IB Physics flashcards · SL and HL · first assessment 2025
A.4 Rigid body mechanics flashcards: IB Physics HL
Revision flashcards for A.4 Rigid body mechanics, written for the IB Diploma Physics course first assessed in 2025. This is a Higher Level topic, so every card is HL only. Use study mode to test yourself one card at a time, or read the full list below with every answer.
Torque, moment of inertia, rotational equilibrium and kinematics, angular momentum and rolling.
- 72 cards
- Definition: 16
- Equation: 12
- Concept/explain: 16
- HL-only cards marked
All 72 A.4 Rigid body mechanics cards
- DefinitionHL only
Define torque (moment of a force).
Show answer
Torque = force × perpendicular distance from the axis to the line of action of the force: τ = Fr sinθ. Vector (direction along axis, clockwise/anticlockwise). SI unit: N m (NOT joule).
- DefinitionHL only
Define moment of inertia.
Show answer
Measure of a body's resistance to change in its rotational motion about a given axis: I = Σmr², the sum over all mass elements of mass × (distance from axis)². Unit kg m². Depends on mass AND how it is distributed about the axis.
- DefinitionHL only
Define angular velocity and angular acceleration.
Show answer
Angular velocity ω = rate of change of angular displacement, Δθ/Δt (rad s⁻¹), ω = 2π/T. Angular acceleration α = rate of change of angular velocity, Δω/Δt (rad s⁻²). Related to linear: v = ωr, a_tangential = αr.
- DefinitionHL only
Define angular momentum.
Show answer
Product of moment of inertia and angular velocity: L = Iω. Vector along the axis of rotation. Unit kg m² s⁻¹ (= N m s). For a point mass: L = mvr.
- DefinitionHL only
State the conditions for a rigid body to be in equilibrium.
Show answer
(1) Translational: resultant force zero, ΣF = 0 in every direction. (2) Rotational: resultant torque zero about ANY axis, Στ = 0 (Σ clockwise = Σ anticlockwise). Both must hold.
- DefinitionHL only
State the law of conservation of angular momentum.
Show answer
The total angular momentum of a system remains constant provided no resultant EXTERNAL torque acts on it: I₁ω₁ = I₂ω₂.
- DefinitionHL only
Define a couple.
Show answer
Two equal and opposite parallel forces whose lines of action do not coincide. Resultant force zero (no translation), but resultant torque = F × d (d = perpendicular separation) about any point, producing pure rotation.
- EquationHL onlyData booklet: Yes
Write the equation for torque and explain θ.
Show answer
τ = Fr sinθ. F = force (N); r = distance from axis to the point of application (m); θ = angle between r and F. r sinθ = perpendicular ('lever arm') distance. Misuse: using the angle between F and the perpendicular; maximum torque when θ = 90°.
- EquationHL onlyData booklet: Yes
Write the general expression for moment of inertia and note how specific shapes are handled in exams.
Show answer
I = Σmr² (booklet). Formulae for specific shapes (e.g. solid sphere 2/5 MR², disc ½MR², hoop MR², rod about end 1/3 ML²) are NOT in the booklet—they are given in the question when needed. Misuse: forgetting to add I of several parts about the same axis.
- EquationHL onlyData booklet: Yes
Write Newton's second law for rotation.
Show answer
τ = Iα. τ = resultant torque (N m); I = moment of inertia (kg m²); α = angular acceleration (rad s⁻²). Analogue of F = ma. Misuse: using the torque of one force instead of the NET torque; forgetting rotational inertia of pulleys.
- EquationHL onlyData booklet: Yes
Write the rotational kinematic equations for constant angular acceleration.
Show answer
ω_f = ω_i + αt; Δθ = ω_i t + ½αt²; ω_f² = ω_i² + 2αΔθ. θ in radians, ω in rad s⁻¹, α in rad s⁻². Direct analogues of suvat. Misuse: mixing revolutions and radians (1 rev = 2π rad).
- EquationHL onlyData booklet: Yes
Write the equations for angular momentum and angular impulse.
Show answer
L = Iω; ΔL = τΔt (angular impulse = area under torque–time graph). Analogues of p = mv, Δp = FΔt. Misuse: using ω in rpm.
- EquationHL onlyData booklet: Yes
Write rotational kinetic energy in two forms.
Show answer
E_k = ½Iω² = L²/(2I). For a rolling body, total KE = ½mv² (translational) + ½Iω² (rotational), with v = ωr for rolling without slipping. Misuse: omitting one of the two KE terms for rolling objects.
- Concept/explainHL only
Explain why a spinning ice skater rotates faster when she pulls her arms in.
Show answer
- No external torque acts (ice friction negligible)
- Angular momentum L = Iω conserved
- Pulling arms in reduces r, so I = Σmr² decreases
- ω must increase to keep Iω constant
- Rotational KE increases (work done by skater's muscles).
- Concept/explainHL only
Explain why a solid sphere rolls down an incline faster than a hollow sphere of the same mass and radius.
Show answer
- Same E_p lost = ½mv² + ½Iω²
- Hollow sphere has larger I (mass further from axis)
- Larger share of energy goes to rotational KE
- Less translational KE, so lower v at bottom
- Result independent of mass and radius (only the shape factor matters).
- Concept/explainHL only
Explain the role of friction when a ball rolls without slipping down a slope.
Show answer
- Static friction acts up the slope at the contact point
- Provides the torque needed to increase ω (τ = Iα)
- Contact point instantaneously at rest, so friction does NO work
- Mechanical energy conserved
- Without friction (frictionless slope) ball would slide, not roll.
- Concept/explainHL only
Explain why the moment of inertia of a body is different about different axes.
Show answer
- I = Σmr² depends on distance of each mass element from the axis
- Choosing an axis far from the mass increases r values
- Same body: I larger about an axis through the end than through the centre
- So a body has no single moment of inertia—axis must be specified.
- Concept/explainHL only
Explain how a couple produces rotation without translation.
Show answer
- Two equal, opposite forces: resultant force zero, so no linear acceleration (N1)
- Lines of action separated by d
- Torques about any point add (both same sense)
- Net torque Fd ≠ 0
- Produces angular acceleration α = Fd/I.
- Worked problemHL onlyData booklet: Yes
A 20 N force acts on a wrench 0.30 m from the nut at 60° to the handle. Find the torque.
Show answer
τ = Fr sinθ = 20 × 0.30 × sin60° = 20 × 0.30 × 0.866 = 5.2 N m.
- Worked problemHL onlyData booklet: Yes
A disc (I = 0.50 kg m²) at rest experiences a constant torque of 2.0 N m for 4.0 s. Find the final angular velocity and the angle turned.
Show answer
α = τ/I = 2.0/0.50 = 4.0 rad s⁻². ω = ω_i + αt = 0 + 4.0 × 4.0 = 16 rad s⁻¹. Δθ = ½αt² = ½ × 4.0 × 16 = 32 rad (≈ 5.1 revolutions).
- Worked problemHL onlyData booklet: Yes
Two 2.0 kg point masses are fixed at each end of a light 1.0 m rod. Find I about the centre and about one end.
Show answer
Centre: I = 2 × (2.0)(0.50²) = 1.0 kg m². One end: I = 2.0(0)² + 2.0(1.0)² = 2.0 kg m². (Larger about the end.)
- Worked problemHL onlyData booklet: Yes
A skater spins at 2.0 rad s⁻¹ with I = 4.0 kg m². She pulls her arms in so I = 1.0 kg m². Find the new ω and the change in KE.
Show answer
I₁ω₁ = I₂ω₂ → ω₂ = 4.0 × 2.0/1.0 = 8.0 rad s⁻¹. KE₁ = ½(4.0)(2.0²) = 8.0 J; KE₂ = ½(1.0)(8.0²) = 32 J. Increase = 24 J, supplied by work done by her muscles.
- Worked problemHL onlyData booklet: Yes
A solid sphere (I = 2/5 mr²) rolls without slipping from rest down a 2.0 m high slope. Find its speed at the bottom.
Show answer
mgh = ½mv² + ½(2/5 mr²)(v/r)² = ½mv² + 1/5 mv² = 7/10 mv². v = √(10gh/7) = √(10 × 9.8 × 2.0/7) = √28 = 5.3 m s⁻¹.
- Worked problemHL only
A uniform 4.0 m beam weighing 200 N rests on a pivot 1.0 m from its left end. What downward force at the left end keeps it horizontal?
Show answer
Weight acts at centre, 1.0 m right of pivot. Torques about pivot: F × 1.0 = 200 × 1.0 → F = 200 N (downward) at the left end. Pivot then supports 400 N.
- Worked problemHL onlyData booklet: Yes
A flywheel decelerates uniformly from 300 rpm to rest in 20 s. Find α and the number of revolutions made.
Show answer
300 rpm = 300 × 2π/60 = 31.4 rad s⁻¹. α = (0 − 31.4)/20 = −1.57 rad s⁻². Δθ = ½(ω_i + ω_f)t = ½ × 31.4 × 20 = 314 rad = 314/2π = 50 revolutions.
- Graph/diagramHL only
What do the gradient and area represent on angular velocity–time and torque–time graphs?
Show answer
ω–t: gradient = angular acceleration α; area = angular displacement Δθ. τ–t: area = angular impulse = change in angular momentum ΔL. Direct analogues of v–t and F–t graphs.
- Graph/diagramHL only
Describe a graph of angular momentum against angular velocity for a rigid body about a fixed axis.
Show answer
Straight line through the origin (L = Iω). Gradient = moment of inertia I. If the mass distribution changes (skater), the point moves along a hyperbola L = constant rather than along the line.
- Exam technique/trapHL only
Trap: which angle goes in τ = Fr sinθ?
Show answer
θ is the angle between the position vector r (axis to point of application) and the force F. If the question gives the angle between F and the PERPENDICULAR to r, use cos of that angle instead. Quick check: force along the line of r gives zero torque.
- Exam technique/trapHL only
Trap: unit conversions in rotational motion.
Show answer
rpm → rad s⁻¹: multiply by 2π/60. Revolutions → radians: × 2π. Degrees → radians: × π/180. Torque unit is N m (not J), even though dimensionally the same. Angular momentum unit kg m² s⁻¹.
- Exam technique/trapHL only
Trap: when is angular momentum conserved, and when is rotational KE conserved?
Show answer
L conserved when net EXTERNAL torque is zero (internal forces, e.g. arms pulled in, do not count). Rotational KE is generally NOT conserved in such changes (work done internally). Both conserved only if I is also unchanged.
- Exam technique/trapHL only
Practical: outline how to measure the moment of inertia of a flywheel.
Show answer
Wrap string with hanging mass m around axle of radius r; release; measure time t for mass to fall height h. a = 2h/t², tension T = m(g − a), torque τ = Tr, α = a/r, so I = τ/α (subtract friction torque by finding minimum mass that just turns wheel). Limitations: friction at bearings, string thickness, timing uncertainty—use light gates or video.
- DefinitionHL only
Define the moment arm (perpendicular distance) of a force about an axis.
Show answer
The moment arm is the perpendicular distance from the axis of rotation to the line of action of the force; torque = force × moment arm. Unit: metre (m); it is a distance (scalar), while torque itself is a vector along the rotation axis. Exam tip: marks are lost for using the distance to the point of application rather than the perpendicular distance to the LINE OF ACTION (the line extended infinitely in both directions through the force arrow). If the force passes through the axis the moment arm is zero, so the torque is zero however large the force. Always draw the extended line of action on the diagram before measuring.
- DefinitionHL only
State what is meant by a rigid body and the modelling assumption this makes in A.4 problems.
Show answer
A rigid body is an extended object in which the separation between any two particles stays constant, so the body does not deform, stretch or compress under the applied forces. Exam tip: the assumption lets you treat every particle as having the same angular velocity ω and angular acceleration α about the axis, so I is a fixed constant for a given axis and the whole body's weight can be taken to act at the centre of gravity. Students lose credit by treating a rope, spring or human body as rigid; a skater who pulls her arms in is NOT a rigid body, which is precisely why her I changes.
- DefinitionHL only
Define the radian and state the link between angular and linear displacement.
Show answer
One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius; θ = s/r, so s = rθ with θ in radians. There are 2π rad in one revolution (1 rad ≈ 57.3°). Exam tip: the radian is dimensionless (m/m), which is why rotational quantities carry units such as rad s⁻¹ that behave as s⁻¹ in algebra. Full marks require the definition in terms of arc length and radius, not "a unit of angle". The relations s = rθ, v = ωr and a = αr are only valid with θ in radians, never degrees or revolutions.
- DefinitionHL onlyData booklet: No – memorise
State the condition for a body to roll without slipping, and what it implies about the contact point.
Show answer
Rolling without slipping requires v = ωr (and a = αr) for the centre of mass, where r is the radius of the rolling body; equivalently the arc length turned equals the distance travelled along the surface. Exam tip: the point of contact is instantaneously at rest relative to the surface, so the friction acting is STATIC friction and it does no work — this is why mechanical energy is conserved for rolling without slipping. Common error: stating that "friction is zero" (it is generally non-zero but does no work), or that kinetic friction acts (it does not unless the body slips). Unit check: m s⁻¹ = (rad s⁻¹)(m).
- DefinitionHL onlyData booklet: Yes
Define angular impulse and state its unit.
Show answer
Angular impulse is the product of the resultant torque and the time for which it acts, ΔL = τΔt, and it equals the change in angular momentum of the body. It is a vector along the axis of rotation. Unit: N m s, equivalently kg m² s⁻¹ (the same unit as angular momentum). Exam tip: for a varying torque the angular impulse is the AREA under a torque–time graph, which is the standard route to full marks in "determine the change in angular momentum" questions. Students lose the mark by writing τ = ΔL/Δt as if it were an instantaneous definition of torque and then multiplying by the wrong time interval.
- DefinitionHL only
Define the centre of gravity of a rigid body and state where it lies for a uniform beam.
Show answer
The centre of gravity is the single point at which the entire weight of the body may be taken to act, so the resultant moment of the distributed weight about that point is zero. For a uniform beam of any length it lies at the geometric centre (midpoint); for a uniform disc or sphere it is at the centre. Exam tip: in a uniform gravitational field the centre of gravity coincides with the centre of mass. Students lose the moments mark by drawing the weight arrow from the pivot or from the end of the beam instead of from its midpoint. If a beam is non-uniform, the position of the centre of gravity is usually the unknown to be found.
- DefinitionHL only
State the principle of moments for a body in rotational equilibrium.
Show answer
For a body in rotational equilibrium the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that same point, i.e. Στ = 0. Exam tip: the phrase "about any point" is the mark-earning part — the resultant torque is zero about EVERY point, so you may choose the most convenient pivot to eliminate an unknown. The most common lost mark is failing to state the point about which moments are taken, or mixing moments taken about two different points in one equation. Rotational equilibrium alone does not imply the body is at rest; it means α = 0, so ω may be constant and non-zero.
- DefinitionHL onlyData booklet: Yes
State what is meant by limiting friction and the condition satisfied when a ladder is on the point of slipping.
Show answer
Limiting friction is the maximum static friction force a surface can supply, f_max = μ_s N, where N is the normal reaction and μ_s the coefficient of static friction. Exam tip: for a ladder "on the point of slipping" (about to slip) the friction has reached this maximum, so you may write f = μ_s N as an equality; at any smaller angle f < μ_s N and the equality is invalid. Students routinely assume f = μN in every ladder question and lose marks; the equality is only justified by the words "just about to slip", "minimum coefficient" or "maximum load". μ_s is a dimensionless ratio with no unit.
- DefinitionHL onlyData booklet: No – derive
Define the angular momentum of a point mass moving in a straight line about an external axis.
Show answer
For a particle of mass m moving with speed v, the angular momentum about a chosen axis is L = mvr⊥, where r⊥ is the perpendicular distance from the axis to the line of motion; equivalently L = mvr sinθ. Unit: kg m² s⁻¹; it is a vector. Exam tip: this is the card that unlocks "child runs tangentially and jumps onto a roundabout" questions — the child has angular momentum about the roundabout's axis BEFORE landing, and it must be included in the conservation equation. Students commonly set the initial angular momentum of the child to zero because "it is not rotating", which loses the whole question.
- EquationHL onlyData booklet: Yes
Write the equations linking linear and angular quantities for a rotating rigid body.
Show answer
s = rθ, v = ωr, a_t = αr, with a_c = ω²r = v²/r for the centripetal component. Symbols: s = arc length (m); θ = angular displacement (rad); v = tangential (linear) speed (m s⁻¹); ω = angular velocity (rad s⁻¹); a_t = tangential acceleration (m s⁻²); α = angular acceleration (rad s⁻²); r = distance from axis (m); a_c = centripetal acceleration (m s⁻²). Valid only with angles in radians, and for a rigid body where ω and α are the same for every particle. Data booklet: v = ωr and a = ω²r appear (topic A.2/A.4). Misuse: applying v = ωr to a wheel that is skidding. Sanity check: a wheel of r = 0.30 m at ω = 10 rad s⁻¹ has rim speed 3.0 m s⁻¹.
- EquationHL onlyData booklet: Yes
Write the total kinetic energy of a body rolling without slipping and show how it splits.
Show answer
E_K,total = ½mv² + ½Iω². Symbols: m = mass (kg); v = speed of centre of mass (m s⁻¹); I = moment of inertia about the centre of mass (kg m²); ω = angular velocity (rad s⁻¹); energies in J. Using v = ωr and writing I = kmr² gives E_K = ½mv²(1 + k), so the fraction in rotation is k/(1 + k). For a solid sphere k = 2/5 (2/7 rotational), a solid cylinder k = ½ (1/3 rotational), a hoop k = 1 (½ rotational). Data booklet: both ½mv² and ½Iω² are given. Misuse: using only ½mv² in an energy conservation line for a rolling object — the standard 2-mark loss. Sanity check: a hoop shares its KE equally between rotation and translation.
- EquationHL onlyData booklet: No – derive
Write the equations for the work done and the power delivered by a torque.
Show answer
W = τΔθ and P = τω. Symbols: W = work done (J); τ = torque (N m); Δθ = angular displacement (rad); P = power (W); ω = angular velocity (rad s⁻¹). Valid for a constant torque about a fixed axis; for a varying torque W is the area under a torque–angle graph. These are the exact rotational analogues of W = Fs and P = Fv. Condition: Δθ must be in radians. Misuse: multiplying torque by the number of revolutions instead of by 2π × revolutions. Sanity check: a motor delivering 50 N m at 20 rad s⁻¹ has an output power of 1.0 kW; the work–energy theorem then gives τΔθ = ½Iω² − ½Iω₀².
- EquationHL onlyData booklet: No – derive
Write the equation for the torque of a couple.
Show answer
τ = Fd, where F = magnitude of each of the two forces (N) and d = perpendicular separation of their lines of action (m); τ is in N m. Conditions: the two forces must be equal in magnitude, opposite in direction and not collinear. The torque of a couple is the same about EVERY point in the plane, which is why a couple produces pure rotation with no translational acceleration (resultant force = 0). Data booklet: not listed separately; it follows from τ = Fr sinθ applied twice. Misuse: using τ = 2Fd (double counting) or using the distance from one force to the axis rather than the separation of the two forces. Sanity check: two 15 N forces 0.40 m apart give a couple of 6.0 N m.
- EquationHL onlyData booklet: No – derive
Write an expression for the acceleration of a body rolling without slipping down an incline.
Show answer
a = g sinθ / (1 + I/mr²), so for I = kmr², a = g sinθ/(1 + k). Symbols: a = acceleration of the centre of mass along the slope (m s⁻²); g = 9.8 m s⁻²; θ = angle of incline to the horizontal; I = moment of inertia about the centre of mass (kg m²); m = mass (kg); r = radius (m). Derived from mg sinθ − f = ma, fr = Iα and a = αr; note it is independent of m and r. Data booklet: no — derive it. Misuse: quoting a = g sinθ (the frictionless sliding result), which is always too large for a rolling body. Sanity check: a solid sphere (k = 2/5) on a 30° slope gives a = 4.9/1.4 = 3.5 m s⁻².
- EquationHL onlyData booklet: Yes
Write the equation used when a rotating body picks up or loses a second body, and state the conditions for its use.
Show answer
I₁ω₁ = I₂ω₂ with I₂ = I₁ + Σmr² when a body of mass m lands and stays at distance r from the axis. Symbols: I = moment of inertia (kg m²); ω = angular velocity (rad s⁻¹); m = added mass (kg); r = its distance from the axis (m). Condition of validity: no resultant EXTERNAL torque about the axis during the interaction — internal forces between the bodies come in equal and opposite pairs and cancel. Data booklet: L = Iω is given; the conservation statement must be quoted. Misuse: also equating the kinetic energies — these "landing" interactions are inelastic and rotational KE always decreases. Sanity check: doubling I halves ω and halves the rotational KE.
- Concept/explainHL only
Explain, using forces and torques, why a ladder resting against a smooth vertical wall does not slip.
Show answer
- Forces: weight W at the midpoint, vertical normal reaction N at the foot, horizontal friction f at the foot towards the wall
- the wall is smooth, so its reaction N_w is purely horizontal with no vertical component
- vertical equilibrium gives N = W and horizontal equilibrium gives f = N_w
- moments about the foot: N_w(L sinθ) = W(L/2)cosθ, so N_w = W/(2 tanθ)
- the friction needed, f = W/(2 tanθ), grows as the ladder is laid more horizontal
- the ladder holds provided f ≤ μN, i.e. μ ≥ 1/(2 tanθ). Exam tip: the incomplete answer is "friction holds it" with no moments taken; the marks come from a moment equation about a stated point and from recognising the wall force is horizontal only because the wall is smooth.
- Concept/explainHL only
Explain why the reactions at the two supports of a loaded beam are generally unequal.
Show answer
- Translational equilibrium gives R₁ + R₂ = total downward load
- rotational equilibrium gives zero resultant torque about any point
- taking moments about one support eliminates that reaction and yields the other directly
- an off-centre load has a larger moment arm about the far support, so the NEAR support must supply the larger reaction
- moving the load towards one support raises that reaction and lowers the other, linearly with position
- a load directly over a support is carried entirely by it, the other taking only its share of the beam's weight. Exam tip: students often use ΣF = 0 alone (one equation, two unknowns) and stall; two independent equations are always needed, and a uniform beam's weight must be included at its midpoint.
- Concept/explainHL only
Explain why the tensions on either side of a pulley of significant mass are not equal.
Show answer
- A massless pulley has I = 0, so zero resultant torque gives any α and the tensions are equal
- a real pulley has I ≠ 0 and must be angularly accelerated by a resultant torque
- that torque comes only from the difference in the rope tensions: (T₁ − T₂)r = Iα
- so T₁ > T₂ whenever the system accelerates, the larger tension being on the descending side
- the string does not slip on the rim, so a = αr links the linear and rotational equations
- at constant velocity (α = 0) the tensions are equal again. Exam tip: classic lost marks are one tension throughout, and forgetting that the pulley's inertia makes the acceleration smaller than the ideal (m₁ − m₂)g/(m₁ + m₂). The effective extra inertia is I/r², in kg.
- Concept/explainHL only
Explain why a yo-yo released from rest falls with an acceleration much less than g.
Show answer
- Two forces act: weight mg down and string tension T up, so ma = mg − T
- the tension also exerts a torque about the axle, Tr = Iα, which spins the yo-yo up
- the string does not slip on the axle, so a = αr
- combining gives a = g/(1 + I/mr²); the axle radius r is tiny while I is set by the much larger disc radius, so I/mr² is large and a is only a few per cent of g
- the tension is therefore almost equal to the weight
- gravitational potential energy converts mostly into ROTATIONAL kinetic energy. Exam tip: incomplete answers invoke air resistance or "friction from the string"; the reasoning wanted is that most energy goes into ½Iω² because ω = v/r is huge for a small axle radius.
- Concept/explainHL only
Explain why a stellar core spins up to millisecond periods when it collapses into a neutron star.
Show answer
- Gravity acts radially towards the centre during collapse, so it exerts no torque about the rotation axis
- with no external torque the angular momentum L = Iω is conserved
- for a sphere I ∝ mR², so collapsing from ~10⁶ km to ~10 km cuts I by a factor of order 10¹⁰
- since Iω is constant, ω rises by the same factor and the period T ∝ R² falls from days to milliseconds
- rotational kinetic energy ½Iω² = L²/2I INCREASES, supplied by gravitational potential energy released in the collapse
- the same argument explains the rapid regular pulses from pulsars. Exam tip: the Nature of Science link — a conservation law applied far outside the laboratory made a prediction pulsar observations later confirmed. Do not write "energy is conserved so it spins faster".
- Concept/explainHL only
Explain what happens to the angular momentum and the kinetic energy when a child jumps onto a rotating roundabout.
Show answer
- There is no resultant external torque about the vertical axis during the landing, so total angular momentum about that axis is conserved
- the child adds moment of inertia mr² to the system, so I increases and ω must decrease in proportion
- rotational kinetic energy ½Iω² = L²/2I therefore DECREASES, since L is fixed and I has increased
- the collision is perfectly inelastic — the child and roundabout end with a common ω
- the "lost" energy is transferred to internal energy and sound by the frictional forces between the child's feet and the platform
- if the child then walks inwards, I falls, ω rises and the child's muscles do work, increasing the KE again. Exam tip: students frequently write ½I₁ω₁² = ½I₂ω₂² alongside I₁ω₁ = I₂ω₂; the two cannot both hold, and quoting energy conservation here loses the marks.
- Concept/explainHL only
Explain qualitatively what changes when a wheel rolls with slipping rather than without slipping.
Show answer
- The contact point is no longer instantaneously at rest, so v ≠ ωr and the linear and rotational equations decouple
- friction becomes KINETIC, f = μ_kN, opposing the relative sliding at the contact
- kinetic friction does negative work, so mechanical energy is NOT conserved — some is dissipated as internal energy (tyre and road heating, skid marks)
- a hard-braked wheel has ωr < v, so friction acts backwards on the car and forwards on the rim, tending to restore rolling
- a spinning drive wheel has ωr > v, so friction acts forwards on the car and slows the spin
- rolling resumes once v = ωr, after which friction is static again. Exam tip: the incomplete answer is "friction is larger when slipping"; the marking point is that the constraint v = ωr fails and energy is dissipated.
- Concept/explainHL only
Explain how the moment of inertia of a composite body is found and why the process works.
Show answer
- I is defined by I = Σmr², a simple sum over all the particles of the body
- sums split, so moments of inertia of the parts about the SAME axis simply add: I_total = I₁ + I₂ + …
- each shape formula given in the question must be used about the correct axis (a rod about its end is ⅓mL², about its centre 1/12 mL²)
- a point mass at distance r contributes mr²
- holes or removed sections are handled by subtracting the missing piece's moment of inertia
- the result holds only about the one axis used throughout. Exam tip: the frequent error is adding values quoted about different axes, or adding masses first and applying one shape formula to the total. Unit check: every term in kg m².
- Concept/explainHL only
Explain why a planet in an elliptical orbit conserves angular momentum although its speed changes.
Show answer
- Gravity on the planet always acts along the line joining it to the star (a central force)
- that line of action passes through the star, so the moment arm is zero and the torque about the star is zero
- with no resultant torque ΔL = τΔt = 0, so L = mvr⊥ about the star is constant
- as r⊥ decreases the speed v must increase, so the planet is fastest at perihelion and slowest at aphelion
- constant L is Kepler's second law: the star–planet line sweeps equal areas in equal times
- kinetic energy and linear momentum are NOT constant, since the force has a component along the velocity. Exam tip: a strong Nature of Science link — an empirical law later explained by a conservation law. Incomplete answers claim "gravity does no work", false in an elliptical orbit.
- Concept/explainHL only
Explain how a body can be in translational equilibrium and yet not be in rotational equilibrium.
Show answer
- Translational equilibrium requires only that the vector sum of the forces is zero, so a_cm = 0
- rotational equilibrium is a separate condition, zero resultant torque about any point, so α = 0
- a couple is the clearest example: two equal, opposite, non-collinear forces give ΣF = 0 but Στ = Fd ≠ 0
- the centre of mass then stays at rest, or moves at constant velocity, while the body angularly accelerates about it
- full equilibrium requires BOTH conditions together
- conversely Στ = 0 with ΣF ≠ 0 is possible, as for a block pushed through its centre of mass, which accelerates without rotating. Exam tip: students commonly answer "is it in equilibrium?" by checking forces only; both conditions must be tested and stated.
- Concept/explainHL only
Explain why a tightrope walker carries a long, heavy, drooping pole.
Show answer
- The pole places mass far from the axis along the rope, so I = Σmr² for walker plus pole is greatly increased
- for a given toppling torque from the walker's weight acting off-line, α = τ/I is therefore much smaller
- the tilt develops slowly, giving far more time to react and shift the centre of gravity back over the rope
- the droop lowers the combined centre of gravity, reducing the toppling torque for a given tilt
- moving the pole gives equal and opposite angular momentum changes in the walker, allowing fine correction
- the pole changes only the RATE of tipping, not the weight. Exam tip: "it lowers the centre of gravity below the rope" is not enough — the marking point is the increase in I and the resulting reduction in angular acceleration.
- Worked problemHL only
A uniform 8.0 m ladder of weight 250 N leans at 60° to the horizontal against a smooth vertical wall. Find the friction at the foot and the minimum coefficient of static friction.
Show answer
Vertical equilibrium: N = W = 250 N. Take moments about the foot (eliminates N and f). Wall force N_w is horizontal, acting at height 8.0 sin60° = 6.93 m. Weight acts at the midpoint, horizontal distance 4.0 cos60° = 2.00 m. Στ = 0: N_w × 6.93 = 250 × 2.00 = 500 N m, so N_w = 72.2 N. Horizontal equilibrium: f = N_w = 72 N (2 s.f.). On the point of slipping f = μN, so μ_min = 72.2/250 = 0.29. Check/Trap: the wall is smooth so its reaction is horizontal only — adding a vertical friction force at the wall makes the problem insoluble; note μ_min = 1/(2 tan60°) = 0.289 confirms the algebra, and a smaller angle needs a larger μ.
- Worked problemHL only
A uniform beam of length 6.0 m and weight 400 N rests on supports at its ends. An 800 N load sits 2.0 m from the left support. Find both reactions.
Show answer
Let R_A (left) and R_B (right). Translational equilibrium: R_A + R_B = 400 + 800 = 1200 N. Take moments about A (eliminates R_A): R_B × 6.0 = 400 × 3.0 + 800 × 2.0 = 1200 + 1600 = 2800 N m, so R_B = 2800/6.0 = 466.7 N ≈ 4.7 × 10² N. Then R_A = 1200 − 467 = 733 N ≈ 7.3 × 10² N. Check/Trap: check with moments about B — R_A × 6.0 = 400 × 3.0 + 800 × 4.0 = 4400, giving R_A = 733 N, consistent. The beam's own 400 N weight must be placed at the 3.0 m midpoint; omitting it is the single most common error, and the nearer support correctly carries the larger reaction.
- Worked problemHL only
Masses of 5.0 kg and 3.0 kg hang from a light string over a pulley of moment of inertia 0.020 kg m² and radius 0.10 m. Find the acceleration and both tensions.
Show answer
For the pulley the effective inertia is I/r² = 0.020/0.10² = 2.0 kg. Newton's laws: 5.0g − T₁ = 5.0a; T₂ − 3.0g = 3.0a; (T₁ − T₂)r = Iα with a = αr. Adding: a = (5.0 − 3.0)(9.8)/(5.0 + 3.0 + 2.0) = 19.6/10.0 = 1.96 ≈ 2.0 m s⁻². Then T₁ = 5.0(9.8 − 1.96) = 39.2 ≈ 39 N and T₂ = 3.0(9.8 + 1.96) = 35.3 ≈ 35 N. Check/Trap: verify the pulley equation — (39.2 − 35.3)(0.10) = 0.39 N m and Iα = 0.020 × (1.96/0.10) = 0.39 N m, so it balances. A massless pulley would have given a = 2.45 m s⁻² and equal tensions; the extra 2.0 kg of effective inertia is what slows it.
- Worked problemHL only
A roundabout (I = 120 kg m²) turns at 1.5 rad s⁻¹. A 30 kg child drops on and sits 1.8 m from the axis. Find the new angular velocity and the kinetic energy lost.
Show answer
No external torque about the vertical axis, so L is conserved: I₁ω₁ = I₂ω₂. Child's contribution: mr² = 30 × 1.8² = 30 × 3.24 = 97.2 kg m². I₂ = 120 + 97.2 = 217.2 kg m². ω₂ = (120 × 1.5)/217.2 = 180/217.2 = 0.829 ≈ 0.83 rad s⁻¹. Energies: E₁ = ½ × 120 × 1.5² = 135 J; E₂ = ½ × 217.2 × 0.829² = 74.6 J. Energy lost = 135 − 75 = 60 J (2 s.f.), transferred to internal energy by friction between the child's feet and the platform. Check/Trap: the child drops vertically, so their initial angular momentum about the axis is zero; energy is NOT conserved here — the check E₂ = L²/2I₂ = 180²/(2 × 217.2) = 74.6 J confirms the arithmetic.
- Worked problemHL only
A solid cylinder (I = ½mr²) rolls without slipping from rest down 3.0 m of a slope inclined at 25°. Find its acceleration and speed at the bottom.
Show answer
Energy method: mgh = ½mv² + ½Iω², with ω = v/r and I = ½mr², so mgh = ¾mv², giving v = √(4gh/3). Height drop h = 3.0 sin25° = 1.27 m, so v = √(4 × 9.8 × 1.27/3) = √16.6 = 4.07 ≈ 4.1 m s⁻¹. Acceleration: a = g sinθ/(1 + k) with k = ½, so a = 9.8 × 0.423/1.5 = 2.76 ≈ 2.8 m s⁻². Check/Trap: v² = 2as = 2 × 2.76 × 3.0 = 16.6, giving v = 4.1 m s⁻¹ — consistent. Note m and r cancel, so the answer is independent of the cylinder's mass and radius; a block sliding without friction would reach 4.99 m s⁻¹, and using ½mv² alone is the standard trap.
- Worked problemHL only
A uniform rod (mass 1.2 kg, length 0.80 m, I = ⅓mL² about one end) is pivoted at one end with a 0.50 kg point mass fixed at the far end. It is released from horizontal. Find the initial angular acceleration.
Show answer
Composite moment of inertia about the pivot: rod ⅓ × 1.2 × 0.80² = ⅓ × 1.2 × 0.64 = 0.256 kg m²; point mass 0.50 × 0.80² = 0.320 kg m². I_total = 0.576 kg m². Torque about the pivot when horizontal: rod's weight acts at 0.40 m, τ = 1.2 × 9.8 × 0.40 = 4.70 N m; point mass, τ = 0.50 × 9.8 × 0.80 = 3.92 N m. Στ = 8.62 N m. α = τ/I = 8.62/0.576 = 15.0 ≈ 15 rad s⁻². Check/Trap: the pivot reaction has zero moment arm about the pivot, so it never enters the torque equation. The rod's weight acts at its centre, not its end; and α is the INITIAL value only — the torque falls as cosθ as the rod swings down.
- Worked problemHL only
A stellar core of radius 7.0 × 10⁸ m rotating once every 25 days collapses to a neutron star of radius 1.0 × 10⁴ m. Estimate the new rotation period.
Show answer
Model both as uniform spheres, I = 2/5 MR², with M unchanged and no external torque, so Iω is constant: R₁²ω₁ = R₂²ω₂. Since ω = 2π/T, this gives T ∝ R², so T₂ = T₁(R₂/R₁)². T₁ = 25 × 24 × 3600 = 2.16 × 10⁶ s. (R₂/R₁)² = (1.0 × 10⁴/7.0 × 10⁸)² = (1.43 × 10⁻⁵)² = 2.04 × 10⁻¹⁰. T₂ = 2.16 × 10⁶ × 2.04 × 10⁻¹⁰ = 4.4 × 10⁻⁴ s (2 s.f.). Check/Trap: the period falls by the SQUARE of the radius ratio, not the ratio itself — the commonest slip. The rotational KE = L²/2I rises enormously, supplied by released gravitational potential energy; real pulsars are slower because mass is ejected and the core is not uniform.
- Graph/diagramHL only
Describe the graph of torque against the angle between the force and the lever arm for a force of fixed magnitude at a fixed distance.
Show answer
Axes: torque τ/N m on the y-axis, angle θ/degrees (or rad) on the x-axis. Shape: a sine curve, τ = Fr sinθ, starting at zero when θ = 0 (force along the arm, line of action through the axis), rising to a maximum of Fr at θ = 90° (force perpendicular to the arm), and returning to zero at θ = 180°. The maximum value of the curve gives the product Fr directly, so with r known the force F can be extracted; equivalently, plotting τ against sinθ linearises the data and gives a straight line through the origin of gradient Fr. If r is doubled the whole curve stretches vertically by a factor of two with the zeros unchanged; if F is halved the amplitude halves. Note the graph is symmetric about θ = 90°, so 30° and 150° give equal torques.
- Graph/diagramHL only
Describe how rotational kinetic energy varies with angular velocity, and how the graph is used to find moment of inertia.
Show answer
Axes: E_K,rot/J against ω/rad s⁻¹. Shape: a parabola through the origin, since E_K = ½Iω², so the energy quadruples when ω doubles. The gradient of this curve is Iω and is not directly useful. To extract I, LINEARISE by plotting E_K against ω²: this gives a straight line through the origin of gradient ½I, so I = 2 × gradient, in kg m². A non-zero positive intercept would indicate a systematic error (for example friction losses or a mis-zeroed energy measurement). Error bars on ω² are found from the fractional uncertainty rule: Δ(ω²)/ω² = 2Δω/ω, so points at large ω carry the widest bars; the uncertainty in I comes from the max/min gradient lines drawn through the error bars. A body with larger I gives a steeper line.
- Graph/diagramHL only
Describe the free-body diagram for a uniform ladder leaning against a smooth wall on rough ground.
Show answer
Draw the ladder as a line at angle θ to the horizontal. Four labelled arrows: weight W acting vertically DOWN from the midpoint of the ladder; normal reaction N acting vertically UP at the foot; friction f acting HORIZONTALLY at the foot, pointing towards the wall; and the wall's reaction N_w acting horizontally AWAY from the wall at the top. Because the wall is smooth there is no vertical force at the top. Reading the diagram: vertical arrows give N = W; horizontal arrows give f = N_w; moments about the foot give N_w(L sinθ) = W(L/2 cosθ). If the wall were rough, a fifth vertical friction arrow appears at the top and the problem becomes statically indeterminate at this level. If a person stands part-way up, add a second downward weight arrow at their position, which increases both N and the friction required.
- Graph/diagramHL only
Describe the angular displacement–time graph for a body with constant angular acceleration from rest.
Show answer
Axes: angular displacement θ/rad against time t/s. Shape: a parabola through the origin, θ = ½αt², with the curve becoming steeper with time. The GRADIENT at any point (drawn as a tangent) gives the instantaneous angular velocity ω in rad s⁻¹; the increasing gradient shows ω is rising. There is no useful area under this graph. To find α, linearise by plotting θ against t²: a straight line through the origin of gradient ½α, so α = 2 × gradient in rad s⁻². A non-zero initial angular velocity adds a linear term, so the curve leaves the origin with a finite gradient ω₀; a negative α (a decelerating flywheel) gives a curve that flattens and then turns over at the moment the body stops. Doubling α doubles the curvature at every time.
- Exam technique/trapHL only
Practical: outline how to determine the moment of inertia of a cylinder by rolling it down an inclined plane.
Show answer
Apparatus: rigid plank on blocks, metre rule, protractor for θ, vernier callipers for r, balance for m, two light gates and timer (or video at known frame rate). Method: set a small θ, release the cylinder from rest at a marked line, time it over a measured distance s, repeat five times, and find a from s = ½at². Controls: same release point, no push, constant θ and surface. Analysis: I = mr²(g sinθ/a − 1); better, vary θ and plot a against sinθ, gradient = g/(1 + I/mr²). Limitations: slipping at large θ, and the cylinder wandering off line. Improvements: light gates remove reaction-time random error, a guide channel keeps it straight, small angles guarantee rolling. "Outline" needs method plus how I is obtained.
- Exam technique/trapHL only
Trap: what goes wrong when students treat the pulley in a connected-body question as massless?
Show answer
The trap is the SL habit of writing one tension throughout and using a = (m₁ − m₂)g/(m₁ + m₂). Students fall for it because every SL pulley is "light and frictionless", so "pulley of moment of inertia I" gets skimmed. It fails because a pulley with I ≠ 0 needs a resultant torque, which can only come from unequal tensions: (T₁ − T₂)r = Iα. Correct approach: a separate Newton's second law equation for EACH hanging mass, a torque equation for the pulley, linked by a = αr, solved simultaneously. The pulley adds effective inertia I/r² (kg) to the denominator, so a is always smaller than the massless value — a quick sanity check. "Show that" requires every equation written out, not just the final formula.
- Exam technique/trapHL only
Trap: choosing the point about which to take moments, and presenting the working for full marks.
Show answer
The trap is taking moments about the centre of a beam or a random point, leaving two unknown reactions in one equation and no way forward. Students fall for it by memorising "take moments about the pivot" without asking which unknowns that removes. Correct approach: since Στ = 0 about EVERY point in equilibrium, choose the point through which the most unknown forces act — usually a support or the ladder's foot — so those have zero moment arm and vanish; then use ΣF = 0 for the rest. Presentation marks: state the point ("moments about A"), include every force's moment plus the body's weight at its centre of gravity, and keep signs consistent. "Determine" needs substituted numbers with units; "show that" needs one extra significant figure.
- Exam technique/trapHL only
Trap: using v = ωr and mechanical energy conservation when a body may be slipping.
Show answer
The trap is applying mgh = ½mv² + ½Iω² with ω = v/r to any object on a slope. Students fall for it because rolling questions are drilled with the constraint always valid. It fails when the surface cannot supply enough static friction — a steep incline, an icy or greased slope, a hard-braked or driven wheel. The contact point then slides, kinetic friction does negative work, energy is dissipated internally, and v ≠ ωr, so both the energy equation and the constraint are invalid. Correct approach: check for the words "rolls without slipping"; if it slips, use ma = ΣF with f = μ_kN and τ = fr = Iα, treating a and α as independent. Uncertainty note: a measured v consistently below the rolling prediction is evidence of slipping — systematic, not random.
Practise this topic with exam-style questions: A.4 Rigid body mechanics questions (HL) · all flashcards
Know it, then write it the way it is marked
One-to-one IB tuition that turns correct physics and maths into full-mark answers.
Book a free consultationThese flashcards are original ExaminerPrep material, written independently. They are not IB documents and do not reproduce IB syllabus text, examination papers or mark schemes; topic references follow the published subject guides. ExaminerPrep is an independent tutoring service. It has been developed independently from and is not endorsed by the International Baccalaureate Organization. "International Baccalaureate", "IB" and "IB Diploma Programme" are registered trademarks of the IBO, used here for descriptive purposes only.