6.1 Rotational Kinetic Energy

Equation for Rotational Kinetic Energy

Rotational kinetic energy represents the energy possessed by an object due to its rotational motion. Just as linear kinetic energy depends on mass and velocity, rotational kinetic energy depends on rotational inertia and angular velocity.

• The equation for rotational kinetic energy is: $K_{\mathrm{rot}}=\frac{1}{2} I \omega^{2}$
• Each term has a specific meaning:
  • $K_{\mathrm{rot}}$ represents the rotational kinetic energy (measured in joules)
  • $I$ represents the rotational inertia (also known as moment of inertia, measured in kg·m²)
  • $\omega$ represents the angular velocity (measured in radians per second)

This equation mirrors the familiar translational kinetic energy equation ($K = \frac{1}{2}mv^2$), highlighting the parallel between linear and rotational motion.

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Equivalence to Translational Energy

The rotational kinetic energy of an object can be directly related to translational energy, showing the fundamental connection between these two forms of motion.

• For a point mass rotating about a fixed axis:
  • The translational speed $v$ relates to angular velocity by $v = r\omega$
  • Substituting this into the rotational energy equation: $K_{\mathrm{rot}} = \frac{1}{2}I\omega^2 = \frac{1}{2}(mr^2)\omega^2 = \frac{1}{2}m(r\omega)^2 = \frac{1}{2}mv^2$
• This demonstrates that for a point mass, rotational kinetic energy equals translational kinetic energy
• For extended objects, we can integrate over all mass elements to find the total rotational energy

Total Kinetic Energy of Systems

Real-world objects often have both rotational and translational motion simultaneously. The total kinetic energy accounts for both types of motion.

• The total kinetic energy is the sum of rotational and translational components: $K_{\mathrm{total}} = K_{\mathrm{rot}} + K_{\mathrm{trans}} = \frac{1}{2}I\omega^2 + \frac{1}{2}Mv_{cm}^2$
• This applies to rigid systems like:
  • A rolling ball (both moving forward and spinning)
  • A thrown frisbee (translating through air while spinning)
  • A car's wheels during normal driving

The separation of kinetic energy into these components allows for clearer analysis of complex mechanical systems.

Rotational Energy with Stationary Center

Many systems exhibit pure rotational motion with their center of mass remaining stationary. These systems still possess significant kinetic energy despite no overall translational movement.

• When the center of mass is stationary ($v_{cm} = 0$):
  • The total kinetic energy equals just the rotational component: $K_{\mathrm{total}} = K_{\mathrm{rot}} = \frac{1}{2}I\omega^2$
  • Individual points within the system still have linear velocities ($v = r\omega$)
• Common examples include:
  • A spinning top maintaining its position
  • A flywheel rotating on a fixed axis
  • Earth rotating on its axis

This concept explains why spinning objects can store substantial energy even when they appear stationary from certain perspectives.

Scalar Nature of Rotational Energy

Rotational kinetic energy, like all forms of energy, is a scalar quantity rather than a vector. This property simplifies many calculations involving energy.

• As a scalar quantity, rotational energy:
  • Has magnitude but no direction
  • Can be added directly without vector components
  • Follows conservation laws in a straightforward manner
• This contrasts with vector quantities in rotational motion:
  • Angular velocity ($\vec{\omega}$)
  • Angular momentum ($\vec{L}$)
  • Torque ($\vec{\tau}$)

The scalar nature allows for simple energy calculations in complex systems with multiple rotating components.

Practice Problem 1: Basic Rotational Kinetic Energy

Solution

First, we need to identify the relevant equation and gather all necessary information:

• The equation for rotational kinetic energy is $K_{\mathrm{rot}} = \frac{1}{2}I\omega^2$
• Mass of the disk: $m = 2.0$ kg
• Radius of the disk: $r = 0.15$ m
• Angular velocity: $\omega = 300$ rpm

We need to convert the angular velocity to radians per second: $\omega = 300 \text{ rpm} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = 31.4 \text{ rad/s}$

For a solid disk, the moment of inertia is: $I = \frac{1}{2}mr^2 = \frac{1}{2} \times 2.0 \text{ kg} \times (0.15 \text{ m})^2 = 0.0225 \text{ kg·m}^2$

Now we can calculate the rotational kinetic energy: $K_{\mathrm{rot}} = \frac{1}{2}I\omega^2 = \frac{1}{2} \times 0.0225 \text{ kg·m}^2 \times (31.4 \text{ rad/s})^2 = 11.1 \text{ J}$

Therefore, the rotational kinetic energy of the disk is 11.1 joules.

Practice Problem 2: Total Kinetic Energy

Solution

For this problem, we need to calculate both the translational and rotational kinetic energy since the sphere is rolling without slipping.

Given information:

• Mass of sphere: $m = 0.5$ kg
• Radius: $r = 0.1$ m
• Translational speed: $v = 2.0$ m/s

For a solid sphere, the moment of inertia is: $I = \frac{2}{5}mr^2 = \frac{2}{5} \times 0.5 \text{ kg} \times (0.1 \text{ m})^2 = 0.002 \text{ kg·m}^2$

When an object rolls without slipping, the relationship between angular velocity and translational velocity is: $v = r\omega$, so $\omega = \frac{v}{r} = \frac{2.0 \text{ m/s}}{0.1 \text{ m}} = 20 \text{ rad/s}$

Now we can calculate:

1. Translational kinetic energy: $K_{\mathrm{trans}} = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.5 \text{ kg} \times (2.0 \text{ m/s})^2 = 1.0 \text{ J}$
2. Rotational kinetic energy: $K_{\mathrm{rot}} = \frac{1}{2}I\omega^2 = \frac{1}{2} \times 0.002 \text{ kg·m}^2 \times (20 \text{ rad/s})^2 = 0.4 \text{ J}$

The total kinetic energy is: $K_{\mathrm{total}} = K_{\mathrm{trans}} + K_{\mathrm{rot}} = 1.0 \text{ J} + 0.4 \text{ J} = 1.4 \text{ J}$

Therefore, the total kinetic energy of the rolling sphere is 1.4 joules.