Centripetal vs Tangential Acceleration: Difference

They are two perpendicular components of one acceleration. Centripetal points at the center of the circle and changes the direction of the velocity. Tangential points along the path and changes its size. Uniform circular motion means the tangential part is zero, not that the acceleration is zero.

AP Physics: Unit 2 (topics 2.9 Circular Motion, 5.2 Connecting Linear and Rotational Motion). Both components are introduced in AP Physics 1 Unit 2, Force and Translational Dynamics, weighted at 18 to 23 percent of the multiple-choice section, at Topic 2.9, Circular Motion, learning objective 2.9.A, describe the motion of an object traveling in a circular path. EK 2.9.A.1 defines centripetal acceleration as the component of an object's acceleration directed toward the center of the object's circular path, with EK 2.9.A.1.i giving its magnitude as the ratio of the object's tangential speed squared to the radius and EK 2.9.A.1.ii restating the inward direction. EK 2.9.A.2 states that the centripetal acceleration can result from a single force, more than one force, or components of forces, with sub-statements on the minimum speed at the top of a vertical loop (derived equation v = sqrt(gr)), on banked surfaces, and on the conical pendulum. EK 2.9.A.3 defines tangential acceleration as the rate at which an object's speed changes, directed tangent to the circular path. EK 2.9.A.4 states that the net acceleration of an object moving in a circle is the vector sum of the centripetal and tangential accelerations. EK 2.9.A.5 introduces period and frequency for the constant-speed case. Topic 2.9 carries two boundary statements: quantitative banked-curve work is limited to the case in which no friction is required, and Kepler's first and second laws are excluded. Suggested skills for Topic 2.9 are 1.B, 2.A, 2.D, 3.A and 3.C. The symbol a_T and the equation a_T = r alpha come from Unit 5, Topic 5.2, Connecting Linear and Rotational Motion, EK 5.2.A.2, with suggested skills 1.C, 2.A, 2.C and 3.B; Unit 5 is weighted at 10 to 15 percent. All four AP equation sheets print a_T = r alpha directly below v = r omega, with a capital T subscript verified on a rendering of the appendix at 400 dots per inch. The AP Physics 1 and AP Physics 2 sheets print a_c = v^2/r; the two Physics C sheets print a_c = v^2/r = r omega^2. No sheet prints the vector sum of the two components.

Same object, same instant, two perpendicular jobs

A velocity can change in two ways: it can turn, or it can get bigger or smaller. Circular motion has one component of acceleration for each of those, and they act at the same time on the same object.

The AP Physics 1 CED defines both as components rather than as separate accelerations, which is the whole idea and is easy to read past.

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 ratio of the object's tangential speed squared to the radius, $a_c = \frac{v^2}{r}$ , and EK 2.9.A.1.ii repeats that it is directed toward the center.
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.

Read those three in order and the structure is plain. There is one acceleration. It has a part pointing at the center and a part pointing along the path, those two directions are always at right angles, and the object's real acceleration is the vector sum.

The division of labour is exact:

a_c = \frac{v^2}{r} \quad \text{turns the velocity} \qquad\qquad a_T = r\alpha \quad \text{resizes the velocity}

Neither can do the other's job, and that is a geometric fact rather than a convention. A component perpendicular to the velocity cannot change the velocity's magnitude to first order, and a component parallel to the velocity cannot change its direction. Split any acceleration into those two directions and you have split it into "how much it turns" and "how much it speeds up".

The coupling between them runs one way and it runs through $v$ . The tangential component changes the speed; the speed then sets the centripetal component through $\frac{v^2}{r}$ . So on an object that is speeding up in a circle, $a_T$ can hold perfectly still while $a_c$ climbs with the square of the speed.

Centripetal vs tangential acceleration, side by side

Property	Centripetal acceleration	Tangential acceleration
Symbol	$a_c$	$a_T$
Direction	Toward the center of the circular path	Along the tangent, parallel to the velocity
What it changes	The direction of the velocity	The magnitude of the velocity
Magnitude	$a_c = \frac{v^2}{r}$ , EK 2.9.A.1.i	$a_T = r\alpha$ , EK 5.2.A.2
Zero when	The object is not moving, or is moving in a straight line	The speed is constant, so $\alpha = 0$
In uniform circular motion	Nonzero, this is the whole acceleration	Zero, by definition of uniform
Sign or sense	Always inward, never outward	Forward when speeding up, backward when slowing
Its angular partner	None. Nothing angular corresponds to it	$\alpha$ , one for one, through $a_T = r\alpha$
Grows with speed how	As $v^2$	Not at all, at fixed $\alpha$
Caused by	The inward component of the net force	The tangential component of the net force
On the AP Physics 1 and 2 sheets	$a_c = \frac{v^2}{r}$	$a_T = r\alpha$
On the two Physics C sheets	$a_c = \frac{v^2}{r} = r\omega^2$	$a_T = r\alpha$
How they combine	Perpendicular components: $a = \sqrt{a_c^2 + a_T^2}$	Perpendicular components, EK 2.9.A.4

Three rows repay the reading.

"Sign or sense." Centripetal acceleration has only one direction available to it. Tangential acceleration reverses when the object slows down, and an object slowing on a circular path still has an inward centripetal component. The two are not a pair of opposites; one is a direction and the other is a signed quantity along a different direction.

"Its angular partner." $\alpha$ maps onto $a_T$ and onto nothing else. There is no angular quantity whose linear counterpart is $a_c$ , because $a_c$ comes from turning rather than from the rate of turning changing. That absence is why angular vs linear acceleration is a separate question from this one.

"Grows with speed how." At a fixed angular acceleration, doubling the speed leaves $a_T$ alone and multiplies $a_c$ by four. On anything that has been speeding up for more than a moment, the centripetal component is the larger by a wide margin, which is worth knowing before you estimate.

The case that separates them: one car, and which component wins

A car enters a circular track of radius $40$ m at $12$ m/s and speeds up at a steady $2.0\ \text{m/s}^2$ along the track. Watch the two components as the lap proceeds.

Moment	Speed $v$	$a_c = \frac{v^2}{r}$	$a_T$	Net acceleration	Tilt from radial
Entering	$12$ m/s	$3.6\ \text{m/s}^2$	$2.0\ \text{m/s}^2$	$4.12\ \text{m/s}^2$	$29.1^\circ$
At $v = 8.94$ m/s, hypothetically	$8.94$ m/s	$2.0\ \text{m/s}^2$	$2.0\ \text{m/s}^2$	$2.83\ \text{m/s}^2$	$45.0^\circ$
After $4.0$ s	$20$ m/s	$10.0\ \text{m/s}^2$	$2.0\ \text{m/s}^2$	$10.2\ \text{m/s}^2$	$11.3^\circ$

Nothing about the car's controls changed between the first row and the third. The driver held the same steady push along the track and the same circle. The acceleration vector rotated from $29^\circ$ off the radial direction to $11^\circ$ off it, purely because the speed grew and $a_c$ grew with its square.

The middle row is the crossover, and it is worth finding for its own sake. The two components are equal when $\frac{v^2}{r} = a_T$ , so at

v = \sqrt{r\,a_T} = \sqrt{(40)(2.0)} = 8.94\ \text{m/s}

Below that speed the tangential part is the larger and the acceleration points mostly along the track. Above it the centripetal part is the larger and the acceleration points mostly inward. Which component dominates is not a property of the situation, it is a property of the moment. That is the sentence to take away, and it is why "is this centripetal or tangential" is usually the wrong question. It is both, and the mix changes.

Worked example one runs this case with the arithmetic shown, including the angular acceleration $\alpha = \frac{a_T}{r} = 0.05\ \text{rad/s}^2$ that a rotational treatment would use instead.

Uniform does not mean unaccelerated

This is the confusion the page exists for, and it comes from the word.

Uniform circular motion means constant speed on a circle. The CED introduces the phrase at EK 2.9.A.5: the revolution of an object traveling in a circular path at a constant speed, which it labels uniform circular motion, can be described using period and frequency. Constant speed means the speed is not changing, and by EK 2.9.A.3 tangential acceleration is precisely the rate at which speed changes. So:

\text{uniform circular motion} \iff a_T = 0

It does not mean $a = 0$ . The velocity is turning the whole time, so $a_c = \frac{v^2}{r}$ is as large as ever, and by EK 2.9.A.4 the net acceleration is the vector sum of the two components, which here is just the centripetal one. An object in uniform circular motion is accelerating, constantly, at $\frac{v^2}{r}$ directed at the center, and never for one instant in a straight line.

The reason the mistake is so durable is that "uniform" in every other kinematics context does mean unaccelerated. Uniform velocity is unaccelerated. Uniform motion is unaccelerated. Uniform circular motion is the exception, because what is uniform is the speed and not the velocity.

Two consequences follow directly and both get tested.

A net force is required to sustain it. EK 2.9.A.2 says centripetal acceleration can result from a single force, more than one force, or components of forces exerted on an object in circular motion. Something must supply that inward net force: tension, friction, gravity, a normal force, or components of several. The centripetal force guide works the routine for identifying it.
The word centripetal never names a new force. It names a direction and the acceleration in that direction. That distinction is the subject of centripetal vs centrifugal force.

Turn it round for the other half of the pair. A tangential acceleration alone, with no centripetal component, is not circular motion at all. It is straight-line motion, since nothing is turning the velocity. So the two components are not symmetric partners: $a_c$ is what makes the path a circle, and $a_T$ is an optional extra that says whether the trip round the circle is speeding up.

Which force produces which component

Both components come from the same net force, split along the same two directions. Newton's second law does not know about circles; you resolve into radial and tangential axes because those are the directions in which the two effects separate.

\Sigma F_{\text{radial}} = ma_c = \frac{mv^2}{r} \qquad\qquad \Sigma F_{\text{tangential}} = ma_T = mr\alpha

The best demonstration is a ball swung in a vertical circle on a string, because there the geometry of one constant force does all the work.

Position of the ball	What gravity does	What the string does
Lowest point	Points straight down, entirely along the radius, so it contributes nothing tangential and $a_T = 0$	Pulls straight up, entirely radial
String horizontal	Points straight down, entirely tangential here, so it supplies the whole of $a_T = g = 9.8\ \text{m/s}^2$	Pulls horizontally, entirely radial, contributes nothing to $a_T$
Highest point	Points straight down, entirely along the radius again, so $a_T = 0$	Pulls straight down, entirely radial

At the bottom of the swing the ball's speed is at its maximum and its tangential acceleration is zero. Those two facts are the same fact: the speed is at a maximum because it has stopped changing at that instant, and it has stopped changing because gravity has no tangential component there. The centripetal acceleration at that moment is the largest it will be all lap, since $a_c$ goes as $v^2$ .

With the string horizontal the tangential acceleration is exactly $g$ , its largest value anywhere on the circle, and the string, no matter how hard it pulls, contributes nothing to it, because a string pulls along its own length and its length is the radius. Worked example two puts numbers on all three positions.

A tension in a string can therefore never change the speed of an object moving in a circle at the end of it. That is a genuinely useful result and it drops straight out of the perpendicularity: the tension is radial, the speed is changed only by the tangential component, and the radial direction has no tangential component. It is also why the work done by a centripetal force on an object in uniform circular motion is zero.

What the subscript on the equation sheet says

The relationship between the tangential component and the rotation is printed on every one of the four AP equation sheets, in the rotational block, immediately under $v = r\omega$ :

a_T = r\alpha

The subscript is a capital T for tangential. It is worth stating plainly because at ordinary reading size the letter is small and italic and can be taken for an $r$ , and $a_r$ would mean the radial component, which is the other one entirely and is not what that line says. Rendered at four hundred dots per inch, the AP Physics 1 sheet and the AP Physics C: Mechanics sheet both show $a_T = r\alpha$ with a capital T.

The sheets differ in one place that matters here, and it is easy to check for yourself on the formulas pages:

AP Physics 1 and AP Physics 2 print $a_c = \frac{v^2}{r}$ in the translational block and $a_T = r\alpha$ in the rotational block.
AP Physics C: Mechanics and AP Physics C: Electricity and Magnetism print $a_c = \frac{v^2}{r} = r\omega^2$ and $a_T = r\alpha$ .

So a Physics C candidate is handed the angular form of the centripetal acceleration and an algebra-based candidate has to build it, which takes one substitution: put $v = r\omega$ into $\frac{v^2}{r}$ and you get $\frac{r^2\omega^2}{r} = r\omega^2$ . That form is often the faster one when a problem gives you a rotation rate rather than a speed.

Two things none of the four sheets prints, and which you therefore have to know:

The vector sum. There is no line reading $a = \sqrt{a_c^2 + a_T^2}$ anywhere. EK 2.9.A.4 states that the net acceleration is the vector sum, and the arithmetic is left to you.
A symbol for the net acceleration of a point on a circle. The symbol keys on the AP Physics 1 and AP Physics 2 sheets list $a$ as acceleration with no qualifier, so the bare $a$ in $\vec{F}_{\text{net}} = m\vec{a}$ and the subscripted $a_T$ and $a_c$ are three different quantities wearing nearly the same letter.

Where the confusion costs a mark

Writing "uniform circular motion, so the acceleration is zero." It is $\frac{v^2}{r}$ toward the center. What is zero is the tangential component.
Adding $a_c$ and $a_T$ as numbers. They are perpendicular, so the magnitude of the net acceleration is $\sqrt{a_c^2 + a_T^2}$ . Adding them overstates the answer, and by a lot when the two are comparable.
Using $a_T = r\alpha$ to find the acceleration of an object in circular motion. It gives one component. If the question asks for the acceleration, it wants the vector sum.
Using $a_c = \frac{v^2}{r}$ with a speed that is changing, without saying which instant. The formula is instantaneous. On an object that is speeding up, $a_c$ has a different value every moment, and using the initial speed for the whole lap is a common way to lose a mark on a multi-part question.
Treating the centripetal direction as outward. It is inward, by EK 2.9.A.1.ii, at every point and in every case.
Assuming tension changes the speed of a ball on a string. It cannot; it is radial and only the tangential component changes the speed.
Reading $a_T$ as $a_r$ . They mean opposite components. Check the subscript before you substitute.
Forgetting that $a_c$ needs the tangential speed. EK 2.9.A.1.i is specific about which speed goes into $\frac{v^2}{r}$ : the object's tangential speed, which for something moving on the circle is simply its speed.
Calling the mix fixed. Which component dominates depends on the current speed, since $a_c$ scales with $v^2$ and $a_T$ need not change at all.
Assigning a force to each component and then counting the force twice. There is one net force. Its radial component produces $a_c$ and its tangential component produces $a_T$ , and free-body diagrams have entries for real forces only, never for $ma_c$ .

When one of them is zero, and why that lulls you

Most first problems are built so that one of the two components is zero, which is a sensible way to teach and a poor way to build intuition.

Uniform circular motion sets $a_T = 0$ . Every conical pendulum, every banked curve at the design speed, every satellite in a circular orbit, every mass on a string swung at constant speed on a frictionless table. In all of them the net force is entirely radial and the acceleration is entirely centripetal, so the word "acceleration" and the phrase "centripetal acceleration" can be used interchangeably without ever going wrong. That habit then transfers to a problem where the object is speeding up, and there it is wrong.

Straight-line motion sets $a_c = 0$ . Every kinematics problem in Unit 1. There $r$ is infinite, $\frac{v^2}{r}$ is zero, and "acceleration" means what tangential acceleration means here. That is the other half of the same habit.

Rotation starting from rest, for the first instant only. At $t = 0$ the speed is zero, so $a_c = \frac{0}{r} = 0$ and the entire acceleration of a rim point is tangential. It stops being true immediately: after even a fraction of a second $a_c$ is climbing as $v^2$ while $a_T$ sits still. A wheel a second or two into spinning up typically has a centripetal component many times the tangential one.

The three places where both are live at once, and where the exam goes:

A vertical circle. Gravity swings from purely radial at the top and bottom to purely tangential at the sides, so the mix changes continuously through one lap even at fixed radius.
Anything spinning up or slowing down. A grinding wheel, a centrifuge, a car pulling out of a bend.
Rolling down an incline. The center of mass accelerates along the slope while every point on the rim also circles the center. Rolling vs sliding works that case.

The habit to build is a two-part answer. When you are asked for the acceleration of something on a circular path, say the centripetal part, say the tangential part, then combine them if the question wants a single number. If the tangential part turns out to be zero, you have lost nothing by checking.

What the CED asks, and how the exam frames it

Both components are introduced in the same place: AP Physics 1 Unit 2, Force and Translational Dynamics, weighted at 18 to 23 percent of the multiple-choice section, at Topic 2.9, Circular Motion. The learning objective is 2.9.A: describe the motion of an object traveling in a circular path.

Its essential knowledge runs in order. EK 2.9.A.1 defines centripetal acceleration as a component directed at the center, with 2.9.A.1.i giving $a_c = \frac{v^2}{r}$ as the tangential speed squared over the radius and 2.9.A.1.ii restating the direction. EK 2.9.A.2 says the centripetal acceleration can result from a single force, more than one force, or components of forces, with three sub-statements: 2.9.A.2.i on the minimum speed at the top of a vertical loop, where the derived equation is $v = \sqrt{gr}$ ; 2.9.A.2.ii on components of static friction and the normal force on a banked surface; and 2.9.A.2.iii on the tension component in a conical pendulum. EK 2.9.A.3 defines tangential acceleration. EK 2.9.A.4 makes the net acceleration the vector sum. EK 2.9.A.5 introduces period and frequency for the constant-speed case, with $T = \frac{1}{f}$ at 2.9.A.5.i and 2.9.A.5.ii, and the derived $T = \frac{2\pi r}{v}$ at 2.9.A.5.iii. Suggested skills for Topic 2.9 are 1.B, 2.A, 2.D, 3.A and 3.C.

The symbol $a_T$ itself comes from elsewhere. Topic 2.9 names tangential acceleration in words; the equation $a_T = r\alpha$ is given in Unit 5, at Topic 5.2, Connecting Linear and Rotational Motion, EK 5.2.A.2, among the derived relationships of linear velocity and the tangential component of acceleration to their angular counterparts. Unit 5 is weighted at 10 to 15 percent, and suggested skills for Topic 5.2 are 1.C, 2.A, 2.C and 3.B. So a full treatment of the two components spans two units, which is part of why the pieces are often learned separately and then not connected.

Topic 2.9 carries two boundary statements. Quantitative work on banked curves is limited to the case in which no friction is required, and Kepler's first and second laws are excluded from the course. Topic 5.2's boundary statement limits descriptions of the direction of rotation to clockwise and counterclockwise with respect to a given axis.

AP Physics 2 has no circular-motion topic in its course framework, whose units run from thermodynamics through modern physics. Its equation sheet still carries the full mechanics and fluids block, including $a_c = \frac{v^2}{r}$ and $a_T = r\alpha$ .

Where to go next: Topic 2.9 has the full CED framing, the centripetal force guide has the force-identification routine, Topic 5.2 has the link to the angular quantities, and the circular motion practice set has problems. The centripetal force calculator will do the radial arithmetic while you concentrate on the tangential side.

A car speeding up on a circular track

A car enters a circular track of radius $40$ m at $12$ m/s and speeds up along the track at a constant $2.0\ \text{m/s}^2$ . Find the centripetal acceleration, the tangential acceleration, the net acceleration and its direction at the moment of entry. Find the speed at which the two components are equal. Then repeat the acceleration calculation $4.0$ s later, and give the angular acceleration of the car about the center.

At entry, centripetal acceleration from EK 2.9.A.1.i: $a_c = \frac{v^2}{r} = \frac{(12)^2}{40} = \frac{144}{40} = 3.6\ \text{m/s}^2$ , directed at the center of the track.
Tangential acceleration at entry: the car speeds up at $2.0\ \text{m/s}^2$ along the track, and EK 2.9.A.3 defines that as the tangential acceleration. $a_T = 2.0\ \text{m/s}^2$ , directed forward along the tangent.
Net acceleration at entry. The components are perpendicular, so by EK 2.9.A.4: $a = \sqrt{a_c^2 + a_T^2} = \sqrt{(3.6)^2 + (2.0)^2} = \sqrt{12.96 + 4.00} = \sqrt{16.96} = 4.12\ \text{m/s}^2$ .
Its direction: $\arctan\frac{a_T}{a_c} = \arctan\frac{2.0}{3.6} = 29.1^\circ$ from the inward radial direction, tilted toward the front of the car.
The crossover speed. Set $\frac{v^2}{r} = a_T$ : $v = \sqrt{r\,a_T} = \sqrt{(40)(2.0)} = \sqrt{80} = 8.94\ \text{m/s}$ . The car entered above this speed, so the centripetal component was already the larger of the two at entry.
Four seconds later. Speed: $v = v_0 + a_T t = 12 + (2.0)(4.0) = 20\ \text{m/s}$ .
Centripetal acceleration now: $a_c = \frac{(20)^2}{40} = \frac{400}{40} = 10.0\ \text{m/s}^2$ . It has grown by a factor of $\left(\frac{20}{12}\right)^2 = 2.78$ while the tangential acceleration has not changed at all.
Net acceleration now: $a = \sqrt{(10.0)^2 + (2.0)^2} = \sqrt{100 + 4} = \sqrt{104} = 10.2\ \text{m/s}^2$ , at $\arctan\frac{2.0}{10.0} = 11.3^\circ$ from radial. The acceleration vector has swung from $29.1^\circ$ off radial to $11.3^\circ$ off it with no change in what the driver is doing.
Angular acceleration about the center, from EK 5.2.A.2 rearranged: $\alpha = \frac{a_T}{r} = \frac{2.0}{40} = 0.050\ \text{rad/s}^2$ . This is constant for the whole manoeuvre, which is exactly the tangential component's behaviour and not the centripetal one's.
As a check on the two descriptions agreeing, the distance covered in those $4.0$ s is $v_0 t + \frac{1}{2}a_T t^2 = (12)(4.0) + \frac{1}{2}(2.0)(16) = 48 + 16 = 64$ m, and the angle swept is $\frac{s}{r} = \frac{64}{40} = 1.6$ rad. Through the angular route: $\theta = \omega_0 t + \frac{1}{2}\alpha t^2$ with $\omega_0 = \frac{12}{40} = 0.30\ \text{rad/s}$ , giving $(0.30)(4.0) + \frac{1}{2}(0.050)(16) = 1.2 + 0.4 = 1.6$ rad. They agree.

At entry: $a_c = 3.6\ \text{m/s}^2$ , $a_T = 2.0\ \text{m/s}^2$ , net $4.12\ \text{m/s}^2$ at $29.1^\circ$ from radial. The two are equal at $8.94$ m/s. After $4.0$ s at $20$ m/s: $a_c = 10.0\ \text{m/s}^2$ , $a_T$ still $2.0\ \text{m/s}^2$ , net $10.2\ \text{m/s}^2$ at $11.3^\circ$ from radial. The angular acceleration is $0.050\ \text{rad/s}^2$ throughout.

A ball on a string in a vertical circle: gravity swaps roles

A ball on a light string of length $0.80$ m is swung in a vertical circle. At the lowest point its speed is $8.0$ m/s. Using $g = 9.8\ \text{m/s}^2$ and ignoring air resistance, find the centripetal and tangential accelerations at the lowest point, at the point where the string is horizontal, and at the highest point. Confirm the string stays taut at the top.

Lowest point, tangential. Gravity points straight down. At the bottom of the circle the radius is vertical, so gravity is entirely along the radius and has no component along the tangent. The string also pulls along the radius. With no tangential force, $a_T = 0$ .
Lowest point, centripetal. $a_c = \frac{v^2}{r} = \frac{(8.0)^2}{0.80} = \frac{64}{0.80} = 80\ \text{m/s}^2$ , directed straight up at the center. Maximum speed and zero tangential acceleration at the same instant, which is not a coincidence: the speed peaks because it has momentarily stopped changing.
String horizontal, speed. Energy conservation from the bottom, rising a height equal to the radius, $h = 0.80$ m: $v^2 = v_0^2 - 2gh = 64 - 2(9.8)(0.80) = 64 - 15.68 = 48.32$ , so $v = 6.95$ m/s.
String horizontal, tangential. The radius is now horizontal, so gravity is entirely tangential. The string is horizontal too and pulls along the radius, contributing nothing tangential. $a_T = g = 9.8\ \text{m/s}^2$ , directed downward along the tangent, slowing the ball on the way up. This is the largest tangential acceleration anywhere on the circle, because it is the one position where all of gravity acts along the path.
String horizontal, centripetal. $a_c = \frac{48.32}{0.80} = 60.4\ \text{m/s}^2$ , directed horizontally at the center. Net acceleration $\sqrt{(60.4)^2 + (9.8)^2} = \sqrt{3648.2 + 96.0} = 61.2\ \text{m/s}^2$ , tilted $\arctan\frac{9.8}{60.4} = 9.2^\circ$ from radial.
Highest point, speed. Rising a height of $2r = 1.6$ m from the bottom: $v^2 = 64 - 2(9.8)(1.6) = 64 - 31.36 = 32.64$ , so $v = 5.71$ m/s.
Highest point, components. The radius is vertical again, so gravity is entirely radial and $a_T = 0$ once more. $a_c = \frac{32.64}{0.80} = 40.8\ \text{m/s}^2$ , directed straight down at the center.
Does the string stay taut? EK 2.9.A.2.i gives the minimum speed at the top of a vertical circular loop as the case where gravity alone supplies the centripetal acceleration, $v = \sqrt{gr} = \sqrt{(9.8)(0.80)} = 2.80$ m/s. The ball is doing $5.71$ m/s there, comfortably above it, so the string is still pulling.
Collect the pattern. $a_T$ went $0$ , then $9.8$ , then $0$ , purely from where gravity points relative to the path. $a_c$ went $80$ , then $60.4$ , then $40.8$ , purely from how fast the ball is going. Two components of one acceleration, driven by two completely different things.

Lowest point: $a_c = 80\ \text{m/s}^2$ upward, $a_T = 0$ . String horizontal, at $6.95$ m/s: $a_c = 60.4\ \text{m/s}^2$ inward, $a_T = 9.8\ \text{m/s}^2$ downward along the path, net $61.2\ \text{m/s}^2$ at $9.2^\circ$ from radial. Highest point, at $5.71$ m/s: $a_c = 40.8\ \text{m/s}^2$ downward, $a_T = 0$ . The minimum speed for a taut string at the top is $2.80$ m/s, so the string stays taut.

A flywheel slowing down: tangential acceleration goes backward

A flywheel of radius $0.25$ m is turning at $\omega = 20\ \text{rad/s}$ and is being braked at a constant $\alpha = 6.0\ \text{rad/s}^2$ opposing the motion. For a point on the rim at that instant, find the tangential acceleration, the centripetal acceleration and the net acceleration, and state the direction of each. Take the direction of rotation as positive.

Declare the sign convention first: the direction of rotation is positive, so the braking gives $\alpha = -6.0\ \text{rad/s}^2$ and the wheel's $\omega = +20\ \text{rad/s}$ .
Tangential acceleration, EK 5.2.A.2: $a_T = r\alpha = (0.25)(-6.0) = -1.5\ \text{m/s}^2$ . The negative sign means it points backward along the tangent, against the rim point's motion. This is the case the word centripetal has no version of: a tangential component can point either way.
Rim speed: $v = r\omega = (0.25)(20) = 5.0$ m/s.
Centripetal acceleration: $a_c = \frac{v^2}{r} = \frac{25}{0.25} = 100\ \text{m/s}^2$ , directed at the axle. Cross-check with the Physics C form $a_c = r\omega^2 = (0.25)(400) = 100\ \text{m/s}^2$ . The two agree, as they must, since one is the other with $v = r\omega$ substituted.
The centripetal component is inward whether the wheel is speeding up, slowing down or holding steady. Braking changes the sign of $a_T$ and does not touch the direction of $a_c$ .
Net acceleration: $a = \sqrt{(100)^2 + (1.5)^2} = \sqrt{10000 + 2.25} = 100.0\ \text{m/s}^2$ , tilted $\arctan\frac{1.5}{100} = 0.86^\circ$ from the inward radial direction, on the trailing side.
A sanity check on the size. At $20\ \text{rad/s}$ the centripetal component is $67$ times the tangential one here. That ratio is $\frac{r\omega^2}{r\lvert\alpha\rvert} = \frac{\omega^2}{\lvert\alpha\rvert} = \frac{400}{6.0} = 66.7$ , and notice that the radius cancels out of it entirely: how lopsided the mix is depends on the rotation rate and the braking rate, not on how big the wheel is.

$a_T = -1.5\ \text{m/s}^2$ , that is $1.5\ \text{m/s}^2$ backward along the tangent, and $a_c = 100\ \text{m/s}^2$ inward. The net acceleration is $100.0\ \text{m/s}^2$ , tilted $0.86^\circ$ from radial. The centripetal component stays inward while the wheel brakes; only the tangential component reverses.

Frequently asked questions

What is the difference between centripetal and tangential acceleration?

They are two perpendicular components of the same acceleration of the same object. Centripetal acceleration points toward the center of the circular path and changes the direction of the velocity; its magnitude is the speed squared divided by the radius. Tangential acceleration points along the path and changes the size of the velocity; its magnitude is the radius times the angular acceleration. The AP Physics 1 CED defines both as components at essential knowledge 2.9.A.1 and 2.9.A.3, and essential knowledge 2.9.A.4 says the net acceleration is their vector sum. Because they are at right angles, combine them as the square root of the sum of their squares.

Is there acceleration in uniform circular motion?

Yes. Uniform circular motion means constant speed, so the tangential component is zero, but the direction of the velocity is changing continuously and the centripetal component is the speed squared over the radius, directed at the center. That is the whole acceleration in that case, and it is not small: an object moving at 12 meters per second on a 40 meter circle has a centripetal acceleration of 3.6 meters per second squared. A net force is required to sustain it, supplied by tension, friction, gravity, a normal force, or components of several, as the CED notes at essential knowledge 2.9.A.2.

Can an object have both centripetal and tangential acceleration at once?

Yes, and any object speeding up or slowing down on a curved path has both. A car entering a 40 meter circular track at 12 meters per second while accelerating along the track at 2.0 meters per second squared has a centripetal component of 3.6 and a tangential component of 2.0, giving a net acceleration of 4.12 meters per second squared tilted about 29 degrees from the inward radial direction. Which component is larger depends on the current speed, because the centripetal one grows with the square of the speed while the tangential one need not change at all.

How do you add centripetal and tangential acceleration?

As perpendicular vector components, never as plain numbers. The magnitude of the net acceleration is the square root of the centripetal component squared plus the tangential component squared, and its direction is tilted from the inward radial direction by the angle whose tangent is the tangential component divided by the centripetal one. Essential knowledge 2.9.A.4 in the AP Physics 1 CED states that the net acceleration of an object moving in a circle is the vector sum of the two. Adding them arithmetically overstates the answer, and by the most when the two are comparable in size.

Does tangential acceleration exist in uniform circular motion?

No. Tangential acceleration is defined as the rate at which an object's speed changes, at essential knowledge 2.9.A.3, and uniform circular motion is motion at constant speed, so the tangential component is exactly zero. The angular acceleration is zero for the same reason, since the two are related by the tangential acceleration equalling the radius times the angular acceleration. What remains is the centripetal component, which is nonzero for any object actually moving on the circle.

Why is the tangential acceleration zero at the bottom of a vertical circle?

Because gravity is the only force with any chance of acting along the path there, and at the lowest point the radius is vertical, so gravity points straight along the radius and has no tangential component at all. The string or rod also pulls along the radius. With nothing pushing the object along its path, its speed is momentarily not changing, which is also why the speed is at its maximum at that point. One quarter turn later, with the string horizontal, gravity is entirely tangential and the tangential acceleration reaches its largest value on the circle, equal to g.

Does the equation sheet print tangential acceleration?

Yes. All four AP equation sheets print the line a with a capital T subscript equals r alpha, in the rotational block directly below v equals r omega. The subscript is a capital T for tangential, not an r for radial, which would mean the other component. The AP Physics 1 and AP Physics 2 sheets print centripetal acceleration as the speed squared over the radius; the AP Physics C: Mechanics and AP Physics C: Electricity and Magnetism sheets print it as the speed squared over the radius and also as r omega squared. None of the four sheets prints the vector sum of the two components, so you supply that yourself.