AP Physics 1 linear and rotational motion sit on a single mathematical scaffold, and the candidates who internalise that scaffold early in their YÖS or TR-YÖS preparation tend to out-score peers who treat the two halves of mechanics as separate worlds. The exam format of AP Physics 1 — both the older pencil-paper sitting and the current digital form — leans on a small set of conversion relations that translate a translational problem into a rotational one, or vice versa. Master those relations, and a question that looks exotic on first read becomes a routine substitution. This article walks through the four conversion equations, the most common question types in which they appear, the rotational inertia tables candidates are expected to memorise, and a four-example problem set built around the kind of phrasing seen in YÖS-style physics modules and AP-style algebra-only items.
The four conversion equations every candidate must memorise
The entire bridge between linear and rotational motion rests on four equations, each of which can be derived by considering an arc length swept by a point on a rotating rigid body. The arc length s equals rθ, where r is the perpendicular distance from the rotation axis and θ is the angle in radians. Differentiate once with respect to time, and s/t becomes v = rω, where ω is the angular velocity. Differentiate again, and the tangential acceleration is at = rα, where α is the angular acceleration. A fourth relation, ac = rω², governs centripetal acceleration, which matters for any object moving in a circle. The AP Physics 1 equation sheet lists these forms, and any equivalent exam that targets university-level physics will either provide them on a formula sheet or test whether the candidate can reconstruct them — a YÖS-style problem on rolling motion, for example, will frequently quote v and ask for ω without printing rω on the page.
Once the four conversions are in muscle memory, the mechanical step in any problem is to identify which variables are linear and which are rotational, decide whether the problem is in disguise, and convert. A wheel of radius 0.30 m rolling at 6.0 m/s has ω = v/r = 6.0/0.30 = 20 rad/s. That single calculation is the engine of roughly a third of rotational kinematics items. Notice that the units work out cleanly: metres per second divided by metres gives 1/second, which is radians per second because radians are dimensionless. Candidates who treat radians as a real SI unit during unit analysis lose marks unnecessarily; the convention is to keep radians in the answer because the radian is the natural angular measure in any formula derived from arc length.
A second anchor: when an object rolls without slipping, the bottom of the object is instantaneously at rest, which means the linear speed of the centre of mass equals rω exactly. YÖS and TR-YÖS candidates preparing for the AP-style mechanics cluster will see this constraint embedded in problems about yo-yos, bicycle wheels, bowling balls, and cylinders on inclines. The rolling-without-slipping condition is not a separate formula so much as a physical requirement that locks v and ω together, and from that lock you can write conservation of energy, conservation of angular momentum, or Newton's second law in either linear or rotational form depending on which is more convenient.
Rotational kinematics versus linear kinematics: matching the template
The rotational equations are structurally identical to the linear equations with three substitutions: x becomes θ, v becomes ω, and a becomes α. The constant-acceleration set for linear motion is x = x₀ + v₀t + ½at², v = v₀ + at, and v² = v₀² + 2a(x − x₀). The rotational equivalents are θ = θ₀ + ω₀t + ½αt², ω = ω₀ + αt, and ω² = ω₀² + 2α(θ − θ₀). Most candidates already know the linear forms; the work is to swap letters and trust the algebra. In YÖS preparation, this is often the moment where a candidate who struggled with rotational questions in high school suddenly finds them tractable, because the underlying template is already known.
A worked example: a disc starts from rest and reaches 30 rad/s in 4.0 s with constant angular acceleration. The angular acceleration is α = (ω − ω₀)/t = 30/4.0 = 7.5 rad/s². The angle swept in that interval is θ = ½αt² = ½(7.5)(16) = 60 rad, which is roughly 9.5 full revolutions. If a point on the rim is 0.20 m from the axis, its linear speed at the end of the 4.0 s interval is v = rω = 0.20 × 30 = 6.0 m/s, and its tangential acceleration is at = rα = 0.20 × 7.5 = 1.5 m/s². This single problem exercises four of the five rotational kinematics skills: identifying α from ω and t, finding θ from α and t, converting ω to v, and converting α to at. Candidates who can run this chain in roughly three minutes have the core machinery locked in.
The trap in rotational kinematics is not the algebra but the geometry. A question will sometimes ask for the linear speed of a point partway out from the axis, and a candidate will read past the phrase "on a wheel of radius R, a point at distance R/2 from the centre." The relevant radius is R/2, not R. Always read the geometry sentence twice. The same caution applies to a problem that gives a wheel rolling at speed v and asks for the speed of a piece of chalk on the rim at the top of the wheel: that speed is 2v, because the top of a rolling wheel moves at twice the speed of the centre. This is a frequently tested edge case in both AP Physics 1 free-response questions and YÖS-style multiple-choice items, and it is the kind of detail that separates a confident 5 from a stuck 3.
Rotational inertia I = Σmr² and the standard table
Rotational inertia, sometimes called the moment of inertia, plays the same role in rotational dynamics that mass plays in linear dynamics. For a discrete set of point masses, I = Σmr², where each mass is squared against its perpendicular distance to the axis. For continuous objects, the sum becomes an integral, but the AP Physics 1 course — and most YÖS-aligned physics modules — reduces the integral to a memorised table. The standard items are: a solid sphere, I = (2/5)MR²; a solid cylinder or disc, I = (2/5)MR² is wrong, it is (1/2)MR²; a hollow sphere or thin spherical shell, I = (2/3)MR²; a thin hoop or thin cylindrical shell, I = MR²; and a slender rod rotating about its centre, I = (1/12)ML². Two of those — solid sphere (2/5)MR² and solid cylinder (1/2)MR² — are by far the most common in exam settings, and the difference between them accounts for a meaningful fraction of the marks lost on rolling-object problems.
The parallel-axis theorem extends the table: I = Icm + Md², where d is the distance from the centre-of-mass axis to the new axis. A thin rod rotating about its end has I = (1/12)ML² + M(L/2)² = (1/3)ML². This is the only parallel-axis shift the AP Physics 1 syllabus typically requires. YÖS-style exam questions occasionally state the parallel-axis theorem implicitly: a door on a hinge is just a thin rod rotating about its end, and the moment of inertia is (1/3)ML². The skill is to recognise the geometry, not to derive the integral each time.
A subtle point: rotational inertia depends on the axis, and the same object can have very different I values about different axes. A solid cylinder about its symmetry axis is (1/2)MR², while the same cylinder about a diameter through its centre is (1/4)MR² + (1/12)ML² for a long cylinder, or (1/4)MR² for a thin disc. The AP Physics 1 exam will not require memorising both, but candidates should be alert to a problem that gives a hint about the axis in the diagram. A cylinder drawn with its circular face on the page, viewed end-on, is rotating about its symmetry axis. A cylinder drawn from the side is rotating about a perpendicular axis through its centre, and the formula is different. Reading the diagram is half the problem.
Newton's second law in rotational form: τ = Iα
The rotational twin of F = ma is τ = Iα, where τ is the net torque about the chosen axis, I is the rotational inertia about that axis, and α is the angular acceleration. Torque is defined as τ = rF sin θ, where θ is the angle between the lever arm vector and the force vector; the perpendicular lever arm r⊥ = r sin θ is often easier to spot in a diagram. Sign conventions matter: pick a rotation direction as positive, then assign positive or negative to each torque based on whether it tries to rotate the system in that direction. Two opposite torques of magnitudes 8 N·m and 5 N·m about the same axis give a net torque of 3 N·m in the direction of the larger one. YÖS candidates often skip the sign step and lose a sign in the answer, which is a quietly devastating habit once it becomes consistent.
Newton's second law for rotation is the engine of the Atwood-style problems that include pulleys. A pulley of mass M, radius R, and moment of inertia I has a tension T₁ on one side and T₂ on the other; the net torque is (T₂ − T₁)R, and α = (T₂ − T₁)R/I. The two hanging masses obey m₁g − T₁ = m₁a and T₂ − m₂g = m₂a, with the constraint a = Rα. The result is a system of three equations in three unknowns. YÖS-style problems of this kind are popular because they test three skills at once: free-body diagrams, rotational dynamics, and the linear-rotational constraint. The same machinery shows up in AP Physics 1 free-response items about reels, yo-yos, and bicycle-chain drives.
A tactical note for both exams: always draw the free-body diagram for the rotating object and label the lever arm for each force. A common error is to assume that the tension on both sides of a pulley is equal, which is only true for a massless, frictionless pulley with no rotational inertia. The moment the pulley has mass, the tensions differ, and the difference is what produces the angular acceleration. AP Physics 1 questions frequently ask the candidate to justify why the tensions are unequal, and YÖS-style questions sometimes give one tension and ask for the other. The formula is just T₂ − T₁ = Iα/R, and the rest is sign discipline.
Rolling objects, energy, and the question of slipping
A rigid object rolling without slipping is a two-degree-of-freedom system: the centre of mass translates with speed v, and the object rotates with angular speed ω = v/R. The total kinetic energy is the sum of translational and rotational parts: KE = ½Mv² + ½Iω². Substituting ω = v/R gives KE = ½Mv²(1 + I/MR²) = ½Mv²(1 + β), where β = I/MR² is a dimensionless shape factor. For a solid sphere, β = 2/5, so the total KE is 7/10 Mv². For a solid cylinder, β = 1/2, so the total KE is 3/4 Mv². For a hollow sphere, β = 2/3, so the total KE is 5/6 Mv². These fractions are useful because they tell you at a glance how much of the gravitational energy went into rotation versus translation when the object rolled down a ramp.
Consider a solid sphere released from rest at height h on a ramp. Conservation of energy gives Mgh = ½Mv²(1 + β), so v = √(2gh/(1 + β)). For a solid sphere, that is v = √(10gh/7), which is slower than the frictionless case v = √(2gh) by the factor √(7/10) ≈ 0.84. A solid cylinder reaches the bottom at v = √(4gh/3), a hollow sphere at v = √(6gh/5). The rank order — solid sphere fastest, then solid cylinder, then hollow sphere — is one of the most testable predictions in the entire rotational cluster, and the AP Physics 1 exam exploits it whenever a question offers three objects and asks which arrives first.
The slipping case is different. A solid sphere sliding down a frictionless ramp has v = √(2gh), the same as a point mass. Adding friction slows the sliding if the friction is kinetic, but the same friction can also start the object rotating, in which case energy is shared between translation and rotation. A problem that explicitly says "sliding without rolling" or "rolling without slipping" is steering you toward one of two branches, and the candidate who misses the word "without" picks the wrong one. This is exactly the kind of phrasing trap that the YÖS and TR-YÖS exams use to differentiate careful readers from confident guessers.
Angular momentum and the conservation principle
Angular momentum about a fixed axis is L = Iω, and the rotational form of Newton's second law is τnet = dL/dt. When the net external torque on a system is zero, the angular momentum is conserved: Iiωi = Ifωf. The textbook example is a figure skater pulling in her arms: reducing I increases ω, and the angular momentum is unchanged. In exam problems, the same principle shows up in a spinning platform onto which a person walks, a disc onto which a ring is dropped, a neutron star that contracts, and a satellite that tethers two modules together.
The direction of angular momentum is perpendicular to the plane of rotation, following the right-hand rule. AP Physics 1 limits the syllabus to two-dimensional situations, so the direction is either into the page or out of the page, and the only relevant question is whether the direction changes. YÖS physics questions occasionally introduce a third dimension in the form of a problem about precession, but those are out of scope for AP Physics 1 and for the typical YÖS physics module. The skill in two dimensions is to track the sign of L, not its vector components, and to recognise that a system with no external torque conserves the signed L.
A worked conservation problem: a disc of mass M, radius R, and moment of inertia (1/2)MR² rotates freely at 30 rad/s. A ring of mass M and radius R, with I = MR², is dropped coaxially onto the disc, and the two reach a common angular speed. Conservation of angular momentum gives (1/2)MR²(30) = ((1/2)MR² + MR²)ωf, so ωf = 30 × (1/2)/(3/2) = 10 rad/s. The kinetic energy before the drop is ½ × (1/2)MR² × 30² = 225 MR². After, the kinetic energy is ½ × (3/2)MR² × 10² = 75 MR². Two-thirds of the kinetic energy is lost to heat and sound during the inelastic interaction, a result worth noticing because conservation of angular momentum does not imply conservation of energy when the system is not isolated in the mechanical sense.
Worked examples in the YÖS-style format
Four examples, in escalating difficulty, build the bridge between AP Physics 1 mechanics and the kind of algebra-only problem a YÖS or TR-YÖS physics module will set. Each example is set up the way a test writer would set it up: short stem, clean numbers, one conversion or one formula in play per step.
Example 1 — direct conversion. A wheel of radius 0.25 m rolls without slipping at 4.0 m/s. Find (a) the angular speed of the wheel, (b) the linear speed of a point on the rim, and (c) the period of one revolution. The angular speed is ω = v/r = 4.0/0.25 = 16 rad/s. A point on the rim has linear speed v = rω = 0.25 × 16 = 4.0 m/s, which is just the centre-of-mass speed because the rim point at the bottom is at rest and the rim point at the top moves at 8.0 m/s. The period is T = 2π/ω = 2π/16 ≈ 0.39 s. This is the entry-level question that opens any rotational unit, and the candidate who cannot finish it in two minutes is not ready for the harder items.
Example 2 — centripetal acceleration. A point mass of 0.50 kg moves in a circle of radius 1.2 m at 6.0 rad/s. Find (a) the linear speed, (b) the centripetal acceleration, and (c) the centripetal force. The linear speed is v = rω = 1.2 × 6.0 = 7.2 m/s. The centripetal acceleration is ac = rω² = 1.2 × 36 = 43.2 m/s², or equivalently v²/r = 51.84/1.2 = 43.2 m/s². The centripetal force is F = ma = 0.50 × 43.2 = 21.6 N, directed toward the centre. The two formulas for centripetal acceleration are interchangeable, and picking the one with the variables you already know is a small but reliable time-saver.
Example 3 — rotational dynamics. A solid cylinder of mass 4.0 kg and radius 0.20 m is mounted on a frictionless axle. A string is wound around the cylinder and pulled with a steady 12 N force. Find (a) the torque, (b) the angular acceleration, and (c) the linear acceleration of the string as it unspools. The torque is τ = Fr = 12 × 0.20 = 2.4 N·m. The moment of inertia is I = (1/2)MR² = (1/2)(4.0)(0.04) = 0.08 kg·m². The angular acceleration is α = τ/I = 2.4/0.08 = 30 rad/s². The linear acceleration of the string is a = rα = 0.20 × 30 = 6.0 m/s². Notice that the answer does not require any reference to gravity because the only force doing work is the applied 12 N pull.
Example 4 — rolling down a ramp. A solid sphere, a solid cylinder, and a hollow sphere, all of the same mass and radius, are released from rest at the same height h on a ramp. Rank them by (a) speed at the bottom, (b) angular speed at the bottom, and (c) total kinetic energy at the bottom. From the energy analysis above, the solid sphere is fastest with v = √(10gh/7), the solid cylinder is next with v = √(4gh/3), and the hollow sphere is slowest with v = √(6gh/5). The angular speed follows the same rank order because ω = v/R is the same conversion for all three. The total kinetic energy is the same for all three objects, equal to Mgh, because they all started at the same height and ended at the bottom. This last result is often surprising to candidates, and the AP-style free-response prompt will sometimes ask the candidate to explain why the translational KE differs even though the total KE is the same.
Common pitfalls and how to avoid them
Five pitfalls account for most of the lost marks in this cluster, and each one is preventable with a small tactical habit. First, treating radians and degrees as if they are interchangeable mid-calculation. Pick radians at the start and never mix. Second, forgetting that v = rω is the tangential speed, not the speed of a specific point on the rim. A point on the rim has an instantaneous velocity that depends on where it is in the rotation: at the bottom of a rolling wheel, it is zero; at the top, it is 2vcm. Third, assuming T₁ = T₂ across a pulley. The tensions differ whenever the pulley has mass. Fourth, mixing up the moment of inertia formulas. A solid sphere is (2/5)MR², a solid cylinder is (1/2)MR², and a hollow sphere is (2/3)MR². Drill the table until it is automatic. Fifth, using ω in degrees per second inside a formula that expects radians per second. The radian is dimensionless, and the formulas on the AP equation sheet assume radians.
| Shape | I about symmetry axis | Shape factor β = I/MR² | Speed factor at bottom of ramp (v/√(2gh)) |
|---|---|---|---|
| Solid sphere | (2/5)MR² | 0.40 | √(10/14) ≈ 0.845 |
| Solid cylinder or disc | (1/2)MR² | 0.50 | √(3/4) ≈ 0.866 |
| Hollow sphere or thin shell | (2/3)MR² | 0.667 | √(5/6) ≈ 0.913 |
| Thin hoop or thin cylindrical shell | MR² | 1.00 | √(1/2) ≈ 0.707 |
| Slender rod about centre | (1/12)ML² | — | — |
The table is a one-page summary that covers roughly 70% of the rotational cluster as it appears in AP Physics 1 and in YÖS-style physics modules. The last column, the speed factor, drops out of the energy analysis developed above. Memorise the moment-of-inertia column, derive the speed factor if asked, and the table does most of the work for you.
How this maps onto the YÖS and TR-YÖS physics module
YÖS and TR-YÖS exams do not test rotational physics in the same depth as AP Physics 1, but the question types that appear in either YÖS-format module have a strong family resemblance to AP-style items. A rolling-wheel problem with a small twist, a pulley-with-mass problem, a conservation-of-angular-momentum problem, and a torque-arm problem are all within range. The scoring benefit of preparing the AP-style cluster for a YÖS exam is that the candidate develops reflexes for identifying which physics is in play, which is more often the bottleneck than the algebra itself. A typical TR-YÖS physics question might combine a 12 N force on a string wrapped around a 0.20 m cylinder with a 3.0 kg hanging mass on the string, and the candidate who can sketch the free-body diagram, write F = ma for the mass, write τ = Iα for the cylinder, and stitch the two together with a = rα has a reliable method regardless of the numbers.
The preparation strategy that pays off most in this cluster is short, repeated, mixed drills. Pick 10 problems: two direct conversions, two centripetal, two pulleys, two rolling, two angular-momentum. Solve them under timed conditions, with the equation sheet open. After each block, mark the misses, group them by skill gap, and pick the next block to address the gap. Within two weeks, the candidate's time-per-problem drops by roughly a third, and the miss rate drops with it. In my experience this works better than reading the textbook, because the cluster is small and the friction is in the execution, not the conceptual understanding.
For candidates balancing YÖS preparation with the AP Physics 1 exam, the order of operations is to learn the linear template first, then translate it letter-for-letter into the rotational template, then layer in the rotational-inertia table, then practice the four conversion equations until they are reflexive. The skill is mechanical and the rewards are concrete. A candidate who has internalised this scaffold can attack a YÖS physics question that mixes translation and rotation in roughly the time a peer spends re-reading the problem statement.
Conclusion and next steps
Linear and rotational motion in AP Physics 1 are the same physics with a small letter substitution, and the candidate who treats them as a single scaffold will find both the AP exam and the YÖS physics module easier than the candidate who treats them as two separate subjects. The four conversion equations, the rotational-inertia table, the rolling-without-slipping energy analysis, and the conservation-of-angular-momentum framework together cover the cluster, and roughly four weeks of focused mixed practice is enough to lock the skills in. Candidates preparing for the TR-YÖS physics module benefit from the same drills, because the underlying problem types overlap heavily with the AP-style items. TestPrep İstanbul's rotational-mechanics diagnostic is a natural starting point for candidates building a sharper preparation plan around v = rω and a = rα.