Section 5.4: Moment of Inertia
The moment of inertia (\(I\)) is the rotational equivalent of mass in linear motion. It measures how difficult it is to change the rotational motion of an object about a specific axis.
\( m_i \) = mass of a particle
\( r_i \) = perpendicular distance of the particle from the axis of rotation
For continuous bodies, the summation is replaced by an integral:
The moment of inertia depends on both the mass distribution and the axis of rotation. Objects with more mass farther from the axis have larger \( I \).
Example: Solid Cylinder
Find the moment of inertia of a solid cylinder of mass 5 kg and radius 0.3 m about its central axis.
Formula for a solid cylinder about its central axis: \( I = \tfrac{1}{2} M R^2 \). \( I = \tfrac{1}{2} (5)(0.3^2) = \tfrac{1}{2} (5)(0.09) = 0.225 \, \text{kg·m}^2 \). The moment of inertia is 0.225 kg·m².
Example: Uniform Rod (Axis Through Center)
A uniform rod of length 2 m and mass 6 kg rotates about its midpoint, perpendicular to its length. Find the moment of inertia.
Formula: \( I = \tfrac{1}{12} M L^2 \). \( I = \tfrac{1}{12}(6)(2^2) = \tfrac{1}{12}(6)(4) = 2 \, \text{kg·m}^2 \). The moment of inertia is 2 kg·m².
Practice Problems
- Find the moment of inertia of a thin ring of mass 2 kg and radius 0.5 m about its central axis.
- A solid sphere of mass 4 kg and radius 0.2 m rotates about its center. Find its \( I \).
- A 3 m long rod of mass 10 kg is pivoted at one end. Find its moment of inertia.
- A hollow cylinder has mass 8 kg and radius 0.4 m. Find its moment of inertia about its central axis.
- Explain why the same object can have different moments of inertia depending on the axis chosen.
Common Formulas for Standard Bodies
- Thin ring (axis through center): \( I = MR^2 \)
- Solid disc/cylinder (axis through center): \( I = \tfrac{1}{2}MR^2 \)
- Hollow cylinder (axis through center): \( I = MR^2 \)
- Solid sphere: \( I = \tfrac{2}{5}MR^2 \)
- Rod (axis through center): \( I = \tfrac{1}{12}ML^2 \)
- Rod (axis through end): \( I = \tfrac{1}{3}ML^2 \)