10.3 Dynamics of Rotational Motion: Rotational Inertia

Rotational motion describes how objects spin around an axis. Where linear motion uses force, mass, and acceleration, rotational motion has its own parallel set of quantities: torque, moment of inertia, and angular acceleration. Understanding how mass is distributed relative to the axis of rotation is the key idea in this section, because that distribution determines how hard it is to speed up or slow down a spinning object.

Rotational Motion and Inertia

Concept of Rotational Inertia

Rotational inertia (also called moment of inertia, $I$) measures how much an object resists changes in its rotational motion. It's the rotational equivalent of mass in linear motion, but with a twist: it depends not just on how much mass an object has, but on where that mass is located relative to the axis of rotation.

Mass farther from the axis contributes more to rotational inertia. Think of a figure skater spinning: with arms extended, mass is far from the axis and rotational inertia is large, so the spin is slow. Pull the arms in, and rotational inertia drops, making the spin faster.

The moment of inertia is calculated as:

$I = \sum m r^2$

For a continuous object, this becomes an integral: $I = \int r^2 \, dm$

• $m$ is the mass of each particle (or differential mass element)
• $r$ is the perpendicular distance from that particle to the axis of rotation

A few important consequences follow from this definition:

• Shape and axis both matter. A cylinder spinning about its central axis has a different $I$ than the same cylinder spinning about one end, because the mass distribution relative to the axis changes.
• Hollow vs. solid objects. A hollow cylinder has more rotational inertia than a solid cylinder of the same total mass and radius. The hollow cylinder's mass is concentrated farther from the axis.
• Parallel axis theorem. If you know the moment of inertia about an axis through the center of mass ($I_{cm}$), you can find it about any parallel axis a distance $d$ away: $I = I_{cm} + Md^2$, where $M$ is the total mass.

Calculation of Torque

Torque ($\tau$) is the rotational equivalent of force. Instead of pushing an object in a straight line, torque causes an object to rotate about an axis.

$\tau = rF\sin\theta$

• $r$ is the distance from the axis of rotation to the point where the force is applied (the lever arm)
• $F$ is the magnitude of the applied force
• $\theta$ is the angle between the force vector and the lever arm vector

You can also write this as $\tau = rF_{\perp}$, where $F_{\perp}$ is the component of force perpendicular to the lever arm. Only the perpendicular component actually produces rotation.

Why does distance from the axis matter? Try pushing a door open near the hinge versus near the handle. The same force produces much more torque (and rotates the door more easily) when applied far from the hinge. That's $r$ doing its job.

Torque has a sign convention: counterclockwise torques are typically positive, and clockwise torques are negative. When multiple torques act on an object, the net torque determines the angular acceleration through Newton's Second Law for rotation:

$\sum \tau = I\alpha$

This is the rotational version of $\sum F = ma$. A larger moment of inertia means you need more torque to achieve the same angular acceleration, just like a larger mass needs more force for the same linear acceleration.

Linear vs. Rotational Motion

Every linear motion quantity has a rotational counterpart. If you already understand linear kinematics and dynamics, you can map that knowledge directly onto rotation:

The constant-acceleration kinematic equations carry over with the same structure:

• Linear: $x = x_0 + v_0 t + \frac{1}{2}at^2$ → Rotational: $\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2$
• Linear: $v = v_0 + at$ → Rotational: $\omega = \omega_0 + \alpha t$
• Linear: $v^2 = v_0^2 + 2a(x - x_0)$ → Rotational: $\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$

Newton's Second Law:

• Linear: $\sum F = ma$
• Rotational: $\sum \tau = I\alpha$

Work-energy theorem:

• Linear: $W = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$
• Rotational: $W = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2$

The pattern is consistent: swap $x$ for $\theta$, $v$ for $\omega$, $a$ for $\alpha$, $m$ for $I$, and $F$ for $\tau$. If you can solve a linear problem, you can solve the rotational version using the same steps with the rotational variables.

Angular Momentum and Conservation

Angular momentum ($L$) is the rotational analog of linear momentum, defined as:

$L = I\omega$

Just as linear momentum is $p = mv$, angular momentum combines the "resistance to change" ($I$) with the "rate of rotation" ($\omega$).

Conservation of angular momentum states that if no net external torque acts on a system, the total angular momentum stays constant:

$L_i = L_f \quad \text{or} \quad I_i \omega_i = I_f \omega_f$

This is why the figure skater example works so well: no external torque is applied, so when $I$ decreases (arms pulled in), $\omega$ must increase to keep $L$ constant.

Rotational kinetic energy is the energy an object has due to its rotation:

$KE_{rot} = \frac{1}{2}I\omega^2$

For an object that's both translating and rotating (like a ball rolling down a ramp), the total kinetic energy is $KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.

One more useful concept: the radius of gyration ($k$) is the distance from the axis at which you could concentrate all of an object's mass and get the same moment of inertia. It's defined by $I = Mk^2$, where $M$ is the total mass. This gives you a quick sense of how "spread out" the mass is relative to the axis.