AP Physics C Mechanics- 6.1 Rotational Kinetic Energy- Study Notes- New Syllabus
AP Physics C Mechanics- 6.1 Rotational Kinetic Energy – Study Notes
AP Physics C Mechanics- 6.1 Rotational Kinetic Energy – Study Notes – per latest Syllabus.
Key Concepts:
- Rotational Kinetic Energy
- Rotational Kinetic Energy with the Center of Mass at Rest
Rotational Kinetic Energy
The rotational kinetic energy of a rigid body is the energy associated with its rotation about a fixed axis. It depends on both the object’s rotational inertia and its angular velocity.
\( \mathrm{K_{rot} = \tfrac{1}{2}I\omega^2} \)
- \( \mathrm{K_{rot}} \): rotational kinetic energy (in joules, J)
- \( \mathrm{I} \): rotational inertia or moment of inertia (in \( \mathrm{kg·m^2} \))
- \( \mathrm{\omega} \): angular velocity (in \( \mathrm{rad/s} \))
Key Idea: Rotational kinetic energy is the rotational equivalent of translational kinetic energy, \( \mathrm{K = \tfrac{1}{2}mv^2} \). The moment of inertia \( \mathrm{I} \) plays the same role in rotational motion as mass \( \mathrm{m} \) does in linear motion.
Equivalence of Rotational and Translational Kinetic Energy
The rotational inertia of an object about a fixed axis can be used to show that the rotational kinetic energy of that object is equivalent to its translational kinetic energy when expressed in terms of the linear velocity of a point on the rim.
From \( \mathrm{v = r\omega} \), substituting into \( \mathrm{K_{rot}} \):
\( \mathrm{K_{rot} = \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}mr^2\left(\dfrac{v}{r}\right)^2 = \tfrac{1}{2}mv^2} \)
This shows that rotational kinetic energy is mathematically analogous to translational kinetic energy.
Total Kinetic Energy of a Rigid System
The total kinetic energy of a rigid system is the sum of:
- Its rotational kinetic energy due to rotation about its center of mass.
- Its translational kinetic energy due to motion of the center of mass.
\( \mathrm{K_{total} = K_{trans} + K_{rot} = \tfrac{1}{2}Mv_{cm}^2 + \tfrac{1}{2}I_{cm}\omega^2} \)
- \( \mathrm{M} \): total mass of the system
- \( \mathrm{v_{cm}} \): velocity of the center of mass
- \( \mathrm{I_{cm}} \): moment of inertia about the center of mass
- \( \mathrm{\omega} \): angular velocity about the center of mass
Key Idea: Objects that both rotate and translate (e.g., rolling objects) possess both forms of kinetic energy. The sum represents the system’s total kinetic energy.
Example
A solid cylinder of mass \( \mathrm{2.0\,kg} \) and radius \( \mathrm{0.20\,m} \) rolls without slipping at a speed of \( \mathrm{3.0\,m/s} \). Find the total kinetic energy of the cylinder.
▶️ Answer / Explanation
Step 1: Write the total kinetic energy formula:
\( \mathrm{K_{total} = \tfrac{1}{2}Mv_{cm}^2 + \tfrac{1}{2}I_{cm}\omega^2} \)
Step 2: For a solid cylinder, \( \mathrm{I_{cm} = \tfrac{1}{2}MR^2} \) and \( \mathrm{\omega = \dfrac{v}{R}} \).
Step 3: Substitute and simplify:
\( \mathrm{K_{total} = \tfrac{1}{2}M v^2 + \tfrac{1}{2}\left(\tfrac{1}{2}MR^2\right)\left(\dfrac{v}{R}\right)^2} \)
\( \mathrm{K_{total} = \tfrac{1}{2}M v^2 + \tfrac{1}{4}M v^2 = \tfrac{3}{4}M v^2} \)
Step 4: Substitute given values:
\( \mathrm{K_{total} = \tfrac{3}{4}(2.0)(3.0)^2 = 13.5\,J} \)
Result: The total kinetic energy of the rolling cylinder is \( \mathrm{13.5\,J} \).
Rotational Kinetic Energy with the Center of Mass at Rest
A rigid system can possess rotational kinetic energy even when its center of mass (CM) is at rest. This occurs because individual points within the system move with linear speeds as the body rotates, and each moving mass element contributes to the system’s total kinetic energy.
Although the CM does not translate, the system still contains energy due to the rotational motion of its parts.
\( \mathrm{K_{rot} = \tfrac{1}{2}I\omega^2} \)
Here, each infinitesimal mass element \( \mathrm{dm} \) contributes \( \mathrm{dK = \tfrac{1}{2}(r\omega)^2\,dm} \) to the total energy, and integrating over the system gives the expression above.
Key Idea:
- Even if the center of mass is stationary (\( \mathrm{v_{cm} = 0} \)), rotation about the CM gives the system kinetic energy.
- Each point on the object has its own linear velocity (\( \mathrm{v = r\omega} \)), which contributes to total energy.
Rotational Kinetic Energy is a Scalar Quantity
Although torque and angular velocity are vector quantities, rotational kinetic energy is a scalar. It has magnitude only and no direction, because energy is a measure of work capacity rather than a vector property.
\( \mathrm{K_{rot} = \tfrac{1}{2}I\omega^2 \geq 0} \)
The value of rotational kinetic energy is always positive, regardless of the direction of rotation (clockwise or counterclockwise).
Example
A ceiling fan rotates with an angular speed of \( \mathrm{20\,rad/s} \). Its center of mass is fixed to the ceiling, so it does not translate. The fan’s moment of inertia is \( \mathrm{0.30\,kg·m^2} \). Find its rotational kinetic energy.
▶️ Answer / Explanation
Step 1: Use the equation for rotational kinetic energy:
\( \mathrm{K_{rot} = \tfrac{1}{2}I\omega^2} \)
Step 2: Substitute values:
\( \mathrm{K_{rot} = \tfrac{1}{2}(0.30)(20)^2 = 0.15(400) = 60\,J} \)
Result: The fan’s center of mass remains stationary, but the fan possesses \( \mathrm{60\,J} \) of rotational kinetic energy due to the motion of its blades.