceleration

rad s^-2

1Choose the fixed rotation axis and keep it for every torque.same axis

2Assign clockwise/anticlockwise signs and add all torques.Στ = τ_anticlockwise - τ_clockwise

3Check whether the signed torque sum is zero.Στ = 0 means α = 0; Στ ≠ 0 means α ≠ 0

4Relate the non-zero torque to angular acceleration.Στ = Iα

Build the formula

A rigid wheel is acted on by torques in opposite rotational senses. Explain what must be true for the wheel to have angular acceleration, and state what happens if the resultant torque is zero.

Common mark losses are discussing only the applied force, ignoring torque direction, or saying zero resultant force means no angular acceleration.

A rigid wheel is acted on by torques in opposite rotational senses. Explain what must be true for the wheel to have angular acceleration, and state what happens if the resultant torque is zero.

Choose

Angular quantities describe rotation itself rather than the linear motion of one point on the object. Use radians for angular displacement because the rotational equations and links such as v = rω assume radian measure. Choose a positive rotational sense, then treat θ, ω, and α as signed quantities. In a rigid body, every point turns through the same angle in the same time, so the angular quantities are common to the body; points farther from the axis travel a larger linear distance.

Angular displacement Δθ is the angle swept out, measured in radians; one full revolution is 2π rad.

Angular velocity ω is the rate of change of angular displacement: ω = Δθ/Δt, with unit rad s^-1.

Angular acceleration α is the rate of change of angular velocity: α = Δω/Δt, with unit rad s^-2.

Clockwise and anticlockwise rotations can be assigned positive or negative signs; keep the sign convention consistent.

For a rigid body rotating about a fixed axis, all points share the same angular displacement, angular velocity, and angular acceleration, even though their linear speeds can differ.

Repair the angular-motion statements.

Spot Errors

Define angular displacement, angular velocity, and angular acceleration for a rigid body rotating about a fixed axis, including their usual units.

Common mark losses are using degrees in equations, calling ω a linear speed, or defining α as the rate of change of angle instead of angular velocity.

Define angular displacement, angular velocity, and angular acceleration for a rigid body rotating about a fixed axis, including their usual units.

Choose

These equations are the rotational versions of constant-acceleration SUVAT. Before choosing an equation, list the known angular quantities and the unknown, then check that α is constant. Use radians for Δθ and rad s^-1 or rad s^-2 for ω and α. Choose the equation that contains the unknown and avoids any quantity you do not know, just as in linear kinematics.

The rotational SUVAT equations apply only when angular acceleration α is constant.

Use angular variables: angular displacement Δθ, initial angular velocity ω0, final angular velocity ω, angular acceleration α, and time t.

Common equations are ω = ω0 + αt, Δθ = ω0t + 1/2αt^2, ω^2 = ω0^2 + 2αΔθ, and Δθ = 1/2(ω0 + ω)t.

Angular displacement must be in radians; convert revolutions using 1 rev = 2π rad.

Keep a consistent positive rotation direction so negative angular velocity or acceleration has physical meaning.

Build the correct angular SUVAT equation for a constant-α situation.

Formula

Target formula ω = ω0 + αt; Δθ = ω0t + 1/2αt^2; ω^2 = ω0^2 + 2αΔθ; Δθ = 1/2(ω0 + ω)t

Δθ

angular displacement

rad

ω0

initial angular velocity

rad s^-1

final angular velocity

rad s^-1

constant angular acceleration

rad s^-2

time interval

1Check whether angular acceleration is constant.α = constant

2Convert any revolutions or degrees to radians.1 rev = 2π rad

3List the known angular quantities and the unknown.Δθ, ω0, ω, α, t

4Choose an equation that contains the unknown and avoids a missing variable.select from angular SUVAT set

5Substitute with the chosen rotational sign convention.positive and negative rotation senses kept consistent

Build the formula

A wheel rotates with constant angular acceleration. Describe how to choose an angular SUVAT equation for an unknown angular quantity, and state two checks needed before substituting values.

Common mark losses are failing to check constant α, using degrees instead of radians, or mixing angular variables with linear SUVAT symbols without defining them.

A wheel rotates with constant angular acceleration. Describe how to choose an angular SUVAT equation for an unknown angular quantity, and state two checks needed before substituting values.

Choose

Moment of inertia plays the role that mass plays in linear motion, but for rotation. A large I means a given resultant torque produces a smaller angular acceleration. The same object can have different moments of inertia about different axes, so always state or identify the axis first. Moving mass away from the axis makes the object harder to spin up or slow down, even if the total mass is unchanged.

Moment of inertia I measures how difficult it is to change an object’s rotational motion about a chosen axis.

It depends on both total mass and how that mass is distributed relative to the axis.

Mass farther from the axis increases I more strongly because distance from the axis matters squared.

The SI unit of moment of inertia is kg m^2.

For IB-style problems, shape-specific formulae for I are normally provided; the key skill is choosing the correct axis and interpreting I.

Repair the moment-of-inertia statements.

Spot Errors

Two wheels have the same mass but one has more mass near the rim. Explain which wheel has the larger moment of inertia about its central axis and state the unit of moment of inertia.

Common mark losses are saying only “more mass” without discussing distance from the axis, or failing to specify the axis of rotation.

Two wheels have the same mass but one has more mass near the rim. Explain which wheel has the larger moment of inertia about its central axis and state the unit of moment of inertia.

Choose

Treat each small mass as concentrated at a point. Choose the rotation axis, measure each point mass’s perpendicular distance from that axis, square the distance, multiply by the mass, then sum all contributions. The result has unit kg m^2 and belongs to that specific axis; changing the axis changes the distances and therefore changes I.

For one point mass, the moment of inertia about an axis is I = mr^2.

For several point masses, add the individual contributions: I = Σm_i r_i^2.

r is the perpendicular distance from the rotation axis to the point mass.

All distances must be measured from the same chosen axis; a mass on the axis contributes zero.

Because r is squared, moving a mass twice as far from the axis makes its contribution four times larger.

Build the point-mass moment of inertia expression.

Formula

Target formula I = Σm_i r_i^2

total moment of inertia about the chosen axis

kg m^2

m_i

one point mass

r_i

perpendicular distance from the axis to that mass

1Choose the rotation axis and keep it fixed.one axis for all r_i

2Measure each perpendicular distance from the axis.r_i

3Square each distance before multiplying.r_i^2

4Multiply each mass by its squared distance.m_i r_i^2

5Add all point-mass contributions.I = Σm_i r_i^2

Build the formula

A set of small masses is fixed to a light rod that rotates about a specified axis. Describe how to calculate the total moment of inertia of the point masses about that axis.

Common mark losses are using distance from the wrong axis, forgetting to square r, or summing distances before calculating each m_i r_i^2 term.

A set of small masses is fixed to a light rod that rotates about a specified axis. Describe how to calculate the total moment of inertia of the point masses about that axis.

Choose

To use the law, choose the rotation axis, calculate every torque about that axis with signs, find the resultant torque, and then use Στ = Iα. The equation is for the resultant torque, not any one applied torque unless it is the only torque. If Στ = 0, then α = 0 and the object is in rotational equilibrium. Keep units consistent: torque in N m, I in kg m^2, and α in rad s^-2.

Rotational Newton’s second law is Στ = Iα for a rigid body about a fixed axis.

Στ is the signed resultant torque about the same axis used to define I.

I is the moment of inertia about that axis; using the wrong axis gives the wrong α.

The angular acceleration has the same sign as the resultant torque under the chosen sign convention.

This is the rotational analogue of F_resultant = ma: torque corresponds to force, I to mass, and α to acceleration.

Build the rotational Newton’s second law calculation.

Formula

Target formula Στ = Iα

Στ

signed resultant torque about the chosen axis

N m

moment of inertia about that same axis

kg m^2

angular acceleration

rad s^-2

1Choose the rotation axis and moment of inertia about that axis.I about chosen axis

2Calculate each torque using its perpendicular lever arm.τ = Fr sinθ

3Add clockwise and anticlockwise torques with signs.Στ

4Use rotational Newton’s second law.α = Στ / I

5Interpret the sign of α as rotational sense.sign of α follows sign of Στ

Build the formula

A rigid body rotates about a fixed axis. It has moment of inertia I about that axis and is acted on by several torques. State how to find its angular acceleration and identify one common condition that must be checked.

Common mark losses are using one applied torque instead of the resultant torque, or using an I value for a different axis.

Choose

Angular momentum is the rotational analogue of linear momentum. For the IB rigid-body model, multiply the moment of inertia about the rotation axis by the angular velocity about that same axis. A larger I at the same ω gives a larger angular momentum, and reversing the rotational sense reverses the sign of L under a signed convention.

For a rigid body rotating about a fixed axis, angular momentum is L = Iω.

I must be the moment of inertia about the same axis as the angular velocity.

Angular momentum has unit kg m^2 s^-1.

Angular momentum depends on both mass distribution and angular velocity; changing either changes L.

Treat direction using the chosen rotational sign convention or axis direction.

Build the angular momentum expression for a rotating rigid body.

Formula

Target formula L = Iω

angular momentum about the chosen axis

kg m^2 s^-1

moment of inertia about that axis

kg m^2

angular velocity about that axis

rad s^-1

1State the rotation axis.chosen axis

2Use moment of inertia about that same axis.I

3Use angular velocity with its rotational sign.ω

4Multiply to find angular momentum.L = Iω

5Attach unit and sign or direction.kg m^2 s^-1 with signed sense

Build the formula

A flywheel has moment of inertia I about its axle and angular velocity ω. State its angular momentum, the unit of angular momentum, and why the axis must be specified.

Common mark losses are using linear momentum notation, omitting the axis, or forgetting the unit kg m^2 s^-1.

A flywheel has moment of inertia I about its axle and angular velocity ω. State its angular momentum, the unit of angular momentum, and why the axis must be specified.

Choose

Start by defining the system and the axis. If the signed resultant external torque about that axis is zero, the total angular momentum about that axis remains constant. This is why a skater spins faster when pulling arms inward: I decreases, so ω increases to keep Iω constant. The condition is about external torque, not just external force, and it must be checked before using a before-after equation.

Angular momentum of a system is conserved when the resultant external torque about the chosen axis is zero.

Use the same system, same axis, and same sign convention before and after.

For a rotating rigid body, conservation can be written L_initial = L_final or I1ω1 = I2ω2 when I changes internally.

Internal torques can redistribute angular momentum within the system, but they do not change the total angular momentum of the system.

Conservation of angular momentum does not automatically mean rotational kinetic energy is conserved.

Build the angular momentum conservation equation and condition.

Formula

Target formula Στ_external = 0 ⇒ L_initial = L_final; I1ω1 = I2ω2

Στ_external

resultant external torque about the chosen axis

N m

L_initial

initial total angular momentum

kg m^2 s^-1

L_final

final total angular momentum

kg m^2 s^-1

moment of inertia about the axis

kg m^2

angular velocity about the axis

rad s^-1

1Define the rotating system and choose the axis.same system, same axis

2Check the resultant external torque about that axis.Στ_external = 0

3Write angular momentum before equals after.L_initial = L_final

4For a rigid-body before-after problem, substitute L = Iω.I1ω1 = I2ω2

5Interpret how changing I affects ω.if I decreases, ω increases for constant L

Build the formula

A skater spinning on low-friction ice pulls their arms inward. Explain why angular velocity increases, and state the condition required for angular momentum conservation.

Common mark losses are saying only “less moment of inertia means faster” without stating zero external torque, or claiming kinetic energy must also be conserved.

A skater spinning on low-friction ice pulls their arms inward. Explain why angular velocity increases, and state the condition required for angular momentum conservation.

Choose

Choose the axis and use the signed resultant torque about that axis. If the torque is constant over the interval, multiply torque by time. If torque varies, find the signed area under the torque-time graph. That angular impulse equals the change in angular momentum, so for a rigid body with constant I it can be written as I(ω_final - ω_initial).

Angular impulse is the time integral of resultant torque about an axis: angular impulse = ∫Στ dt.

Angular impulse changes angular momentum: ∫Στ dt = ΔL.

For constant or average resultant torque, ΔL = Στ Δt.

The area under a torque-time graph gives angular impulse, with sign showing rotational sense.

Units of angular impulse are N m s, equivalent to kg m^2 s^-1.

Build the angular impulse relation.

Formula

Target formula ∫Στ dt = ΔL; for constant torque, ΔL = ΣτΔt

Στ

signed resultant torque about the chosen axis

N m

Δt

time interval

ΔL

change in angular momentum

kg m^2 s^-1

∫Στ dt

angular impulse or torque-time area

N m s

1Choose the axis and sign convention.same axis for torque and L

2Use the resultant torque about that axis.Στ

3Use area under τ-t if torque varies, or product if torque is constant.∫Στ dt or ΣτΔt

4Set angular impulse equal to change in angular momentum.∫Στ dt = ΔL

5For constant I, connect ΔL to angular velocity change.ΔL = I(ω_f - ω_i)

Build the formula

A varying resultant torque acts on a wheel for a short time. Explain how to determine the change in angular momentum from a torque-time graph, and state the constant-torque special case.

Common mark losses are using force-time area, omitting the sign of torque, or using τΔt when the torque is not constant without first finding an average torque.

A varying resultant torque acts on a wheel for a short time. Explain how to determine the change in angular momentum from a torque-time graph, and state the constant-torque special case.

Choose

Use rotational kinetic energy when the object’s mass distribution rotates about an axis. The expression mirrors linear kinetic energy, with I replacing m and ω replacing v. For a wheel rolling along a surface, the centre of mass translates while the wheel also rotates, so the total kinetic energy is the sum of translational and rotational kinetic energies. Do not use only 1/2mv^2 when rotational energy is present.

A rigid body rotating about an axis has rotational kinetic energy E_rot = 1/2 Iω^2.

I must be the moment of inertia about the same axis as the angular speed ω.

Rotational kinetic energy is a scalar energy in joules; the sign of ω does not affect E_rot.

If an object is both translating and rotating, include both terms: E_k,total = 1/2mv^2 + 1/2Iω^2.

For rolling without slipping, v = ωr can connect the translational and rotational parts.

Build the rotational and rolling kinetic energy expressions.

Formula

Target formula E_rot = 1/2 Iω^2; E_k,total = 1/2mv^2 + 1/2Iω^2

E_rot

rotational kinetic energy

moment of inertia about the rotation axis

kg m^2

angular speed

rad s^-1

mass of the object

speed of the centre of mass

m s^-1

1Decide whether the object is only rotating or both translating and rotating.fixed-axis rotation or rolling

2Use moment of inertia about the same axis as ω.I with ω

3Calculate rotational kinetic energy.E_rot = 1/2Iω^2

4If the centre of mass moves, add translational kinetic energy.1/2mv^2 + 1/2Iω^2

5For rolling without slipping, connect v and ω if needed.v = ωr

Build the formula

A wheel rolls without slipping. State the expression for its total kinetic energy and explain why a translational kinetic energy term alone is incomplete.

Common mark losses are using only 1/2mv^2 for rolling motion, using the wrong axis for I, or treating kinetic energy as directional.

A wheel rolls without slipping. State the expression for its total kinetic energy and explain why a translational kinetic energy term alone is incomplete.

Choose

A.4 questions become much safer when you begin by naming the system and rotation axis. Then decide whether the prompt is asking for a turning effect, equilibrium, angular kinematics, rotational dynamics, angular momentum, angular impulse, or rotational energy. Each model has a condition: same axis, signed torques, radians, constant angular acceleration, zero external torque, or a rolling/non-rolling energy choice. State that condition before substituting numbers.

Torque is a signed turning effect about an axis: τ = Fr sinθ = Fd_perpendicular.

Rotational equilibrium requires Στ = 0 about the chosen axis; complete static equilibrium also needs ΣF = 0.

Angular kinematics uses θ, ω, and α in radians; angular SUVAT equations require constant α.

Moment of inertia depends on mass distribution about the axis; point masses use I = Σm_i r_i^2.

Rotational dynamics uses Στ = Iα with resultant torque and I about the same axis.

Angular momentum is L = Iω; it is conserved only when resultant external torque about the axis is zero.

Angular impulse is signed torque-time area: ∫Στ dt = ΔL.

Rotational kinetic energy is E_rot = 1/2Iω^2; rolling motion also has translational kinetic energy.

Match each A.4 retrieval cue to the physics move it should trigger.

Match

Reasons

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IB Physics HL A.4 Rigid body mechanics Notes

Turn Force into Torque

Use Rotational equilibrium

Turn Force into Torque

Use Angular motion quantities

Use Angular motion equations

Moment of inertia

Point-mass moment of inertia

Use Rotational Newton’s second law

Use Angular momentum

Use Angular momentum conservation

Use Angular impulse

Use Rotational kinetic energy

Retrieve the A.4 Rigid body mechanics Model