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.

The formula for rotational inertia is I = mr^2, where m is mass and r is distance from the axis. For multiple objects, you add up their individual inertias. The parallel axis theorem helps calculate inertia about different axes.

Rotational inertia of rigid systems

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
• Systems with mass concentrated farther from the axis rotate more slowly when given the same torque
• The distribution of mass matters more than total mass alone
• A hollow cylinder spins differently than a solid one of the same mass

For an object 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 multiple objects, sum their individual inertias: $I_{tot} = \sum I = \sum mr^2$
• This explains why figure skaters pull in their arms to spin faster, reducing their rotational inertia

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.

• Spinning objects rotate most efficiently around their center of mass
• This principle explains why balancing a spinning object makes it more stable
• A baton twirler can spin the baton faster when grasping it at its center of mass
• Gyroscopes and tops are designed with this principle in mind

The parallel axis theorem allows us to calculate rotational inertia around any axis parallel to one through the center of mass.

• The equation is $I' = I_{cm} + Md^2$ where:
  • $I'$ is rotational inertia about the parallel axis (kg⋅m²)
  • $I_{cm}$ is rotational inertia about the center of mass axis (kg⋅m²)
  • $M$ is the total system mass (kg)
  • $d$ is the perpendicular distance between axes (m)
• This theorem saves time by letting us calculate rotational inertia for any parallel axis once we know it for the center of mass
• Understanding this relationship helps explain stability in rotating objects

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:

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

Practice Problem 3: Real-World Application

This problem involves the conservation of angular momentum:

1. Angular momentum is conserved: $L = I\omega = \text{constant}$

• Initial: $L = I_1\omega_1 = 4.8 \times 2 = 9.6 \text{ kg}\cdot\text{m}^2/\text{s}$

1. Solve for final angular velocity:

• $\omega_2 = \frac{L}{I_2} = \frac{9.6}{2.4} = 4 \text{ revolutions/second}$

1. The figure skater doubles her rotation rate by halving her rotational inertia.