Rigid body mechanics is one of the highest-yield HL-only topics in the new IB Physics syllabus. Topic A.4 takes everything you learned about translation in A.1–A.3 and rebuilds it for rotation: torque replaces force, moment of inertia replaces mass, angular velocity replaces velocity, and angular momentum replaces linear momentum. The structural parallels make A.4 surprisingly learnable — once you've memorised the linear-to-rotational analogy table, half the topic falls out for free. The other half is the geometry: identifying lever arms, perpendicular distances, and the right moment of inertia for the right shape.
This cheatsheet condenses every formula, trick and trap from Topic A.4 Rigid Body Mechanics into one page. It covers torque and rotational equilibrium, angular SUVAT (with the linking equations $v = r\omega$, $a_t = r\alpha$), the standard moment-of-inertia table, $\tau = I\alpha$ for pulleys and rolling objects, conservation of angular momentum, rotational KE and the energy-conservation method for rolling down inclines. Scroll to the bottom for the printable PDF and the full Photon Academy library.
§1 — Torque A.4 HL
Torque:$\tau = F r \sin\theta$
Equivalent form:Torque $=$ force $\times$ perpendicular distance from pivot to line of action
Rotational equilibrium:$\sum \tau = 0$
TrickThe perpendicular distance from the pivot to the line of action of the force is $r\sin\theta$ — often the most direct route. Draw the line of action and drop a perpendicular from the pivot.
Trap$\theta$ is the angle between the force vector and the lever arm (line from pivot to point of application), not the angle between the force and any surface.
NoteMaximum torque occurs at $\theta = 90°$ (lever arm perpendicular to force, $\tau = Fr$). Zero torque when $\theta = 0°$ — the line of action passes through the pivot.
Taking moments — equilibrium
Choose a pivot at an unknown force to eliminate it from the equation. Then $\sum \tau_{\text{CCW}} = \sum \tau_{\text{CW}}$.
§2 — Angular Kinematics (Rotational SUVAT) A.4 HL
| Linear | Rotational | Variable missing |
|---|
| $v = u + at$ | $\omega_f = \omega_i + \alpha t$ | $\Delta\theta$ |
| $s = \tfrac{1}{2}(u + v)t$ | $\Delta\theta = \tfrac{1}{2}(\omega_i + \omega_f)t$ | $\alpha$ |
| $s = ut + \tfrac{1}{2}at^2$ | $\Delta\theta = \omega_i t + \tfrac{1}{2}\alpha t^2$ | $\omega_f$ |
| $v^2 = u^2 + 2as$ | $\omega_f^2 = \omega_i^2 + 2\alpha\Delta\theta$ | $t$ |
Linking equations: $v = r\omega$, $a_t = r\alpha$, $a_c = r\omega^2$.
Trap$\Delta\theta$ must be in radians. If the problem gives revolutions $n$, convert: $\Delta\theta = 2\pi n$.
TrickWhen time is not given and not asked for, jump straight to $\omega_f^2 = \omega_i^2 + 2\alpha\Delta\theta$.
§3 — Moment of Inertia A.4 HL
Definition: $I = \sum m_i r_i^2$ — sum of mass × (distance from axis)² for every particle.
| Shape | Axis | $I$ |
|---|
| Solid disc / cylinder | Central, along axis of symmetry | $\tfrac{1}{2} M R^2$ |
| Thin ring (hoop) | Central, along axis of symmetry | $M R^2$ |
| Solid sphere | Through centre | $\tfrac{2}{5} M R^2$ |
| Thin rod | Through centre, perpendicular | $\tfrac{1}{12} M L^2$ |
| Thin rod | Through end, perpendicular | $\tfrac{1}{3} M L^2$ |
Note$I$ depends on both total mass and how the mass is distributed relative to the axis. More mass further from the axis means larger $I$.
Trap$I$ changes if the axis changes, even for the same object — compare the two thin-rod values above. Always state which axis is being used.
TrickFor a system of discrete masses, just sum $m r^2$ for each piece: $I = m_1 r_1^2 + m_2 r_2^2 + \ldots$
§4 — Newton's Second Law for Rotation A.4 HL
$$\tau_{\text{net}} = I \alpha \quad \Longleftrightarrow \quad F = m a$$
Angular acceleration: $\alpha = \dfrac{\tau_{\text{net}}}{I}$.
TrickFor a pulley problem: write $F = ma$ for each block, $\tau = I\alpha$ for the pulley, and use $a = r\alpha$ to link them. Solve simultaneously — you'll get $a$, $\alpha$ and the tensions.
Trap"Net torque is zero" does not mean "no torques act" — it means the torques balance. Always use the net torque in $\tau = I\alpha$.
§5 — Angular Momentum & Angular Impulse A.4 HL
Angular momentum:$L = I \omega$
Conservation:$I_1 \omega_1 = I_2 \omega_2$ (no external torque)
Angular impulse:$\Delta L = \tau \, \Delta t$
TrickFor conservation problems: write $I_1 \omega_1 = I_2 \omega_2$, list what is given, what changes, and solve for the unknown $\omega$.
TrapAngular impulse units are N m s (newton metre seconds), NOT N m (torque) and NOT N s (linear impulse). This is a common Paper 1 MCQ trap.
NoteWhen a skater pulls in their arms, $I$ decreases ⇒ $\omega$ increases. Rotational KE also increases — the extra energy comes from the work done by the skater's muscles.
§6 — Rotational Kinetic Energy & Rolling A.4 HL
Rotational KE:$E_K = \tfrac{1}{2} I \omega^2 = \dfrac{L^2}{2 I}$
Rolling (no slip):$v = r\omega$ ⇒ $E_{K,\text{total}} = \tfrac{1}{2} M v^2 + \tfrac{1}{2} I \omega^2$
Solid cylinder rolling:$E_{K,\text{total}} = \tfrac{3}{4} M v^2$ (translational : rotational $= 2:1$)
TrickFor rolling down a slope, use energy conservation $Mgh = \tfrac{1}{2} M v^2 + \tfrac{1}{2} I \omega^2$, substitute $\omega = v/r$, and solve for $v$. No torque equations needed.
TrapA hollow object (ring, hollow sphere) has greater $I$ than a solid one of the same mass and radius, so it rolls more slowly down a slope (more energy goes to rotation, less to translation).
Note$E_K = L^2 / (2I)$ is extremely useful when comparing objects with the same angular momentum or same rotational KE.
§7 — Linear ↔ Rotational Analogy & Exam Attack Plan A.4 — All sections
| Linear | Rotational |
|---|
| $F$ (force, N) | $\tau$ (torque, N m) |
| $m$ (mass, kg) | $I$ (moment of inertia, kg m²) |
| $a$ (acceleration, m s⁻²) | $\alpha$ (angular acceleration, rad s⁻²) |
| $v$ (velocity, m s⁻¹) | $\omega$ (angular velocity, rad s⁻¹) |
| $p = mv$ (momentum, N s) | $L = I\omega$ (angular momentum, N m s) |
| $F\,\Delta t = \Delta p$ | $\tau\,\Delta t = \Delta L$ |
| $E_K = \tfrac{1}{2} m v^2$ | $E_K = \tfrac{1}{2} I \omega^2$ |
| $\sum F = 0$ (translational equilibrium) | $\sum \tau = 0$ (rotational equilibrium) |
| Question trigger | Reach for |
|---|
| "Beam balanced on a pivot, find unknown force" | $\sum \tau = 0$ — pick a pivot at an unknown |
| "Wheel spins up from rest, find $\omega$" | Angular SUVAT — convert revs to radians first |
| "Solid sphere / cylinder / ring, find $I$" | Look up $I$ from the standard table |
| "Pulley with mass attached" | $F = ma$ for the block + $\tau = I\alpha$ for the pulley + $a = r\alpha$ |
| "Skater pulls in arms" | Conservation of $L$: $I_1\omega_1 = I_2\omega_2$ |
| "Object rolls down a slope, find $v$ at bottom" | $Mgh = \tfrac{1}{2} M v^2 + \tfrac{1}{2} I \omega^2$, $\omega = v/r$ |
| "Compare hollow vs solid roll-down" | Larger $I$ ⇒ slower at bottom |
Worked Example — IB-Style Rolling Down a Slope
Question (HL Paper 2 style — 7 marks)
A solid uniform sphere of mass 0.80 kg and radius 0.050 m is released from rest at the top of an inclined plane and rolls (without slipping) down a vertical drop of 0.60 m. Take $g = 9.81$ m s⁻² and $I_{\text{sphere}} = \tfrac{2}{5} M R^2$. Calculate (a) the linear speed of the sphere at the bottom of the slope, and (b) the fraction of the kinetic energy that is rotational.
Solution
- Energy conservation: $M g h = \tfrac{1}{2} M v^2 + \tfrac{1}{2} I \omega^2$. (M1)
- Rolling without slipping: $\omega = v/R$. Substitute $I = \tfrac{2}{5} M R^2$:
$\tfrac{1}{2} I \omega^2 = \tfrac{1}{2} \cdot \tfrac{2}{5} M R^2 \cdot \dfrac{v^2}{R^2} = \tfrac{1}{5} M v^2$. (M1)
- Total KE: $\tfrac{1}{2} M v^2 + \tfrac{1}{5} M v^2 = \tfrac{7}{10} M v^2$. So $Mgh = \tfrac{7}{10} M v^2$ ⇒ $v = \sqrt{\dfrac{10 g h}{7}}$. (M1)(A1)
- $v = \sqrt{\dfrac{10 \times 9.81 \times 0.60}{7}} = \sqrt{8.41} = 2.90$ m s⁻¹. (A1) [part (a)]
- Fraction rotational $= \dfrac{\tfrac{1}{5} M v^2}{\tfrac{7}{10} M v^2} = \dfrac{2}{7} \approx 0.286 \;(28.6\%)$. (M1)(A1) [part (b)]
Examiner's note: Notice that mass and radius both cancel — the speed at the bottom only depends on $g$, $h$ and the shape (through the $\tfrac{2}{5}$ factor). For comparison, a solid cylinder ($I = \tfrac{1}{2}MR^2$) gives $v = \sqrt{4gh/3}$ (slower than the sphere), and a hollow ring ($I = MR^2$) gives $v = \sqrt{gh}$ (slowest). The sphere always wins.
Common Student Questions
Which angle do I use in $\tau = F r \sin\theta$?
The angle between the force vector and the lever arm — the line from the pivot to the point where the force is applied. NOT the angle the force makes with any surface or the ground. Maximum torque is at $\theta = 90°$ (force perpendicular to lever arm). Zero torque is at $\theta = 0°$ (line of action passes through the pivot). The fastest method in IB exams is $\tau = F \times $ perpendicular distance from pivot to line of action of force.
Why must $\Delta\theta$ be in radians for angular SUVAT?
Because the linking equations $v = r\omega$ and $a_t = r\alpha$ assume $\theta$ is in radians (their derivation uses arc length $s = r\theta$, which only works in radians). If a problem gives revolutions $n$, convert immediately: $\Delta\theta = 2\pi n$ radians. Putting degrees into $\omega_f^2 = \omega_i^2 + 2\alpha\Delta\theta$ is one of the fastest ways to lose marks in A.4 — it can change your answer by a factor of $(180/\pi)^2 \approx 3300$.
Why does a hollow sphere roll down a slope slower than a solid one?
Because the hollow sphere has a larger moment of inertia for the same mass and radius. Energy conservation gives $Mgh = \tfrac{1}{2} M v^2 + \tfrac{1}{2} I \omega^2$. With $\omega = v/r$ and a larger $I$, more of the released gravitational PE goes into rotational KE and less into translational KE — so $v$ is smaller at the bottom. Order from fastest to slowest: solid sphere ($\tfrac{2}{5} MR^2$), solid cylinder ($\tfrac{1}{2} MR^2$), thin ring ($MR^2$).
What are the units of angular momentum?
N m s (newton metre seconds) — NOT N m (which is torque) and NOT N s (which is linear impulse). This is a frequent Paper 1 MCQ trap. Equivalent units: kg m² s⁻¹. Memorise: $L = I\omega$, so $[L] = [\text{kg m}^2][\text{rad s}^{-1}] = $ kg m² s⁻¹ $=$ N m s. Angular impulse $\tau \, \Delta t$ has the same units, because angular impulse equals $\Delta L$.
Why does an ice skater spin faster when she pulls in her arms?
Conservation of angular momentum. With no external torque, $L = I\omega$ is constant. Pulling arms in reduces $I$ (mass moves closer to the spin axis), so $\omega$ must increase. Note that her rotational KE also increases — the extra energy comes from the work she does with her muscles to pull her arms in against the centrifugal-like effect. This is a classic full-marks "explain" question.
What's NOT in this cheatsheet
This page gives you the formulas and the traps. The full Photon Academy A.4 Rigid Body Mechanics library (only available to enrolled students or via the resource library subscription) adds:
- Notes PDF — every concept worked through in full, with proofs and exam-style commentary on pulleys, ladders, and rolling problems.
- Tutorial booklet — IB-style questions sequenced from torque equilibrium through angular momentum conservation and rolling.
- Tutorial Solutions — full mark-scheme-style worked solutions with M1/A1/B1 annotations.
- Practice Solutions — extra past-paper-style problems with detailed walk-throughs (including P1A-style HL multiple choice).
- Cheatsheet PDF — print-ready, brand-formatted, the same one our students take into mock exams.
Get the printable PDF version
Same cheatsheet, formatted for A4 print — keep it next to your study desk. Free for signed-in users.
