Angular vs Linear Acceleration: The Difference

Angular acceleration is how fast a rotation rate changes, in radians per second squared, and every point on a rigid body shares one value of it. Linear acceleration belongs to one point and is in meters per second squared. They link by a equals r alpha, which gives only the tangential part.

AP Physics: Unit 5 (topics 2.9 Circular Motion, 5.1 Rotational Kinematics, 5.2 Connecting Linear and Rotational Motion, 6.5 Rolling). Angular acceleration is defined in AP Physics 1 Unit 5, Torque and Rotational Dynamics, weighted at 10 to 15 percent of the multiple-choice section. Topic 5.1, Rotational Kinematics, carries one learning objective, 5.1.A, describe the rotation of a system with respect to time using angular displacement, angular velocity, and angular acceleration; EK 5.1.A.3 defines average angular acceleration as the average rate at which the angular velocity changes with respect to time, and EK 5.1.A.4 states that the angular quantities around one axis are analogous to the linear ones in one dimension and demonstrate the same mathematical relationships. Topic 5.2, Connecting Linear and Rotational Motion, learning objective 5.2.A, carries the bridge: EK 5.2.A.2 gives the derived relationships s = r theta, v = r omega and a_T = r alpha for the linear velocity and the tangential component of acceleration, and EK 5.2.A.3 states that all points within a rigid system have the same angular velocity and angular acceleration. The two-component picture of a point's linear acceleration comes from Unit 2, Topic 2.9, Circular Motion: EK 2.9.A.1 defines centripetal acceleration as the component directed toward the center, EK 2.9.A.3 defines tangential acceleration as the rate at which speed changes, and EK 2.9.A.4 makes the net acceleration their vector sum. The rolling relationship a of the center of mass equals r alpha is separate, at EK 6.5.B.1 in Unit 6, Topic 6.5, and EK 6.5.C.1 states that when slipping the center-of-mass motion and the rotational motion cannot be directly related. Both Topic 5.1 and Topic 5.2 carry the same boundary statement: 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. Suggested skills are 1.B, 2.A, 2.D, 3.A and 3.C for Topic 5.1, 1.C, 2.A, 2.C and 3.B for Topic 5.2, 1.B, 2.A, 2.D, 3.A and 3.C for Topic 2.9, and 1.A, 2.A, 2.C and 3.C for Topic 6.5. All four AP equation sheets print v = r omega, a_T = r alpha and delta x_cm = r delta theta; none prints v_cm = r omega or a_cm = r alpha, which live only in the CED at EK 6.5.B.1.

Different quantities, not one quantity in two costumes

Put a mark on the rim of a bicycle wheel and a second mark halfway to the hub, then spin the wheel up. Ask how fast the rotation is speeding up and there is one answer for the whole wheel. Ask how fast a mark is speeding up and you have to say which mark.

That asymmetry is written into the AP Physics 1 CED. EK 5.2.A.3: for a rigid system, all points within that system have the same angular velocity and angular acceleration. Linear acceleration carries no such guarantee, and the CED's own rigid-system definition at EK 5.1.A.1.i says why: a rigid system holds its shape but different points on the system move in different directions during rotation, and a rigid system cannot be modeled as an object.

The two definitions sit in different units and different units are the fastest way to see that they are different quantities.

Angular acceleration, EK 5.1.A.3: the average rate at which the angular velocity changes with respect to time, $\alpha_{\text{avg}} = \frac{\Delta\omega}{\Delta t}$ . Radians per second squared.
Linear acceleration, EK 1.2.B.2 and its neighbours: the rate at which a velocity changes, $\vec{a} = \frac{\Delta\vec{v}}{\Delta t}$ . Meters per second squared.

The bridge, EK 5.2.A.2, is one line:

a_T = r\alpha

Read the subscript. It says tangential, and the restriction it carries is the subject of most of this page: $r\alpha$ is not the linear acceleration of the point, it is one perpendicular component of it.

The unit conversion looks suspiciously free, and that is worth naming early. Multiply $\text{rad/s}^2$ by meters and you get $\text{m/s}^2$ , because the radian is an arc length divided by a radius, a ratio of two lengths, and so carries no dimension at all. Nothing in the algebra warns you that $r\alpha$ answers a narrower question than $\vec{a}$ does. You have to know.

Angular vs linear acceleration, side by side

Property	Angular acceleration	Linear acceleration
Symbol	$\alpha$	$\vec{a}$
SI unit	$\text{rad/s}^2$	$\text{m/s}^2$
Defined as	$\frac{\Delta\omega}{\Delta t}$ , EK 5.1.A.3	$\frac{\Delta\vec{v}}{\Delta t}$
What it belongs to	The whole rigid system	One point, or one object
Same at every radius	Yes, EK 5.2.A.3	No, it scales with $r$
Value on the rotation axis	Unchanged, the system still spins up	$a_T = 0$ there, since $r = 0$
Number of independent parts	One, a magnitude with a rotational sense	Two for circular motion: tangential and centripetal
Zero when the rate is steady	Yes, $\alpha = 0$	No, $a_c$ survives, EK 2.9.A.4
Direction handling in AP Physics 1	Clockwise or counterclockwise about a stated axis	A full vector in the plane
Its dynamics law	$\alpha_{\text{sys}} = \frac{\Sigma\tau}{I_{\text{sys}}}$	$\vec{a}_{\text{sys}} = \frac{\Sigma\vec{F}}{m_{\text{sys}}}$
Angular unit required	Radians whenever it multiplies a radius	Not applicable
Printed link between them	$a_T = r\alpha$ , on all four AP equation sheets	$\alpha = \frac{a_T}{r}$ , the same line rearranged

Two rows deserve a second look.

"Value on the rotation axis." A point sitting on the axis of a wheel that is spinning up has $a_T = (0)(\alpha) = 0$ , while the wheel's angular acceleration is whatever it is. One of the two quantities is zero and the other is not, in the same rigid body at the same instant, which settles the question of whether they are the same thing.

"Number of independent parts." This is the row that does the real damage on an exam, and it gets a section of its own below. $\alpha$ is one number. The linear acceleration of a point going round a circle is two perpendicular numbers, and $r\alpha$ is only one of them.

The case that separates them: one alpha, four different linear accelerations

A wheel of radius $0.30$ m is spinning up at a constant $\alpha = 8.0\ \text{rad/s}^2$ . Look at four points on it at the same instant.

Point	Distance from axis	Angular acceleration	Tangential acceleration $a_T = r\alpha$
Rim	$0.30$ m	$8.0\ \text{rad/s}^2$	$2.4\ \text{m/s}^2$
Two thirds out	$0.20$ m	$8.0\ \text{rad/s}^2$	$1.6\ \text{m/s}^2$
One third out	$0.10$ m	$8.0\ \text{rad/s}^2$	$0.80\ \text{m/s}^2$
On the axle	$0$ m	$8.0\ \text{rad/s}^2$	$0\ \text{m/s}^2$

One column is constant down the table and the other spans every value from zero to $2.4\ \text{m/s}^2$ . That is the single most useful fact on this page: a rotating rigid body has one angular acceleration and as many linear accelerations as it has points.

The practical consequence is a reading habit. When a problem gives you "the acceleration" of something rotating, the number is unusable until you know which point it belongs to, and the giveaway is the unit. A figure in $\text{rad/s}^2$ describes the whole body and needs no further address. A figure in $\text{m/s}^2$ describes one location and is wrong everywhere else.

Worked example one runs this table and then shows what happens when you feed the same $\alpha$ in degrees per second squared into $a_T = r\alpha$ .

a equals r alpha is only the tangential half

Here is the error that survives longest, because the equation people remember is true and the sentence they build out of it is not.

$a_T = r\alpha$ gives the tangential component of a point's linear acceleration. The point also has a centripetal component, and the CED names all three pieces separately.

EK 2.9.A.1: centripetal acceleration is the component of an object's acceleration directed toward the center of the object's circular path. EK 2.9.A.1.i gives its magnitude as the tangential speed squared over the radius, $a_c = \frac{v^2}{r}$ .
EK 2.9.A.3: tangential acceleration is the rate at which an object's speed changes and is directed tangent to the object's circular path.
EK 2.9.A.4: the net acceleration of an object moving in a circle is the vector sum of the centripetal acceleration and tangential acceleration.

So the mapping between the two sides of this page is one to two:

Angular quantity	What it produces at a point at distance $r$
$\omega$	Tangential speed $v = r\omega$
$\alpha$	Tangential acceleration $a_T = r\alpha$ , and nothing else
Nothing angular	Centripetal acceleration $a_c = \frac{v^2}{r}$ , which exists whenever the point moves

Read the bottom row. No angular quantity corresponds to $a_c$ , because centripetal acceleration comes from the velocity changing direction rather than from the rotation rate changing. A wheel turning at a steady rate has $\alpha = 0$ and $a_T = 0$ at every point, and every point off the axis still has a nonzero linear acceleration. "Constant angular velocity" and "no acceleration" are not the same statement, and the gap between them is the whole content of centripetal vs tangential acceleration.

One asymmetry in the printing is worth knowing. The two Physics C sheets print centripetal acceleration in an angular form as well, $a_c = \frac{v^2}{r} = r\omega^2$ , while the AP Physics 1 and AP Physics 2 sheets print only $a_c = \frac{v^2}{r}$ . The physics is the same on both, since substituting $v = r\omega$ gets you from one to the other in a line, but a C candidate has $r\omega^2$ handed to them and an AP Physics 1 candidate has to build it.

Rolling prints a third equation that looks identical and is not

There is a second relationship of the form "linear equals $r$ times angular", and it is a different statement from $a_T = r\alpha$ even though the symbols nearly match.

EK 6.5.B.1: 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 the equations

\Delta x_{\text{cm}} = r\Delta\theta \qquad v_{\text{cm}} = r\omega \qquad a_{\text{cm}} = r\alpha

The difference is what the left-hand side names.

$a_T = r\alpha$ is about a point on the body, at distance $r$ from the axis, and it gives the tangential part of that point's acceleration. The axis can be fixed in space; the body need not go anywhere.
$a_{\text{cm}} = r\alpha$ is about the center of mass of the whole body, it requires the rolling-without-slipping condition, and $r$ is specifically the radius at which the body touches the ground.

They coincide numerically for a rolling wheel because the contact radius is the wheel's radius and the rotation is about the center, but the claims are not interchangeable. Break the rolling condition and the second equation fails while the first still holds: EK 6.5.C.1 says that when slipping, the motion of a system's center of mass and the system's rotational motion cannot be directly related. A car wheel on ice still has a perfectly good $a_T = r\alpha$ at its rim, and the car's own acceleration has stopped following it. That case is rolling vs sliding.

The equation sheets make the distinction visible by what they print. All four sheets print $a_T = r\alpha$ and $\Delta x_{\text{cm}} = r\Delta\theta$ . None of the four prints $v_{\text{cm}} = r\omega$ or $a_{\text{cm}} = r\alpha$ : those two live in the CED at EK 6.5.B.1 and not on the sheet, so you carry them yourself.

The dynamics on each side, term for term

Each acceleration has a law that produces it, and comparing the two laws is the cleanest way to see which quantities are analogous.

\vec{a}_{\text{sys}} = \frac{\Sigma\vec{F}}{m_{\text{sys}}} = \frac{\vec{F}_{\text{net}}}{m_{\text{sys}}} \qquad\qquad \alpha_{\text{sys}} = \frac{\Sigma\tau}{I_{\text{sys}}} = \frac{\tau_{\text{net}}}{I_{\text{sys}}}

Both are printed on all four AP equation sheets, side by side in the same block. Reading across them:

Translational	Rotational	Why the swap is not automatic
Force $\vec{F}$	Torque $\tau = r_{\perp}F = rF\sin\theta$	A torque needs a chosen axis; a force does not
Mass $m$	Rotational inertia $I = \sum m_i r_i^2$	$m$ is a property of the object, $I$ is a property of the object and the axis together
$\vec{a}$ in $\text{m/s}^2$	$\alpha$ in $\text{rad/s}^2$	Different dimensions, so the numbers are never directly comparable

The middle row is the one that breaks the symbol-swapping reflex, and it has a page of its own at mass vs rotational inertia. The practical version: the same net torque applied to the same object about two different axes produces two different angular accelerations, while the same net force on the same object always produces the same linear acceleration.

A useful consistency check falls out of these two laws. Because $\tau = rF$ for a force applied tangentially at radius $r$ , and $I = mr^2$ for a single particle at that radius, $\alpha = \frac{rF}{mr^2} = \frac{F}{mr} = \frac{a}{r}$ , which is $a = r\alpha$ again. The two laws are not independent statements about the world; the rotational one is the translational one with the geometry of a chosen axis folded in.

Where the confusion costs a mark

Every item below is a specific way of losing points, not a general warning.

Reporting an $\alpha$ in $\text{m/s}^2$ or an $a$ in $\text{rad/s}^2$ . The units are not interchangeable and the radian's dimensionlessness is what makes the slip invisible in the algebra. Check the unit against the question: does it ask about the body or about a point?
Using $a = r\alpha$ for the full linear acceleration. It is the tangential component only. EK 2.9.A.4 makes the net acceleration the vector sum of the tangential and centripetal parts.
Adding $a_T$ and $a_c$ as plain numbers. They are perpendicular, so combine them with $\sqrt{a_T^2 + a_c^2}$ .
Concluding that $\alpha = 0$ means the acceleration is zero. It means the tangential part is zero. Every point off the axis is still accelerating toward the axis.
Feeding degrees per second squared into $a_T = r\alpha$ . The result is too large by $57.2958$ . The radian requirement bites the moment an angular quantity multiplies a length, which is exactly what this equation does.
Saying the outer point has a larger angular acceleration. EK 5.2.A.3 says every point shares $\alpha$ . What the outer point has is a larger $a_T$ .
Using $a_{\text{cm}} = r\alpha$ for an object that is slipping. EK 6.5.C.1 rules it out, and the CED's Topic 6.5 boundary statement says the precise relationships while rolling and slipping are beyond AP Physics 1 and 2 and will not be asked for quantitatively.
Swapping $m$ for $I$ without changing the axis discussion. Rotational inertia depends on the axis, so an $\alpha$ quoted without an axis is incomplete in a way an $a$ never is.
Letting the sign convention drift. EK 5.1.A.1.ii lets you pick which rotational sense counts as positive. Pick once, write it down, and hold it. An $\alpha$ that changes sign because you changed your mind about clockwise is a defect, not a deceleration.

When they track each other, and why that lulls you

Three situations let you slide between the two without noticing.

One point, one fixed radius. If the entire question concerns the rim of a single wheel, $r$ never changes, so $a_T$ is $\alpha$ times a constant. Every proportional claim about one is true of the other and the distinction never surfaces.

Starting from rest with $\alpha$ constant. Then $\omega$ is small early on, so $a_c = r\omega^2$ is small early on, and the net linear acceleration really is close to $r\alpha$ for the first moments. It stops being close very quickly: $a_c$ grows with the square of $\omega$ while $a_T$ holds still, so a wheel that is a few seconds into spinning up has a centripetal component many times the tangential one.

A radius of one meter. Then $a_T$ and $\alpha$ share a numeric value and your working looks the same in either column. This is a coincidence of the meter, not a piece of physics.

The distinction comes back in four exam-shaped places.

Two radii in one problem. Gears meshing, a belt over two pulleys, a person walking inward on a rotating platform.
A rope leaving a spool. The rope's linear acceleration equals $r\alpha$ at the radius where it leaves, so a spool whose radius changes as it unwinds changes the relationship as it goes.
Rolling. The bridge is $a_{\text{cm}} = r\alpha$ and it holds only while the contact point is not sliding.
Anything at constant angular velocity. $\alpha = 0$ and the linear acceleration is entirely centripetal, which is uniform circular motion and the case where treating $r\alpha$ as the acceleration returns exactly zero and is exactly wrong.

The reflex worth building: before you write $r\alpha$ , say out loud whether you want the whole acceleration of a point or only the part that changes its speed. If it is the whole thing, you have another component to find.

What the CED asks, and what the sheets print

Angular acceleration is defined in AP Physics 1 Unit 5, Torque and Rotational Dynamics, weighted at 10 to 15 percent of the multiple-choice section. Topic 5.1, Rotational Kinematics, carries one learning objective, 5.1.A: describe the rotation of a system with respect to time using angular displacement, angular velocity, and angular acceleration. EK 5.1.A.3 defines average angular acceleration as the average rate at which the angular velocity changes with respect to time. EK 5.1.A.4 states the analogy in the CED's own words: angular displacement, angular velocity, and angular acceleration around one axis are analogous to linear displacement, velocity, and acceleration in one dimension and demonstrate the same mathematical relationships. EK 5.1.A.4.i lists the three constant-angular-acceleration equations and EK 5.1.A.4.ii covers reading them off graphs. Suggested skills for Topic 5.1 are 1.B, 2.A, 2.D, 3.A and 3.C.

The bridge is Topic 5.2, Connecting Linear and Rotational Motion, 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. EK 5.2.A.1 gives $\Delta s = r\Delta\theta$ , EK 5.2.A.2 gives $s = r\theta$ , $v = r\omega$ and $a_T = r\alpha$ as derived relationships of linear velocity and of the tangential component of acceleration to their angular counterparts, and EK 5.2.A.3 gives the shared- $\alpha$ statement. Suggested skills for Topic 5.2 are 1.C, 2.A, 2.C and 3.B.

The two-component picture comes from Unit 2, Topic 2.9, Circular Motion, at EK 2.9.A.1, 2.9.A.3 and 2.9.A.4. The rolling relationships come from Unit 6, Topic 6.5, Rolling, at EK 6.5.B.1, with EK 6.5.C.1 marking where they stop applying.

Both Topic 5.1 and Topic 5.2 carry the same single boundary statement: 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. So $\alpha$ is handled with a sign rather than as a vector along the axis, and the right-hand rule for angular acceleration is outside AP Physics 1.

On the sheets, checked line by line: all four AP equation sheets print $v = r\omega$ and $a_T = r\alpha$ next to each other, plus $\Delta x_{\text{cm}} = r\Delta\theta$ and the three constant-angular-acceleration equations. The AP Physics 1 and AP Physics 2 sheets write angular acceleration only through those relationships; the two Physics C sheets add the calculus definitions $\omega = \frac{d\theta}{dt}$ and $\alpha = \frac{d\omega}{dt}$ , and they print centripetal acceleration as $a_c = \frac{v^2}{r} = r\omega^2$ where the algebra-based sheets print $a_c = \frac{v^2}{r}$ alone. The symbol keys on the AP Physics 1 and 2 sheets read $\alpha$ as angular acceleration and $a$ as acceleration, with no note that the $a$ in $a_T = r\alpha$ is a component.

For the routines, rotational kinematics works the three angular equations and the kinematic equations guide works their linear originals. The CED framing sits on Topic 5.1, Topic 5.2 and Topic 2.9, and the torque and rotational motion practice set has problems to work. The AP Physics 1 equation sheet shows the rotational block in full.

One angular acceleration, four points, four linear accelerations

A wheel of radius $0.30$ m starts from rest and spins up at a constant $\alpha = 8.0\ \text{rad/s}^2$ . Find the angular acceleration and the tangential acceleration of points at $r = 0.30$ , $0.20$ , $0.10$ and $0$ m. Then find the angular velocity and the centripetal acceleration of the rim point $1.5$ s later, and the net linear acceleration there. Take the direction of rotation as positive.

Angular acceleration at all four points. EK 5.2.A.3 settles this without arithmetic: for a rigid system, all points within that system have the same angular velocity and angular acceleration. All four points have $\alpha = 8.0\ \text{rad/s}^2$ .
Tangential acceleration at the rim, from EK 5.2.A.2: $a_T = r\alpha = (0.30)(8.0) = 2.4\ \text{m/s}^2$ .
At $r = 0.20$ m: $a_T = (0.20)(8.0) = 1.6\ \text{m/s}^2$ . At $r = 0.10$ m: $a_T = (0.10)(8.0) = 0.80\ \text{m/s}^2$ . At $r = 0$ : $a_T = (0)(8.0) = 0\ \text{m/s}^2$ .
Collect the comparison. One value of $\alpha$ , four values of $a_T$ : $2.4$ , $1.6$ , $0.80$ and $0\ \text{m/s}^2$ . The ratios $2.4 : 1.6 : 0.80 : 0$ are exactly $0.30 : 0.20 : 0.10 : 0$ , which is what $a_T = r\alpha$ requires with $\alpha$ shared.
Angular velocity after $1.5$ s, from EK 5.1.A.4.i: $\omega = \omega_0 + \alpha t = 0 + (8.0)(1.5) = 12\ \text{rad/s}$ .
Rim speed: $v = r\omega = (0.30)(12) = 3.6\ \text{m/s}$ . Centripetal acceleration, from EK 2.9.A.1.i: $a_c = \frac{v^2}{r} = \frac{(3.6)^2}{0.30} = \frac{12.96}{0.30} = 43.2\ \text{m/s}^2$ , directed at the axle.
Net linear acceleration at the rim. The two components are perpendicular, so by EK 2.9.A.4 combine them as vectors: $a = \sqrt{(43.2)^2 + (2.4)^2} = \sqrt{1866.24 + 5.76} = \sqrt{1872} = 43.3\ \text{m/s}^2$ , tilted $\arctan\frac{2.4}{43.2} = 3.18^\circ$ from the inward radial direction.
The size of the gap is the point. At this instant $r\alpha$ is $2.4\ \text{m/s}^2$ and the rim point's actual acceleration is $43.3\ \text{m/s}^2$ , eighteen times larger. Calling $r\alpha$ the acceleration of the point would be wrong by that factor, and the error grows as the wheel speeds up because $a_c$ scales with $\omega^2$ while $a_T$ does not move.
Now the degrees test. Convert: $\alpha = 8.0\ \text{rad/s}^2 \times 57.2958 = 458\ \text{deg/s}^2$ . Substituting into $a_T = r\alpha$ gives $(0.30)(458) = 137\ \text{m/s}^2$ for a bicycle wheel spinning up gently. Wrong by the factor $57.2958$ , because the radian's definition as arc length over radius is what makes $a_T = r\alpha$ true at all.

All four points share $\alpha = 8.0\ \text{rad/s}^2$ . Their tangential accelerations are $2.4$ , $1.6$ , $0.80$ and $0\ \text{m/s}^2$ at $r = 0.30$ , $0.20$ , $0.10$ and $0$ m. After $1.5$ s the rim has $\omega = 12\ \text{rad/s}$ , $v = 3.6$ m/s and $a_c = 43.2\ \text{m/s}^2$ , giving a net linear acceleration of $43.3\ \text{m/s}^2$ tilted $3.18^\circ$ from radial. Using $458\ \text{deg/s}^2$ in $a_T = r\alpha$ returns $137\ \text{m/s}^2$ , too large by $57.2958$ .

A car speeding up: which acceleration is which

A car accelerates from rest along a straight road at a constant $2.6\ \text{m/s}^2$ for $5.0$ s. Its tires have radius $0.32$ m and roll without slipping. Find the angular acceleration of a tire, its angular velocity and the number of revolutions it turns in that time, and check the distance travelled two ways. Then state which quantity in the problem $a_T = r\alpha$ would give you.

Identify which relationship applies. The car's $2.6\ \text{m/s}^2$ is the acceleration of the tire's center of mass, so the relevant statement is EK 6.5.B.1 for rolling without slipping: $a_{\text{cm}} = r\alpha$ . This is the rolling bridge, not the point-on-a-body relationship.
Angular acceleration: $\alpha = \frac{a_{\text{cm}}}{r} = \frac{2.6}{0.32} = 8.125\ \text{rad/s}^2$ .
Angular velocity after $5.0$ s: $\omega = \omega_0 + \alpha t = 0 + (8.125)(5.0) = 40.6\ \text{rad/s}$ . Cross-check through the linear side: $v = a_{\text{cm}}t = (2.6)(5.0) = 13\ \text{m/s}$ , and $\omega = \frac{v}{r} = \frac{13}{0.32} = 40.6\ \text{rad/s}$ . The two routes agree.
Angular displacement: $\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 0 + \frac{1}{2}(8.125)(5.0)^2 = \frac{1}{2}(8.125)(25) = 101.6\ \text{rad}$ .
Revolutions: $\frac{101.56}{2\pi} = 16.2$ revolutions.
Distance, first route, straight linear kinematics: $x = \frac{1}{2}a t^2 = \frac{1}{2}(2.6)(25) = 32.5$ m.
Distance, second route, through the rolling relationship printed on all four sheets: $\Delta x_{\text{cm}} = r\Delta\theta = (0.32)(101.5625) = 32.5$ m. The agreement is the check that the rolling condition was applied consistently.
Now the question the example is really about. $a_T = r\alpha$ at the tire's rim gives $(0.32)(8.125) = 2.6\ \text{m/s}^2$ , numerically identical to the car's acceleration, and it is a different quantity: it is the tangential component of the acceleration of a rim point measured about the wheel's own center. That rim point is also being carried forward with the car and swung round the axle at $a_c = r\omega^2 = (0.32)(40.625)^2 = 528\ \text{m/s}^2$ at the five-second mark. Its actual acceleration in the road frame is nothing like $2.6\ \text{m/s}^2$ . The numbers matching is what makes this confusion so durable.

$\alpha = 8.125\ \text{rad/s}^2$ , $\omega = 40.6\ \text{rad/s}$ after $5.0$ s, and the tire turns $101.6$ rad, which is $16.2$ revolutions. The distance is $32.5$ m by linear kinematics and $32.5$ m by $\Delta x_{\text{cm}} = r\Delta\theta$ . The car's $2.6\ \text{m/s}^2$ is $a_{\text{cm}}$ , not the acceleration of any particular point on the tire: a rim point has a centripetal component of about $528\ \text{m/s}^2$ about the axle at that moment.

Constant angular velocity, and the acceleration that does not vanish

A turntable of radius $0.15$ m rotates at a steady $\omega = 12\ \text{rad/s}$ . Find the angular acceleration and the tangential acceleration of a coin on the rim, then find the coin's actual linear acceleration. Repeat for a coin at $0.05$ m.

Angular acceleration. The rate is steady, so $\Delta\omega = 0$ and by EK 5.1.A.3, $\alpha = \frac{\Delta\omega}{\Delta t} = 0\ \text{rad/s}^2$ .
Tangential acceleration at the rim: $a_T = r\alpha = (0.15)(0) = 0\ \text{m/s}^2$ . The coin's speed is not changing, which is what EK 2.9.A.3 says the tangential component measures.
Rim speed: $v = r\omega = (0.15)(12) = 1.8\ \text{m/s}$ .
Centripetal acceleration at the rim: $a_c = \frac{v^2}{r} = \frac{(1.8)^2}{0.15} = \frac{3.24}{0.15} = 21.6\ \text{m/s}^2$ , directed at the center. The Physics C form gives the same number in one step: $a_c = r\omega^2 = (0.15)(144) = 21.6\ \text{m/s}^2$ .
Net linear acceleration at the rim, by EK 2.9.A.4: the tangential part is zero, so the vector sum is just the centripetal part, $21.6\ \text{m/s}^2$ pointed at the center. Angular acceleration zero, linear acceleration $21.6\ \text{m/s}^2$ , same object, same instant.
The inner coin at $r = 0.05$ m: same $\alpha = 0$ , same $a_T = 0$ , speed $v = (0.05)(12) = 0.60\ \text{m/s}$ , and $a_c = \frac{0.36}{0.05} = 7.2\ \text{m/s}^2$ . One third the radius, one third the centripetal acceleration, since $a_c = r\omega^2$ is linear in $r$ at fixed $\omega$ .
State the trap in one line. $\alpha = 0$ tells you the rotation rate is holding steady. It tells you nothing about whether points on the body are accelerating, and here every point off the axis is, by as much as $21.6\ \text{m/s}^2$ .

$\alpha = 0\ \text{rad/s}^2$ and $a_T = 0$ everywhere on the turntable. The rim coin still has a linear acceleration of $21.6\ \text{m/s}^2$ directed at the center, and the coin at $0.05$ m has $7.2\ \text{m/s}^2$ . Zero angular acceleration does not mean zero linear acceleration.

Frequently asked questions

What is the difference between angular and linear acceleration?

Angular acceleration is the rate at which a rotation rate changes, measured in radians per second squared, and every point on a rigid body shares one value of it: the AP Physics 1 CED states this at essential knowledge 5.2.A.3. Linear acceleration is the rate at which a velocity changes, measured in meters per second squared, and it belongs to one particular point, so a rim point and a point near the hub have different values at the same instant. They are linked by a subscript T equals r alpha, which gives only the tangential component of the point's acceleration, not the whole of it.

Does a equal r alpha give the total linear acceleration?

No. It gives the tangential component only, which is why the AP equation sheets write it with a subscript T. A point moving on a circle also has a centripetal component equal to its speed squared over the radius, directed at the center, and essential knowledge 2.9.A.4 says the net acceleration is the vector sum of the two. Because they are perpendicular, combine them as the square root of the sum of their squares rather than by adding. For a wheel that has been speeding up for a while the centripetal part is usually much the larger of the two, since it grows with the square of the angular velocity while the tangential part stays fixed.

Can angular acceleration be zero while linear acceleration is not?

Yes, and this is the most common version of the confusion. A wheel turning at a constant rate has zero angular acceleration, so the tangential acceleration of every point is zero, but every point off the rotation axis is still changing the direction of its velocity and therefore still accelerating toward the axis. A turntable of radius 0.15 m spinning at a steady 12 radians per second has zero angular acceleration and a rim acceleration of 21.6 meters per second squared. Uniform circular motion and zero acceleration are different conditions.

Do all points on a rotating object have the same angular acceleration?

Yes for a rigid body, and the AP Physics 1 CED requires it at essential knowledge 5.2.A.3: for a rigid system, all points within that system have the same angular velocity and angular acceleration. Their tangential accelerations differ, because tangential acceleration is r times the shared angular acceleration and each point sits at a different r. A point on the rotation axis has zero tangential acceleration while sharing the body's angular acceleration, which is the cleanest demonstration that the two are different quantities rather than one quantity in two units.

What are the units of angular acceleration?

Radians per second squared. Linear acceleration is in meters per second squared, so the two are never directly comparable and quoting one in the other's unit is an error. The relationship a subscript T equals r alpha converts between them because the radian is defined as an arc length divided by a radius, which makes it a ratio of two lengths and therefore dimensionless. That is also why the unit check does not catch a mistake here: meters times radians per second squared simplifies to meters per second squared with nothing left over to warn you.

How do you convert angular acceleration to linear acceleration?

Multiply by the distance from the rotation axis to the point you care about, using radians per second squared: the tangential acceleration is r times alpha. If the angular acceleration is given in degrees per second squared, convert first by dividing by 57.2958, because the relationship only holds in radians. Remember that the result is the tangential component alone. For a body rolling without slipping there is a separate relationship, a of the center of mass equals r alpha, given in the CED at essential knowledge 6.5.B.1, and it applies to the center of mass rather than to a point on the rim.

Is angular acceleration a vector in AP Physics 1?

Not in the full three-dimensional sense. The boundary statement carried by both Topic 5.1 and Topic 5.2 says 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, so you handle its direction with a sign rather than with a vector along the axis. Essential knowledge 5.1.A.1.ii describes the convention: one rotational sense is typically taken as mathematically positive and the other as negative. Choose which at the start of a problem and hold that choice to the end.