Topic 6.5: Rolling

What Topic 6.5 requires

Topic 6.5 carries three learning objectives, tying Topic 6.3 for the most in Unit 6, and five essential-knowledge statements.

6.5.A: describe the kinetic energy of a system that has translational and rotational motion. 6.5.B: describe the motion of a system that is rolling without slipping. 6.5.C: describe the motion of a system that is rolling while slipping.

Boundary statement, quoted whole, because it is one sentence: "Rolling friction is beyond the scope of AP Physics C: Mechanics."

This is the only boundary statement anywhere in Unit 6. Topics 6.1, 6.2, 6.3, 6.4 and 6.6 print none.

The suggested skills are 1.B (create quantitative graphs with appropriate scales and units, including plotting data), 2.A (derive a symbolic expression), 2.D (predict new values or factors of change using functional dependence between variables), 3.A (create experimental procedures appropriate for a given scientific question) and 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim). Five skills, more than any other topic in Unit 6, which lists four apiece for 6.1 through 6.4 and three for 6.6. Checking the unit's suggested-skill table topic by topic, 1.B, 2.D and 3.A appear only under Topic 6.5 within this unit.

What the framework actually says about friction while rolling

This is worth getting exactly right, because the usual textbook phrasing is stronger than the framework's.

Essential knowledge 6.5.B.2, in full, is: "For ideal cases, rolling without slipping implies that the frictional force does not dissipate any energy from the rolling system."

Read the two hedges. "For ideal cases" is doing work, and so is "does not dissipate any energy", which is a claim about energy leaving the system, not the broader claim that the friction force does no work in every sense. The framework does not say "static friction does no work." If you want that phrasing on a free-response answer, derive it rather than cite it.

The derivation is short and uses only what the framework supplies.

Why the friction is static. Rolling without slipping means $v_{\text{cm}} = r\omega$, per 6.5.B.1. The material point of the body that is momentarily touching the ground has two velocity contributions: $v_{\text{cm}}$ forward from the translation, and $r\omega$ backward from the rotation about the centre of mass. Those are equal in size, so the contact point is instantaneously at rest relative to the surface. No relative sliding at the interface is what distinguishes static friction from kinetic friction, so the friction here is static. The equation sheet prints it in its inequality form, $\lvert \vec{F}_f \rvert \leq \lvert \mu \vec{F}_N \rvert$, which is the static statement: the friction takes whatever value the constraint requires, up to a maximum.

Why nothing is dissipated. Statement 6.5.C.2 gives the framework's own mechanism for dissipation in the slipping case: the point of application of the kinetic friction force moves with respect to the surface, so energy is dissipated. Reverse that reasoning for the rolling case. The point of application does not move with respect to the surface, so there is no sliding for the friction force to grind against, and no energy leaves the system. That is 6.5.B.2, arrived at from 6.5.C.2 rather than asserted.

Why "ideal cases". Real wheels and real surfaces deform at the contact, and the energy lost to that deformation is rolling friction. The unit's boundary statement removes it: "Rolling friction is beyond the scope of AP Physics C: Mechanics." So on this exam the ideal case is the case, and the hedge in 6.5.B.2 is telling you what the idealisation is rather than warning you off it.

One thing 6.5.B.2 emphatically does not say: that there is no friction force. There is, and on a ramp it is the force supplying the torque that spins the object up. Set it to zero and the object slides down without rotating at all.

Friction transfers energy between the two buckets without removing any

The sharper version of 6.5.B.2, and the version that survives a follow-up question, is this: static friction on a body rolling without slipping moves energy from the translational store to the rotational store and takes none out of the total.

Work it out for an object rolling down a ramp, with $f$ the static friction force acting up the slope and $R$ the radius.

• Work on the translational motion: the friction opposes the centre of mass, so it contributes $-f\,\Delta x_{\text{cm}}$.
• Work on the rotational motion: the friction acts at the rim, exerting a torque $fR$ about the centre of mass in the sense that spins the object up, so by Topic 6.2 it contributes $+\int fR\, d\theta = +fR\,\Delta\theta$.
• The rolling constraint says $\Delta x_{\text{cm}} = R\Delta\theta$, so the two are equal in size and opposite in sign, and they sum to zero.

So the friction force removes energy from the translational bucket at exactly the rate it adds it to the rotational bucket. Nothing is lost, and the total mechanical energy is still conserved, which is what makes the energy route to a ramp problem legitimate. Worked example 2 puts numbers to it: $-5.52$ J and $+5.52$ J on a two-metre ramp.

It also answers the next question. If friction does no net work, why does a rolling object reach the bottom slower than a sliding one? Because friction diverted part of the gravitational energy into rotation, so less is available as translational kinetic energy. The energy is not missing, it is spinning.

And it explains why mass and radius drop out of the final speed. All that survives is the dimensionless ratio

which is $\tfrac{1}{2}$ for a uniform solid cylinder or disk, $1$ for a hoop or thin cylindrical shell, $\tfrac{2}{5}$ for a uniform solid sphere and $\tfrac{2}{3}$ for a thin spherical shell. None of those values is printed on the equation sheet, so a question that needs one either supplies it in the stem or expects you to produce it from $I = \int r^2\, dm$.

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

Statement 6.5.B.1 gives three equations, and it is worth noticing what the equation sheet prints of them.

Those subscript differences are not cosmetic, and mixing them up is a real defect. On the sheet, $v = r\omega$ and $a_T = r\alpha$ are general statements about a point at distance $r$ from a rotation axis: $r$ is that point's distance from the axis and $a_T$ is a tangential acceleration. In 6.5.B.1 they are specialised to rolling: $r$ is the radius of the rolling object, and the quantities on the left describe its centre of mass, which is not on the rim.

So the rolling case is the one place in Unit 6 where $r$ is unambiguously the radius. Everywhere else, $r$ is a distance from an axis or a reference point, and getting into the habit of asking "distance from what?" is worth more than memorising which equation uses which letter.

The constraint also has a consequence that trips people up: the contact point is instantaneously at rest, the centre moves at $v_{\text{cm}}$, and the top of the wheel moves at $2v_{\text{cm}}$. The framework's vocabulary appendix makes the underlying point about a wheel: "At any given instant in time, Point A is traveling with a greater translational speed and in a different direction than Point B."

Statement 6.5.C.1 marks the boundary of the constraint in one line: when slipping, the motion of a system's centre of mass and the system's rotational motion cannot be directly related. Read that as a licence and a warning. Once slipping starts, you have two independent variables and you need two independent equations, one translational and one rotational, and $v_{\text{cm}} = r\omega$ is not available until the moment slipping stops.

The energy split, and the ratio that decides a race

Statement 6.5.A.1 gives the total as a sum, $K_{\text{tot}} = K_{\text{trans}} + K_{\text{rot}}$. That exact equation is not printed on the equation sheet. The sheet prints $K = \frac{1}{2}mv^2$ and $K_{\text{rot}} = \frac{1}{2}I\omega^2$ on separate lines; adding them is the step 6.5.A.1 is asking for.

Combine the sum with the rolling constraint and everything about a ramp race falls out at once. Substituting $\omega = v_{\text{cm}}/R$:

with $\beta = I_{\text{cm}}/MR^2$ as above. Set that equal to $Mgh$ and the mass cancels:

Three consequences that are exactly what skill 2.D, functional dependence, is asking you to see:

• Mass does not matter. It cancels. A heavy hoop and a light hoop tie.
• Radius does not matter either, because $\beta$ is dimensionless and independent of $R$ for any shape of uniform composition.
• Only the shape matters, through $\beta$. Smaller $\beta$ means a smaller share of the energy goes into rotation and a bigger share into translation, so the object arrives faster. Solid sphere ($\beta = \tfrac{2}{5}$) beats solid cylinder ($\tfrac{1}{2}$) beats spherical shell ($\tfrac{2}{3}$) beats hoop ($1$).

That is the reasoning the framework's sample free-response Question 4 asks for. Its published example response says that more rotational inertia means more gravitational potential energy converted into rotational kinetic energy and less left as translational kinetic energy, so the object with the least translational kinetic energy moves slowest and reaches the bottom last, and it concludes that "the hoop will move the slowest and the solid sphere the fastest".

The unit's fifth sample instructional activity is the classroom version: roll a hoop and a disk of equal mass and radius down identical ramps, then explain why the disk reached the bottom in less time using energy bar charts and to-scale free-body diagrams.

Two routes down the ramp, and the framework credits both

A rolling-down-a-ramp question can be attacked with energy or with forces and torques, and the AP Physics C: Mechanics scoring guidelines for sample free-response Question 4 credit both explicitly.

The forces route, as the published example response runs it. Take torques about the contact point, so that friction has no lever arm and gravity does. The scoring guidelines write it as

where $I + MR^2$ is the rotational inertia about the contact point, from the printed parallel-axis theorem $I' = I_{\text{cm}} + Md^2$ with $d = R$. Substituting $\alpha = a/R$ gives

The forces route, the other way. The guidelines also credit taking torques about the centre of mass instead, writing the pair $F_f R = I\alpha$ and $Mg\sin\theta - F_f = Ma$ and solving simultaneously. Same acceleration, and it hands you the friction force as a bonus: $F_f = \frac{\beta}{1 + \beta}Mg\sin\theta$. The energy route. The scoring note attached to that item says outright that a derivation which correctly applies conservation of energy can earn the same points; that is the $v_{\text{cm}} = \sqrt{2gh/(1+\beta)}$ of the previous section.

The friction expression is the one to carry away, because it answers the question the energy route cannot: how rough does the surface have to be for the object to roll rather than slide? Since static friction is capped at $\mu_s F_N = \mu_s Mg\cos\theta$, rolling without slipping requires

Mass and radius drop out again. Steeper ramps and larger $\beta$ both demand more friction. Worked example 1 computes it for a cylinder on a 25 degree ramp and gets $\mu_s \geq 0.155$.

Rolling while slipping (6.5.C)

Objective 6.5.C is a whole learning objective, so this is not an optional extra. Two statements define it.

6.5.C.1: when slipping, the motion of a system's centre of mass and the system's rotational motion cannot be directly related. In practice: drop the constraint. Track $v_{\text{cm}}(t)$ and $\omega(t)$ as two separate functions of time, with kinetic friction acting.

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.

That second statement is the exact mirror of the rolling case and the reason the two are separate objectives. It also gives you the quantitative handle: the energy dissipated is the friction force times the relative sliding distance, which is the difference between how far the centre of mass travelled and how far the rim rolled, $\Delta x_{\text{cm}} - R\Delta\theta$. That difference is zero when rolling, which is 6.5.B.2 again.

The standard problem is an object launched onto a surface spinning too slowly, or not at all, for its speed. Kinetic friction then does two things at once: it decelerates the centre of mass and it spins the object up. Both continue until $v_{\text{cm}} = R\omega$, at which instant the slipping stops, the friction switches to static, and the object rolls the rest of the way. Worked example 3 runs that transition and checks the dissipated energy against the sliding distance to the digit.

An elegant shortcut worth knowing for this class of problem: angular momentum about a point on the surface is constant throughout the slipping phase, because the friction force acts along the surface line and therefore has no lever arm about a point on it, gravity and the normal force cancel, and the normal force also has no torque about a point directly below. That gets you the final speed in one line without touching the time dependence. Worked example 3 checks it both ways.

What is genuinely the same as AP Physics 1, and what is not

Being honest here is more useful than manufacturing a difference. On Topic 6.5 the two courses overlap more than anywhere else in Unit 6.

Identical between the two courses:

• The three rolling-constraint equations of 6.5.B.1, word for word.
• The energy split of 6.5.A.1 as a sum of two terms.
• The friction statement 6.5.B.2, which is the same sentence in both frameworks.
• The parallel-axis theorem, $I' = I_{\text{cm}} + Md^2$, printed on both sheets.
• The $\beta$ ranking that decides a ramp race, and the fact that mass and radius cancel.

Genuinely different in AP Physics C: Mechanics:

• $I = \int r^2\, dm$ is printed here and not there, so a rolling object with a non-uniform mass distribution is answerable here and is not there.
• Skill 2.A, deriving a symbolic expression, carries 25 to 30% of the multiple-choice section. The framework's sample Question 4 part B asks for $t$ in terms of $I$, and its published answer is $t = \sqrt{\dfrac{2L(I + MR^2)}{MgR^2\sin\theta}}$.
• Skill 3.A, creating experimental procedures, appears in Unit 6 only under this topic, and both of the unit's experiment activities are rolling activities.
• The slipping-to-rolling transition is tractable here as a two-variable problem in time, and the angular-momentum shortcut for it follows from the cross product in Topic 6.3.

If you want the algebra-based treatment, the [AP Physics 1 Topic 6.5 page](/ap-physics-1/unit-6-energy-and-momentum-of-rotating-systems/6-5-rolling) is for students in that course. This page is for students in AP Physics C: Mechanics. On the constraint and the energy split the two say the same thing, because the frameworks say the same thing. The difference is the derivation-first framing and the rotational inertia integral.

How Topic 6.5 is tested

The framework's sample free-response Question 4 is a Qualitative/Quantitative Translation question worth 8 points, tagged with skills 3.B, 3.C, 2.A and 2.D, and aligned to learning objectives 1.3.A, 3.4.B, 5.4.A, 5.6.A, 5.4.B and 6.1.A. It places a hollow sphere, a uniform solid sphere and a hoop of the same mass $M$ and radius $R$ at the top of a ramp of length $L$ at angle $\theta$, releases all three from rest, and rolls them down without slipping.

Its three parts are a fair template for what this topic asks. Part A: rank the times from greatest to least and justify, with the instruction that the justification may reference equations but must include conceptual reasoning beyond algebraic solutions. Part B: derive the relationship between the time and the rotational inertia, beginning from a fundamental physics principle or an equation from the reference information. Part C: a new sphere is a thin shell filled with liquid that does not rotate with the shell, and you must say whether it reaches the bottom in more, less or the same time as the solid sphere and justify using the part B equation. The published answer to part C is less time, because less mass rotates so the rotational inertia is smaller.

Two lessons from the scoring guidelines. The instruction to begin from a fundamental principle or a reference-sheet equation is not decoration: part B's first point is awarded for a multistep derivation that uses Newton's second law in rotational form. And the unit's Preparing for the AP Exam note uses a rolling example to make its point about justifications, saying that stating one disk is rolling faster than another because of "conservation of energy" is not a complete enough answer to earn credit.

Both of the unit's laboratory-flavoured sample activities sit under 6.5. One releases a yo-yo down a ramp and uses a meterstick and stopwatch to find its final velocity and release height, then its outer radius and mass, and determines its rotational inertia using energy concepts. The other is the hoop-and-disk race with energy bar charts and to-scale free-body diagrams.

Where it connects: Topic 6.1 supplies the two energy terms, Topic 6.2 supplies the work done by the friction torque, and Topic 6.4 supplies the shortcut for the slipping phase. The inclined plane guide owns the ramp geometry, the static compared with kinetic friction guide owns the distinction this topic depends on, and the Unit 6 hub lists all six topics.