AP Physics C Mechanics- 5.4 Rotational Inertia- Study Notes- New Syllabus
AP Physics C Mechanics- 5.4 Rotational Inertia – Study Notes
AP Physics C Mechanics- 5.4 Rotational Inertia – Study Notes – per latest Syllabus.
Key Concepts:
- Rotational Inertia (Moment of Inertia)
- Minimum Rotational Inertia and the Center of Mass
- Parallel Axis Theorem
Definition of Rotational Inertia
Rotational inertia (also called the moment of inertia) measures a rigid system’s resistance to changes in its rotational motion. It depends both on the mass of the system and on how that mass is distributed relative to the axis of rotation.
- A system with more mass farther from the axis has greater rotational inertia.
- It plays the same role in rotational dynamics that mass plays in linear dynamics.
Key Idea: The greater the rotational inertia, the harder it is to change the object’s rotational speed.
Example
Two solid disks have the same total mass. Disk A has more of its mass concentrated near the center, while Disk B has mass distributed farther from the axis. Which has the greater rotational inertia?
▶️ Answer / Explanation
Step 1: Rotational inertia increases with mass distribution farther from the axis.
Step 2: Disk B’s mass is farther from the axis, so it resists angular acceleration more.
Result: Disk B has the greater rotational inertia.
Rotational Inertia of a Point Mass
The rotational inertia of a single point mass \( \mathrm{m} \) located a perpendicular distance \( \mathrm{r} \) from the axis of rotation is given by:
\( \mathrm{I = mr^2} \)
- \( \mathrm{I} \): rotational inertia (in \( \mathrm{kg·m^2} \))
- \( \mathrm{m} \): mass of the object (in \( \mathrm{kg} \))
- \( \mathrm{r} \): perpendicular distance from the rotation axis (in \( \mathrm{m} \))
Key Idea: The farther the mass is from the axis, the larger its rotational inertia. Even a small mass can have a large \( \mathrm{I} \) if \( \mathrm{r} \) is large.
Example
A \( \mathrm{2.0\,kg} \) mass is attached to the end of a \( \mathrm{0.5\,m} \) rod and rotated about one end. Find the rotational inertia of the mass about the axis.
▶️ Answer / Explanation
Step 1: Use the formula \( \mathrm{I = mr^2} \).
\( \mathrm{I = (2.0)(0.5)^2 = (2.0)(0.25) = 0.50\,kg·m^2} \)
Result: The rotational inertia of the mass is \( \mathrm{0.50\,kg·m^2} \).
Rotational Inertia of a System of Point Masses
The total rotational inertia of a system of multiple particles (or discrete masses) about a single axis is the sum of the rotational inertias of each individual mass about that axis:
\( \mathrm{I_{tot} = \sum_i m_i r_i^2} \)
- \( \mathrm{m_i} \): mass of the \( \mathrm{i^{th}} \) particle
- \( \mathrm{r_i} \): perpendicular distance of each mass from the rotation axis
- \( \mathrm{I_{tot}} \): total rotational inertia about the chosen axis
Key Idea: Each mass contributes to the total rotational inertia in proportion to both its mass and the square of its distance from the axis.
Example
Three point masses are located at distances \( \mathrm{r_1 = 0.10\,m} \), \( \mathrm{r_2 = 0.20\,m} \), and \( \mathrm{r_3 = 0.30\,m} \) from a rotation axis, with respective masses \( \mathrm{m_1 = 1.0\,kg} \), \( \mathrm{m_2 = 2.0\,kg} \), and \( \mathrm{m_3 = 3.0\,kg} \). Find the total rotational inertia.
▶️ Answer / Explanation
Step 1: Apply \( \mathrm{I_{tot} = \sum m_i r_i^2} \).
\( \mathrm{I_{tot} = (1.0)(0.10)^2 + (2.0)(0.20)^2 + (3.0)(0.30)^2} \)
\( \mathrm{I_{tot} = 0.01 + 0.08 + 0.27 = 0.36\,kg·m^2} \)
Result: The total rotational inertia is \( \mathrm{0.36\,kg·m^2} \).
Rotational Inertia of Continuous Bodies
For solid objects (continuous mass distributions), rotational inertia can be calculated by summing the contributions of all infinitesimal mass elements \( \mathrm{dm} \). The total rotational inertia is given by the integral:
\( \mathrm{I = \displaystyle \int r^2\,dm} \)
- \( \mathrm{r} \): perpendicular distance from the axis of rotation to the differential mass element
- \( \mathrm{dm} \): infinitesimal element of mass
- The integral adds up all such contributions across the entire solid.
Key Idea: For irregular or continuous shapes (e.g., rods, disks, spheres), the rotational inertia must be found by integration using mass density and geometry.
Example
Find the moment of inertia of a uniform thin rod of mass \( \mathrm{M} \) and length \( \mathrm{L} \) about an axis through one end, perpendicular to its length.
▶️ Answer / Explanation
Step 1: Write the general formula \( \mathrm{I = \int r^2\,dm} \).
Since \( \mathrm{dm = \dfrac{M}{L}\,dr} \), the linear mass density \( \mathrm{\lambda = M/L} \).
Step 2: Substitute and integrate:
\( \mathrm{I = \int_0^L r^2 \dfrac{M}{L} dr = \dfrac{M}{L}\int_0^L r^2 dr} \)
\( \mathrm{I = \dfrac{M}{L} \left[\dfrac{r^3}{3}\right]_0^L = \dfrac{ML^2}{3}} \)
Result: The rotational inertia of a uniform rod about one end is \( \mathrm{I = \dfrac{1}{3}ML^2} \).
Minimum Rotational Inertia and the Center of Mass
A rigid system’s rotational inertia about a given plane is at its minimum when the axis of rotation passes through the system’s center of mass (CM).
- The center of mass is the point where the total mass of the system can be considered to act for rotational motion.
- If the axis of rotation is shifted away from the center of mass, the rotational inertia increases because the mass elements are, on average, farther from the new axis.
- This concept forms the foundation of the parallel axis theorem.
Key Idea: Rotation through the center of mass always minimizes \( \mathrm{I} \). Any parallel axis farther away adds an additional term due to translational separation.
Example
Two identical rods rotate about different axes—Rod A about its center and Rod B about one end. Which has the smaller rotational inertia and why?
▶️ Answer / Explanation
Step 1: The rotational inertia about the center is \( \mathrm{I_{cm} = \dfrac{1}{12}ML^2} \).
Step 2: The rotational inertia about one end (using the parallel axis theorem) is \( \mathrm{I_{end} = I_{cm} + M\left(\dfrac{L}{2}\right)^2 = \dfrac{1}{12}ML^2 + \dfrac{1}{4}ML^2 = \dfrac{1}{3}ML^2} \).
Result: Rod A (rotating about its center) has the smaller rotational inertia. The rotational inertia is always minimum about the center of mass.
The Parallel Axis Theorem
The Parallel Axis Theorem relates the rotational inertia of a rigid body about any axis to the rotational inertia about a parallel axis through its center of mass.
\( \mathrm{I’ = I_{cm} + Md^2} \)
- \( \mathrm{I’} \): rotational inertia about the new axis
- \( \mathrm{I_{cm}} \): rotational inertia about an axis through the center of mass
- \( \mathrm{M} \): total mass of the object
- \( \mathrm{d} \): perpendicular distance between the two axes
Key Idea: Moving the axis of rotation a distance \( \mathrm{d} \) from the center of mass adds an extra term \( \mathrm{Md^2} \) to account for the mass being farther from the new axis.
Example
A uniform thin rod of mass \( \mathrm{M = 2.0\,kg} \) and length \( \mathrm{L = 1.2\,m} \) has a rotational inertia about its center given by \( \mathrm{I_{cm} = \tfrac{1}{12}ML^2} \). Find its rotational inertia about one end using the parallel axis theorem.
▶️ Answer / Explanation
Step 1: Identify known quantities:
\( \mathrm{M = 2.0\,kg}, \; \mathrm{L = 1.2\,m}, \; \mathrm{I_{cm} = \tfrac{1}{12}ML^2}, \; \mathrm{d = \tfrac{L}{2} = 0.6\,m}} \)
Step 2: Apply the parallel axis theorem:
\( \mathrm{I’ = I_{cm} + Md^2 = \tfrac{1}{12}(2.0)(1.2)^2 + (2.0)(0.6)^2} \)
Step 3: Simplify:
\( \mathrm{I’ = (0.24) + (0.72) = 0.96\,kg·m^2} \)
Result: The rotational inertia of the rod about one end is \( \mathrm{0.96\,kg·m^2} \).