6.5 Kinetic Energy of a System with Translational and Rotational Motion

When an object both moves and spins, its total kinetic energy is the sum of translational energy ($\frac{1}{2}mv {cm}^2$) and rotational energy ($\frac{1}{2}I\omega^2$). For rolling without slipping, the no-slip condition $v {cm} = r\omega$ links the two motions, and static friction does no work, so mechanical energy is conserved.

Why This Matters for the AP Physics C: Mechanics Exam

Rolling problems pull together ideas from energy, rotation, and friction, so they are common in both multiple-choice and free-response questions. You will often need to combine conservation of energy with the no-slip condition to find a speed, acceleration, or energy partition. Free-response answers usually ask you to explain your reasoning, not just write an equation, so being able to say why static friction does no work or why energy is conserved is just as important as the math. Unit 6 carries a 10 to 15 percent weighting on the exam, and rolling sits right in the middle of it.

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Key Takeaways

• Total kinetic energy of a rolling object: $K_{tot} = K_{trans} + K_{rot} = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I\omega^2$.
• Rolling without slipping gives three constraints: $\Delta x_{cm} = r\Delta\theta$, $v_{cm} = r\omega$, and $a_{cm} = r\alpha$.
• In ideal rolling without slipping, static friction is a constraint force that does no work, so mechanical energy is conserved.
• When slipping, $v_{cm} = r\omega$ no longer holds, and kinetic friction dissipates mechanical energy as thermal energy.
• The energy split between translational and rotational depends on the moment of inertia, so objects with more mass farther from the axis roll slower down a ramp.
• Rolling friction is outside the scope of this course.

Total Kinetic Energy

When an object both moves linearly and rotates, you have to account for both forms of energy. The total kinetic energy is the complete energy of motion for the system, found by adding the two pieces:

$K_{\text{tot}} = K_{\text{trans}} + K_{\text{rot}}$

Where:

• $K_{\text{tot}}$ is the total kinetic energy of the system
• $K_{\text{trans}}$ is the translational kinetic energy, $\frac{1}{2}mv_{cm}^2$ (with $m$ the mass and $v_{cm}$ the center-of-mass speed)
• $K_{\text{rot}}$ is the rotational kinetic energy about the center of mass, $\frac{1}{2}I\omega^2$ (with $I$ the moment of inertia and $\omega$ the angular velocity)

This combined approach lets you analyze motions like a rolling ball, where both the moving center of mass and the spinning contribute to the overall energy. Both terms are scalars, so you simply add them.

Rolling Without Slipping

Key Relationships

Rolling without slipping is the ideal case where an object rolls along a surface with no relative motion at the point of contact. The contact point is instantaneously at rest, which ties the linear motion of the center of mass directly to the rotation.

These equations connect translational and rotational motion:

• $\Delta x_{\mathrm{cm}} = r \Delta \theta$ : linear displacement equals radius times angular displacement
• $v_{\mathrm{cm}} = r \omega$ : linear velocity equals radius times angular velocity
• $a_{\mathrm{cm}} = r \alpha$ : linear acceleration equals radius times angular acceleration

These relationships show up in everyday scenarios such as:

• A car wheel rolling on a road
• A ball rolling down a ramp
• A cylinder rolling on a flat surface

Friction in Ideal Rolling

In perfect rolling without slipping, static friction matters but works in a special way.

Static friction in rolling motion:

• Acts as a constraint force rather than a dissipative force
• Keeps the contact point momentarily at rest, preventing slipping
• Does no work on the system, because there is no displacement at the point where the force is applied
• Lets mechanical energy be conserved, unlike kinetic friction

Because no energy is lost, you can use conservation of mechanical energy to solve rolling-without-slipping problems.

Rolling with Slipping

Center of Mass vs Rotational Motion

When slipping occurs, the clean link between linear and angular motion breaks down.

In rolling with slipping:

• The equations $\Delta x_{\mathrm{cm}} = r \Delta \theta$, $v_{\mathrm{cm}} = r \omega$, and $a_{\mathrm{cm}} = r \alpha$ no longer apply
• The center-of-mass motion and the rotational motion become decoupled
• You must track translational and rotational components separately
• The contact point has real motion relative to the surface

This happens when a vehicle's wheels spin on ice or when a ball is thrown with backspin onto a surface.

Kinetic Friction in Slipping

When slipping occurs, kinetic friction takes over and changes the energy picture.

Kinetic friction during slipping:

• Converts mechanical energy into thermal energy
• Does negative work on the system, lowering its total mechanical energy
• Often drives the system toward eventual rolling without slipping
• Acts at a contact point that moves relative to the surface, which is why it dissipates energy

Track this energy loss when you predict the final state of an object that starts out slipping.

How to Use This on the AP Physics C: Mechanics Exam

Problem Solving

For a rolling-without-slipping object, the fastest path is usually conservation of energy. Set the change in gravitational potential energy equal to the gain in total kinetic energy, then substitute $\omega = v_{cm}/r$ so everything is in terms of $v_{cm}$:

$mg\Delta h = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I\omega^2$

The moment of inertia controls how the energy splits. A larger $I$ (more mass farther from the axis) sends more energy into rotation, leaving less for translation, so that object reaches a lower speed at the bottom of a ramp.

Free Response

Stating "by conservation of energy" or "because of friction" is not enough to support a stronger score. Walk through the steps: name the principle, state why it applies (for example, static friction does no work in pure rolling so mechanical energy is conserved), then connect that to your claim. If two objects roll down the same ramp, explain the role of $I$ in the energy partition before declaring which is faster.

Common Trap

Watch the friction type. Static friction in pure rolling does no work, so energy is conserved. Kinetic friction during slipping does negative work and removes mechanical energy. Mixing these up will break your energy bookkeeping.

Common Misconceptions

• Friction always removes energy. In rolling without slipping, static friction does no work because the contact point is instantaneously at rest. Only kinetic friction during slipping dissipates energy.
• $v_{cm} = r\omega$ always works. This holds only while rolling without slipping. Once an object slips, the center-of-mass and rotational motions are decoupled and you cannot use it.
• All round objects roll down a ramp at the same speed. The energy split depends on $I$. A hoop ($I = mr^2$) ends up slower than a solid sphere ($I = \frac{2}{5}mr^2$) of the same mass and radius, because the hoop puts more energy into rotation.
• A rolling object's energy is purely translational. You must add the rotational term. Using only $\frac{1}{2}mv^2$ undercounts the total kinetic energy.
• Heavier objects always roll faster. When you solve down an incline, the mass cancels out. The final speed depends on the shape (through $I/mr^2$) and the height, not the mass.

Practice Problem 1: Total Kinetic Energy

Solution

Find both the translational and rotational pieces, then add them:

1. Translational kinetic energy: $K_{trans} = \frac{1}{2}mv^2 = \frac{1}{2} \times 2.0 \text{ kg} \times (5.0 \text{ m/s})^2 = 25.0 \text{ J}$

2. Since the sphere rolls without slipping, use $v_{cm} = r\omega$ to get the angular velocity: $\omega = \frac{v_{cm}}{r} = \frac{5.0 \text{ m/s}}{0.10 \text{ m}} = 50.0 \text{ rad/s}$

3. Rotational kinetic energy: $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} \times \frac{2}{5}mr^2 \times \omega^2$ $K_{rot} = \frac{1}{2} \times \frac{2}{5} \times 2.0 \text{ kg} \times (0.10 \text{ m})^2 \times (50.0 \text{ rad/s})^2$ $K_{rot} = \frac{1}{2} \times \frac{2}{5} \times 2.0 \text{ kg} \times 0.01 \text{ m}^2 \times 2500 \text{ rad}^2/\text{s}^2$ $K_{rot} = 10.0 \text{ J}$

4. 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 Without Slipping

Solution

Because the cylinder rolls without slipping, mechanical energy is conserved, so use conservation of energy:

1. The initial energy is gravitational potential energy: $E_i = mgh_i$

2. The final energy is translational plus rotational kinetic energy (taking the final height as the reference): $E_f = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

3. Height change from the distance along the incline: $\Delta h = 3.0 \text{ m} \times \sin(30°) = 3.0 \text{ m} \times 0.5 = 1.5 \text{ m}$

4. By conservation of energy: $mg\Delta h = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

5. For rolling without slipping, $\omega = \frac{v}{r}$, so: $mg\Delta h = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{v}{r}\right)^2$ $mg\Delta h = \frac{1}{2}mv^2 + \frac{1}{4}mv^2$ $mg\Delta h = \frac{3}{4}mv^2$

6. Solving for $v$: $v^2 = \frac{4g\Delta h}{3} = \frac{4 \times 9.8 \text{ m/s}^2 \times 1.5 \text{ m}}{3} = 19.6 \text{ m}^2/\text{s}^2$ $v = 4.43 \text{ m/s}$

Notice the mass cancels out, so the final speed depends only on the shape of the object and the height dropped.

Related AP Physics C: Mechanics Guides

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