AP Physics 1- 6.1 Rotational Kinetic Energy- Study Notes- New Syllabus
AP Physics 1-6.1 Rotational Kinetic Energy – Study Notes
AP Physics 1-6.1 Rotational Kinetic Energy – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Rotational Kinetic Energy
Rotational Kinetic Energy
Rotational kinetic energy is the energy of an object due to its rotational motion about an axis.
- It is analogous to translational kinetic energy but depends on the object’s moment of inertia and angular velocity.
- It is a scalar quantity.
Relevant Equation
\( KE_\text{rot} = \dfrac{1}{2} I \omega^2 \)
Where:
- \( I \) = moment of inertia of the rotating object (depends on mass distribution and axis of rotation)
- \( \omega \) = angular velocity (rad/s)
- \( I \) = moment of inertia of the rotating object (depends on mass distribution and axis of rotation)
Key Points
- Moment of inertia plays the same role in rotational motion as mass does in linear motion.
- An object rolling without slipping has both translational kinetic energy and rotational kinetic energy.
- Total kinetic energy of a rolling object: \( KE_\text{total} = \dfrac{1}{2} m v^2 + \dfrac{1}{2} I \omega^2 \)
- Rotational KE depends not only on the object’s mass but also on how that mass is distributed relative to the axis of rotation.
Example:
A solid disk of mass 2 kg and radius 0.5 m is rotating at 10 rad/s about its central axis. Find its rotational kinetic energy.
▶️Answer/Explanation
Step 1: Moment of inertia of a solid disk about its central axis: \( I = \dfrac{1}{2} m r^2 = \dfrac{1}{2} \cdot 2 \cdot (0.5)^2 = 0.25 \, \text{kg·m}^2 \)
Step 2: Apply rotational KE formula: \( KE_\text{rot} = \dfrac{1}{2} I \omega^2 = \dfrac{1}{2} \cdot 0.25 \cdot (10^2) \)
Step 3: Calculate → \( KE_\text{rot} = 12.5 \, \text{J} \)
Final Answer: The rotational kinetic energy of the disk is 12.5 J.
Example:
A solid sphere of mass 3 kg and radius 0.2 m is rolling without slipping at a linear speed of 4 m/s. Find its rotational kinetic energy.
▶️Answer/Explanation
Step 1: For a solid sphere, moment of inertia about central axis: \( I = \dfrac{2}{5} m r^2 = \dfrac{2}{5} \cdot 3 \cdot (0.2)^2 = 0.048 \, \text{kg·m}^2 \)
Step 2: Since rolling without slipping, \( v = r\omega \) → \( \omega = \dfrac{v}{r} = \dfrac{4}{0.2} = 20 \, \text{rad/s} \)
Step 3: Rotational KE: \( KE_\text{rot} = \dfrac{1}{2} I \omega^2 = \dfrac{1}{2} \cdot 0.048 \cdot (20^2) \)
= \( 9.6 \, \text{J} \)
Final Answer: The rotational kinetic energy of the rolling sphere is 9.6 J.
Example:
A uniform rod of length 1.5 m and mass 4 kg rotates about an axis at one end perpendicular to its length with angular velocity 6 rad/s. Find its rotational kinetic energy.
▶️Answer/Explanation
Step 1: Moment of inertia of a rod about one end: \( I = \dfrac{1}{3} m L^2 = \dfrac{1}{3} \cdot 4 \cdot (1.5^2) \)
= \( \dfrac{1}{3} \cdot 4 \cdot 2.25 = 3 \, \text{kg·m}^2 \)
Step 2: Rotational KE: \( KE_\text{rot} = \dfrac{1}{2} I \omega^2 = \dfrac{1}{2} \cdot 3 \cdot (6^2) \)
= \( 54 \, \text{J} \)
Final Answer: The rotational kinetic energy of the rod is 54 J.