Topic 6.5: Rolling

What Topic 6.5 requires

Topic 6.5 carries three learning objectives and five essential knowledge statements. Here is every one of them, counted off the CED page.

• 6.5.A Describe the kinetic energy of a system that has translational and rotational motion.
• 6.5.A.1 The total kinetic energy of a system is the sum of the system's translational and rotational kinetic energies. The CED gives the relevant equation $K_{\text{tot}} = K_{\text{trans}} + K_{\text{rot}}$.
• 6.5.B Describe the motion of a system that is rolling without slipping.
• 6.5.B.1 While rolling without slipping, the translational motion of a system's center of mass is related to the rotational motion of the system itself with the equations $\Delta x_{\text{cm}} = r\Delta\theta$, $v_{\text{cm}} = r\omega$, and $a_{\text{cm}} = r\alpha$.
• 6.5.B.2 For ideal cases, rolling without slipping implies that the frictional force does not dissipate any energy from the rolling system.
• 6.5.C Describe the motion of a system that is rolling while slipping.
• 6.5.C.1 When slipping, the motion of a system's center of mass and the system's rotational motion cannot be directly related.
• 6.5.C.2 When a rotating system is slipping relative to another surface, the point of application of the force of kinetic friction exerted on the system moves with respect to the surface, so the force of kinetic friction will dissipate energy from the system.

The two motion objectives are a matched pair: 6.5.B locks linear and angular quantities together, 6.5.C pulls them apart. Almost every question here is asking which of the two you are in.

Two boundary statements govern the topic, and the second has an exception clause that must travel with it every time it is quoted.

• Boundary statement 1. Rolling friction is beyond the scope of AP Physics 1.
• Boundary statement 2. The precise mathematical relationships between linear and angular quantities while a rigid body is rolling while slipping are beyond the scope of AP Physics 1 and 2, and students will not be expected to model those relationships quantitatively. However, students are expected to qualitatively explain the changes to linear and angular quantities while a rigid body is rolling while slipping.

That second boundary is routinely half-quoted as "rolling while slipping is not on the exam." Its closing sentence says the opposite about words: you can be asked to explain, in prose, what happens to the linear speed and the angular speed of a wheel that is spinning and skidding at once. What is excluded is the calculation, not the concept.

The CED lists four suggested skills for Topic 6.5: 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws. No calculation skill is listed.

Topic 6.5 sits in Unit 6, weighted 5 to 8 percent of the multiple-choice section across roughly 8 to 14 class periods.

Total kinetic energy is a sum, not a choice

Statement 6.5.A.1 settles the most common wrong instinct on this topic. A rolling object is not either translating or rotating. It does both, and its kinetic energy is the sum:

Two details in that expanded form matter more than they look. The speed in the first term is the speed of the center of mass, not the rim and not the contact point, whose speed is zero. The rotational inertia in the second is taken about the center of mass, because the object spins about its own center while that center travels. That is the decomposition Topic 6.1 sets up in statement 6.1.A.1.ii.

Now check the AP Physics 1 equation sheet against the CED, because the two are not the same document. The sheet prints $K = \frac{1}{2}mv^2$ in the translational group and $K = \frac{1}{2}I\omega^2$ in the rotational group, as separate entries. It does not print $K_{\text{tot}} = K_{\text{trans}} + K_{\text{rot}}$ anywhere. That equation appears only in the CED, attached to 6.5.A.1, so it is content you are expected to know rather than to copy off the reference sheet mid-problem. Kinetic energy is a scalar, so this is ordinary addition, with no components and no vector diagram.

The rolling condition, and the one place r really is the radius

Statement 6.5.B.1 gives three equations that say the same thing at three levels of derivative:

The middle one is the rolling condition: a system rolling without slipping has its center of mass speed and its angular speed locked together by the radius.

**Here $r$ is genuinely the radius of the rolling object**, which is the exception rather than the rule. In Topic 5.2, the $r$ in $v = r\omega$ means the distance from the axis to whichever point you are asking about, and one spinning disk has a different $r$ for every point on it. Rolling is the case where geometry forces a single value, because the axle sits exactly one radius above the contact patch. Carry "$r$ is always the object's radius" out of this topic and you will get Topic 5.2 wrong; carry "$r$ is never the object's radius" and you will get this one wrong.

The physical statement is simpler than the algebra. Rolling without slipping means the contact point is instantaneously at rest relative to the surface. Every centimeter of circumference that comes around is laid on the ground exactly once, so the arc that unrolls equals the distance the axle advances. That is $\Delta x_{\text{cm}} = r\Delta\theta$ read as a sentence. It also means the top of a rolling wheel moves at $2v_{\text{cm}}$ while the bottom moves at zero.

The equation-sheet check here is fussy but it decides derivation credit. The sheet prints $\Delta x_{\text{cm}} = r\Delta\theta$ in the rotational group, exactly as the CED gives it. It also prints $v = r\omega$ and $a_T = r\alpha$, but those are the general point-on-a-rotating-body relations, not the center-of-mass versions. If a free-response part says to begin from an equation on the reference information, $\Delta x_{\text{cm}} = r\Delta\theta$ is the one unambiguously there for rolling.

The contact point is at rest, so static friction dissipates nothing

Because the contact point is instantaneously at rest relative to the ground, the friction acting there is static, not kinetic. Topic 2.7 supplies the sheet's $|\vec{F}_f| \leq |\mu \vec{F}_N|$, an inequality because the static force takes whatever value the situation demands, up to a ceiling.

Statement 6.5.B.2 makes the energy claim: for ideal cases, rolling without slipping implies that the frictional force does not dissipate any energy from the rolling system. The reason is the one 6.5.C.2 gives in the negative. Kinetic friction dissipates energy because its point of application moves with respect to the surface, grinding one material across the other. In rolling, the point of application does not move with respect to the surface at all, so nothing grinds and no energy leaves as thermal energy.

Now the exact bookkeeping, because "friction does no work" is true here for a specific reason and false in general. Take a ball rolling down a ramp, with the positive direction taken down the slope and held there for the rest of this page. Static friction points up the slope, and it hits the two energy accounts like this:

• On the translational motion it does negative work, $-f d$, where $d$ is how far the center of mass travels.
• On the rotational motion it exerts a torque $fR$ about the center of mass, doing work $\tau \Delta\theta = fR\Delta\theta = f d$, using the rolling condition $d = R\Delta\theta$.

The two are equal and opposite, so the total work static friction does on the rolling object is exactly zero. What it decides is the split: it moves energy from the translational account into the rotational one at exactly the rate that keeps $v_{\text{cm}} = r\omega$ true. That is why the total mechanical energy of a system rolling without slipping down a ramp is constant, and why you may write a conservation of energy equation for it with no dissipation term.

Two warnings follow from the reasoning rather than from the slogan. Static friction is not always workless: a crate on the bed of an accelerating truck is held by static friction and is not sliding, yet that force does positive work on it. What is special about rolling is not that the friction is static, it is that the point of application is instantaneously at rest in the ground frame. And no friction means no rolling: friction is the only force in the standard ramp problem with a torque about the center of mass, so on a frictionless ramp an object cannot spin up at all. It slides down without rotating, and $v_{\text{cm}} = r\omega$ never applies to it.

The race down the ramp: the ratio I over MR squared decides it

Which object wins a race down a ramp has a clean answer with a clean derivation, which is why suggested skill 2.A hangs on this topic. Release a symmetric object from rest at height $h$ and let it roll without slipping to the bottom. Write its rotational inertia about its center as a dimensionless number times $MR^2$, so $I_{\text{cm}} = \beta M R^2$. Static friction dissipates nothing, so mechanical energy is constant:

Substitute $\omega = v_{\text{cm}}/R$ and $I_{\text{cm}} = \beta MR^2$. The rotational term becomes $\frac{1}{2}\beta MR^2 \cdot v_{\text{cm}}^2/R^2 = \frac{1}{2}\beta M v_{\text{cm}}^2$, and the radius cancels itself out:

**Mass cancels. Radius cancels. Only $\beta$ survives.** The winner is whichever object has the smallest rotational inertia relative to $MR^2$, meaning the one with its mass concentrated nearest its axis. The same algebra fixes the energy split at the bottom as a fraction depending on nothing but $\beta$:

Before you memorize that column of $\beta$ values, read the boundary statement that governs them, which lives in Topic 5.4 rather than here: students do not need to know the rotational inertia of extended rigid systems, as these will be provided within the exam, and students should have a qualitative understanding of the factors that affect rotational inertia. A rolling problem hands you the rotational inertia. What you supply is the reasoning and the ranking, which you can build from $I = \sum m_i r_i^2$ with no standard result at all: a hoop puts every gram at the full radius, a solid disk puts most of its mass closer in, so the hoop has the larger $I$ for the same $M$ and $R$.

The CED puts this race in front of teachers three times, in optional activities rather than required content. Two Unit 6 sample instructional activities race a ring, a disk and a loaded low-friction cart of equal mass down identical inclines (the cart wins, then the disk, then the ring), and roll a hoop against a disk to be explained with energy bar charts and to-scale free-body diagrams. A third, in the Instructional Approaches section, rolls a hollow sphere against a solid one. None of that promises an exam scenario, but three appearances tell you what the writers think the topic is for.

The forces version: acceleration, and the friction rolling requires

The energy route gives the speed at the bottom, not the acceleration or the friction force, and the CED's sample activity asks for the force explanation as well as the energy one. Keep the positive direction pointing down the slope.

Draw the free-body diagram first, which is suggested skill 1.A. Three forces act on an object rolling down an incline of angle $\theta$: gravity at the center of mass, the normal force at the contact patch, and static friction at the contact patch pointing up the slope. Take torques about the center of mass and notice which forces drop out.

• Gravity acts at the center of mass, so its lever arm about that point is zero.
• The normal force acts at the contact point, and for a symmetric round object it points straight at the center, so its line of action passes through the center of mass and its torque is zero too. This is where the first boundary statement earns its keep: in reality the contact patch deforms and the normal force shifts slightly forward, which is what rolling friction is, and the CED puts rolling friction beyond the scope of AP Physics 1.
• Static friction is the only one left with a torque about the center of mass, of magnitude $fR$.

So the two equations are Newton's second law along the slope and Newton's second law in rotational form about the center:

The rolling condition $a_{\text{cm}} = R\alpha$ ties them together. Substituting $\alpha = a_{\text{cm}}/R$ gives $f = \beta M a_{\text{cm}}$, and putting that back into the force equation gives

The same $1+\beta$ that appeared in the energy result appears here, which is a good consistency check. The friction result also answers a question the energy route cannot: will it actually roll? Static friction is capped at $\mu_s N = \mu_s M g \cos\theta$, so rolling without slipping is only possible when

Steepen or polish the ramp past that and the object slips, at which point everything above stops applying and you are in Topic 6.5.C. Note that the mass cancels out of that requirement, so a heavy ball and a light ball of the same shape need the same coefficient.

Rolling while slipping, and what you are expected to say about it

Objective 6.5.C covers the case where the surface cannot supply enough friction, or where an object arrives already spinning at the wrong rate. Statement 6.5.C.1 is blunt: when slipping, the motion of a system's center of mass and the system's rotational motion cannot be directly related. All three of $\Delta x_{\text{cm}} = r\Delta\theta$, $v_{\text{cm}} = r\omega$, and $a_{\text{cm}} = r\alpha$ are off the table.

Statement 6.5.C.2 supplies the energy consequence, worth reading for its mechanism rather than its conclusion: when a rotating system is slipping relative to another surface, the point of application of the force of kinetic friction exerted on the system moves with respect to the surface, so the force of kinetic friction will dissipate energy from the system. Sliding contact is what dissipates, the same criterion that made the rolling case lossless, applied the other way round.

The boundary statement tells you how far to take it, and here it is in full because the half-quoted version means something different: the precise mathematical relationships between linear and angular quantities while a rigid body is rolling while slipping are beyond the scope of AP Physics 1 and 2, and students will not be expected to model those relationships quantitatively. However, students are expected to qualitatively explain the changes to linear and angular quantities while a rigid body is rolling while slipping.

So learn the two standard cases as stories.

A ball thrown down a lane with no spin. It starts with $v_{\text{cm}} > 0$ and $\omega = 0$. The contact point slides forward relative to the lane, so kinetic friction acts backward: it slows the center of mass, and its torque about the center spins the ball up. So $v_{\text{cm}}$ falls and $\omega$ rises until $v_{\text{cm}} = r\omega$. From that instant the ball rolls, the friction goes static, and the dissipation stops.

A wheel spun up and dropped. Now $v_{\text{cm}} = 0$ and $\omega$ is large. The contact point slides backward relative to the ground, so kinetic friction acts forward. The center of mass speeds up, the torque opposes the spin so $\omega$ falls, and again they meet at $v_{\text{cm}} = r\omega$. This is a drag car leaving the line.

In both, the answer to "what changes?" is the same: the linear and angular speeds move toward each other, kinetic friction is doing it, and mechanical energy is lost the whole time the surfaces slide.

The CED offers teachers a ranking task on exactly this, with its own answer key. A wheel rolls down an incline from rest and across a flat surface in three cases: tracks rough enough that there is no slipping, tracks with some friction but with slipping, and tracks with negligible friction. Ranking the final translational kinetic energies, the final rotational kinetic energies, and the final total mechanical energies gives $K_{T3} > K_{T2} > K_{T1}$, $K_{R1} > K_{R2} > K_{R3}$, and $E_1 = E_3 > E_2$. All three follow from this page. Case 3 has no friction, so nothing spins the wheel up and nothing dissipates. Case 1 rolls, so it puts a fixed fraction into rotation and still loses nothing. Case 2 sits between them on both splits and is the only one that dissipates.

The first boundary statement matters here too. Rolling friction is out of scope, so in AP Physics 1 an object rolling without slipping along a level surface keeps rolling at constant speed indefinitely. If a question shows a rolling object slowing on the flat, something else is doing it.

How Topic 6.5 is tested, and what goes wrong

The four suggested skills work as a checklist of question types. Skill 1.A means diagrams: free-body diagrams of a rolling object, energy bar charts for the top and bottom of a ramp. Skill 2.A means symbolic derivations, most often $v_{\text{cm}} = \sqrt{2gh/(1+\beta)}$ or $a_{\text{cm}} = g\sin\theta/(1+\beta)$ with a rotational inertia supplied in the stem. Skill 2.C means comparisons. Skill 3.C means writing the justification out.

Unit 6's own exam-preparation note is aimed at that last one, and it uses a rolling example: when writing justifications for claims, simply referencing an equation, law, or physical principle is not sufficient, and stating that one disk is rolling faster than another because of "conservation of energy" is not a complete enough answer to earn credit on the free-response section. The version that earns credit names the chain: both disks release the same $Mgh$ per kilogram, the one with the smaller $I/MR^2$ sends a smaller fraction into rotation, so a larger fraction stays translational, so its center of mass is faster.

The errors that cost points cluster tightly.

• Dropping the rotational term. Setting $Mgh = \frac{1}{2}Mv^2$ for a rolling object gives $v = \sqrt{2gh}$, the answer for a frictionless slide, and it is always too fast.
• Double counting. You may compute the total as $\frac{1}{2}Mv_{\text{cm}}^2 + \frac{1}{2}I_{\text{cm}}\omega^2$, or as $\frac{1}{2}I_{\text{contact}}\omega^2$ using the sheet's parallel axis theorem $I' = I_{\text{cm}} + Md^2$ with $d = R$. Both give the same number. What you may not do is use the contact-point rotational inertia and add a translational term as well.
• **Using $v = r\omega$ while the object slips.** Statement 6.5.C.1 forbids it, and a question that mentions ice, a polished ramp, or a spinning wheel dropped onto the ground is telling you which case you are in.
• Thinking mass or radius decides the race. They cancel.
• Treating the friction in the rolling case as a loss. Subtracting a $\mu M g \cos\theta \cdot d$ term from a rolling energy equation is a common and expensive slip.
• **Feeding degrees into $\omega$.** Every rotational relation here needs radians, for the reason Topic 5.1 sets out. Degrees belong in $\sin\theta$ for the incline.
• Assuming rolling without checking. If a problem gives $\mu_s$ as well as the ramp angle, it is probably asking whether $\mu_s \geq \beta\tan\theta/(1+\beta)$ holds.

If the incline geometry is the shaky part rather than the rotation, work through the inclined plane problems guide first, and use the kinetic energy calculator to check the translational half of any split you compute here.