Rolling vs Sliding: What Is the Difference?

One condition, and the rest is consequences

A wheel can move forward and it can turn. Rolling without slipping is the case where those two motions are locked together in a fixed ratio, and sliding is any case where they are not.

The AP Physics 1 CED writes the lock as three equations. EK 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

They are one statement written three ways. The wheel lays out exactly as much ground as it unrolls arc, so the distance travelled is $r$ times the angle turned, and differentiating that gets you the other two.

Then the CED says what happens when the lock fails. EK 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. Not "are related by a harder formula". Cannot be directly related. $v_{\text{cm}}$ and $\omega$ become two independent numbers, and a problem that gives you one tells you nothing about the other.

Three cases are worth naming before going further, because "sliding" is used loosely.

• Rolling without slipping. $v_{\text{cm}} = r\omega$ holds. The contact point is instantaneously at rest against the surface, so any friction there is static.
• Sliding without rotating. A block, or a wheel with the brakes fully locked. $\omega = 0$ while $v_{\text{cm}}$ is not, so $v_{\text{cm}} \neq r\omega$ by the widest possible margin.
• Rolling while slipping. Both motions are happening and they do not match, which is a wheel spinning on ice or a bowling ball in the first meters of its throw. The contact point is moving against the surface, so the friction is kinetic.

This page is mostly about the first two, because that is the comparison the exam builds problems on. The third is the one the CED deliberately keeps qualitative, and there is a section on that below.

Rolling vs sliding, side by side

The row to notice is the one about energy. Static friction at a rolling contact does no work. Its point of application is not moving, so there is no displacement for the force to act through. That is the technical reason a rolling object on a ramp can be treated with pure energy conservation, and it is the reason the whole ramp problem is tractable. The CED states the conclusion at EK 6.5.B.2: for ideal cases, rolling without slipping implies that the frictional force does not dissipate any energy from the rolling system.

The contrasting statement is EK 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. Note that the CED gives the reason, not just the rule: the point of application moves. That is the whole distinction between the two frictions in this context, and it is the same reasoning as static vs kinetic friction applied to a contact that keeps being renewed.

The case that separates them: the race down the ramp

Release four objects from rest at the same height $h = 1.2$ m and let them reach the bottom. One is a block on a frictionless surface, so it slides without rotating. The other three roll without slipping.

Every one of the rolling objects arrives with the same total kinetic energy as the sliding block. Each converted the same $mgh$ and none of them lost any to friction, because in every rolling case the friction was static and did no work. What differs is the split. The block put all of it into translation. The hoop put half of it into spinning, so only half was available to make it move, and it arrives slowest.

That is the payoff of this comparison and it is worth stating as a sentence: a rolling object is slower not because it lost energy but because it had to spend some of it turning.

The ordering is fixed by $\beta$ alone, the shape factor, which measures how far the mass sits from the axis in units of the radius. Mass close to the axis means a small $\beta$, a small rotational share and a fast arrival. A hoop has all of its mass out at the rim, so $\beta = 1$, the split is exactly even and of the four objects listed it comes last.

Two things this race does not depend on, which the next section proves:

• How heavy the object is. A bowling ball and a marble of the same shape arrive together.
• How big it is. A large sphere and a small sphere arrive together, provided both roll.

The moments of inertia in the table above are not printed on any AP equation sheet. None of the four sheets carries a table of rotational inertias for common shapes: what they print is the definition $I = \sum m_i r_i^2$ and the parallel axis theorem $I' = I_{\text{cm}} + Md^2$. On an exam the $I$ for a shape is given to you in the question, so read the stem before you assume a $\frac{1}{2}$.

Why the mass and the radius drop out

This is worth doing rather than remembering, because the cancellation is quick and it tells you which quantities actually matter.

Write the rotational inertia of any round object about its own center as $I = \beta m r^2$, where $\beta$ is a pure number set by the shape. Conserve energy from a height $h$, using the rolling condition $\omega = \frac{v}{r}$ from EK 6.5.B.1 and the total kinetic energy from EK 6.5.A.1:

The $r^2$ in the rotational inertia and the $r^2$ in $\omega^2$ cancel against each other:

Now $m$ divides out of both sides:

No $m$, no $r$. The radius went because the rolling condition ties $\omega$ to $\frac{v}{r}$ while the rotational inertia carries an $r^2$, and the two dependences are exact inverses. The mass went for the same reason it always goes in an energy problem about gravity: it appears once on each side.

The same structure appears in the acceleration. On an incline of angle $\theta$, with static friction $f$ up the slope, Newton's second law along the slope gives $ma_{\text{cm}} = mg\sin\theta - f$, and torque about the center gives $fr = I\alpha = \beta m r^2 \frac{a_{\text{cm}}}{r}$, so $f = \beta m a_{\text{cm}}$. Substituting:

Again no $r$ anywhere, and no $m$ except in the friction force itself, which is proportional to weight as friction always is. The last expression is the smallest coefficient of static friction that will keep the object rolling: exceed the slope it can handle and it starts to slip, and every equation above it stops applying.

Two corollaries worth carrying:

• A steeper ramp needs more friction to keep something rolling, since $\mu_{s,\text{min}}$ rises with $\tan\theta$.
• A hoop needs more friction than a sphere on the same slope, since $\frac{\beta}{1+\beta}$ rises with $\beta$. At $25^\circ$ a solid sphere needs $\mu_s \geq 0.133$, a solid cylinder $0.155$ and a hoop $0.233$.

The contact point is standing still

The single mental picture that makes rolling behave is this one. At the instant of contact, the bottom of a rolling wheel is not moving.

Take the velocity of any point on a rolling wheel as the sum of two pieces: the whole wheel translating at $v_{\text{cm}}$, and the point circling the center at $r\omega$. At the contact point those two are equal in size and opposite in direction, because the rolling condition says $v_{\text{cm}} = r\omega$ and the bottom of the wheel is moving backward relative to the center. They cancel exactly.

The top of a rolling wheel moves at twice the speed of the car it is attached to. That is a genuine and checkable prediction of the rolling condition, and it is why the spokes of a moving bicycle wheel blur at the top and look sharp at the bottom.

Three consequences follow, and each of them is a place students lose marks by reasoning from the sliding picture instead.

• The friction is static, not kinetic. Static friction is the friction between surfaces that are not moving relative to one another, and here they are not. That is why $\mu_s$ and not $\mu_k$ sets the limit, and why the friction force takes whatever value the rolling requires rather than the fixed $\mu_k N$.
• That friction does no work. Work needs the point of application to move through a displacement. This one does not move, so no energy is dissipated, which is exactly EK 6.5.B.2. Each instant a new bit of the wheel becomes the contact point, and each of them is at rest while it holds the job.
• A rolling object can therefore be treated with straightforward energy conservation, which is the only reason the ramp race above is a one-line calculation.

Now break the condition. If a wheel is spinning faster than it is travelling, $r\omega > v_{\text{cm}}$, the contact point is moving backward against the ground, and kinetic friction acts forward on the wheel. If it is travelling faster than it is spinning, $r\omega < v_{\text{cm}}$, the contact point is sliding forward and friction acts backward. In both cases the friction pushes the system back toward the rolling condition, which is why a bowling ball thrown without spin eventually rolls, and why a car whose wheels are spinning on ice eventually grips.

What the CED will and will not ask about slipping

Topic 6.5 carries two boundary statements and they are unusually specific about where the quantitative work stops.

Boundary statement one: rolling friction is beyond the scope of AP Physics 1. So the slow loss that eventually stops a real ball rolling on a flat floor is not modelled. Treat an ideal rolling contact as lossless, which is what EK 6.5.B.2 licenses.

Boundary statement two: 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 is an unusually clear instruction and it tells you exactly what to prepare. You will not be asked to compute the moment a slipping ball starts to roll. You will be asked to say which way $v_{\text{cm}}$ and $\omega$ move and why. The qualitative account is short:

In both rows the two quantities move toward each other and the process dissipates energy the whole time, by EK 6.5.C.2, so the final rolling speed is always less than a lossless analysis would give.

The second boundary statement also names AP Physics 2, which is worth flagging: AP Physics 2's course framework does not include rotational mechanics, but the AP Physics 2 equation sheet carries the full mechanics block, so the same rotational lines are available there.

Where the confusion costs a mark

• Using $v_{\text{cm}} = r\omega$ when the object is slipping. EK 6.5.C.1 says the two motions cannot be directly related then. Check the condition before you use the equation.
• Writing $K = \frac{1}{2}mv^2$ for a rolling object. By EK 6.5.A.1 the total kinetic energy is $K_{\text{trans}} + K_{\text{rot}}$, and for a solid cylinder the rotational term is half as big again as you would otherwise have. Missing it makes the object arrive too fast.
• Subtracting a friction loss on a rolling ramp. In the ideal case there is none, by EK 6.5.B.2, because the static friction acts at a point that is not moving.
• Using $\mu_k N$ for the friction on a rolling object. The contact is static, so the friction takes whatever value rolling demands, up to a maximum of $\mu_s N$. On an incline that value is $\frac{\beta m g \sin\theta}{1+\beta}$, which is generally well below the maximum.
• Assuming the heavier object wins the ramp race. Mass cancels. So does radius. Only $\beta$, the shape factor, decides.
• Assuming the bigger object wins. Same cancellation. A large hoop and a small hoop arrive together, and both lose to any solid sphere.
• Quoting a rotational inertia from memory. No AP equation sheet prints a table of them. The exam gives you the $I$ for the shape in question, and if it has not, you are meant to build it from $I = \sum m_i r_i^2$ or the parallel axis theorem.
• Saying a rolling object loses energy to friction and therefore arrives slower. It arrives slower and it loses no energy. Those are two separate facts and running them together is a scoring error even though the conclusion is right.
• Forgetting that a rolling object still has translational kinetic energy. Both terms are live at once, which is the subject of rotational vs translational kinetic energy.
• Attempting a quantitative rolling-while-slipping calculation. The Topic 6.5 boundary statement says you will not be asked for one. A qualitative account is what earns the mark.

What the two have in common, and why that lulls you

The two motions look alike from a distance and several common setups make them behave alike.

On a flat surface at constant speed, they are indistinguishable from the outside. A puck sliding at $6.0$ m/s and a wheel rolling at $6.0$ m/s cover the same ground in the same time. Nothing about the trajectory gives the difference away, and the distinction only appears when you ask about energy or about what happens when something changes.

The center of mass obeys the same law in both cases. $\vec{a}_{\text{cm}} = \frac{\vec{F}_{\text{net}}}{m}$ holds whether the object rolls, slides or tumbles. That is why the translational half of a rolling problem looks exactly like a block problem, and why it is so natural to stop there.

The linear momentum is the same for both. $\vec{p} = m\vec{v}_{\text{cm}}$ takes no notice of whether the object is spinning, which is one of the differences worked in angular vs linear momentum.

A very small $\beta$ makes the difference numerically small. A thin-walled tube filled with a dense fluid that does not rotate with it, or any object with most of its mass at the axis, rolls down a ramp almost as fast as it would slide. Nothing is wrong with the physics; the rotational share is just small.

The difference shows up in four exam-shaped places.

• Any energy question. The split between translation and rotation is the answer.
• Any race or ordering question. The ranking is by $\beta$ and it is worth having the four common values to hand.
• Any question about friction. Static and doing no work, against kinetic and dissipating.
• Any question about the contact point. Rolling puts it at rest, and sliding does not, and almost every conceptual multiple-choice question about rolling is really a question about that point.

The habit: before writing any equation, check $v_{\text{cm}}$ against $r\omega$. If they match, you have energy conservation and static friction and the whole toolkit. If they do not, you have kinetic friction and a dissipative problem, and the CED expects words rather than numbers.

What the CED asks, and what the sheets print

Rolling is AP Physics 1 Unit 6, Energy and Momentum of Rotating Systems, weighted at 5 to 8 percent of the multiple-choice section, at Topic 6.5, Rolling. It is the last topic of the unit and it assumes both Unit 5's rotational dynamics and Unit 3's energy machinery.

Topic 6.5 carries three learning objectives.

• 6.5.A: describe the kinetic energy of a system that has translational and rotational motion. EK 6.5.A.1 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. EK 6.5.B.1 gives $\Delta x_{\text{cm}} = r\Delta\theta$, $v_{\text{cm}} = r\omega$ and $a_{\text{cm}} = r\alpha$. EK 6.5.B.2 says that 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. EK 6.5.C.1 says that when slipping, the motion of a system's center of mass and the system's rotational motion cannot be directly related. EK 6.5.C.2 explains that the point of application of the kinetic friction force moves with respect to the surface, so that force dissipates energy from the system.

Suggested skills for Topic 6.5 are 1.A, 2.A, 2.C and 3.C. The two boundary statements are quoted in full in the section above.

On the equation sheets, checked line by line across all four:

• All four print $\Delta x_{\text{cm}} = r\Delta\theta$, the position form of the rolling condition, in the rotational block.
• None of the four prints $v_{\text{cm}} = r\omega$ or $a_{\text{cm}} = r\alpha$. Those two live in the CED at EK 6.5.B.1 and nowhere on the sheet, so you supply them. What the sheets do print is $v = r\omega$ and $a_T = r\alpha$, which are the statements about a point on a rotating rigid system rather than about the center of mass of a rolling one. The equations look the same and the claims are not, which is the trap covered in angular vs linear acceleration.
• All four print $K = \frac{1}{2}mv^2$ and a rotational kinetic energy. The AP Physics 1 and AP Physics 2 sheets write the second one as $K = \frac{1}{2}I\omega^2$ with no subscript, so the two energies share a symbol; the two Physics C sheets write $K_{\text{rot}} = \frac{1}{2}I\omega^2$ and keep them apart.
• None of the four prints a table of rotational inertias for common shapes. They print $I = \sum m_i r_i^2$ and $I' = I_{\text{cm}} + Md^2$, with the Physics C sheets adding $I = \int r^2\,dm$. The $\beta$ values used on this page come from the shapes themselves and would be given in an exam question.
• The friction limit $\lvert \vec{F}_f \rvert \leq \lvert \mu \vec{F}_N \rvert$ is on all four, with one symbol $\mu$ rather than separate static and kinetic coefficients.

For the surrounding material: Topic 6.5 has the full CED framing, Topic 6.1 covers the rotational energy term, conservation of energy has the energy routine, inclined plane problems has the ramp geometry, and static vs kinetic friction covers the friction distinction this page leans on. The torque and rotational motion practice set has problems to work.