Moment of Inertia

Newton's second law for linear motion states $F = ma$. For rotation about a fixed axis, the analogous law is $\tau = I\alpha$, where $\tau$ (tau) is the net torque, $I$ is the moment of inertia, and $\alpha$ (alpha) is the angular acceleration. Comparing the two equations shows that $I$ plays the role of mass: a large $I$ means a large torque is needed to produce a given angular acceleration.

For a single point mass $m$ located at perpendicular distance $r$ from the rotation axis, the moment of inertia is simply $I = mr^2$. The units are $\text{kg}\cdot\text{m}^2$. Notice the $r^2$ dependence: doubling the distance from the axis quadruples the contribution to $I$. This is why mass far from the axis matters far more than mass close to it.

For a collection of point masses, the total moment of inertia is the sum of individual contributions: $I = \sum_i m_i r_i^2$. Each mass $m_i$ is at perpendicular distance $r_i$ from the axis. This additive property means we can build up the moment of inertia of any object by imagining it sliced into small pieces and summing their contributions.

For a solid object with continuously distributed mass, the sum becomes an integral: $I = \int r^2 \, dm$, where $r$ is the perpendicular distance from each infinitesimal mass element $dm$ to the rotation axis. Evaluating this integral for common shapes yields the standard formulas used in physics: for example, a solid sphere of mass $M$ and radius $R$ rotating about a diameter has $I = \frac{2}{5}MR^2$, while a thin rod of mass $M$ and length $L$ rotating about its center has $I = \frac{1}{12}ML^2$.

If you know the moment of inertia $I_{\text{cm}}$ of an object about an axis through its center of mass, you can find the moment of inertia about any parallel axis using the parallel axis theorem: $I = I_{\text{cm}} + Md^2$, where $M$ is the total mass and $d$ is the distance between the two parallel axes. This powerful shortcut avoids recomputing integrals from scratch. For example, the same thin rod rotating about one end has $I = \frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{3}ML^2$.

A rotating body possesses kinetic energy given by $KE_{\text{rot}} = \frac{1}{2}I\omega^2$, where $\omega$ is the angular velocity in radians per second. Compare this with translational kinetic energy $KE_{\text{trans}} = \frac{1}{2}mv^2$: the parallel between $I$ and $m$, and between $\omega$ and $v$, is exact. A rolling object (like a ball rolling down a ramp) has both forms simultaneously: $KE_{\text{total}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.

$I_{\text{cm}} = 0.24\,\text{kg}\cdot\text{m}^2$, $\quad I_{\text{end}} = 0.96\,\text{kg}\cdot\text{m}^2$

• Torque and Rotational Force — Torque is the rotational analogue of force. The equation $\tau = I\alpha$ directly connects torque to moment of inertia, making torque the essential companion concept.
• Angular Momentum — Angular momentum $L = I\omega$ depends directly on moment of inertia. Conservation of angular momentum explains phenomena like the spinning skater, where changing $I$ changes $\omega$.
• Rotational Kinetic Energy — Rotational kinetic energy $KE = \frac{1}{2}I\omega^2$ requires knowledge of moment of inertia and is essential for energy conservation problems involving rolling or spinning objects.
• Mass and Inertia — Moment of inertia is the direct rotational analogue of mass. Understanding why mass resists linear acceleration provides the intuition for why $I$ resists angular acceleration.
• Angular Kinematics — Angular acceleration $\alpha$ is the quantity that moment of inertia resists. Familiarity with angular displacement, velocity, and acceleration is prerequisite to applying $\tau = I\alpha$.