AP Physics 1- 5.4 Rotational Inertia- Study Notes- New Syllabus
AP Physics 1-5.4 Rotational Inertia – Study Notes
AP Physics 1-5.4 Rotational Inertia – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Rotational Inertia
- Rotational Inertia about an Axis Not Through the Center of Mass
Rotational Inertia
Rotational inertia (also called moment of inertia) is the rotational analog of mass in linear motion.
It quantifies how difficult it is to change the rotational state of an object about a given axis.
The formula is:
\( I = \sum m_i r_i^2 \)
where \( m_i \) is the mass of each particle and \( r_i \) is its perpendicular distance from the axis of rotation.
- The larger the mass and the farther it is distributed from the axis, the larger the rotational inertia.
- It depends on both mass distribution and axis of rotation.
Common Rotational Inertia Formulas
| Object | Axis of Rotation | Moment of Inertia \( I \) |
|---|---|---|
| Solid disk / cylinder | Through center | \( I = \dfrac{1}{2} M R^2 \) |
| Thin hoop / ring | Through center | \( I = M R^2 \) |
| Solid sphere | Through center | \( I = \dfrac{2}{5} M R^2 \) |
| Thin rod | Through center | \( I = \dfrac{1}{12} M L^2 \) |
| Thin rod | Through end | \( I = \dfrac{1}{3} M L^2 \) |
Example :
Find the moment of inertia of a thin rod of length \( 2 \, \text{m} \) and mass \( 4 \, \text{kg} \) about an axis through its center and perpendicular to its length.
▶️Answer/Explanation
\( I = \dfrac{1}{12} M L^2 = \dfrac{1}{12} (4)(2^2) = \dfrac{16}{12} = 1.33 \, \text{kg·m}^2 \)
Answer: \( I = 1.33 \, \text{kg·m}^2 \)
Example :
Calculate the moment of inertia of a solid sphere of mass \( 2 \, \text{kg} \) and radius \( 0.5 \, \text{m} \) about an axis through its center.
▶️Answer/Explanation
\( I = \dfrac{2}{5} M R^2 = \dfrac{2}{5} (2)(0.5^2) = \dfrac{2}{5}(0.5) = 0.2 \, \text{kg·m}^2 \)
Answer: \( I = 0.2 \, \text{kg·m}^2 \)
Example :
A hoop of radius \( 0.3 \, \text{m} \) and mass \( 1.5 \, \text{kg} \) rotates about its central axis. Find its moment of inertia.
▶️Answer/Explanation
\( I = M R^2 = (1.5)(0.3^2) = 1.5 \times 0.09 = 0.135 \, \text{kg·m}^2 \)
Answer: \( I = 0.135 \, \text{kg·m}^2 \)
Rotational Inertia about an Axis Not Through the Center of Mass
The rotational inertia of a rigid system depends not only on its mass and shape but also on the choice of the axis of rotation.
If the axis of rotation does not pass through the center of mass (COM), the Parallel Axis Theorem is used.
Parallel Axis Theorem:
\( I = I_{CM} + M d^2 \)
- \( I \) = moment of inertia about the new axis
- \( I_{CM} \) = moment of inertia about a parallel axis through the center of mass
- \( M \) = total mass of the body
- \( d \) = perpendicular distance between the two axes
This theorem shows that shifting the axis away from the center of mass always increases the moment of inertia.
Example :
A thin rod of length \( 2 \, \text{m} \) and mass \( 3 \, \text{kg} \) has a moment of inertia about its center: \( I_{CM} = \dfrac{1}{12} M L^2 \). Find the moment of inertia about an axis through one of its ends, perpendicular to its length.
▶️Answer/Explanation
\( I_{CM} = \dfrac{1}{12} M L^2 = \dfrac{1}{12} (3)(2^2) = 1 \, \text{kg·m}^2 \)
Distance from COM to end: \( d = \dfrac{L}{2} = 1 \, \text{m} \)
\( I = I_{CM} + M d^2 = 1 + 3(1^2) = 4 \, \text{kg·m}^2 \)
Answer: \( I = 4 \, \text{kg·m}^2 \)
Example :
A solid disk of radius \( 0.5 \, \text{m} \) and mass \( 2 \, \text{kg} \) has rotational inertia about its center: \( I_{CM} = \dfrac{1}{2} M R^2 \). Find its moment of inertia about a parallel axis tangent to its rim.
▶️Answer/Explanation
\( I_{CM} = \dfrac{1}{2} M R^2 = \dfrac{1}{2}(2)(0.5^2) = 0.25 \, \text{kg·m}^2 \)
Distance from COM to tangent axis: \( d = R = 0.5 \, \text{m} \)
\( I = I_{CM} + M d^2 = 0.25 + (2)(0.25) = 0.75 \, \text{kg·m}^2 \)
Answer: \( I = 0.75 \, \text{kg·m}^2 \)
Example :
A solid sphere of radius \( 0.2 \, \text{m} \) and mass \( 4 \, \text{kg} \) has \( I_{CM} = \dfrac{2}{5} M R^2 \). Find its moment of inertia about a tangent axis.
▶️Answer/Explanation
\( I_{CM} = \dfrac{2}{5} (4)(0.2^2) = \dfrac{2}{5}(0.16) = 0.128 \, \text{kg·m}^2 \)
Distance from COM to tangent: \( d = R = 0.2 \, \text{m} \)
\( I = I_{CM} + M d^2 = 0.128 + 4(0.04) = 0.288 \, \text{kg·m}^2 \)
Answer: \( I = 0.288 \, \text{kg·m}^2 \)