Summary to follow. 12 syllabus statements · 27 questions · about twenty minutes.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank ·
Specialist review in progress
· How these pages are made
Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Physics guide (first assessment 2025, updated November 2023).
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In preparation: 0 of 12 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
Torque (τ) HL
Torque (τ)
The turning effect of a force about an axis, τ = Fr sin θ, where F is the magnitude of the force, r is the distance from the axis to the point where the force is applied, and θ is the angle between the line of action of the force and the line from the axis to that point. SI unit: N m. The torque is greatest when the force acts at right angles to that line (θ = 90°) and zero when its line of action passes through the axis (θ = 0° or 180°).
Perpendicular distance of the line of action
The quantity r sin θ in τ = Fr sin θ: the shortest distance from the axis to the line of action of the force. The torque equals the force multiplied by this perpendicular distance, so moving the point of application along the line of action does not change the torque.
Sense of a torque
Whether a torque tends to turn a body clockwise or counter-clockwise about the axis, as seen from one chosen side. Torques in opposite senses are given opposite signs when they are combined, so a resultant torque is found by subtracting the total torque in one sense from the total in the other. The vector nature of torque is not required.
Students often think The angle in the torque equation is used with cosine (or measured from the perpendicular to the lever), so τ = Fr cos θ, as in the equation for work. In fact θ is the angle between the line of action of the force and the line joining the axis to the point where the force is applied. The torque is greatest at θ = 90°.
Students often think Torque is simply force × distance from the axis, whatever direction the force acts in. In fact Yes. τ = Fr sin θ: the same force at the same point produces less torque the more closely it points along the line to the axis.
Rotational equilibrium HL
Rotational equilibrium
The state of a body whose resultant torque is zero, so it has no angular acceleration: it is either not rotating or rotating at a constant angular speed. For a body at rest the total clockwise torque about any axis equals the total counter-clockwise torque about the same axis. Rotational equilibrium is independent of translational equilibrium (zero resultant force); a body in full equilibrium needs both.
Centre of mass
The point of an extended body at which, for linear motion, its whole mass may be taken as concentrated; the weight of the body can be taken to act there. For a uniform symmetrical body such as a uniform plank it is at the geometric centre. It does not replace the mass distribution when rotation is considered: the moment of inertia depends on where all the mass is. Calculating the position of the centre of mass is not required.
Students often think A body in rotational equilibrium is not rotating at all. In fact Yes. Rotational equilibrium means zero resultant torque and therefore zero angular acceleration; the body may be at rest or rotating at a constant angular speed.
Students often think A beam balances when the downward forces on each side of the pivot are equal in size. In fact Not in general. It balances when the total clockwise torque equals the total counter-clockwise torque; each force's distance from the pivot matters.
Rigid body HL
Rigid body
An extended body whose shape and size do not change when forces act on it, so the distance between any two of its particles is fixed. Every particle of a rigid body rotating about a fixed axis turns through the same angle in the same time, so the whole body has one angular displacement, one angular velocity and one angular acceleration.
Unbalanced (resultant) torque
A non-zero sum of the torques acting on a body about an axis, with torques of opposite sense subtracted. An unbalanced torque on an extended rigid body causes angular acceleration, just as an unbalanced force causes linear acceleration. A body can experience an unbalanced torque when the resultant force on it is zero, for example two equal and opposite forces whose lines of action do not coincide.
Students often think A torque is needed to keep a body rotating: a steady torque keeps it turning at a steady angular speed, and when the torque is removed the rotation dies away. In fact No. With zero resultant torque the angular speed stays constant; a torque is needed only to change it. A constant resultant torque gives a constant angular acceleration, so the angular speed increases steadily while it acts.
Students often think A torque gives the body a store of 'turning force' that keeps speeding it up until it is used up. In fact No. Angular acceleration exists only while an unbalanced torque acts (α = τ/I). Once the torque is removed, α = 0 immediately.
Angular position and angular displacement (θ, Δθ) HL
Angular position and angular displacement (θ, Δθ)
Angular position θ is the angle of a line fixed in the body measured from a reference direction; angular displacement Δθ is the change in angular position. Both are measured in radians (rad): one revolution is 2π rad, and an arc of length s at radius r subtends Δθ = s/r.
Angular velocity (ω)
The rate of change of angular position, ω = Δθ/Δt; its magnitude is the angular speed. SI unit: rad s⁻¹. All points of a rigid body share the same ω, while the linear speed of a point at distance r from the axis is v = ωr. A body making f revolutions per second has ω = 2πf. A formal vector treatment is not required.
Angular acceleration (α)
The rate of change of angular velocity, α = Δω/Δt. SI unit: rad s⁻². It is caused by an unbalanced torque, and is zero whenever the body rotates at a constant angular speed.
Rolling without slipping
Motion of a round body along a surface in which the point in contact with the surface is instantaneously at rest, so the speed of the centre of mass v and the angular speed ω about the centre are linked by v = ωR, where R is the radius. It is the only combination of rotational and translational motion considered in the course.
Students often think Every point of a rigid body moves together, so all points have the same linear speed. In fact No. They share the same angular velocity, but the linear speed v = ωr is greater for points further from the axis.
Students often think Angular speed and linear speed are the same thing, so a point that moves faster must be rotating faster, and a wheel moving at 7 m s⁻¹ rotates at 7 rad s⁻¹. In fact No. Angular speed ω (rad s⁻¹) is the rate of turning and is the same for every point of a rigid body; linear speed v (m s⁻¹) is the rate of travel of a point and equals ωr.
Equations for uniform angular acceleration HL
Equations for uniform angular acceleration
For constant α: Δθ = ((ω_f + ω_i)/2)t; ω_f = ω_i + αt; Δθ = ω_i t + ½αt²; ω_f² = ω_i² + 2αΔθ. They are the rotational counterparts of the equations for uniform linear acceleration, with Δθ, ω and α replacing s, v and a. Angles must be in radians and a slowing body has α of opposite sign to ω.
Students often think The angular speed can be treated as constant, so Δθ = ωt, and a body turns through equal angles in equal times. In fact No. The angular speed changes, so Δθ = ((ω_f + ω_i)/2)t or Δθ = ω_i t + ½αt² must be used; the body does not turn through equal angles in equal times.
Students often think Angular acceleration is always positive, so the magnitude of α is substituted as a positive number even when the wheel slows. In fact Opposite to the sign of ω. Taking the direction of rotation as positive, a slowing wheel has negative α.
Moment of inertia (I) HL
Moment of inertia (I)
The property of a body that measures its resistance to angular acceleration about a particular axis, playing the role in rotation that mass plays in linear motion. SI unit: kg m². It depends on the mass of the body and on how that mass is distributed about the axis: mass further from the axis contributes more. The same body has different moments of inertia about different axes. Formulas for specific mass distributions are provided when needed.
Students often think Moment of inertia is just the rotational name for mass, so bodies of equal mass have equal moments of inertia, and anything that changes the mass changes I in proportion. In fact No. It depends on the mass and on how that mass is distributed about the axis. Bodies of equal mass can have very different moments of inertia, and adding mass on the axis adds nothing.
Students often think A body that rotates faster has a larger moment of inertia, because it is harder to stop. In fact No. For a rigid body about a given axis, I depends only on the mass distribution, not on the angular speed.
Moment of inertia of a system of point masses HL
Moment of inertia of a system of point masses
I = Σmr², the sum over all the masses of each mass multiplied by the square of its perpendicular distance from the axis. Because r is squared, doubling the distance of a mass from the axis quadruples its contribution, while a mass on the axis contributes nothing.
Students often think Moment of inertia is proportional to distance from the axis, so r is not squared (I = Σmr, and doubling the distance doubles I). In fact On r²: I = Σmr². Doubling the distance of a mass from the axis quadruples its contribution.
Students often think The distances in I = Σmr² can be measured from the end of the rod where the positions are given. In fact From the axis of rotation. Positions given from one end of a rod must be converted to distances from the axis first.
Newton’s second law for rotation HL
Newton’s second law for rotation
τ = Iα, where τ is the average resultant torque on a body about an axis, I is its moment of inertia about that axis and α its angular acceleration. For the same resultant torque, a body with a larger moment of inertia has a smaller angular acceleration.
Students often think The driving torque alone determines the angular acceleration; friction torque can be ignored in τ = Iα. In fact No. τ is the resultant torque: the driving torque minus any opposing torque such as friction at the bearing.
Students often think Force and torque are interchangeable, so a tangential force F can be used as the torque in τ = Iα. In fact No. The torque of a force applied at distance r at right angles to the radius is τ = Fr. The force must be multiplied by its distance from the axis.
Angular momentum (L) HL
Angular momentum (L)
For an extended body rotating about an axis, L = Iω. SI unit: kg m² s⁻¹ (equivalent to N m s). It depends on both the moment of inertia (so on the mass distribution) and the angular speed. The vector nature of angular momentum is not required.
Students often think Angular momentum is the ordinary momentum mv of the body, so it depends on its mass and on how fast the body (or its centre of mass) is moving. In fact No. Angular momentum L = Iω depends on the moment of inertia and angular speed. A wheel spinning on a fixed axle has zero linear momentum (its centre of mass is at rest) but non-zero angular momentum.
Students often think A body has large angular momentum when a large torque is acting on it, and no angular momentum once the torque stops. In fact No. L = Iω depends on the present moment of inertia and angular speed. Torque determines how fast L is changing (τ = ΔL/Δt), not its value.
Conservation of angular momentum HL
Conservation of angular momentum
The angular momentum of a body or system stays constant unless a resultant external torque acts on it. When the moment of inertia of an extended body changes (a skater pulling in her arms) or two bodies become coupled (a disc dropped onto a rotating turntable), Iω before equals Iω after, so the angular speed changes. Kinetic energy is not necessarily conserved in these processes.
Students often think With no external torque the angular speed stays the same, even when the body changes shape or another body joins it. In fact Not if the moment of inertia changes. It is angular momentum Iω that is conserved; if I decreases, ω increases in proportion.
Students often think Kinetic energy is conserved whenever angular momentum is conserved, so ½I₁ω₁² = ½I₂ω₂² can be used to find the new angular speed. In fact No. Angular momentum is conserved, but the rotational kinetic energy E_k = L²/2I changes. It increases when I decreases (internal work is done) and decreases when bodies couple by friction (energy is dissipated).
Angular impulse (ΔL) HL
Angular impulse (ΔL)
The product of the average resultant torque and the time for which it acts, ΔL = τΔt, equal to the change in angular momentum Δ(Iω). SI unit: N m s (equivalent to kg m² s⁻¹). On a torque–time graph it is the area under the graph. Torques of opposite sense produce angular impulses of opposite sign.
Students often think The change in angular momentum is equal to the torque, so the time for which the torque acts does not matter. In fact No. Angular impulse is torque multiplied by the time for which it acts, ΔL = τΔt (unit N m s); torque (N m) is the rate of change of angular momentum.
Students often think The angular impulse gives the final angular momentum (or the final angular speed) directly. In fact Only if the body starts from rest. In general ΔL = L_final − L_initial, so the initial angular momentum must be included.
Rotational kinetic energy HL
Rotational kinetic energy
The kinetic energy of a body due to its rotation, E_k = ½Iω² = L²/2I. SI unit: J. For a fixed angular momentum, the rotational kinetic energy is inversely proportional to the moment of inertia, so a body that reduces its moment of inertia while L stays constant gains rotational kinetic energy from work done by internal forces.
Kinetic energy of a rolling body
A body rolling without slipping has translational kinetic energy ½Mv² (its mass treated as concentrated at the centre of mass, moving at v) and rotational kinetic energy ½Iω² about the centre of mass, with ω = v/R. The total is ½Mv² + ½Iω². On a slope, the larger the fraction of energy that goes into rotation, the smaller the speed gained for a given loss of gravitational potential energy.
Students often think The kinetic energy of a moving body is ½Mv² of its centre of mass, so rotation adds nothing, and any work done on the body appears as ½Mv². In fact No. That is only its translational kinetic energy. If the body also rotates it has rotational kinetic energy ½Iω² as well, so a rolling body has E_k = ½Mv² + ½Iω².
Students often think Rotational kinetic energy is Iω², without the ½. In fact ½Iω², the rotational counterpart of ½mv².
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 A rigid body is free to turn about a fixed axis. One of the forces acting on it lies in a plane perpendicular to the axis. Under which condition is the torque of this force about the axis zero? HL
Answer and reasoning
It acts perpendicular to the line from the axis to where it acts. — A student who uses cos θ in the torque equation thinks a perpendicular force gives zero torque. In fact θ = 90° gives sin θ = 1, the maximum torque for that force and distance.
The body remains at rest while this force acts on it. — A student who thinks a torque exists only when the body turns picks this. A body can stay at rest while this force exerts a torque, if other torques balance it; the force's own torque is still Fr sin θ.
An equal force acts on the other side, at the same distance. — A student who thinks a balancing force removes this force's own torque picks this. An equal force on the other side at the same distance can make the RESULTANT torque zero, but the torque of this force is still Fr sin θ, which other forces do not change.
Its line of action passes through the axis of rotation. — τ = Fr sin θ. If the line of action passes through the axis, θ = 0° or 180°, so sin θ = 0; equivalently the perpendicular distance from the axis to the line of action is zero.
2 A rigid body is in rotational equilibrium. Which statement about the body must be true? HL
Answer and reasoning
The body is not rotating about any axis. — A student who equates equilibrium with being at rest picks this. A body with zero resultant torque has constant angular velocity, which need not be zero: a wheel spinning steadily on a frictionless axle is in rotational equilibrium.
The forces on each side of its axis are equal. — A student who judges balance by equal forces picks this. The torques must balance, and a small force far from the axis can balance a large force close to it.
The resultant torque acting on the body is zero. — Rotational equilibrium is defined by zero resultant torque, so there is no angular acceleration. The body may be at rest or rotating at a constant angular speed.
The resultant force acting on the body is zero. — A student who merges the two equilibrium conditions picks this. Zero resultant torque does not require zero resultant force: a box pulled along by a force through its centre of mass accelerates without rotating.
3 A wheel is mounted on a frictionless axle. Starting from rest, a constant resultant torque acts on it for a few seconds and is then removed. How does the wheel move after the torque is removed? HL
Answer and reasoning
It keeps rotating at a constant angular speed. — With no resultant torque there is no angular acceleration (α = τ/I = 0), so the wheel keeps the angular speed it had when the torque was removed. This is the rotational form of Newton's first law.
It slows down steadily and then stops rotating. — A student who thinks a torque is needed to keep a body rotating picks this. Real wheels slow because friction exerts a torque; on a frictionless axle nothing changes the angular speed.
Its angular speed rises until the torque is used up. — A student who pictures the torque as stored in the wheel picks this. Angular acceleration exists only while a torque acts: once the torque is removed, α = 0 at once.
It stops, as its angular acceleration is now zero. — A student who confuses zero angular acceleration with zero angular velocity picks this. α = 0 means the angular velocity is constant, at the non-zero value reached when the torque stopped.
4 A rigid disc rotates at a steady rate about a fixed axis through its centre. Point P on the disc is 0.10 m from the axis and point Q is 0.20 m from the axis. How do the motions of P and Q compare? HL
Answer and reasoning
Q has twice the angular velocity and twice the linear speed. — A student who does not separate angular from linear speed thinks the faster-moving point Q must also rotate faster. Both points complete a revolution in the same time, so their angular velocities are equal.
Both have the same angular velocity and the same linear speed. — A student who thinks every part of a rigid body moves at the same speed picks this. In rotation Q travels round a circle twice as large in the same time, so its linear speed v = ωr is twice P's.
Same angular velocity; Q has twice the linear speed of P. — Every point of a rigid body turns through the same angle in the same time, so P and Q share one angular velocity. Linear speed v = ωr, so Q, at twice the distance, has twice the linear speed.
Q has half the angular velocity of P, because ω = v/r. — A student who reads ω = v/r with v held constant picks this. On a rigid body ω is the same everywhere; it is v that is proportional to r.
5 A flywheel rotating at 1500 revolutions per minute is brought to rest with uniform angular deceleration in 20 s. Through what angular displacement does it turn while stopping? (1 revolution = 2π rad.) HL
3.14 × 10³ rad — A student who treats the angular speed as constant uses Δθ = ω_i t = 157 × 20 = 3.14 × 10³ rad. The flywheel slows uniformly, so its average angular speed is half the initial value.
1.50 × 10⁴ rad — A student who uses 1500 rev min⁻¹ as if it were 1500 rad s⁻¹ gets (1500/2) × 20 = 1.50 × 10⁴ rad. The rate must first be converted: 1500 × 2π/60 = 157 rad s⁻¹.
4.71 × 10³ rad — A student who takes the deceleration as positive uses Δθ = ω_i t + ½αt² with α = +7.85 rad s⁻², getting 3142 + 1571 = 4.71 × 10³ rad. For a slowing wheel α is negative, which gives 3142 − 1571 = 1.57 × 10³ rad.
Working ω_i = 1500 rev min⁻¹ × 2π rad/rev ÷ 60 s min⁻¹ = 157.1 rad s⁻¹; ω_f = 0. Δθ = ((ω_f + ω_i)/2)t = (157.1 rad s⁻¹/2)(20 s) = 1571 rad ≈ 1.57 × 10³ rad. (Check: α = −157.1/20 = −7.85 rad s⁻²; Δθ = ω_i t + ½αt² = 3142 − 1571 = 1571 rad.)
6 A system of small masses rotates about a fixed axis. Which single change doubles its moment of inertia about that axis? HL
Answer and reasoning
Doubling each mass, with all distances kept unchanged — I = Σmr². Doubling every m with every r unchanged doubles every term, so I doubles.
Doubling each distance from the axis, with all masses unchanged — A student who thinks I is proportional to r picks this. Because I = Σmr², doubling every distance multiplies I by 2² = 4.
Doubling the angular speed at which the system is rotating — A student who thinks moment of inertia depends on how fast the body spins picks this. I depends only on the masses and their distances from the axis; doubling ω doubles L, not I.
Adding a mass equal to the total mass, placed on the axis — A student who treats moment of inertia as mass alone picks this. A mass on the axis has r = 0, so it adds nothing to Σmr²; I is unchanged.
7 A flywheel of moment of inertia 0.50 kg m² is driven by a belt that exerts a tangential force of 40 N on its rim, 0.30 m from the axis. A frictional torque of 2.0 N m acts at the bearing, opposing the rotation. What is the angular acceleration of the flywheel? HL
Answer and reasoning
24 rad s⁻² — A student who uses the driving torque alone gets 12/0.50 = 24 rad s⁻². τ in τ = Iα is the resultant torque, so the 2.0 N m friction torque must be subtracted.
20 rad s⁻² — Driving torque = Fr = 40 × 0.30 = 12 N m. The friction torque has the opposite sense, so the resultant torque is 12 − 2.0 = 10 N m, and α = τ/I = 10/0.50 = 20 rad s⁻².
28 rad s⁻² — A student who adds the torques regardless of sense gets (12 + 2.0)/0.50 = 28 rad s⁻². The friction torque opposes the rotation, so it reduces the resultant torque.
76 rad s⁻² — A student who uses the belt force as if it were a torque gets (40 − 2.0)/0.50 = 76 rad s⁻². The force must be multiplied by its distance from the axis: τ = 40 × 0.30 = 12 N m.
Working Driving torque τ₁ = Fr = (40 N)(0.30 m) = 12 N m. Friction torque 2.0 N m in the opposite sense. Resultant τ = 12 − 2.0 = 10 N m. α = τ/I = (10 N m)/(0.50 kg m²) = 20 rad s⁻².
8 A rigid body rotates about a fixed axis. Its angular momentum about that axis is determined by which quantities? HL
Answer and reasoning
Its mass and angular speed, however the mass is arranged — A student who treats moment of inertia as mass alone picks this. Two bodies of the same mass spinning at the same rate have different angular momenta if their mass is at different distances from the axis.
Its mass and the linear speed of its centre of mass — A student who confuses angular momentum with linear momentum mv picks this. A wheel spinning on a fixed axle has a stationary centre of mass but a non-zero angular momentum Iω.
The resultant torque that is acting on it at that instant — A student who links the amount of rotation to the torque currently acting picks this. Torque sets the RATE of change of angular momentum; a freely spinning body with no torque still has L = Iω.
Its moment of inertia about that axis and its angular speed — L = Iω. The moment of inertia carries the information about the mass and how it is distributed about the axis; the angular speed says how fast the body turns.
9 A diver leaves a springboard rotating slowly with her body straight. In the air she pulls into a tight tuck and rotates much faster. Air resistance is negligible. Why does her angular speed increase? HL
Answer and reasoning
Her moment of inertia decreases while her angular momentum stays constant. — In the air no external torque acts about her centre of mass, so her angular momentum Iω is constant. Tucking brings mass closer to the axis, reducing I, so ω increases.
Pulling in her limbs exerts an external torque on her, which speeds her up. — A student who thinks any increase in angular speed needs an external torque picks this. The forces that pull her limbs in are internal; they cannot change her total angular momentum. The spin rate rises because I falls.
Her weight exerts a torque about her centre of mass as she falls. — A student who links the fall to the faster spin picks this. Her weight acts at her centre of mass, so it has no torque about that point; it accelerates her centre of mass downward but does not change her spin.
Her angular momentum increases as she falls and her speed increases. — A student who confuses angular momentum with linear momentum picks this. Her linear momentum increases as she falls, but her angular momentum about her centre of mass is constant because no torque acts about it.
10 A wheel of moment of inertia 0.50 kg m² rotates counter-clockwise at 12 rad s⁻¹. A torque acts on it for 2.0 s, after which the wheel rotates clockwise at 4.0 rad s⁻¹. What is the magnitude of the average torque on the wheel? HL
Answer and reasoning
2.0 N m — A student who takes the change in angular speed as the larger value minus the smaller gets 0.50 × (12 − 4.0)/2.0 = 2.0 N m. The wheel reverses: with counter-clockwise positive, Δω = (−4.0) − (+12) = −16 rad s⁻¹, so the average torque has magnitude 0.50 × 16/2.0 = 4.0 N m.
1.0 N m — A student who takes the change in angular momentum to be the final angular momentum uses 0.50 × 4.0 = 2.0 N m s and gets 1.0 N m. The initial angular momentum must be subtracted: ΔL = L_final − L_initial.
4.0 N m — Taking counter-clockwise as positive, ΔL = I(ω_f − ω_i) = 0.50 × (−4.0 − 12) = −8.0 N m s. τ = ΔL/Δt = −8.0/2.0, a torque of 4.0 N m in the clockwise sense.
8.0 N m — A student who equates the torque with the angular impulse gives 0.50 × 16 = 8.0 as the torque. 8.0 N m s is the angular impulse; dividing by the 2.0 s gives the average torque, 4.0 N m.
Working Take counter-clockwise as positive: ω_i = +12 rad s⁻¹, ω_f = −4.0 rad s⁻¹. ΔL = I(ω_f − ω_i) = (0.50 kg m²)(−16 rad s⁻¹) = −8.0 N m s. τ = ΔL/Δt = −8.0/2.0 = −4.0 N m, i.e. 4.0 N m clockwise.
Read the ones marked not yet in Learn, then Verify.
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17 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 A mechanic tightens a nut with a spanner. She applies a force of 45 N to the handle at a point 0.30 m from the axis of the nut. The force makes an angle of 70° with the line joining the axis to the point where the force is applied. What is the magnitude of the torque of this force about the axis? HL
Answer and reasoning
4.62 N m — A student who uses cos θ, as in the equation for work, gets 45 × 0.30 × cos 70° = 4.62 N m. The turning component of the force is the one perpendicular to the line from the axis, F sin θ.
12.7 N m — τ = Fr sin θ, where θ is the angle between the force and the line from the axis to the point of application: τ = 45 × 0.30 × sin 70° = 13.5 × 0.940 = 12.7 N m.
13.5 N m — A student who multiplies force by distance and ignores the angle gets 45 × 0.30 = 13.5 N m. That is the torque only when the force is at 90° to the line from the axis; at 70° it is smaller.
10.4 N m — A student whose calculator is in radian mode gets sin 70 = 0.774 and 13.5 × 0.774 = 10.4 N m. The angle is in degrees; sin 70° = 0.940.
Working τ = Fr sin θ = (45 N)(0.30 m)(sin 70°) = 13.5 N m × 0.940 = 12.7 N m.
2 A uniform plank of length 4.0 m and weight 150 N rests on a pivot 1.2 m from end A. A load of weight 90 N hangs from the other end, B. The weight of the plank acts at its midpoint. What weight must hang from end A for the plank to rest horizontally in equilibrium? HL
Answer and reasoning
310 N — Take torques about the pivot. The plank's weight acts 0.80 m from the pivot on the B side and the load at B is 2.8 m from it: 150 × 0.80 + 90 × 2.8 = 372 N m in one sense. W × 1.2 = 372, so W = 310 N.
210 N — A student who ignores the plank's own weight gets W × 1.2 = 90 × 2.8, so W = 210 N. The plank's weight acts at its midpoint, 0.80 m from the pivot, and has a torque of 120 N m that must also be balanced.
240 N — A student who balances forces rather than torques makes the weight at A equal to 150 + 90 = 240 N. Equilibrium needs the torques about the pivot to balance, and each force's distance from the pivot matters.
550 N — A student who uses positions measured from end A as the distances takes 2.0 m for the plank's weight and 4.0 m for the load at B: 1.2W = 150 × 2.0 + 90 × 4.0 = 660 N m, so W = 550 N. Distances must be measured from the pivot: 0.80 m for the plank's weight and 2.8 m for the load.
Working Take torques about the pivot (the pivot force then has no torque). Midpoint is 2.0 m from A, i.e. 0.80 m from the pivot on the B side; B is 2.8 m from the pivot. Torque in the B sense: (150 N)(0.80 m) + (90 N)(2.8 m) = 120 + 252 = 372 N m. Torque in the A sense: W(1.2 m). Rotational equilibrium: 1.2W = 372, so W = 310 N.
3 A uniform bar lies at rest on frictionless horizontal ice. Two horizontal forces of equal, constant magnitude are then applied, one at each end of the bar, in opposite directions; each force is kept perpendicular to the bar at all times. Air resistance is negligible. Which statement about the bar while the forces act is correct? HL
Answer and reasoning
It stays at rest and does not turn. — A student who treats zero resultant force as full equilibrium picks this. Zero resultant force keeps the centre of mass at rest, but the two forces act along different lines and turn the bar in the same sense, so there is a resultant torque and the bar gains angular speed.
Its angular speed increases steadily. — The forces act along different lines and turn the bar in the same sense, so the resultant torque is not zero; it stays constant because each force stays perpendicular to the bar, so the angular acceleration α = τ/I is constant and the angular speed increases steadily. The forces are equal and opposite, so the resultant force is zero and the centre of mass stays at rest.
It rotates at a constant angular speed. — A student who thinks a steady torque keeps a body turning at a steady rate picks this. A constant resultant torque gives a constant angular acceleration (α = τ/I), so the angular speed keeps increasing for as long as the forces act.
Its centre of mass accelerates. — A student who thinks off-centre forces move a body as well as turn it picks this. The motion of the centre of mass depends only on the resultant force, a_cm = F/M, and here the equal and opposite forces give F = 0, so the centre of mass stays at rest; the forces only make the bar turn.
4 A bicycle travels along a level road at a constant 7.0 m s⁻¹. Its wheels, each of diameter 0.70 m, roll without slipping. What is the angular speed of each wheel? HL
Answer and reasoning
10.0 rad s⁻¹ — A student who uses the diameter as the radius gets 7.0/0.70 = 10.0 rad s⁻¹. The distance from the axle to the rim is the radius, 0.35 m.
3.18 rad s⁻¹ — A student who divides by the circumference gets 7.0/(2π × 0.35) = 3.18. That is the number of revolutions per second, not the angular speed in rad s⁻¹; multiply by 2π to get 20.0 rad s⁻¹.
7.00 rad s⁻¹ — A student who treats linear and angular speed as the same quantity gives 7.00 rad s⁻¹. They are linked by v = ωr, so ω = v/r = 20.0 rad s⁻¹.
20.0 rad s⁻¹ — For rolling without slipping v = ωr, with r = 0.70/2 = 0.35 m. ω = v/r = 7.0/0.35 = 20.0 rad s⁻¹.
Working Rolling without slipping: v = ωr, r = 0.70 m/2 = 0.35 m. ω = v/r = (7.0 m s⁻¹)/(0.35 m) = 20.0 rad s⁻¹.
5 A wheel starts from rest and turns with a uniform angular acceleration of 3.0 rad s⁻². What is its angular speed at the moment it completes 8.0 revolutions? HL
Answer and reasoning
6.93 rad s⁻¹ — A student who uses 8.0 revolutions as 8.0 rad gets ω_f = √(2 × 3.0 × 8.0) = 6.93 rad s⁻¹. Each revolution is 2π rad, so Δθ = 50.3 rad.
12.3 rad s⁻¹ — A student who drops the factor 2 uses ω_f² = αΔθ = 3.0 × 50.3 and gets 12.3 rad s⁻¹. The equation is ω_f² = ω_i² + 2αΔθ.
8.68 rad s⁻¹ — A student who finds the time (t = √(2Δθ/α) = 5.79 s) and then divides the angle by the time gets 50.3/5.79 = 8.68 rad s⁻¹. That is the average angular speed; starting from rest, the final value is twice as large.
6 A turntable starts from rest and rotates with uniform angular acceleration. During the first 2.0 s it turns through an angle θ. Through what angle does it turn during the next 2.0 s? HL
Answer and reasoning
θ, as equal time intervals give equal angles turned — A student who treats the angular speed as constant picks this. The turntable is accelerating, so it turns faster in the second interval and covers a larger angle.
2θ, as its angular speed at the end has doubled — A student who scales by the final angular speed of each interval picks this. The angle depends on the AVERAGE angular speed in each interval: ½ω₁ in the first and 1½ω₁ in the second, a ratio of 3.
3θ, as the angle turned from rest reaches 4θ after 4.0 s — From rest, Δθ = ½αt², so the angle turned from the start is proportional to t². After 4.0 s it is (4.0/2.0)² × θ = 4θ. The angle turned in the second interval is 4θ − θ = 3θ.
4θ, as the angle is proportional to t², and t doubles — A student who confuses the angle turned from the start with the angle turned in the interval picks this. 4θ is the angular position after 4.0 s; θ of that was turned in the first 2.0 s, leaving 3θ.
Working From rest with uniform α: θ(t) = ½αt². θ(2.0 s) = 2α = θ; θ(4.0 s) = 8α = 4θ. Angle turned between 2.0 s and 4.0 s = 4θ − θ = 3θ.
7 A thin hoop and a uniform solid disc have the same mass and the same radius. Each rotates about an axis through its centre, perpendicular to its plane. Which has the greater moment of inertia about this axis? HL
Answer and reasoning
Neither, because bodies of equal mass have equal moments of inertia. — A student who treats moment of inertia as the same thing as mass picks this. Moment of inertia depends on the distribution of mass about the axis, not just the amount.
The hoop, as all its mass is at the maximum distance from the axis. — Moment of inertia depends on how far the mass is from the axis. All of the hoop's mass is at distance R (I = MR²); much of the disc's mass is nearer the axis (I = ½MR²).
The disc, since its mass is spread out over its whole area. — A student who equates 'spread out' with 'covering more area' picks this. What matters is distance from the axis: the disc has much of its mass close to the centre, so its moment of inertia is smaller.
Whichever of the two is rotating at the greater angular speed. — A student who thinks moment of inertia depends on how fast a body spins picks this. For a rigid body about a given axis, I depends only on the mass distribution.
8 A uniform rod of mass M and length L can rotate either about an axis through its centre (I = ML²/12) or about an axis through one end (I = ML²/3), both perpendicular to the rod. The same resultant torque acts in each case. What is the ratio (angular acceleration about the centre)/(angular acceleration about the end)? HL
Answer and reasoning
2.00 — A student who takes I proportional to distance rather than distance squared argues that the mass is on average L/4 from a central axis and L/2 from an end axis, giving a ratio of 2. Because I depends on r², the ratio is 2² = 4.
0.25 — A student who thinks a larger moment of inertia gives a larger angular acceleration writes α ∝ I and gets (1/12)/(1/3) = 0.25. In fact α = τ/I, so the body with the smaller I has the larger α.
1.00 — A student who thinks moment of inertia depends only on mass concludes that the same rod has the same I about any axis. The formulas show I about the end is four times I about the centre.
4.00 — α = τ/I and τ is the same, so α_centre/α_end = I_end/I_centre = (ML²/3)/(ML²/12) = 4.00. Rotating about the centre puts the mass closer to the axis, so I is smaller and α larger.
Working α = τ/I; same τ, so α_centre/α_end = I_end/I_centre = (ML²/3)/(ML²/12) = 12/3 = 4.00.
9 A light rod of length 0.80 m rotates about an axis perpendicular to it through its midpoint. Small masses are fixed to the rod: 0.50 kg at end A, 0.20 kg at 0.20 m from end A, and 0.30 kg at end B. The centre of mass of the system is 0.12 m from the axis. What is the moment of inertia of the system about the axis? HL
Answer and reasoning
3.6 × 10⁻¹ kg m² — A student who does not square the distances gets 0.50(0.40) + 0.20(0.20) + 0.30(0.40) = 0.360. That quantity has unit kg m, not kg m²; each distance must be squared.
2.0 × 10⁻¹ kg m² — A student who measures distances from end A gets 0.50(0)² + 0.20(0.20)² + 0.30(0.80)² = 0.200 kg m². That is the moment of inertia about an axis through A; here the axis is at the midpoint, so r = 0.40, 0.20 and 0.40 m.
1.4 × 10⁻¹ kg m² — Distances from the axis are 0.40 m, 0.20 m and 0.40 m. I = Σmr² = 0.50(0.40)² + 0.20(0.20)² + 0.30(0.40)² = 0.080 + 0.008 + 0.048 = 0.136 kg m². The centre-of-mass position is not needed.
1.4 × 10⁻² kg m² — A student who places all 1.0 kg at the centre of mass gets 1.0 × (0.12)² = 0.0144 kg m². The centre-of-mass model applies to linear motion; for rotation each mass contributes mr² at its own distance.
Working Distances from the axis at the midpoint: A at 0.40 m, the 0.20 kg mass at 0.40 − 0.20 = 0.20 m, B at 0.40 m. I = Σmr² = (0.50 kg)(0.40 m)² + (0.20 kg)(0.20 m)² + (0.30 kg)(0.40 m)² = 0.080 + 0.008 + 0.048 = 0.136 kg m² = 1.4 × 10⁻¹ kg m².
10 A uniform disc of mass 3.0 kg and radius 0.20 m rotates at 240 revolutions per minute about an axis through its centre, perpendicular to its plane. For this axis, I = ½MR². What is the angular momentum of the disc? HL
Answer and reasoning
3.02 kg m² s⁻¹ — A student who uses the point-mass formula I = MR² for the disc gets I = 0.12 kg m² and L = 3.02 kg m² s⁻¹. Much of the disc's mass is nearer the axis than R; the given formula is I = ½MR².
14.4 kg m² s⁻¹ — A student who uses 240 rev min⁻¹ as if it were 240 rad s⁻¹ gets 0.060 × 240 = 14.4. Convert first: ω = 240 × 2π/60 = 25.1 rad s⁻¹.
1.51 kg m² s⁻¹ — I = ½ × 3.0 × 0.20² = 0.060 kg m² and ω = 240 × 2π/60 = 25.1 rad s⁻¹. L = Iω = 0.060 × 25.1 = 1.51 kg m² s⁻¹.
7.54 kg m² s⁻¹ — A student who does not square the radius gets I = ½ × 3.0 × 0.20 = 0.30 and L = 0.30 × 25.1 = 7.54. The formula is I = ½MR², so R must be squared.
Working I = ½MR² = ½(3.0 kg)(0.20 m)² = 0.060 kg m². ω = 240 × 2π/60 = 25.13 rad s⁻¹. L = Iω = (0.060 kg m²)(25.13 rad s⁻¹) = 1.51 kg m² s⁻¹.
11 A turntable of moment of inertia 0.020 kg m² rotates freely at 3.0 rad s⁻¹ on a frictionless bearing. A disc of moment of inertia 0.010 kg m², not rotating, is dropped onto it so that both share the same axis. Friction between them soon makes them rotate together. What is their common angular speed? HL
Answer and reasoning
2.0 rad s⁻¹ — No external torque acts about the axis, so angular momentum is conserved: 0.020 × 3.0 = (0.020 + 0.010)ω, giving ω = 0.060/0.030 = 2.0 rad s⁻¹.
3.0 rad s⁻¹ — A student who thinks the angular speed is conserved when there is no external torque picks this. The moment of inertia has increased, so to keep Iω constant the angular speed must fall.
1.5 rad s⁻¹ — A student who averages the two angular speeds, (3.0 + 0)/2, gets 1.5 rad s⁻¹. The speeds must be weighted by the moments of inertia: the turntable has twice the disc's I.
2.4 rad s⁻¹ — A student who conserves kinetic energy, ½(0.020)(3.0)² = ½(0.030)ω², gets 2.4 rad s⁻¹. Friction between the discs dissipates energy, so kinetic energy is not conserved; angular momentum is.
Working No external torque about the axis: L conserved. I₁ω₁ + I₂ω₂ = (I₁ + I₂)ω → (0.020)(3.0) + (0.010)(0) = (0.030)ω → ω = 0.060/0.030 = 2.0 rad s⁻¹.
12 A student sits on a stool that rotates freely about a vertical axis. She holds a 2.5 kg mass in each hand at 0.80 m from the axis and rotates at 2.0 rad s⁻¹. She then pulls both masses in to 0.20 m from the axis. Take the moment of inertia of the student and stool, without the masses, as a constant 2.0 kg m², and treat the masses as point masses. What is her new angular speed? HL
Answer and reasoning
4.0 rad s⁻¹ — A student who does not square the distances gets I_i = 2.0 + 5.0 × 0.80 = 6.0 and I_f = 2.0 + 5.0 × 0.20 = 3.0, so ω_f = 4.0 rad s⁻¹. Each mass contributes mr².
4.7 rad s⁻¹ — I_i = 2.0 + 2 × 2.5 × 0.80² = 5.2 kg m²; I_f = 2.0 + 2 × 2.5 × 0.20² = 2.2 kg m². With no external torque, I_iω_i = I_fω_f, so ω_f = 5.2 × 2.0/2.2 = 4.7 rad s⁻¹.
3.1 rad s⁻¹ — A student who conserves kinetic energy, ½I_iω_i² = ½I_fω_f², gets 3.1 rad s⁻¹. The student does work pulling the masses in, so kinetic energy increases; the conserved quantity is angular momentum.
2.0 rad s⁻¹ — A student who thinks angular speed stays constant without an external torque picks this. The moment of inertia has fallen from 5.2 to 2.2 kg m², so ω must rise to keep Iω constant.
Working I_i = 2.0 + 2(2.5 kg)(0.80 m)² = 2.0 + 3.2 = 5.2 kg m². I_f = 2.0 + 2(2.5 kg)(0.20 m)² = 2.0 + 0.20 = 2.2 kg m². No external torque about the axis: I_iω_i = I_fω_f → ω_f = (5.2)(2.0)/2.2 = 4.7 rad s⁻¹.
13 A wheel of moment of inertia 0.25 kg m² rotates counter-clockwise at 40 rad s⁻¹. A clockwise torque is then applied. A graph of the magnitude of this torque against time rises in a straight line from 0 at t = 0 to 4.0 N m at t = 0.50 s, stays at 4.0 N m until t = 1.50 s, then falls in a straight line to 0 at t = 2.0 s. No other torque acts. What is the angular speed of the wheel at t = 2.0 s? HL
Answer and reasoning
64.0 rad s⁻¹ — A student who ignores the sense of the torque adds the angular impulse: (10 + 6.0)/0.25 = 64.0 rad s⁻¹. The torque is clockwise and the wheel turns counter-clockwise, so the torque reduces the angular momentum.
24.0 rad s⁻¹ — A student who takes the angular impulse as the final angular momentum gets 6.0/0.25 = 24.0 rad s⁻¹. 24.0 rad s⁻¹ is the CHANGE in angular speed; the final value is 40 − 24 = 16.0 rad s⁻¹.
8.00 rad s⁻¹ — A student who multiplies the peak torque by the whole time uses 4.0 × 2.0 = 8.0 N m s and gets (10 − 8.0)/0.25 = 8.00 rad s⁻¹. The average torque is less than the peak; the angular impulse is the area under the graph, 6.0 N m s.
16.0 rad s⁻¹ — Angular impulse = area under the graph = ½(0.50)(4.0) + (1.0)(4.0) + ½(0.50)(4.0) = 6.0 N m s, clockwise. Initial L = 0.25 × 40 = 10 kg m² s⁻¹ counter-clockwise, so final L = 10 − 6.0 = 4.0 and ω = 4.0/0.25 = 16.0 rad s⁻¹, still counter-clockwise.
Working Angular impulse = area under τ–t graph = ½(0.50 s)(4.0 N m) + (1.00 s)(4.0 N m) + ½(0.50 s)(4.0 N m) = 1.0 + 4.0 + 1.0 = 6.0 N m s (clockwise). Take counter-clockwise as positive: L_i = (0.25)(40) = 10 kg m² s⁻¹; L_f = 10 − 6.0 = 4.0 kg m² s⁻¹; ω_f = 4.0/0.25 = 16.0 rad s⁻¹, counter-clockwise.
14 A solid cylinder of mass 2.0 kg and radius 0.10 m is released from rest and rolls without slipping down a slope, descending a vertical height of 0.90 m. For the cylinder, I = ½MR². Take g = 9.8 m s⁻² and ignore energy dissipation. What is the speed of its centre of mass at the bottom of the slope? HL
Answer and reasoning
4.2 m s⁻¹ — A student who ignores rotational kinetic energy uses Mgh = ½Mv² and gets v = √(2gh) = 4.2 m s⁻¹. That is the speed of a body sliding without friction; a rolling cylinder puts a third of its kinetic energy into rotation.
3.4 m s⁻¹ — Mgh = ½Mv² + ½Iω² with ω = v/R and I = ½MR², so Mgh = ½Mv² + ¼Mv² = ¾Mv². v = √(4gh/3) = √(4 × 9.8 × 0.90/3) = 3.4 m s⁻¹.
3.0 m s⁻¹ — A student who writes rotational kinetic energy as Iω² (no ½) gets Mgh = ½Mv² + ½Mv², so v = √(gh) = 3.0 m s⁻¹. Rotational kinetic energy is ½Iω² = ¼Mv² here.
1.7 m s⁻¹ — A student who does not square R in I = ½MR² gets I = 0.10 and ½Iω² = 5v², so 17.64 = v² + 5v² and v = 1.7 m s⁻¹. With R squared, I = 0.010 kg m² and ½Iω² = 0.50v².
Working E_p lost = Mgh = (2.0 kg)(9.8 m s⁻²)(0.90 m) = 17.64 J. Rolling without slipping: ω = v/R. E_k = ½Mv² + ½(½MR²)(v/R)² = ¾Mv² = 1.5v² (in J, v in m s⁻¹). 1.5v² = 17.64 → v² = 11.76 → v = 3.4 m s⁻¹.
15 A skater spinning on ice with negligible friction pulls her arms in towards her body and spins faster. How do her angular momentum L and her rotational kinetic energy E_k change? HL
Answer and reasoning
L is unchanged; E_k is unchanged, since no external work is done. — A student who assumes kinetic energy is conserved whenever angular momentum is picks this. E_k = L²/2I: with L fixed and I smaller, E_k must rise. Internal forces (her muscles) do the work.
L increases, because her mass is the same and she spins faster; E_k increases. — A student who thinks the moment of inertia depends only on mass assumes I is unchanged, so faster spin means more L. Pulling her arms in reduces I; L = Iω stays constant.
L is unchanged; E_k increases, as she does work pulling her arms in. — No external torque acts, so L = Iω is constant. E_k = L²/2I, and I decreases, so E_k increases. The extra energy comes from the work her muscles do pulling her arms inward.
L increases, as pulling her arms in exerts a torque on her; E_k increases. — A student who thinks any spin-up needs an external torque picks this. The forces pulling her arms in are internal and cannot change her total angular momentum; ω rises because I falls.
16 A thin hoop of mass 3.0 kg (I = MR²) and a solid cylinder of mass 1.0 kg (I = ½MR²), with equal radii, are released together from rest at the top of the same slope. Both roll without slipping. Which reaches the bottom first? HL
Answer and reasoning
The cylinder, as a smaller fraction of its E_k is rotational. — For rolling without slipping, E_k = ½Mv²(1 + I/MR²). The cylinder (I/MR² = ½) has v² = 4gh/3 at each height; the hoop (I/MR² = 1) has v² = gh. The cylinder is faster at every point and arrives first; the masses cancel.
The hoop, because it is the heavier and so it accelerates faster. — A student who believes heavier objects move down faster picks this. A larger weight comes with a proportionally larger inertia, so mass cancels; only the mass distribution (I/MR²) matters.
They arrive together, as the acceleration does not depend on mass. — A student who over-generalises free fall picks this. The mass does cancel, but the shapes differ: the hoop puts half its kinetic energy into rotation, the cylinder only a third.
The hoop, because its larger moment of inertia keeps it rolling faster. — A student who reads 'inertia' as a tendency to keep going faster picks this. A larger I/MR² means more of the lost E_p goes into rotation, so the hoop gains LESS speed.
17 A thin hoop (I = MR²) and a uniform solid disc (I = ½MR²) have the same mass M and the same radius R. Each is mounted on a frictionless axle through its centre, perpendicular to its plane, and is initially at rest. The same constant resultant torque τ is then applied to each. Which statement correctly compares their subsequent motion? HL
Answer and reasoning
The hoop has twice the angular acceleration of the disc, because its larger moment of inertia responds more to the torque. — A student who treats moment of inertia as a measure of how strongly a body responds to a torque picks this. Moment of inertia is resistance to angular acceleration: α = τ/I, so the larger I of the hoop gives it the smaller angular acceleration, half that of the disc.
After equal times the disc is turning faster, because its angular acceleration is twice that of the hoop. — From τ = Iα, α = τ/I. The disc's moment of inertia is half the hoop's, so for the same torque its angular acceleration is twice as large: α_disc = 2τ/MR², α_hoop = τ/MR². Starting from rest, ω = αt, so after any equal time the disc has the greater angular speed.
The two angular accelerations are equal, because the bodies have equal mass and the same torque acts on each. — A student who takes moment of inertia to be just the rotational name for mass picks this. Equal masses do not give equal moments of inertia: I depends on how the mass is distributed about the axis, and the hoop's mass is all at radius R, giving it twice the disc's I and half its angular acceleration.
Each body turns at a constant angular speed, and the two speeds are equal because the torques are equal. — A student who thinks a steady torque produces a steady angular speed picks this. A resultant torque produces angular acceleration, not a fixed angular speed: while τ acts, each body's angular speed keeps increasing, at a rate α = τ/I that is different for the two bodies.
That was your twenty minutes. Real practice on A.4 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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