Linear vs Angular Velocity: The Difference

Angular velocity is how fast something turns, in radians per second, and every point on a rigid system shares one value of it. Linear velocity is how fast a particular point moves through space, in meters per second, and it grows with distance from the axis. They are linked by v equals r omega.

AP Physics: Unit 5 (topics 1.2 Displacement, Velocity, and Acceleration, 5.1 Rotational Kinematics, 5.2 Connecting Linear and Rotational Motion). Linear velocity is defined in AP Physics 1 Unit 1, Kinematics, at Topic 1.2, learning objective 1.2.B, describe the average velocity and acceleration of an object, with EK 1.2.B.2 giving average velocity as displacement divided by the time interval. Angular velocity is defined in Unit 5, Torque and Rotational Dynamics, weighted at 10 to 15 percent of the multiple-choice section over about 15 to 20 class periods. Topic 5.1 has one learning objective, 5.1.A, describe the rotation of a system with respect to time using angular displacement, angular velocity, and angular acceleration, carrying EK 5.1.A.1 through 5.1.A.4; EK 5.1.A.1 defines angular displacement as the measurement of the angle, in radians, through which a point on a rigid system rotates about a specified axis, and EK 5.1.A.4.i gives the three constant-angular-acceleration equations. Topic 5.2 is the bridge, 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, carrying EK 5.2.A.1, the arc-length relation, EK 5.2.A.2, the derived relationships s = r theta, v = r omega and a_T = r alpha, and EK 5.2.A.3, which states that all points within a rigid system have the same angular velocity and angular acceleration. 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. The radian requirement applies to the equations in which an angle multiplies a length, not to the constant-angular-acceleration equations, which are consistent in any single angular unit. Suggested skills are 1.C, 2.B, 2.C and 3.C for Topic 1.2, 1.B, 2.A, 2.D, 3.A and 3.C for Topic 5.1, and 1.C, 2.A, 2.C and 3.B for Topic 5.2.

One number for the whole system, one number per point

Spin a bicycle wheel. Ask how fast it is turning and there is one answer. Ask how fast it is moving and the question is incomplete, because you have to say which part.

The AP Physics 1 CED makes the first half of that a required statement. EK 5.2.A.3: for a rigid system, all points within that system have the same angular velocity and angular acceleration. One number, shared by every point from the hub to the rim.

Linear velocity has no such property, and the CED's own appendix spells out why. Its definition of a rigid system reads: a system that does not change shape, but different points within the system may move in different directions and with different speeds. The discussion that follows marks two points A and B on a rotating wheel and says that at any given instant Point A is traveling with a greater translational speed and in a different direction than Point B. Different speeds, different directions, same wheel, same instant.

EK 5.1.A.1.i says the same thing from the other side: a rigid system is one that holds its shape but in which different points on the system move in different directions during rotation, and a rigid system cannot be modeled as an object. That last clause is the reason angular velocity exists as a separate quantity. If you could collapse the wheel to a point at its center, its linear velocity would be one number and you would not need a second quantity.

So the two answer different questions:

Angular velocity answers how fast the arrangement is rotating. EK 5.1.A.2 defines the average version: average angular velocity is the average rate at which angular position changes with respect to time, $\omega_{\text{avg}} = \frac{\Delta\theta}{\Delta t}$ .
Linear velocity answers how fast one chosen point is moving through space. EK 1.2.B.2 defines the average version: average velocity is the displacement of an object divided by the interval of time in which that displacement occurs, $\vec{v}_{\text{avg}} = \frac{\Delta\vec{x}}{\Delta t}$ .

EK 5.2.A.2 links them for a point at distance $r$ from the axis:

v = r\omega

Read $r$ as the part of that equation that makes the linear velocity a property of a point rather than of the system. Change $r$ and $v$ changes; $\omega$ does not.

Linear vs angular velocity, side by side

Question you are asking	Linear velocity	Angular velocity
Symbol	$\vec{v}$	$\omega$
SI unit	$\text{m/s}$	$\text{rad/s}$
What it belongs to	One point, or one object	The whole rigid system
Same for every point on a spinning wheel	No, it scales with $r$	Yes, EK 5.2.A.3
Definition	$\frac{\Delta\vec{x}}{\Delta t}$ , EK 1.2.B.2	$\frac{\Delta\theta}{\Delta t}$ , EK 5.1.A.2
Direction	Tangent to the path, changing continuously	Clockwise or counterclockwise about a stated axis
Zero at the axis	Yes, a point on the axis has $v = 0$	No, the system still turns
Link between them	$v = r\omega$ , EK 5.2.A.2	$\omega = \frac{v}{r}$ for a point at distance $r$
Its acceleration counterpart	$a_T = r\alpha$ , tangential only	$\alpha = \frac{\Delta\omega}{\Delta t}$ , EK 5.1.A.3
Its kinetic energy	$K = \frac{1}{2}mv^2$	$K = \frac{1}{2}I\omega^2$
Angular unit required	Not applicable	Radians whenever it multiplies a radius
CED essential knowledge	1.2.B.2, 1.2.B.4	5.1.A.1, 5.1.A.2, 5.2.A.3

The two rows to sit with:

"Zero at the axis." A point on the rotation axis of a spinning wheel has zero linear velocity while the wheel turns at full speed. That single fact makes it impossible for the two quantities to be the same thing described two ways. One of them is zero and the other is not, in the same object at the same instant.

"Direction." These behave completely differently and the CED restricts the angular case deliberately. The boundary statement under Topic 5.1, repeated verbatim under 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 angular velocity gets two possible directions and a sign convention, which EK 5.1.A.1.ii describes: one direction of angular displacement about an axis of rotation, clockwise or counterclockwise, is typically indicated as mathematically positive, with the other direction becoming mathematically negative. Linear velocity, meanwhile, points along the tangent and so sweeps continuously through every direction in the plane during one revolution.

The case that separates them: two points on one wheel

A wheel of radius $0.34$ m turns at a steady $\omega = 8.0\ \text{rad/s}$ . Compare a point on the rim with a point $0.12$ m from the axis, at the same instant.

	Rim point, $r = 0.34$ m	Inner point, $r = 0.12$ m
Angular velocity	$8.0\ \text{rad/s}$	$8.0\ \text{rad/s}$
Period of one revolution	$0.785$ s	$0.785$ s
Linear speed, $v = r\omega$	$2.72\ \text{m/s}$	$0.96\ \text{m/s}$
Distance traveled per revolution	$2.14$ m	$0.754$ m
Direction of the linear velocity	Tangent at that point	Tangent at that point, a different direction

Both points complete a revolution in the same $0.785$ s, because they turn through the same $2\pi$ radians at the same rate. The rim point covers $2.83$ times the distance in that time, because its circle is $2.83$ times bigger, so its speed is $2.83$ times larger. That ratio is exactly $\frac{0.34}{0.12}$ , which is the whole content of $v = r\omega$ .

The two quantities disagree by a factor you get to choose by picking a point, and they agree only when $r$ happens to be $1$ m. There is nothing physical about that coincidence; it is a numerical accident of the meter.

The same experiment run at the axis gives the cleanest version. A point at $r = 0$ has $v = 0$ while the wheel turns at $8.0\ \text{rad/s}$ . A rider on a merry-go-round standing at the center is not moving through space and is very definitely rotating.

Worked example one runs these numbers and worked example three shows that a single angular acceleration produces two different linear accelerations at the same point, in perpendicular directions.

Radians, and exactly where the factor bites

This is the part that produces wrong answers, and the rule is narrower than it is usually stated. Getting it half right is worse than not knowing it.

The correct rule: the radian requirement applies the moment an angle multiplies a length. Nowhere else.

Equations that require radians, because an angular quantity is multiplied by a radius:

Equation	Source
$\Delta s = r\Delta\theta$	EK 5.2.A.1
$s = r\theta$	EK 5.2.A.2, and the sheet's geometry table
$v = r\omega$	EK 5.2.A.2
$a_T = r\alpha$	EK 5.2.A.2
$\Delta x_{\text{cm}} = r\Delta\theta$	AP Physics 1 equation sheet, rotational block
$K = \frac{1}{2}I\omega^2$ and $L = I\omega$	Both derive from $v = r\omega$

The reason is the definition of the radian: it is an arc length divided by a radius, a ratio of two lengths, so $s = r\theta$ holds as an identity only in radians. Substitute degrees and every one of those equations is wrong by the conversion factor, $1\ \text{rad} = \frac{180^\circ}{\pi} = 57.2958^\circ$ . The CED writes the requirement into the definition itself, at EK 5.1.A.1: angular displacement is the measurement of the angle, in radians, through which a point on a rigid system rotates about a specified axis.

Equations that do not care which angular unit you use, as long as you use one unit consistently:

\omega = \omega_0 + \alpha t

\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2

\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)

Those are the three constant-angular-acceleration relationships from EK 5.1.A.4.i, and degrees work in all three. No length appears in any of them, so there is nothing for the radian's definition to be needed for. Every term in the first equation carries one factor of the angular unit, every term in the second carries one, and every term in the third carries two, so a change of unit multiplies both sides of each equation by the same factor and cancels. The same is true of $\omega_{\text{avg}} = \frac{\Delta\theta}{\Delta t}$ and $\alpha_{\text{avg}} = \frac{\Delta\omega}{\Delta t}$ : degrees per second and degrees per second squared are perfectly valid there.

Worked example two solves a turntable problem twice, once entirely in radians and once entirely in degrees, gets consistent answers from all three kinematics equations both times, and then computes an arc length where only the radian version survives.

So the practical procedure is: work the angular kinematics in whatever unit the problem gave you, then convert to radians before the first equation that contains an $r$ . In practice, converting to radians at the start is simpler and always safe, and it is what the equation sheet's symbol key assumes when it labels $\omega$ as angular speed.

One genuine complication worth knowing about, since it looks like a contradiction. The AP Physics 1 equation sheet prints its trigonometric values in degrees, for $0^\circ$ , $30^\circ$ , $37^\circ$ , $45^\circ$ , $53^\circ$ , $60^\circ$ and $90^\circ$ . That is not inconsistent. Those angles are geometric angles inside a $\sin$ or $\cos$ , such as the $\theta$ in $\tau = rF\sin\theta$ or in $W = Fd\cos\theta$ , and a trigonometric function returns the same number whichever unit you feed it in, provided your calculator is set to match. Geometric angles in degrees, rotational angles in radians, and the dividing line is whether the angle multiplies a radius.

One angular quantity, several linear ones

The mapping between the two is not one to one, and that is worth setting out because it catches people who have learned to swap symbols mechanically.

A point moving on a circle has an acceleration with two perpendicular parts, and the CED names both.

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, with $a_c = \frac{v^2}{r}$ .
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, and EK 5.2.A.2 gives it as $a_T = r\alpha$ .
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.

Meanwhile there is exactly one angular acceleration, $\alpha$ , a single number with a clockwise or counterclockwise sense. So:

Angular quantity	Linear counterparts at a point at distance $r$
$\theta$	Arc length $s = r\theta$ , and a position on the circle
$\omega$	Speed $v = r\omega$ , tangential, direction changing constantly
$\alpha$	Tangential acceleration $a_T = r\alpha$ , and separately the centripetal acceleration $a_c = \frac{v^2}{r}$ , which exists even when $\alpha = 0$

The bottom-right cell is where the analogy stops being a substitution rule. A wheel turning at constant angular velocity has $\alpha = 0$ and $a_T = 0$ , and every point on it still has a nonzero centripetal acceleration. Nothing angular corresponds to $a_c$ , because it comes from the direction of the velocity changing rather than from the rotation rate changing.

Worked example three makes the size of the gap concrete: on a grinding wheel spinning up, the centripetal acceleration is about $67$ times the tangential one, and one of them grows as the wheel speeds up while the other holds still.

One more asymmetry. Angular velocity does not depend on where you measure it, but it does depend on which axis you named. EK 5.1.A.1 says angular displacement is measured about a specified axis, and EK 5.1.A.1.iii adds that if the rotation of a system about an axis may be well described using the motion of the system's center of mass, the system may be treated as a single object, giving the example that the rotation of Earth about its axis may be considered negligible when considering the revolution of Earth about the center of mass of the Earth and Sun system. Same planet, two rotations, two angular velocities, and which one you mean depends entirely on the axis you named.

Where the confusion costs a mark

Each of these is a specific scoring error.

Using $v = r\omega$ with $\omega$ in degrees per second. Wrong by $57.2958$ , and the answer is so large that it usually survives no sanity check at all. Convert to radians per second first.
Assuming degrees are wrong in the angular kinematics equations too. They are not. $\omega = \omega_0 + \alpha t$ , $\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2$ and $\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$ hold in any consistent angular unit, and needlessly converting is a source of arithmetic slips rather than a safeguard. What matters is converting before an equation that contains an $r$ .
Using revolutions or revolutions per minute directly in $v = r\omega$ . One revolution is $2\pi$ radians, so an rpm figure needs two conversions: divide by $60$ for per second, then multiply by $2\pi$ .
Quoting a single linear speed for a rotating rigid system. Ask which point. The CED's appendix says different points may move with different speeds, and EK 5.1.A.1.i says a rigid system cannot be modeled as an object.
Saying the outer point has a larger angular velocity. EK 5.2.A.3 says all points within a rigid system have the same angular velocity and angular acceleration. The outer point has a larger linear speed, which is a different claim.
Saying a point at the axis is not rotating. It has $v = 0$ and shares the system's $\omega$ .
Forgetting the centripetal acceleration when $\alpha = 0$ . Constant angular velocity means $a_T = 0$ , not $a = 0$ . EK 2.9.A.4 makes the net acceleration the vector sum of the two parts.
Adding $a_c$ and $a_T$ as numbers. They are perpendicular, so combine them as vector components.
Reporting an angular velocity without naming the axis. EK 5.1.A.1 measures angular displacement about a specified axis, and the Topic 5.1 boundary statement limits direction descriptions to clockwise and counterclockwise with respect to a given axis of rotation.
Letting the sign convention drift. EK 5.1.A.1.ii lets you choose which sense is positive. Choose once, write it down, and hold it to the end of the problem.

When they track each other, and why that lulls you

Three situations let you treat the two as interchangeable, and they cover most first problems.

A single point at a fixed radius. If the whole question concerns the rim of one wheel, $r$ is a constant and $v$ is just $\omega$ multiplied by that constant. Every proportional statement you make about one is true of the other, so the distinction never surfaces.

A question phrased entirely in one language. "The wheel spins at $8.0\ \text{rad/s}$ ; find the angular displacement in $3.0$ s" never leaves the angular column, and "the car travels at $24$ m/s; how far in $3.0$ s" never leaves the linear one. No conversion, no radian rule, no chance to slip.

A radius of one meter. Then $v$ and $\omega$ share a numeric value and the two columns of your working look identical. This is a trap rather than a simplification, because it teaches that the numbers should match.

The distinction reappears in four exam-shaped places.

Two points at different radii. Gears, pulleys of different sizes, a person on a merry-go-round moving inward or outward.
A belt or a rope connecting two wheels. The belt does not stretch, so the two rims share a linear speed while their angular velocities differ inversely with their radii. This is the case people get backwards.
Rolling. Topic 6.5 links the translational motion of the center of mass to the rotation, and the sheet's $\Delta x_{\text{cm}} = r\Delta\theta$ is the bridge, which is another equation with an $r$ in it and therefore another place radians are compulsory. Topic 6.5, Rolling covers it.
Energy. $K = \frac{1}{2}mv^2$ for the center of mass and $K = \frac{1}{2}I\omega^2$ for the rotation are separate terms, and a rolling object has both, which is Topic 6.1.

The reflex: when a problem mentions two radii, decide first which of the two quantities is shared, then convert. Points on one rigid body share $\omega$ . Points linked by an inextensible belt or by rolling contact share $v$ .

What the CED asks, and how the exam frames it

The two quantities are defined in two different units, and the bridge between them has a topic of its own.

Linear velocity is defined in Unit 1, Kinematics, at Topic 1.2, Displacement, Velocity, and Acceleration, learning objective 1.2.B: describe the average velocity and acceleration of an object. EK 1.2.B.2 gives $\vec{v}_{\text{avg}} = \frac{\Delta\vec{x}}{\Delta t}$ , EK 1.2.B.4 says an object is accelerating if the magnitude and/or direction of the object's velocity are changing, and EK 1.2.B.5 says calculating an average over a very small time interval yields a value very close to the instantaneous one. Suggested skills for Topic 1.2 are 1.C, 2.B, 2.C and 3.C.

Angular velocity is defined in Unit 5, Torque and Rotational Dynamics, weighted at 10 to 15 percent of the multiple-choice section over about 15 to 20 class periods. Topic 5.1, Rotational Kinematics, has one learning objective, 5.1.A: describe the rotation of a system with respect to time using angular displacement, angular velocity, and angular acceleration. It carries EK 5.1.A.1 through 5.1.A.4, including the radian requirement in 5.1.A.1, the rigid-system definition in 5.1.A.1.i, the sign convention in 5.1.A.1.ii, the three constant-acceleration equations in 5.1.A.4.i, and the graph reading in 5.1.A.4.ii. 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. Three pieces of essential knowledge sit under it: 5.2.A.1 gives $\Delta s = r\Delta\theta$ , 5.2.A.2 gives the derived relationships $s = r\theta$ , $v = r\omega$ and $a_T = r\alpha$ , and 5.2.A.3 says all points within a rigid system share the angular velocity and angular acceleration. Suggested skills for Topic 5.2 are 1.C, 2.A, 2.C and 3.B.

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 you will not be asked for an angular velocity as a vector along the axis, and the right-hand rule for $\vec{\omega}$ is outside AP Physics 1.

EK 5.1.A.4 is worth reading as the exam's own summary of the analogy: 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. The mathematics transfers; the ownership does not, which is what this page is about.

On the AP Physics 1 equation sheet, the rotational block opens with the three constant-angular-acceleration equations and then prints $v = r\omega$ and $a_T = r\alpha$ next to each other, followed later by $\Delta x_{\text{cm}} = r\Delta\theta$ . The symbol key labels $\omega$ as angular speed, $\theta$ as angle or angular position, and $v$ as velocity or speed. The geometry table on the preceding page prints $s = r\theta$ with a figure of an arc, which is the identity the radian rule comes from.

For the routines, rotational kinematics has the three angular equations worked, and the kinematic equations guide has their linear originals. The CED framing is on Topic 5.1, Topic 5.2 and Topic 1.2, and the torque and rotational motion practice set has problems to work.

Two points on one wheel, same angular velocity, different speeds

A wheel of radius $0.34$ m turns at a constant $\omega = 8.0\ \text{rad/s}$ . Compare a point on the rim with a point $0.12$ m from the axis: find the angular velocity, linear speed, period, and distance traveled per revolution for each. Then show what happens if you use degrees per second in $v = r\omega$ .

Angular velocity, both points. EK 5.2.A.3 settles it without a calculation: for a rigid system, all points within that system have the same angular velocity. Both points turn at $8.0\ \text{rad/s}$ .
Linear speed of the rim point, from EK 5.2.A.2: $v = r\omega = (0.34\ \text{m})(8.0\ \text{rad/s}) = 2.72\ \text{m/s}$ . The radian is a ratio of two lengths and so carries no dimension, which is why $\text{m} \times \text{rad/s}$ comes out as $\text{m/s}$ .
Linear speed of the inner point: $v = (0.12)(8.0) = 0.96\ \text{m/s}$ . The ratio of the two speeds is $\frac{2.72}{0.96} = 2.83$ , which is exactly $\frac{0.34}{0.12}$ , as $v = r\omega$ requires with $\omega$ shared.
Period, both points: $T = \frac{2\pi}{\omega} = \frac{2\pi}{8.0} = 0.785\ \text{s}$ . The same for both, because both sweep the same $2\pi$ radians at the same rate. A stopwatch cannot tell the two points apart.
Distance traveled per revolution. Rim: $2\pi r = 2\pi(0.34) = 2.14\ \text{m}$ . Inner: $2\pi(0.12) = 0.754\ \text{m}$ . Ratio $2.83$ again, and this is the physical reason the speeds differ: same time, different distances.
Now the degrees test. Convert: $\omega = 8.0\ \text{rad/s} \times 57.2958\ \frac{\text{deg}}{\text{rad}} = 458\ \text{deg/s}$ . Substituting that into $v = r\omega$ gives $(0.34)(458) = 156$ , which would be $156$ m/s for a bicycle wheel, about $560$ km/h. Wrong by the factor $57.2958$ , and wrong in a way that a moment's thought about the size of the answer catches.
One more comparison to fix the ownership. Take a third point, on the axis itself, at $r = 0$ . Its linear speed is $(0)(8.0) = 0\ \text{m/s}$ and its angular velocity is $8.0\ \text{rad/s}$ , the same as everywhere else. Three points on one rigid wheel: three linear speeds, $2.72$ , $0.96$ and $0$ m/s, and one angular velocity.

Both points share $\omega = 8.0\ \text{rad/s}$ and a period of $0.785$ s. The rim point moves at $2.72$ m/s and covers $2.14$ m per revolution; the inner point moves at $0.96$ m/s and covers $0.754$ m, a ratio of $2.83$ in both cases. A point on the axis has $v = 0$ and the same $\omega$ . Feeding $458$ deg/s into $v = r\omega$ returns $156$ m/s, too large by $57.2958$ .

The same turntable in radians and in degrees

A turntable starts from rest and accelerates uniformly at $3.0\ \text{rad/s}^2$ for $4.0$ s. Find the final angular velocity and the angular displacement using all three constant-angular-acceleration equations. Then repeat the whole calculation working only in degrees. Finally find the arc length traveled and the final linear speed of a point $0.15$ m from the axis, in both unit systems, and say which answers are valid.

Radians, first equation. $\omega = \omega_0 + \alpha t = 0 + (3.0)(4.0) = 12\ \text{rad/s}$ .
Radians, second equation. $\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2 = 0 + 0 + \frac{1}{2}(3.0)(4.0)^2 = \frac{1}{2}(3.0)(16) = 24\ \text{rad}$ .
Radians, third equation, as a check. $\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0) = 0 + 2(3.0)(24) = 144$ , so $\omega = 12\ \text{rad/s}$ , agreeing with the first equation.
Now convert the acceleration to degrees and start again. $\alpha = 3.0\ \text{rad/s}^2 \times 57.2958 = 171.9\ \text{deg/s}^2$ .
Degrees, first equation. $\omega = 0 + (171.887)(4.0) = 687.5\ \text{deg/s}$ . Convert back: $\frac{687.55}{57.2958} = 12.0\ \text{rad/s}$ , the same physical answer.
Degrees, second equation. $\theta = \frac{1}{2}(171.887)(16) = 1375\ \text{deg}$ . Convert back: $\frac{1375.1}{57.2958} = 24.0\ \text{rad}$ , the same physical answer. Note $1375^\circ$ is a little under four full turns, and $24$ rad is $3.82$ turns, so the two descriptions agree.
Degrees, third equation. $\omega^2 = 2(171.887)(1375.1) = 4.727 \times 10^5$ , so $\omega = 687.5\ \text{deg/s}$ , matching the degrees version of the first equation exactly. All three constant-angular-acceleration equations worked perfectly in degrees. Nothing was lost, because no length appears in any of them.
Arc length, where the unit stops being free. At $r = 0.15$ m, using EK 5.2.A.1 with radians: $\Delta s = r\Delta\theta = (0.15)(24) = 3.6\ \text{m}$ . Using the degree figure instead: $(0.15)(1375.1) = 206\ \text{m}$ . The ratio of those two is $\frac{206.26}{3.6} = 57.2958$ , the conversion factor, and only $3.6$ m is correct.
Linear speed, same story. With radians: $v = r\omega = (0.15)(12) = 1.8\ \text{m/s}$ . With degrees per second: $(0.15)(687.55) = 103\ \text{m/s}$ , wrong by the same factor.
State the rule the two halves of this example establish. The angular kinematics equations are unit-agnostic: use degrees throughout and every answer converts back correctly. The factor of $57.2958$ appears the moment an angle multiplies a radius, which is $\Delta s = r\Delta\theta$ , $v = r\omega$ , $a_T = r\alpha$ and $\Delta x_{\text{cm}} = r\Delta\theta$ . Converting to radians at the start is the safe habit precisely because you cannot then forget the boundary.

$\omega = 12\ \text{rad/s}$ and $\theta = 24$ rad, equivalently $687.5$ deg/s and $1375$ deg. All three constant-angular-acceleration equations give consistent answers in either unit. The arc length at $r = 0.15$ m is $3.6$ m and the linear speed is $1.8$ m/s, and both require radians: the degree figures return $206$ m and $103$ m/s, too large by $57.2958$ .

One angular acceleration, two linear accelerations

A grinding wheel of radius $0.18$ m is turning at $20\ \text{rad/s}$ and speeding up at a constant $6.0\ \text{rad/s}^2$ . For a point on the rim, find the linear speed, the tangential acceleration and the centripetal acceleration at that instant, and the magnitude of the net linear acceleration. Then find the same quantities $2.0$ s later, and the arc length the point travels in that time. Take the direction of rotation as positive.

Linear speed at the start, from EK 5.2.A.2: $v = r\omega = (0.18)(20) = 3.6\ \text{m/s}$ , directed along the tangent.
Tangential acceleration, also EK 5.2.A.2: $a_T = r\alpha = (0.18)(6.0) = 1.08\ \text{m/s}^2$ , directed along the tangent in the direction of motion, since the wheel is speeding up. EK 2.9.A.3 names this the rate at which the object's speed changes.
Centripetal acceleration, from EK 2.9.A.1.i: $a_c = \frac{v^2}{r} = \frac{(3.6)^2}{0.18} = \frac{12.96}{0.18} = 72\ \text{m/s}^2$ , directed toward the axis.
Net linear acceleration. The two components are perpendicular, so by EK 2.9.A.4 combine them as vectors: $a = \sqrt{(72)^2 + (1.08)^2} = \sqrt{5184 + 1.17} = 72.0\ \text{m/s}^2$ , tilted $\arctan\frac{1.08}{72} = 0.86^\circ$ from the inward radial direction. The centripetal part is $67$ times the tangential part here, so the net acceleration points almost straight at the axis.
Now $2.0$ s later. Angular velocity from the first kinematics equation: $\omega = 20 + (6.0)(2.0) = 32\ \text{rad/s}$ . Linear speed: $v = (0.18)(32) = 5.76\ \text{m/s}$ .
Tangential acceleration: still $a_T = r\alpha = 1.08\ \text{m/s}^2$ , because $\alpha$ is constant and $r$ has not moved. Centripetal acceleration: $a_c = \frac{(5.76)^2}{0.18} = \frac{33.18}{0.18} = 184\ \text{m/s}^2$ , which has grown by a factor of $\left(\frac{32}{20}\right)^2 = 2.56$ .
Angular displacement over those $2.0$ s: $\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = (20)(2.0) + \frac{1}{2}(6.0)(2.0)^2 = 40 + 12 = 52\ \text{rad}$ , which is $\frac{52}{2\pi} = 8.28$ revolutions.
Arc length traveled by the rim point: $\Delta s = r\Delta\theta = (0.18)(52) = 9.36\ \text{m}$ . Radians were compulsory in that line and optional in the line before it, which is the boundary this page keeps returning to.
Collect the comparison. Angular quantities across those two seconds: one $\alpha$ , unchanged at $6.0\ \text{rad/s}^2$ . Linear quantities at the rim: $v$ went from $3.6$ to $5.76$ m/s, $a_c$ went from $72$ to $184\ \text{m/s}^2$ , and $a_T$ held still at $1.08\ \text{m/s}^2$ . One angular number, three linear behaviors.

At the start: $v = 3.6$ m/s, $a_T = 1.08\ \text{m/s}^2$ , $a_c = 72\ \text{m/s}^2$ , net linear acceleration $72.0\ \text{m/s}^2$ tilted $0.86^\circ$ from radial. After $2.0$ s: $\omega = 32\ \text{rad/s}$ , $v = 5.76$ m/s, $a_T$ still $1.08\ \text{m/s}^2$ , $a_c = 184\ \text{m/s}^2$ . The point swept $52$ rad, that is $8.28$ revolutions, traveling $9.36$ m of arc.

Frequently asked questions

What is the difference between linear and angular velocity?

Angular velocity is how fast something rotates, measured in radians per second, and every point on a rigid system shares the same value: the AP Physics 1 CED states this at essential knowledge 5.2.A.3. Linear velocity is how fast one particular point moves through space, measured in meters per second, and it grows with distance from the axis, so a rim point moves faster than a point near the hub while both turn at the same rate. A point on the axis has zero linear velocity while the system rotates. The two are linked by v equals r omega, where r is the point's distance from the axis.

Does v equals r omega require radians?

Yes. The radian is defined as an arc length divided by a radius, so the identity arc length equals r times angle holds only in radians, and v equals r omega follows from it. Using degrees per second makes the answer too large by the conversion factor, one radian equals 180 over pi degrees, about 57.2958. The same requirement applies to every equation in which an angle or an angular rate multiplies a length: arc length equals r delta theta, tangential acceleration equals r alpha, and the center-of-mass displacement of a rolling object equals r delta theta. The AP Physics 1 CED writes the requirement into the definition at essential knowledge 5.1.A.1, which specifies angular displacement in radians.

Can you use degrees in the angular kinematics equations?

Yes, provided you use one angular unit consistently. The three constant-angular-acceleration relationships in essential knowledge 5.1.A.4.i contain no length at all, so a change of unit multiplies both sides of each by the same factor and cancels. Work a problem entirely in degrees per second and degrees per second squared and every answer converts back correctly. The same is true of average angular velocity as delta theta over delta t and average angular acceleration as delta omega over delta t. The factor of 57.2958 only bites the moment an angle multiplies a radius, so convert to radians before the first equation containing an r. Converting at the start is the safest habit because it removes the boundary from consideration.

Do all points on a rotating wheel have the same angular velocity?

Yes, 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. They also all complete a revolution in the same time, so they share a period and a frequency. What differs is the linear velocity, because each point travels a circle of a different circumference in that same time. The CED's appendix marks two points on a rotating wheel and notes that at any given instant the outer one is traveling with a greater translational speed and in a different direction than the inner one.

How do you convert rpm to radians per second?

Divide by 60 to get revolutions per second, then multiply by 2 pi because one revolution is 2 pi radians. So 600 rpm is 10 revolutions per second, which is about 62.8 radians per second. Doing only one of the two steps is a common error and it leaves the answer wrong by a factor of 60 or of about 6.28. Convert before using v equals r omega, tangential acceleration equals r alpha, rotational kinetic energy as one half I omega squared, or angular momentum as I omega, since every one of those needs radians.

Why does a point at the center of a merry-go-round have zero speed?

Because its distance from the axis is zero, and linear speed is that distance multiplied by the angular velocity. Standing at the exact center you rotate through a full turn with everyone else, so your angular velocity is the same as theirs, but you trace out no circle and cover no distance, so your linear speed is zero. This is the cleanest demonstration that linear and angular velocity are different quantities rather than one quantity in two units: at that point one of them is zero and the other is not, in the same rigid system at the same instant.

Is angular velocity a vector in AP Physics 1?

Not in the full three-dimensional sense. The boundary statement under 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 direction with a sign rather than with a vector along the axis, and the right-hand rule for angular velocity is outside the course. Essential knowledge 5.1.A.1.ii describes the convention: one direction of angular displacement about an axis is typically indicated as mathematically positive, with the other becoming mathematically negative. Choose which sense is positive at the start of a problem and hold it throughout.