5.4 Rotational Inertia

Rotational inertia measures how hard it is to change an object's spinning motion. It depends on the object's mass and how that mass is spread out from the spin axis. Objects with more mass farther from the axis have higher rotational inertia.

For a point mass, or for one object in a system treated as concentrated at a distance $r$ from the axis, the rotational inertia is $I = mr^2$. For a system of several discrete objects, the total rotational inertia is $I_{tot} = \sum m_i r_i^2$. In AP Physics 1, students are expected to calculate rotational inertia mainly for systems of five or fewer objects arranged in two dimensions; for extended rigid objects, the needed rotational inertia values will be provided on the exam.

The parallel axis theorem helps calculate inertia about different axes.

Rotational inertia of rigid systems

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Resistance to rotational changes

Rotational inertia quantifies the opposition a rigid system exhibits when subjected to changes in its rotational motion. This concept is fundamental to understanding how objects resist rotational acceleration.

• Mass distribution plays a critical role in determining rotational inertia
• For the same applied net torque, systems with more mass concentrated farther from the axis have greater rotational inertia and therefore undergo smaller angular acceleration.
• The distribution of mass matters more than total mass alone
• A hollow cylinder spins differently than a solid one of the same mass

For a point mass (or an object treated as concentrated at a single point) rotating at a perpendicular distance r from an axis, rotational inertia follows a simple equation.

• The formula is $I = mr^2$ where:
  • $I$ represents rotational inertia (kg⋅m²)
  • $m$ is mass (kg)
  • $r$ is perpendicular distance to the rotation axis (m)
• For a system of multiple discrete objects, sum their individual inertias: $I_{tot} = \sum I_i = \sum m_i r_i^2$
• A figure skater pulling in her arms decreases her rotational inertia because more of her mass moves closer to the axis.

Rotational inertia off-center

Minimum rotational inertia

The rotational inertia of a rigid system reaches its minimum value when rotation occurs around an axis through the center of mass. Any other axis parallel to this results in larger rotational inertia.

• A rigid system has the smallest rotational inertia when the axis passes through its center of mass.
• Moving the axis farther from the center of mass increases rotational inertia if the new axis is parallel.
• This is described quantitatively by the parallel axis theorem: $I' = I_{cm} + Md^2$.

Parallel axis theorem

• The parallel axis theorem relates the rotational inertia of a rigid system about an axis passing through its center of mass ($I_{cm}$) to the rotational inertia about any axis parallel to it ($I'$)
• The equation for the parallel axis theorem is $I' = I_{cm} + Md^2$
  • $I'$ is the rotational inertia about the parallel axis (kg⋅m²)
  • $I_{cm}$ is the rotational inertia about the axis through the center of mass (kg⋅m²)
  • $M$ is the total mass of the system (kg)
  • $d$ is the perpendicular distance between the parallel axes (m)
• Enables calculation of a system's rotational inertia about any parallel axis if the rotational inertia about the center of mass axis is known
• Example: A 4 kg rod has a rotational inertia of 0.8 kg⋅m² about an axis through its center. To find the rotational inertia about a new axis 0.2 m from the center axis, use $I' = 0.8\text{ kg}\cdot\text{m}^2 + (4\text{ kg})(0.2\text{ m})^2 = 0.96\text{ kg}\cdot\text{m}^2$

Practice Problem 1: Rotational Inertia Calculation

To solve this problem, we use the basic rotational inertia formula and the parallel axis theorem:

1. For the initial situation, calculate using $I_{tot} = \sum mr^2$:

• $I_{tot} = (3 \text{ kg})(0.4 \text{ m})^2 + (2 \text{ kg})(0.7 \text{ m})^2$
• $I_{tot} = 3 \times 0.16 + 2 \times 0.49 = 0.48 + 0.98 = 1.46 \text{ kg}\cdot\text{m}^2$

1. For the second scenario with the axis through the 3 kg mass:

• The 3 kg mass now has $r = 0$, so its contribution is zero
• The 2 kg mass is now 1.1 m from the axis (0.7 + 0.4)
• $I_{tot} = 0 + (2 \text{ kg})(1.1 \text{ m})^2 = 2 \times 1.21 = 2.42 \text{ kg}\cdot\text{m}^2$

Practice Problem 2: Parallel Axis Theorem Application

To solve this problem:

1. Use the parallel axis theorem: $I' = I_{cm} + Md^2$

• $I_{cm} = 0.225 \text{ kg}\cdot\text{m}^2$
• $M = 5 \text{ kg}$
• $d = 0.3 \text{ m}$ (distance from center to edge)

1. Calculate the new rotational inertia:

• $I' = 0.225 + 5 \times (0.3)^2$
• $I' = 0.225 + 5 \times 0.09$
• $I' = 0.225 + 0.45 = 0.675 \text{ kg}\cdot\text{m}^2$