6.5 Rolling

What is rolling motion in AP Physics 1?

Rolling motion combines straight-line (translational) motion of the center of mass with rotation about that center. The key idea is that an object rolling without slipping has its linear and angular motion locked together by $v_{cm} = r\omega$, and its total kinetic energy is the sum of translational and rotational parts. When an object slips, that link breaks and kinetic friction starts draining energy as heat.

more resources to help you study

practice multiple choiceFRQ practice & scoringcheatsheetsscore calculator

Why This Matters for the AP Physics 1 Exam

Rolling pulls together energy, torque, friction, and rotational motion from earlier in Unit 6 and from Units 3 through 5, so it shows up in both the multiple-choice and free-response sections. You will often need to decide whether a quantity increases, decreases, or stays the same when something changes, and then explain your reasoning step by step.

A big skill here is justification. On the free-response section, saying an object speeds up "because of conservation of energy" is not enough. You have to walk from the principle (like total kinetic energy splitting into translational and rotational pieces) to your actual claim. Rolling problems give you a clean place to practice that kind of careful reasoning with both equations and concepts.

Key Takeaways

• Total kinetic energy of a rolling object is $K_{tot} = K_{trans} + K_{rot} = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I\omega^2$.
• Rolling without slipping links linear and angular motion: $v_{cm} = r\omega$, $a_{cm} = r\alpha$, and $\Delta x_{cm} = r\Delta\theta$.
• In ideal rolling without slipping, static friction acts at the contact point but does not dissipate energy, so mechanical energy is conserved.
• The contact point of a rolling-without-slipping object is momentarily at rest relative to the surface.
• When an object slips, $v_{cm} = r\omega$ no longer holds, and kinetic friction converts mechanical energy into thermal energy.
• The moment of inertia $I$ changes the energy split, so different shapes (sphere, cylinder, hoop) reach different speeds from the same drop.

Kinetic Energy of Translational and Rotational Motion

A rolling object moves in two ways at once: its center of mass slides forward, and the whole body spins. The total kinetic energy adds both.

• Total kinetic energy is the sum of the two parts: $K_{tot} = K_{trans} + K_{rot}$
• Translational kinetic energy depends on mass and center-of-mass speed: $K_{trans} = \frac{1}{2}mv_{cm}^2$
• Rotational kinetic energy depends on moment of inertia and angular velocity: $K_{rot} = \frac{1}{2}I\omega^2$
• For a solid sphere rolling without slipping, this becomes $K_{tot} = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}\left(\frac{2}{5}mR^2\right)\omega^2$

Because $I$ depends on shape, two objects with the same mass and speed can carry different amounts of rotational energy.

Rolling Without Slipping

When an object rolls without slipping, the point touching the surface is momentarily at rest while the rest of the object rotates around it. That single condition ties linear and angular motion together.

• Linear velocity relates to angular velocity: $v_{cm} = r\omega$
• Linear acceleration relates to angular acceleration: $a_{cm} = r\alpha$
• Linear displacement relates to angular displacement: $\Delta x_{cm} = r\Delta\theta$
• The instantaneous velocity at the contact point is zero.

Friction in Ideal Rolling

Static friction is what makes rolling without slipping possible, but it does not drain energy from the system.

• The contact point is momentarily at rest, so there is no sliding there.
• Static friction supplies the force that keeps the object rolling rather than skidding.
• This friction can create a torque about the center of mass.
• Unlike kinetic friction, static friction in ideal rolling does not reduce the system's mechanical energy, so you can use conservation of energy.

Rolling While Slipping

When an object slips, it slides across the surface while spinning, and the clean link between linear and angular motion breaks.

• Linear velocity and angular velocity are no longer tied together: $v_{cm} \neq r\omega$
• The contact point now has a nonzero velocity relative to the surface.
• Linear and rotational motion have to be analyzed separately using Newton's laws and the rotational form of Newton's second law.
• Kinetic friction often acts to bring the object toward rolling without slipping over time.

Energy Dissipation During Slipping

Once an object slips, friction turns some mechanical energy into thermal energy.

• Kinetic friction acts at the contact point while the object slides relative to the surface.
• Because the point where kinetic friction is applied moves relative to the surface, that friction dissipates energy as heat.
• During slipping, both translational and rotational motion change, but they cannot be connected by $v_{cm} = r\omega$ or $a_{cm} = r\alpha$.
• Slipping may decrease until the object rolls without slipping, but you are only expected to describe this qualitatively.

How to Use This on the AP Physics 1 Exam

Problem Solving

For rolling-without-slipping problems, the move is almost always to write total kinetic energy as $K_{trans} + K_{rot}$ and then use $v_{cm} = r\omega$ to get everything in terms of one variable. On an incline, set the lost gravitational potential energy equal to the gained total kinetic energy. The radius usually cancels, which is why the shape (through $I$) matters more than the size.

Free Response

When you justify a claim, connect the principle to the result. For example, if you claim a hoop rolls down slower than a solid cylinder from the same height, explain that the hoop has a larger $I/(mR^2)$, so more of the same gravitational energy goes into rotation, leaving less for translation, which means a smaller $v_{cm}$ at the bottom. Naming "conservation of energy" alone will not support a stronger score.

Common Trap

Watch the friction type. Static friction in rolling without slipping does no work and lets you use energy conservation. Kinetic friction during slipping does negative work and removes mechanical energy. Mixing these up is one of the fastest ways to set up a rolling problem wrong.

Practice Problem 1: Kinetic Energy of Rolling Objects

Solution:

1. Identify the known quantities:

• Mass: $m = 2.0$ kg
• Radius: $r = 0.10$ m
• Linear velocity: $v_{cm} = 5.0$ m/s

1. Since the sphere is rolling without slipping, we can relate linear and angular velocities: $v_{cm} = r\omega$ $\omega = \frac{v_{cm}}{r} = \frac{5.0\text{ m/s}}{0.10\text{ m}} = 50\text{ rad/s}$

2. Calculate the translational kinetic energy: $K_{trans} = \frac{1}{2}mv_{cm}^2 = \frac{1}{2}(2.0\text{ kg})(5.0\text{ m/s})^2 = 25.0\text{ J}$

3. Calculate the rotational kinetic energy using the moment of inertia for a solid sphere: $I = \frac{2}{5}mr^2 = \frac{2}{5}(2.0\text{ kg})(0.10\text{ m})^2 = 0.008\text{ kg}\cdot\text{m}^2$ $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(0.008\text{ kg}\cdot\text{m}^2)(50\text{ rad/s})^2 = 10.0\text{ J}$

4. Calculate the total kinetic energy: $K_{tot} = K_{trans} + K_{rot} = 25.0\text{ J} + 10.0\text{ J} = 35.0\text{ J}$

Practice Problem 2: Rolling Down an Incline

Solution:

1. Identify the known quantities:

• Mass: $m = 5.0$ kg
• Radius: $r = 0.20$ m
• Initial velocity: $v_i = 0$ m/s
• Vertical displacement: $h = 2.0$ m

1. Apply conservation of energy. Initially, the cylinder has only gravitational potential energy. $E_i = mgh$

2. At the bottom, the energy is distributed between translational and rotational kinetic energy: $E_f = \frac{1}{2}mv_f^2 + \frac{1}{2}I\omega_f^2$

3. For a solid cylinder, the moment of inertia is: $I = \frac{1}{2}mr^2$

4. Since the cylinder rolls without slipping: $\omega_f = \frac{v_f}{r}$

5. Substitute into conservation of energy equation: $mgh = \frac{1}{2}mv_f^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v_f}{r})^2$ $mgh = \frac{1}{2}mv_f^2 + \frac{1}{4}mv_f^2$ $mgh = \frac{3}{4}mv_f^2$ $v_f = \sqrt{\frac{4gh}{3}} = \sqrt{\frac{4(9.8\text{ m/s}^2)(2.0\text{ m})}{3}} = 5.1\text{ m/s}$

Practice Problem 3: Friction Force in Rolling

Solution:

1. Identify the known quantities:

• Mass: $m = 7.2$ kg
• Radius: $r = 0.108$ m
• Applied force: $F = 10$ N

1. Apply Newton's Second Law to find the linear acceleration: $F - f_s = ma_{cm}$

2. Apply the rotational version of Newton's Second Law: $\tau = I\alpha$ $f_s \cdot r = \frac{2}{5}mr^2 \cdot \alpha$

3. For rolling without slipping: $a_{cm} = r\alpha$

4. Solve for the friction force by combining these equations: $a_{cm}=\frac{F-f_s}{m}$ From torque and the rolling condition, $f_s r = \frac{2}{5}mr^2\alpha, \quad a_{cm}=r\alpha \Rightarrow f_s = \frac{2}{5}m a_{cm}$ Substitute $a_{cm}=\frac{F-f_s}{m}$: $f_s = \frac{2}{5}(F-f_s)$ $5f_s = 2F - 2f_s$ $7f_s = 2F$ $f_s = \frac{2}{7}F = \frac{2}{7}(10\text{ N}) = 2.86\text{ N}$

Common Misconceptions

• "Friction always removes energy." In rolling without slipping, static friction does no work because the contact point is not sliding, so mechanical energy is conserved. Only kinetic friction during slipping dissipates energy.
• "$v_{cm} = r\omega$ works for everything that spins and moves." This relationship only holds for rolling without slipping. Once an object slips, you cannot use it.
• "Heavier objects roll down faster." For rolling without slipping, mass and radius usually cancel out. What changes the speed is the shape through the moment of inertia, since that sets how the energy splits between translation and rotation.
• "Rotational kinetic energy is separate and you only need translational energy." A rolling object stores energy in both, and ignoring the rotational part gives a wrong total and a wrong final speed.
• "The contact point moves with the object." For rolling without slipping, the contact point is momentarily at rest relative to the surface, which is exactly why static friction acts there.
• "Rolling means the object is rotating but not translating." A rolling object's center of mass is moving forward and the object is spinning at the same time, and the total kinetic energy includes both.

Related AP Physics 1 Guides

• 6.3 Angular Momentum and Angular Impulse
• 6.4 Conservation of Angular Momentum
• 6.6 Motion of Orbiting Satellites
• 6.1 Rotational Kinetic Energy
• 6.2 Torque and Work