AP Physics C: Mechanics · Topic 5.2

Topic 5.2: Connecting Linear and Rotational Motion

Unit 5: Torque and Rotational Dynamics10-15% of the multiple-choice section

Topic 5.2 relates a point's linear motion to the rotation of the rigid system it belongs to: arc length is radius times angle, speed is radius times angular velocity, tangential acceleration is radius times angular acceleration. The required content is identical in AP Physics C and AP Physics 1.

AP Physics: Unit 5 (topics 5.2 Connecting Linear and Rotational Motion). AP Physics C: Mechanics Unit 5, Topic 5.2. One learning objective, 5.2.A: describe the linear motion of a point on a rotating rigid system that corresponds to the rotational motion of that point, and vice versa. Three essential-knowledge statements: 5.2.A.1 (for a point at distance r from a fixed axis, the linear distance s traveled as the system rotates through an angle delta-theta is delta-s = r delta-theta); 5.2.A.2 (derived relationships of linear velocity and of the tangential component of acceleration to their respective angular quantities are s = r theta, v = r omega, a_T = r alpha); 5.2.A.3 (for a rigid system, all points within that system have the same angular velocity and angular acceleration). Two-paragraph boundary statement: 'AP Physics C: Mechanics expects students to be able to mathematically manipulate the magnitudes of angular displacement, angular velocity, and angular acceleration using vector conventions. However, the directions of the vectors will not be assessed on the exam.' and 'Descriptions of the directions of rotational kinematics quantities for a point or rigid body are limited to clockwise and counterclockwise with respect to a given axis of rotation.' Suggested skills 1.C, 2.A, 2.C, 3.B. All three essential-knowledge statements, their equations and all four suggested skills are printed identically in the AP Physics 1 CED under the same numbering, making 5.2 the only Unit 5 topic with no difference in required content between the two courses; AP Physics 1 prints only the second paragraph of the boundary statement, worded 'for a point or object'. Printed on the C: Mechanics sheet: v = r omega and a_T = r alpha in the rotational column, s = r theta in the geometry box, a_c = v^2/r = r omega^2 in the translational column. Not printed on either sheet: delta-s = r delta-theta. The AP Physics 1 sheet prints a_c = v^2/r only, without the r omega^2 form, and prints s = r theta in its own geometry box.

What Topic 5.2 requires, and the honest headline

Topic 5.2 carries one learning objective, 5.2.A: describe the linear motion of a point on a rotating rigid system that corresponds to the rotational motion of that point, and vice versa. The CED suggests four skills for it: 1.C, 2.A, 2.C and 3.B.

Three essential-knowledge statements sit under it.

5.2.A.1 says that for a point at a distance $r$ from a fixed axis of rotation, the linear distance $s$ traveled by the point as the system rotates through an angle $\Delta\theta$ is given by

\Delta s = r\Delta\theta

5.2.A.2 says derived relationships of linear velocity and of the tangential component of acceleration to their respective angular quantities are given by the following equations:

s = r\theta \qquad v = r\omega \qquad a_T = r\alpha

5.2.A.3 says that for a rigid system, all points within that system have the same angular velocity and angular acceleration.

Now the headline, because a student deciding what to read next deserves it in one sentence. Every one of those three statements is printed word for word in the AP Physics 1 course and exam description, with the same numbering, the same equations, and the same wording. The four suggested skills are the same four. Topic 5.2 is the only topic in Unit 5 where both of those hold at once. The two frameworks separate at Topic 5.1 (derivatives against averages), at 5.3 (the direction of torque), at 5.4 (the rotational-inertia integral), and, for Topics 5.5 and 5.6, in the suggested skills even though the required content matches. At 5.2 they do not separate at all.

The rest of this page is about the three things that do change: one paragraph of the boundary statement, one line on the equation sheet, and what the word "derived" is allowed to mean once you have calculus.

The boundary statement, and the one paragraph AP Physics 1 does not print

Topic 5.2 in AP Physics C: Mechanics carries a two-paragraph boundary statement. Quoted whole:

"AP Physics C: Mechanics expects students to be able to mathematically manipulate the magnitudes of angular displacement, angular velocity, and angular acceleration using vector conventions. However, the directions of the vectors will not be assessed on the exam."

"Descriptions of the directions of rotational kinematics quantities for a point or rigid body are limited to clockwise and counterclockwise with respect to a given axis of rotation."

AP Physics 1's Topic 5.2 prints only the second paragraph, and words it "for a point or object" rather than "for a point or rigid body".

This is the same pair of paragraphs the CED prints under Topic 5.1, with one word changed: 5.1 says "the directions of said vectors" and 5.2 says "the directions of the vectors". Nothing turns on that.

What does turn on it is the scope. The restriction covers the rotational kinematics quantities: angular displacement, angular velocity, angular acceleration. It says nothing about torque, and Topic 5.3 in this same unit prints no boundary statement and does give you a right-hand rule. So within Unit 5 you work with clockwise and counterclockwise for $\theta$ , $\omega$ and $\alpha$ , and with a genuine vector direction for $\vec{\tau}$ . That asymmetry is deliberate and it is worth carrying into every problem you set up.

What "derived relationships" means when you actually have calculus

Statement 5.2.A.2 calls $s = r\theta$ , $v = r\omega$ and $a_T = r\alpha$ derived relationships in both courses. Only in this course are you handed the tools to derive them in two lines, and only in this course is $\omega$ defined as a derivative in the first place.

Start from 5.2.A.1 and note the one physical fact that makes everything work: for a point fixed in a rigid system, $r$ is a constant. It does not change as the system turns. So differentiating $s = r\theta$ with respect to time gives

\frac{ds}{dt} = r\frac{d\theta}{dt} \quad \Longrightarrow \quad v = r\omega

and differentiating once more gives

\frac{dv}{dt} = r\frac{d\omega}{dt} \quad \Longrightarrow \quad a_T = r\alpha

using the two definitions from Topic 5.1. That is the whole derivation. The constancy of $r$ is the hinge: if $r$ changed with time, the product rule would add a term and neither result would hold. That is exactly what happens on a spool whose radius of wrapped string shrinks, and it is why such a problem is harder than it looks.

Skill 2.A, deriving a symbolic expression by selecting and following a logical mathematical pathway, is a suggested skill for this topic in both courses. What differs is the weight it carries downstream: on the AP Physics C: Mechanics multiple-choice section, 2.A is weighted at 25 to 30%, the highest band the CED gives any single skill there, ahead of 2.B at 20 to 25%.

There is one more consequence of the derivative definitions, and it is the reason worked example 1 is set up the way it is. Since $v = r\omega$ holds at every instant and $r$ is constant, any statement about $\omega(t)$ transfers immediately to $v(t)$ by multiplying by $r$ . A non-constant angular acceleration produces a non-constant tangential acceleration in exactly the same shape.

The tangential component is only one of two

Read the subscript in 5.2.A.2. The statement says "the tangential component of acceleration", and the sheet prints $a_T$ with a capital T. It is not the whole acceleration of the point.

A point moving on a circle of radius $r$ has two perpendicular acceleration components at every instant:

$a_T = r\alpha$ , tangent to the circle, which changes the point's speed. It is zero whenever $\alpha$ is zero, which is to say whenever the system turns at a constant rate.
$a_c = v^2/r = r\omega^2$ , directed along the radius toward the axis, which changes the point's direction. It is zero only when $\omega$ is zero, that is, only at an instant when the system is not turning at all.

The magnitude of the total acceleration is $\sqrt{a_T^2 + a_c^2}$ , and its angle away from the inward radial direction is $\arctan(a_T/a_c)$ . Worked example 2 computes both, and finds a case where the tangential part is 1 part in 18 of the radial part even though the wheel is speeding up hard.

Here is the one line of the equation sheet that differs between the two courses on this topic. The AP Physics C: Mechanics sheet prints

a_c = \frac{v^2}{r} = r\omega^2

with both forms. The AP Physics 1 sheet prints only $a_c = v^2/r$ . The angular form is the one you want in a Topic 5.2 problem, because $\omega$ is usually what you are given, and on this exam you do not have to reconstruct it.

Centripetal acceleration is Unit 2 content in both courses: Topic 2.10 in AP Physics C: Mechanics Unit 2 and Topic 2.9 in AP Physics 1. Topic 5.2 is where it meets the angular variables.

Same angular velocity, different linear speeds

Statement 5.2.A.3 is one sentence and it is the sentence most Topic 5.2 questions are built on: for a rigid system, all points within that system have the same angular velocity and angular acceleration.

What is shared and what is not:

Shared across the whole rigid system: $\theta$ , $\omega$ , $\alpha$ . Every point sweeps the same angle in the same time.
Not shared: $s$ , $v$ , $a_T$ , $a_c$ . Each of these scales with that point's own $r$ , measured perpendicular from the axis.

So two points on one turntable at radii $r_1$ and $r_2$ have speeds in the ratio $r_2/r_1$ and tangential accelerations in the same ratio, while their centripetal accelerations are in that ratio too, since $a_c = r\omega^2$ with $\omega$ shared. Worked example 3 works a stepped pulley on exactly this principle.

The CED's own Topic 5.1 sample instructional activity is this idea as a demonstration: put stickers near the center and near the edge of a rotating turntable and have students predict and explain the relationships between linear and rotational position, displacement, speed, and acceleration.

Two cautions about $r$ :

$r$ is the perpendicular distance from the axis to the point, not the distance from the center of mass and not the length of the object. A point on the axis has $r = 0$ and therefore zero speed no matter how fast the system spins.
When two separate rigid bodies are coupled by a belt or by rolling contact, it is $v$ that is shared at the contact, not $\omega$ . Statement 5.2.A.3 is about points within one rigid system. Two pulleys joined by a belt are two systems.

Which Topic 5.2 equations are printed, on which sheet

Equation	CED statement	C: Mechanics sheet	AP Physics 1 sheet
$\Delta s = r\Delta\theta$	5.2.A.1	no	no
$s = r\theta$	5.2.A.2	yes, geometry box	yes, geometry box
$v = r\omega$	5.2.A.2	yes	yes
$a_T = r\alpha$	5.2.A.2	yes	yes
$a_c = \frac{v^2}{r} = r\omega^2$	Unit 2	yes, both forms	$\frac{v^2}{r}$ only
$\Delta x_{\text{cm}} = r\Delta\theta$	Unit 6 rolling	yes	yes

Three rows repay a second look.

$s = r\theta$ is printed, but not where you would look for it. It sits in the GEOMETRY AND TRIGONOMETRY box under Circle, alongside $A = \pi r^2$ and $C = 2\pi r$ , on both sheets. It is easy to conclude it is missing by scanning only the mechanics table, and that conclusion is wrong.

$\Delta s = r\Delta\theta$ is not printed in that form on either sheet. What both sheets print is $\Delta x_{\text{cm}} = r\Delta\theta$ , with a center-of-mass subscript. That is the rolling relation and it belongs to Unit 6, not here. For Topic 5.2 use the geometry-box $s = r\theta$ and take differences yourself.

The $a_T$ subscript is a capital T for tangential. Both sheets print it that way. It is not a radial component, and reading it as one inverts the physics of every problem it appears in.

Traps that cost points on this topic

Degrees anywhere in these equations. Every relation on this page assumes radians, because the radian is defined as the ratio of arc length to radius. Feeding degrees into $s = r\theta$ makes the arc length too large by a factor of $180/\pi$ . Convert first.

Using the wrong $r$ . It is the perpendicular distance from the rotation axis to the specific point you are asked about. On a stepped pulley there are two of them, and the problem will use both.

Reporting $a_T$ as the acceleration. It is one component. If the question asks for the magnitude of the acceleration of a point, you owe it $\sqrt{a_T^2 + a_c^2}$ . This is the single most common way to lose a point on a Topic 5.2 calculation.

Assuming $\omega$ is shared between coupled bodies. Belt-driven pulleys share the belt's linear speed. A rolling wheel shares the contact point's linear speed with the ground. Only points inside one rigid body share $\omega$ .

Assuming $r$ is constant when the problem says it is not. The derivation of $v = r\omega$ differentiates $s = r\theta$ treating $r$ as a constant. On a spool that is unwinding, or a mass on a string being pulled through a hole, $r$ changes and that step is invalid.

Treating zero angular acceleration as zero acceleration. A point on a system turning at a steady rate has $a_T = 0$ and $a_c = r\omega^2$ , which is usually the larger of the two anyway. Worked example 1 gets a centripetal acceleration above two thousand times $g$ from a disk that is not speeding up at all.

If you are taking AP Physics 1, this is not your page

Because the required content of Topic 5.2 is identical in the two frameworks, the honest advice here is different from the advice on the other five pages of this unit: for the physics of Topic 5.2 itself, the [AP Physics 1 Topic 5.2 page](/ap-physics-1/unit-5-torque-and-rotational-dynamics/5-2-connecting-linear-and-rotational-motion) covers the same three essential-knowledge statements, and it is the page written for AP Physics 1 students. This page is for AP Physics C: Mechanics students, and what it adds is the derivation from the derivative definitions, the second form of $a_c$ that only the C sheet prints, and the symbolic style that skill 2.A is weighted for on this exam.

AP Physics C: Mechanics is a calculus-based introductory college-level course equivalent to the first course in a college calculus-based physics sequence, and its only stated prerequisite is that students have taken or are concurrently taking calculus. That prerequisite does no work at all in Topic 5.2. It does a great deal in Topic 5.1, where the two definitions become derivatives, and in Topic 5.4, where rotational inertia becomes an integral. If you are looking for what makes this course different, read those two.

The Unit 5 hub lists all six topics with their objectives and boundary statements, and the rotational kinematics guide owns the step-by-step routine for the angular equations.

Deriving the two relations, then a point on a hard-disk platter

A rigid disk spins about its central axis at a steady 7200 revolutions per minute. Consider a point on it at $r = 3.5$ cm from the axis. First derive $v = r\omega$ and $a_T = r\alpha$ from $s = r\theta$ , then find (a) the angular velocity in $\mathrm{rad/s}$ , (b) the linear speed of the point, (c) its tangential acceleration, and (d) its centripetal acceleration, expressed as a multiple of $g = 9.8\ \mathrm{m/s^2}$ .

Derivation first, because 5.2.A.2 calls these derived relationships and skill 2.A is what the exam weights most heavily. The point is fixed in the rigid disk, so $r$ is a constant. Differentiating $s = r\theta$ with respect to time and using $\omega = d\theta/dt$ from 5.1.A.2 gives $v = ds/dt = r\, d\theta/dt = r\omega$ .
Differentiate once more, using $\alpha = d\omega/dt$ from 5.1.A.3: $a_T = dv/dt = r\, d\omega/dt = r\alpha$ . Both results depend on $r$ being constant, and that is the only assumption used.
(a) Convert to radians per second before anything else. $\omega = 7200\ \frac{\text{rev}}{\text{min}} \times \frac{2\pi\ \text{rad}}{1\ \text{rev}} \times \frac{1\ \text{min}}{60\ \text{s}} = 240\pi = 754\ \mathrm{rad/s}$ .
(b) $v = r\omega = (0.035)(753.98) = 26.4\ \mathrm{m/s}$ . Note the radius converted to metres first; 3.5 cm is 0.035 m.
(c) The disk spins at a steady rate, so $\alpha = 0$ and therefore $a_T = r\alpha = 0$ . The point's speed is not changing.
(d) The point's acceleration is entirely centripetal. The C: Mechanics sheet prints $a_c = v^2/r = r\omega^2$ , so use the angular form directly: $a_c = (0.035)(753.98)^2 = (0.035)(568{,}490) = 1.99 \times 10^4\ \mathrm{m/s^2}$ .
As a multiple of $g$ : $19{,}897/9.8 = 2.03 \times 10^3$ , about 2000 times the acceleration due to gravity at Earth's surface.
Cross-check with the other printed form: $v^2/r = (26.389)^2/0.035 = 696.4/0.035 = 1.99 \times 10^4\ \mathrm{m/s^2}$ . The two agree, as they must, since $v = r\omega$ is what turns one into the other.
Read the result. The tangential acceleration is exactly zero and the total acceleration is two thousand $g$ . A steady rotation rate does not mean a small acceleration; it means no tangential component.

$v = r\omega$ and $a_T = r\alpha$ follow from differentiating $s = r\theta$ twice with $r$ constant. (a) $\omega = 240\pi = 754\ \mathrm{rad/s}$ . (b) $v = 26.4\ \mathrm{m/s}$ . (c) $a_T = 0$ , because the rate is steady. (d) $a_c = 1.99 \times 10^4\ \mathrm{m/s^2}$ , about $2.0 \times 10^3$ times $g$ .

Both components at once, and the instant when they are equal

A grinding wheel of radius $R = 0.125$ m starts from rest and speeds up at a constant angular acceleration $\alpha = 8.0\ \mathrm{rad/s^2}$ . For a point on the rim, find (a) the tangential acceleration, (b) the angular velocity, centripetal acceleration, total acceleration magnitude and direction at $t = 1.5$ s, and (c) a symbolic expression for the time at which the two components are equal in magnitude.

(a) From 5.2.A.2, $a_T = R\alpha = (0.125)(8.0) = 1.0\ \mathrm{m/s^2}$ . Because $\alpha$ is constant and $R$ is constant, this value holds at every instant.
(b) The angular acceleration is constant, so 5.1.A.4.i applies: $\omega = \omega_0 + \alpha t = 0 + (8.0)(1.5) = 12\ \mathrm{rad/s}$ .
$a_c = R\omega^2 = (0.125)(12)^2 = (0.125)(144) = 18.0\ \mathrm{m/s^2}$ .
The two components are perpendicular, so the magnitude is $\sqrt{a_T^2 + a_c^2} = \sqrt{(1.0)^2 + (18.0)^2} = \sqrt{325} = 18.0\ \mathrm{m/s^2}$ to three figures.
Its direction is $\arctan(a_T/a_c) = \arctan(1.0/18.0) = 3.2^\circ$ away from the inward radial direction, tilted toward the direction of motion because the wheel is speeding up. The total acceleration points very nearly straight at the axis.
(c) Set $a_c = a_T$ : $R\omega^2 = R\alpha$ . The radius cancels immediately, leaving $\omega^2 = \alpha$ . From rest at constant $\alpha$ , $\omega = \alpha t$ , so $\alpha^2 t^2 = \alpha$ and $t = 1/\sqrt{\alpha}$ .
That result is independent of $R$ , so the two components cross over at the same instant everywhere on the wheel, from the hub to the rim. This is the kind of functional-dependence answer skill 2.C and skill 2.D are written for.
Check it numerically: $t = 1/\sqrt{8.0} = 0.354$ s. Then $\omega = (8.0)(0.35355) = 2.83\ \mathrm{rad/s}$ , giving $a_c = (0.125)(2.8284)^2 = (0.125)(8.0) = 1.0\ \mathrm{m/s^2}$ , which equals $a_T = 1.0\ \mathrm{m/s^2}$ exactly.
Sanity check on the ordering: before $t = 0.354$ s the tangential part dominates, after it the centripetal part does, and by $t = 1.5$ s the centripetal part is 18 times larger. Centripetal acceleration grows as $t^2$ while tangential acceleration is flat, so it always wins eventually.

(a) $a_T = 1.0\ \mathrm{m/s^2}$ , constant. (b) At $t = 1.5$ s: $\omega = 12\ \mathrm{rad/s}$ , $a_c = 18.0\ \mathrm{m/s^2}$ , total $18.0\ \mathrm{m/s^2}$ at $3.2^\circ$ from the inward radius. (c) $t = 1/\sqrt{\alpha} = 0.354$ s, independent of the radius.

A stepped pulley: one angular velocity, two linear speeds

A stepped pulley is a single rigid body with two coaxial radii, $r_1 = 0.040$ m and $r_2 = 0.100$ m. A cord wrapped around the inner step is pulled so that the cord's speed is $0.60\ \mathrm{m/s}$ and increasing at $1.5\ \mathrm{m/s^2}$ . Find (a) the pulley's angular velocity and angular acceleration, (b) the linear speed and tangential acceleration of a point on the outer rim, and (c) the total acceleration magnitude of that outer-rim point.

Identify the system. The two steps are machined from one piece, so this is one rigid system, and 5.2.A.3 applies: every point on it shares one $\omega$ and one $\alpha$ . What differs from point to point is $r$ .
(a) The cord leaves the inner step tangentially, so the cord's speed equals the inner surface's linear speed. From $v = r\omega$ , $\omega = v_1/r_1 = 0.60/0.040 = 15\ \mathrm{rad/s}$ .
From $a_T = r\alpha$ , $\alpha = a_{T,1}/r_1 = 1.5/0.040 = 37.5\ \mathrm{rad/s^2}$ .
(b) Apply the same two relations at the outer radius with the same $\omega$ and $\alpha$ . $v_2 = r_2\omega = (0.100)(15) = 1.5\ \mathrm{m/s}$ , and $a_{T,2} = r_2\alpha = (0.100)(37.5) = 3.75\ \mathrm{m/s^2}$ .
Check the structure rather than just the numbers. Both linear quantities scale by $r_2/r_1 = 0.100/0.040 = 2.5$ : $1.5/0.60 = 2.5$ and $3.75/1.5 = 2.5$ . That factor is the whole content of 5.2.A.3.
(c) The outer point also has $a_c = r_2\omega^2 = (0.100)(15)^2 = (0.100)(225) = 22.5\ \mathrm{m/s^2}$ , directed at the axis.
Total magnitude: $\sqrt{(3.75)^2 + (22.5)^2} = \sqrt{14.06 + 506.25} = \sqrt{520.3} = 22.8\ \mathrm{m/s^2}$ .
Compare with the inner step, where $a_c = r_1\omega^2 = (0.040)(225) = 9.0\ \mathrm{m/s^2}$ . The centripetal accelerations are in the ratio 2.5 as well, since $a_c = r\omega^2$ with $\omega$ shared.
One thing this problem is not. If the two radii belonged to two separate pulleys joined by a belt, the shared quantity would be the belt's linear speed, not $\omega$ , and the two angular velocities would then be in the inverse ratio. Statement 5.2.A.3 is about points within one rigid system.

(a) $\omega = 15\ \mathrm{rad/s}$ and $\alpha = 37.5\ \mathrm{rad/s^2}$ , shared by the whole pulley. (b) $v_2 = 1.5\ \mathrm{m/s}$ and $a_{T,2} = 3.75\ \mathrm{m/s^2}$ , both larger than the inner values by the factor $r_2/r_1 = 2.5$ . (c) The outer point's total acceleration is $22.8\ \mathrm{m/s^2}$ , dominated by its $22.5\ \mathrm{m/s^2}$ centripetal component.

Frequently asked questions

Is Topic 5.2 different in AP Physics C Mechanics and AP Physics 1?

The required course content is not. Essential knowledge 5.2.A.1, 5.2.A.2 and 5.2.A.3 appear in both course and exam descriptions with the same numbering, the same wording and the same equations: arc length equals radius times angular displacement, then arc length equals radius times angle, linear speed equals radius times angular velocity, and tangential acceleration equals radius times angular acceleration, and finally all points in a rigid system share one angular velocity and one angular acceleration. The four suggested skills, 1.C, 2.A, 2.C and 3.B, are also the same in both. Three things do differ: AP Physics C: Mechanics adds a paragraph to the boundary statement about manipulating magnitudes using vector conventions, its equation sheet prints centripetal acceleration in the angular form as well as the speed form, and its Topic 5.1 defines angular velocity as a derivative, so a Topic 5.2 question can involve a non-constant angular acceleration.

Is s equals r theta on the AP Physics C equation sheet?

Yes, but not in the mechanics table where most people look for it. It is printed in the GEOMETRY AND TRIGONOMETRY box of the Table of Information, under Circle, next to the area and circumference formulas, and it is printed there on the AP Physics 1 sheet as well. Concluding that it is absent because it does not appear among the mechanics equations is a mistake worth avoiding. Two related equations are worth separating from it: the mechanics table prints the change in center-of-mass position as radius times change in angle, which is the rolling relation from Unit 6, and neither sheet prints the Topic 5.2 form giving the change in arc length as radius times change in angle. For Topic 5.2 work from the geometry-box relation and take differences yourself.

What is the difference between tangential and centripetal acceleration?

They are the two perpendicular components of the acceleration of a point moving on a circle, and they do different jobs. Tangential acceleration points along the direction of motion and equals the radius times the angular acceleration; it changes how fast the point is going, and it is zero whenever the system turns at a steady rate. Centripetal acceleration points inward along the radius and equals the square of the speed divided by the radius, which is the same as the radius times the square of the angular velocity; it changes the direction of motion, and it is zero only at an instant when the system is not turning at all. The magnitude of the total acceleration is the square root of the sum of their squares. Essential knowledge 5.2.A.2 covers the tangential component only, which is why the equation sheet writes it with a capital T subscript.

Why do all points on a rotating rigid body have the same angular velocity?

Because a rigid system holds its shape. Essential knowledge 5.1.A.1.i defines a rigid system as one that holds its shape but in which different points move in different directions during rotation. If two points on it swept different angles in the same time, the angle between them would change, and the body would have deformed. Statement 5.2.A.3 states the consequence directly: for a rigid system, all points within that system have the same angular velocity and angular acceleration. Their linear quantities are not shared, because arc length, speed and tangential acceleration each scale with that point's own perpendicular distance from the axis. A point twice as far from the axis moves twice as fast. Note the restriction to one rigid system: two pulleys joined by a belt share the belt's linear speed, not an angular velocity.

Do I have to use radians in v equals r omega?

Yes. The relation is derived from the definition of the radian as the ratio of arc length to radius, so it only produces the correct linear quantity when the angular quantity is in radians. Using degrees makes the answer too large by a factor of about 57.3, and using revolutions makes it too large by a factor of about 6.28. Convert at the top of the solution: one revolution is 2 pi radians, and revolutions per minute become radians per second by multiplying by 2 pi and dividing by 60. The same applies to the arc-length relation and to the tangential-acceleration relation. The radian is dimensionless precisely because it is a ratio of two lengths, which is why radius in metres times angular velocity in radians per second gives metres per second with no leftover unit.

How do I find the total acceleration of a point on a rotating object?

Compute both components and combine them with the Pythagorean theorem, because they are perpendicular. The tangential component is the radius times the angular acceleration, and the centripetal component is the radius times the square of the angular velocity, which the AP Physics C: Mechanics equation sheet prints alongside the speed-squared-over-radius form. The magnitude of the total acceleration is the square root of the sum of their squares, and its direction sits at an angle equal to the inverse tangent of the tangential component divided by the centripetal component, measured from the inward radial direction toward the direction of motion. A common error is to report the tangential component alone when the question asks for the acceleration of the point; on a wheel spinning at any appreciable rate the centripetal part is usually much the larger of the two.

Can a point on a rotating object have acceleration when the rotation rate is constant?

Yes, and it is usually large. A constant angular velocity means the angular acceleration is zero, so the tangential component of the point's acceleration is zero and its speed is not changing. But its direction is changing continuously, and that requires a centripetal acceleration equal to the radius times the square of the angular velocity, directed toward the axis. A point 3.5 cm from the axis of a disk turning at 7200 revolutions per minute has zero tangential acceleration and a centripetal acceleration of about 2 times 10 to the fourth metres per second squared, roughly 2000 times the acceleration due to gravity at Earth's surface. Constant angular velocity rules out one component of the acceleration, not the acceleration itself.