6.1 Rotational Kinetic Energy

Rotational kinetic energy is a crucial concept in physics, describing the energy of rotating objects. It's calculated using the equation K = 1/2Iω², where I is rotational inertia and ω is angular velocity.

Understanding rotational kinetic energy helps explain how objects store energy through rotation. It's essential for analyzing spinning systems, from flywheels to planets, and complements translational kinetic energy in describing motion.

Rotational kinetic energy of rigid systems

Rotating objects store energy in their spinning motion. This energy depends on both how fast the object rotates and how its mass is distributed.

• Calculated using the formula $K=\frac{1}{2}I\omega^2$ where I is rotational inertia and ω is angular velocity
• Larger objects with mass distributed far from the rotation axis have greater rotational inertia
• Faster rotation (higher angular velocity) increases rotational energy exponentially due to the squared term

Equation for rotational kinetic energy

The fundamental equation $K=\frac{1}{2}I\omega^2$ mirrors the formula for translational kinetic energy but uses rotational equivalents.

• The formula structure mimics translational kinetic energy ($K=\frac{1}{2}mv^2$) with rotational inertia replacing mass and angular velocity replacing linear velocity
• Rotational inertia (I) depends on both the object's mass and its distribution relative to the rotation axis
• Angular velocity (ω) measures how quickly the object rotates in radians per second

Rotational vs translational kinetic energy

A rigid object can possess rotational kinetic energy even when its center of mass isn't moving through space.

• A spinning basketball on your fingertip has rotational energy but no translational energy
• Each particle in a rotating object follows a circular path, giving it individual linear velocity despite the stationary center of mass
• The sum of all these individual kinetic energies equals the rotational kinetic energy

Total kinetic energy of rigid systems

Real-world objects often combine both rotational and translational motion, requiring us to calculate their total energy.

• Total kinetic energy equals the sum of translational and rotational components: $K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
• A rolling ball has both types of energy - it's moving forward and spinning
• For a perfectly rolling object without slipping, these energies maintain a fixed ratio based on the object's geometry

Rotational energy with stationary center of mass

Understanding how objects can have energy while appearing stationary helps explain many physical phenomena.

• A spinning top demonstrates rotational energy despite its fixed position on a surface
• A gyroscope maintains stability through conservation of angular momentum while its center remains stationary
• The Earth rotates on its axis, giving it enormous rotational energy even though its center of mass follows a nearly fixed path around the Sun

Scalar nature of rotational energy

Energy quantities like rotational kinetic energy have magnitude but no direction, making them scalar values.

• You can add energies directly without considering direction
• This differs from vector quantities like angular velocity which require directional components
• The scalar nature allows for direct energy conservation calculations and simplifies energy transfer analysis

Practice Problem 1: Rotational Kinetic Energy

To solve this problem:

1. Identify the given values:

• Mass (m) = 2.0 kg
• Radius (r) = 0.15 m
• Angular velocity (ω) = 5.0 rad/s
• Moment of inertia formula: $I = \frac{2}{5}mr^2$

1. Calculate the moment of inertia: $I = \frac{2}{5}mr^2 = \frac{2}{5} \times 2.0 \times (0.15)^2 = \frac{2}{5} \times 2.0 \times 0.0225 = 0.018 \text{ kg} \cdot \text{m}^2$

2. Calculate the rotational kinetic energy: $K = \frac{1}{2}I\omega^2 = \frac{1}{2} \times 0.018 \times (5.0)^2 = 0.225 \text{ J}$

The rotational kinetic energy of the sphere is 0.225 joules.

Practice Problem 2: Combined Kinetic Energy

To solve this problem:

1. Identify the given values:

• Mass (m) = 5.0 kg
• Radius (r) = 0.10 m
• Linear speed (v) = 3.0 m/s
• Moment of inertia formula: $I = \frac{1}{2}mr^2$

1. For rolling without slipping, the relationship between linear and angular speed is: $v = r\omega$ or $\omega = \frac{v}{r} = \frac{3.0}{0.10} = 30 \text{ rad/s}$

2. Calculate the moment of inertia: $I = \frac{1}{2}mr^2 = \frac{1}{2} \times 5.0 \times (0.10)^2 = 0.025 \text{ kg} \cdot \text{m}^2$

3. Calculate the translational kinetic energy: $K_{trans} = \frac{1}{2}mv^2 = \frac{1}{2} \times 5.0 \times (3.0)^2 = 22.5 \text{ J}$

4. Calculate the rotational kinetic energy: $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} \times 0.025 \times (30)^2 = 11.25 \text{ J}$

5. Calculate the total kinetic energy: $K_{total} = K_{trans} + K_{rot} = 22.5 + 11.25 = 33.75 \text{ J}$

The total kinetic energy of the rolling cylinder is 33.75 joules.

Practice Problem 3: Energy Conservation

To solve this problem:

1. Identify the given values:

• Mass (m) = 1.0 kg
• Radius (r) = 0.20 m
• Height (h) = 0.50 m
• Moment of inertia formula: $I = \frac{2}{5}mr^2$

1. Apply conservation of energy: Gravitational potential energy at the top equals total kinetic energy at the bottom. $mgh = K_{total} = K_{trans} + K_{rot}$

2. For a solid sphere rolling without slipping, the relationship between translational and rotational kinetic energy is: $K_{rot} = \frac{I\omega^2}{2} = \frac{I}{2} \times \frac{v^2}{r^2} = \frac{I}{2r^2} \times v^2$

3. Substitute the moment of inertia: $K_{rot} = \frac{\frac{2}{5}mr^2}{2r^2} \times v^2 = \frac{1}{5}mv^2$

4. Therefore: $K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$

5. From conservation of energy: $mgh = \frac{7}{10}mv^2$

6. Solve for v: $v^2 = \frac{10gh}{7} = \frac{10 \times 9.8 \times 0.50}{7} = 7.0$ $v = 2.65 \text{ m/s}$

7. Calculate the total kinetic energy: $K_{total} = \frac{7}{10}mv^2 = \frac{7}{10} \times 1.0 \times 7.0 = 4.9 \text{ J}$

The total kinetic energy of the sphere at the bottom of the ramp is 4.9 joules.