6.5 Kinetic Energy of a System with Translational and Rotational Motion

Total Kinetic Energy

When analyzing objects that both move linearly and rotate, we must account for both forms of energy. The total kinetic energy represents the complete energy of motion for the system.

The total kinetic energy is the sum of translational and rotational energies:

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

Where:

• $K_{\text{tot}}$ is the total kinetic energy of the system 🌀
• $K_{\text{trans}}$ is the translational kinetic energy, calculated as $\frac{1}{2}mv^2$ (where $m$ is mass, $v$ is velocity)
• $K_{\text{rot}}$ is the rotational kinetic energy, calculated as $\frac{1}{2}I\omega^2$ (where $I$ is moment of inertia, $\omega$ is angular velocity)

This combined approach allows us to analyze complex motions like a rolling ball, where both types of motion contribute to the overall energy.

more resources to help you study

practice questionscheatsheetscore calculator

Rolling Without Slipping

Key Relationships

Rolling without slipping represents an ideal case where an object rolls along a surface with no relative motion at the point of contact. This creates a direct relationship between linear and angular motion.

The following 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 apply to many 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 the case of perfect rolling without slipping, static friction plays a crucial role but in a unique way.

Static friction in rolling motion:

• Acts as a constraint force rather than a dissipative force
• Prevents slipping by keeping the point of contact momentarily at rest
• Does not perform work on the system since there is no displacement at the point of application 🔋
• Allows mechanical energy to be conserved (unlike kinetic friction)

This idealized model simplifies our analysis since we don't need to account for energy losses due to friction.

Rolling with Slipping

Center of Mass vs Rotational Motion

When slipping occurs, the elegant relationship 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 rotational motion become partially decoupled
• We must analyze translational and rotational components separately
• The point of contact has relative motion with respect to the surface

This situation commonly occurs 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 fundamentally changes the energy dynamics of the system.

Kinetic friction during slipping:

• Converts mechanical energy into thermal energy (heat)
• Performs negative work on the system, reducing its total mechanical energy 🔧
• Gradually brings the system toward a state of rolling without slipping
• The magnitude of energy loss depends on the coefficient of kinetic friction and the slipping distance

Understanding this energy dissipation is essential for accurately predicting the final state of a rolling object that initially slips.

Practice Problem 1: Total Kinetic Energy

Solution

To find the total kinetic energy, we need to calculate both the translational and rotational components:

1. First, calculate the 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. For the rotational kinetic energy, we need the angular velocity. Since the sphere is rolling without slipping, we can use the relationship $v_{cm} = r\omega$: $\omega = \frac{v_{cm}}{r} = \frac{5.0 \text{ m/s}}{0.10 \text{ m}} = 50.0 \text{ rad/s}$

3. Now calculate the 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. The total kinetic energy is the sum: $K_{tot} = K_{trans} + K_{rot} = 25.0 \text{ J} + 10.0 \text{ J} = 35.0 \text{ J}$

Practice Problem 2: Rolling Without Slipping

Solution

Since the cylinder rolls without slipping, we can use conservation of energy to solve this problem:

1. The initial energy is purely gravitational potential energy: $E_i = mgh = mg(h_i)$

2. The final energy consists of translational kinetic energy, rotational kinetic energy, and the remaining gravitational potential energy: $E_f = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 + mg(h_f)$

3. The height change can be calculated from the distance traveled 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: $mgh_i = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 + mgh_f$

5. Rearranging: $mg\Delta h = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

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

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