Topic 5.2: Connecting Linear and Rotational Motion

What Topic 5.2 requires

Topic 5.2 carries one learning objective, 5.2.A: describe the linear motion of a point on a rotating rigid system that corresponds to the rotational motion of that point, and vice versa. Note the two-way phrasing. You are expected to go from angular quantities to linear ones and back again.

Three essential knowledge statements support it.

• 5.2.A.1 states that for a point at a distance $r$ from a fixed axis of rotation, the linear distance $s$ traveled by the point as the system rotates through an angle $\Delta\theta$ is given by the equation $\Delta s = r\Delta\theta$.
• 5.2.A.2 states that derived relationships of linear velocity and of the tangential component of acceleration to their respective angular quantities are given by three equations: $s = r\theta$, $v = r\omega$, and $a_T = r\alpha$.
• 5.2.A.3 states that for a rigid system, all points within that system have the same angular velocity and angular acceleration.

The word derived in 5.2.A.2 is a hint about how the topic is assessed. These are not new laws of physics; they follow from the definition of the radian, and suggested skill 2.A asks you to walk that derivation.

One boundary statement applies, the same sentence that closes Topic 5.1: descriptions of the directions of rotation for a point or object are limited to clockwise and counterclockwise with respect to a given axis of rotation.

The suggested skills are 1.C, create qualitative sketches of graphs that represent features of a model or the behavior of a physical system; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim. Skill 2.C is the giveaway: comparing two locations in a single scenario is exactly what "two riders at different radii on the same merry-go-round" asks for, and that comparison is what suggested skill 2.C is listed for here.

Unit 5 is weighted at 10 to 15 percent of the multiple-choice section, with roughly 15 to 20 class periods suggested for the unit's six topics.

Everything here is in radians

Before any of these equations mean anything, the angle has to be in radians. That is not a stylistic preference; it is where the equations come from.

A radian is defined as arc length divided by radius. Turn that definition around and you get $s = r\theta$ directly, with $\theta$ in radians. Divide both sides by an elapsed time and you get $v = r\omega$. Divide the change in $v$ by an elapsed time and you get $a_T = r\alpha$. That three-line chain is the derivation skill 2.A wants, and every link in it depends on $\theta$ being the arc-length-over-radius ratio rather than a count of degrees.

Because a radian is one length divided by another, it has no physical dimension, which is why the units work out with nothing left over. Meters times radians gives meters, and meters times radians per second gives meters per second. The rad label simply drops out, and that vanishing act is a feature of the definition rather than sloppy bookkeeping.

Substitute degrees instead and every answer is wrong by a factor of $180/\pi$, about 57.3. A wheel of radius 0.50 m turned through a quarter circle carries a point along an arc of $(0.50)(\pi/2) = 0.785$ m, while feeding in 90 would claim 45 m. Convert revolutions and degrees to radians on the first line of your work.

Two conversions handle the usual cases: one revolution is $2\pi$ rad, and one rpm is $2\pi/60$ rad/s, about 0.105 rad/s. Topic 5.1 has the fuller conversion table.

What r means, and what it does not mean

Read essential knowledge 5.2.A.1 slowly: for **a point at a distance $r$ from a fixed axis of rotation**. The $r$ in these equations is the distance from the axis to the specific point you are asking about. It is not a property of the object.

This matters because the same letter $r$ elsewhere in physics often does mean the radius of something. On a rotating disk of radius 0.60 m, a point on the rim has $r = 0.60$ m, a point halfway out has $r = 0.30$ m, and a point on the axis itself has $r = 0$. Three different values of $r$ on one object, all at the same instant, all correct for their own point.

Three consequences worth stating on their own:

A point on the axis does not move. With $r = 0$, both $v = r\omega$ and $a_T = r\alpha$ give zero no matter how fast the system spins. The center of a spinning wheel goes nowhere.

Linear speed is proportional to distance from the axis. Double $r$ and you double $v$, provided $\omega$ is the same. That proportionality is the comparison suggested skill 2.C keeps asking you to make.

The object's outer radius only enters when the point of interest is on the rim. A pulley of radius 0.12 m has a rope over its rim, so the rope's speed uses $r = 0.12$ m. A dot painted at 0.06 m from the axle on that same pulley moves at half the rope's speed. Ask which point first, then choose $r$.

Rolling is the case where this is easiest to get wrong in the other direction. A wheel rolling on the ground has its contact point exactly one radius from the axle, so the wheel's own radius is the $r$ that converts its spin into how far it travels. That gets its own section below. It is still the distance-to-a-point rule, just applied to a point that happens to sit on the rim. That gets its own section below, and it is a special case rather than the definition.

Arc length, and what the equation sheet prints

Essential knowledge 5.2.A.1 gives the distance relation in change form:

Statement 5.2.A.2 then lists $s = r\theta$ alongside $v = r\omega$ and $a_T = r\alpha$. Both forms say the same thing, and the change form is the one to trust, because it never depends on where you called the angle zero.

Now check the AP Physics 1 formula sheet and note exactly which of these are printed. It prints $v = r\omega$ and $a_T = r\alpha$. It does not print $s = r\theta$ or $\Delta s = r\Delta\theta$. What it does print, in the rotational group, is

which is the same algebra applied to the center of mass of a rolling system. So the arc-length relation for a general point is one you carry in your head, even though a near-identical equation is sitting on the sheet in front of you. Knowing which is which stops you from mis-citing the sheet in a derivation, and a free-response derivation can ask you to begin from an equation on the reference information.

A quick unit check on the arc-length relation: meters equals meters times radians, and since radians are dimensionless, that balances. Try the same check with degrees and it fails, which is another way of seeing why the unit is not optional.

The arc length $\Delta s$ is a distance along a curved path, not a straight-line displacement. A point on the rim of a wheel that makes one full turn has traveled $\Delta s = 2\pi r$ of arc but has a displacement of zero, since it is back where it started. Distance and displacement part company here exactly as they do in Topic 1.2.

v = r omega, and which velocity it gives you

The $v$ here is the speed of the point along its circular path, and its direction is tangent to the circle, perpendicular to the line joining the point to the axis. That is what makes it the tangential velocity. For a point moving in a circle it is the whole velocity, because a point on a rigid rotating system has no radial velocity at all: its distance from the axis is fixed, which follows from essential knowledge 5.1.A.1.i saying a rigid system holds its shape.

The direction changes continuously even when the speed does not. A point on a wheel spinning at a steady rate has a constant $v$ and a constantly turning velocity vector, which is exactly the uniform circular motion set up in Topic 2.9.

Deriving it is a three-step job, and it is worth being able to produce it on demand for a skill 2.A question:

1. Start from the definition of the radian, in the change form 5.2.A.1 gives: $\Delta s = r\Delta\theta$.
2. Divide both sides by $\Delta t$: $\frac{\Delta s}{\Delta t} = r\frac{\Delta\theta}{\Delta t}$.
3. The left side is the average speed of the point and the right side has the average angular velocity from 5.1.A.2, so $v = r\omega$. Because $r$ is constant for a rigid system, the relation holds instant by instant, not just on average.

Two consistency checks worth carrying. First, if two points on the same rigid system have different speeds, their radii must differ in the same ratio. Second, if the same point on a system doubles its speed, $\omega$ must have doubled, because $r$ cannot change on a rigid system.

Tangential acceleration, and the acceleration that was already there

The subscript is load-bearing. Essential knowledge 5.2.A.2 is careful to say this is the relationship for the tangential component of acceleration, not for the acceleration of the point. A point on a rotating system generally has two perpendicular acceleration components:

The two answer different questions. Tangential acceleration says the point is speeding up or slowing down along its path. Centripetal acceleration says the point's direction of travel is turning. Topic 2.9 covers the centripetal side in full, including where the inward force comes from, so use that page for it rather than rebuilding it here.

What Topic 5.2 adds is that both components can now be written from the angular quantities. Substituting $v = r\omega$ into $a_c = v^2/r$ gives

That form is genuinely useful, and it is worth knowing that the AP Physics 1 sheet prints only $a_c = v^2/r$, not $a_c = r\omega^2$. The second form is a two-line derivation from the first, which is precisely the kind of thing suggested skill 2.A rewards, but do not go looking for it on the reference information.

Three cases, so the distinction sticks:

• Wheel spinning at a steady rate: $\alpha = 0$, so $a_T = 0$, but $a_c$ is not zero for any point off the axis. The point accelerates even though its speed never changes.
• Wheel speeding up: both components are non-zero, and the total acceleration of a point is the vector sum, $\sqrt{a_T^2 + a_c^2}$, tilted away from the inward radial line toward the direction of travel.
• Wheel momentarily at rest but still angularly accelerating: $v = 0$ so $a_c = 0$, while $a_T = r\alpha$ is not zero. This is the instant a swinging rod reaches the top of its arc.

Same angular velocity, different linear speeds

Essential knowledge 5.2.A.3 is one sentence and it does a lot of work: for a rigid system, all points within that system have the same angular velocity and angular acceleration.

Notice what is not in that list. Points on a rigid system share $\omega$ and $\alpha$. They do not generally share $v$, $a_T$, $a_c$, or arc length, because each of those carries a factor of $r$, and points at different distances from the axis have different $r$. Two riders on the same merry-go-round complete a turn in the same time, and the outer rider covers more ground doing it.

Written as a comparison between two points on one system, which is the skill 2.C form:

The centripetal accelerations scale the same way when written as $a_c = r\omega^2$, since $\omega$ is shared. That is a good self-check on a two-point problem: if $r$ doubles, then $v$, $a_T$, and $a_c$ all double, and the total acceleration vector at the two points points in the same direction relative to the radius even though it has a different magnitude.

One caution about the word rigid. Statement 5.2.A.3 covers points within one rigid system. Two objects that touch or are linked by a belt are not one rigid system, so they generally do not share $\omega$. What they share is the linear speed at the contact point or along the belt, and you convert back to each object's own $\omega$ using its own radius. Getting that backwards, and assuming two meshed gears or belted pulleys turn at the same rate, is an easy way to lose a question.

Skill 1.C asks for qualitative sketches, and the sketch that shows up here is $v$ against $r$ at a fixed instant: a straight line through the origin with slope $\omega$. If a system speeds up, that line keeps passing through the origin and gets steeper. A curved line, or a line with an intercept, contradicts $v = r\omega$.

Rolling without slipping, and where the CED puts it

Rolling is the headline application of everything above, and it is worth being precise about where it sits in the course. The CED does not list rolling under Topic 5.2. It is Topic 6.5 in Unit 6, and Unit 6's own introduction says students will use the content and skills of both Units 5 and 6 to study rolling without slipping. Topic 5.2 is where the tools get built; Topic 6.5 is where the CED asks you to use them.

Essential knowledge 6.5.B.1 states that while rolling without slipping, the translational motion of a system's center of mass is related to the rotational motion of the system itself with three equations:

Those are the same algebra as the Topic 5.2 relations, but they come from a different place. The center of mass sits on the axis, so $v = r\omega$ does not apply to it directly. What makes them true is the no-slip constraint: the contact point is instantaneously at rest, so the axle advances by exactly the arc length that unrolls onto the ground, and $r$ is the wheel's own radius because that is the radius doing the unrolling. The first of the three is printed on the AP Physics 1 equation sheet in exactly that form. The other two are not, though the sheet's $v = r\omega$ and $a_T = r\alpha$ are the same algebra with a general point in place of the center of mass.

The physical condition behind them: rolling without slipping means the contact point is instantaneously at rest relative to the surface. The wheel is not sliding, so the distance the axle advances equals the arc length that has unrolled onto the ground. A wheel that spins on ice covers arc length while going nowhere, and none of the three equations applies.

Statement 6.5.B.2 adds that for ideal cases, rolling without slipping implies that the frictional force does not dissipate any energy from the rolling system. Statement 6.5.C.1 covers the other side: when slipping, the motion of a system's center of mass and the system's rotational motion cannot be directly related.

Two boundary statements govern this material, and the second one has an exception clause that has to travel with it. The first: rolling friction is beyond the scope of AP Physics 1. The second: the precise mathematical relationships between linear and angular quantities while a rigid body is rolling while slipping are beyond the scope of AP Physics 1 and 2, and students will not be expected to model those relationships quantitatively; however, students are expected to qualitatively explain the changes to linear and angular quantities while a rigid body is rolling while slipping. So a question can ask you to describe in words what happens to a wheel that is spinning and skidding at once. It just will not ask you to compute it.

How Topic 5.2 is tested, and what goes wrong

Unit 5's AP Classroom Progress Check runs about 18 multiple-choice questions and 4 free-response questions across Mathematical Routines, Translation Between Representations, Experimental Design and Analysis, and Qualitative or Quantitative Translation. Topic 5.2's suggested skills point at derivations (2.A) and graph sketches (1.C) especially.

The failures worth guarding against:

Using the object's radius when the point is not on the rim. Essential knowledge 5.2.A.1 says a point at a distance $r$ from the axis. Identify the point, then measure to it.

Assuming linked objects share an angular velocity. Statement 5.2.A.3 applies within one rigid system. Belted pulleys and meshed gears share the linear speed at contact, and each converts that to its own $\omega$ through its own radius.

**Calling $a_T = r\alpha$ the acceleration of the point.** It is one perpendicular component of it. Whenever the point is moving, a centripetal component is present as well, and a question asking for the magnitude of the acceleration wants both.

Substituting degrees or revolutions. Every relation on this page assumes radians.

Applying the rolling equations to a slipping wheel. Statement 6.5.C.1 says the two motions cannot be directly related once slipping starts, and the boundary statement puts the quantitative version out of scope for AP Physics 1 and 2 while still expecting a qualitative explanation.

Where this leads: Topic 5.3 and Topic 5.4 supply the cause of the angular acceleration that this topic converts into linear terms, and Topic 5.6 puts them together. The rotational kinematics guide walks the angular problem-solving routine end to end, and the centripetal force guide handles the inward-force half of the acceleration story.