The Timpson Lyceum · University Physics I (2025)

Rotation of Rigid Bodies and the Parallel-Axis Theorem – Lecture 9 (University Physics I)

Overview

This lecture moves from treating objects as point particles to treating them as rigid bodies — extended objects with mass spread over a volume that cannot bend, stretch, or deform. The goal is to describe how such bodies rotate and how much energy lives in that rotation. The first half builds the kinematics of rotation (angular position, angular velocity, angular acceleration) and shows that each rotational quantity maps onto a familiar linear one through the radius $r$ . The second half introduces the moment of inertia — the rotational analog of mass — and uses it to compute rotational kinetic energy and to derive the moments of inertia of several standard shapes by integration. By the end you should be able to estimate, for example, the energy stored in the rotation of the Earth.

Key Concepts

Rigid body

An idealized extended object whose mass is distributed over a region and whose shape never changes (no bending, warping, or deformation). Real objects only approximate this, but it is far closer to everyday objects than a single point.

Rotational kinematics (angular quantities)

Angular position $\theta$ : the angle of a point on a circle, measured counterclockwise from the $+x$ -axis, in radians. All angles are measured from a common starting reference.
Arc length / linear position relates to angle through the radius: $x = r\theta$ , so a displacement $\Delta x = r\,\Delta\theta$ .
Angular velocity (speed) $\omega$ $ω$ : the rate of change of angle.
- Average: $\omega_{\text{avg}} = \dfrac{\Delta\theta}{\Delta t}$ ; instantaneous: $\omega = \dfrac{d\theta}{dt}$ . Units: rad/s (radians are dimensionless).
- Sign carries direction (counterclockwise positive); a full vector treatment uses the right-hand rule, but a sign suffices for these problems.
Angular acceleration $\alpha$ : the rate of change of angular velocity. Average $\alpha_{\text{avg}} = \dfrac{\Delta\omega}{\Delta t}$ ; instantaneous $\alpha = \dfrac{d\omega}{dt}$ . Units: rad/s².

Linking angular and linear motion (valid in radians)

$v = r\omega$ — linear (tangential) speed grows with distance from the axis. At the center ( $r=0$ ) the speed is zero; farther out means faster.
$a_{\text{tan}} = r\alpha$ — tangential acceleration, how fast the speed changes (speeding up / slowing down).
$a_{\text{rad}} = \dfrac{v^2}{r} = \omega^2 r$ — radial (centripetal) acceleration, always pointing toward the center; this is what changes the direction of motion.
The total acceleration is the vector sum of these two perpendicular pieces.

Constant- $\alpha$ rotational kinematic equations (parallels of the linear ones; only valid when $\alpha$ is constant)

\begin{aligned} \omega &= \omega_0 + \alpha t \\ \theta &= \theta_0 + \omega_0 t + \tfrac{1}{2}\alpha t^2 \\ \theta - \theta_0 &= \tfrac{1}{2}(\omega_0 + \omega)\,t \\ \omega^2 &= \omega_0^2 + 2\alpha(\theta - \theta_0) \end{aligned}

If $\theta(t)$ is not quadratic in $t$ (so $\alpha$ is not constant), these do not apply — differentiate directly: $\omega = d\theta/dt$ , $\alpha = d\omega/dt$ . The "midpoint = average of endpoints" shortcut only works for constant $\alpha$ .

Moment of inertia $I$

The rotational analog of mass: an object's resistance to angular acceleration. Unlike mass, it is not a property of the object alone — it depends on the chosen axis of rotation and on how the mass is distributed relative to that axis.
Discrete masses: $I = \sum_i m_i r_i^2$ , where $r_i$ is each mass's perpendicular distance from the axis.
Continuous body: $I = \displaystyle\int r^2\,dm$ .
Because distance enters squared, doubling a mass's distance from the axis quadruples its contribution. Mass on the axis ( $r=0$ ) contributes nothing. Units: kg·m².

Rotational kinetic energy

Summing $\tfrac12 m_i v_i^2$ over all pieces, substituting $v_i = \omega r_i$ and factoring out $\tfrac12\omega^2$ , gives $K_{\text{rot}} = \tfrac{1}{2}I\omega^2$
This is the exact rotational counterpart of $K = \tfrac12 mv^2$ , with $I$ playing the role of $m$ and $\omega$ the role of $v$ .

Moments of inertia by integration (uniform density)

For a uniform body, set up $dm$ from the mass-per-unit-length/area/volume and choose a slicing that keeps $r$ constant over each piece.
Thin rod, axis through center: $I = \tfrac{1}{12}ML^2$ .
Thin rod, axis through one end: $I = \tfrac{1}{3}ML^2$ — four times larger, because the mass sits farther from the axis (the $r^2$ dependence).
Hollow (thick-walled) cylinder, axis along its center, using cylindrical shells: $I = \tfrac{1}{2}M\!\left(R_1^2 + R_2^2\right)$ $I = \frac{1}{2} M (R_{1}^{2} + R_{2}^{2})$ .
- Solid cylinder/disk (set $R_1 = 0$ ): $I = \tfrac{1}{2}MR^2$ .
- Thin cylindrical shell ( $R_1 \approx R_2 = R$ ): $I = MR^2$ (all mass at one radius).
Solid sphere, axis through a diameter, using disk slices: $I = \tfrac{2}{5}MR^2$ .
Thin spherical shell: $I = \tfrac{2}{3}MR^2$ .

Worked Example

A discus thrower's hand. An athlete's arm has length (radius) $r = 0.800\ \text{m}$ . At the instant of interest the arm rotates at $\omega = 10.0\ \text{rad/s}$ and is speeding up at $\alpha = 50.0\ \text{rad/s}^2$ . Find the radial acceleration, the tangential acceleration, and the magnitude of the total acceleration of the discus.

Radial (centripetal) acceleration:

\begin{aligned} a_{\text{rad}} &= \omega^2 r \\ &= (10.0)^2 (0.800) = 80.0\ \text{m/s}^2 \end{aligned}

Tangential acceleration:

\begin{aligned} a_{\text{tan}} &= \alpha r \\ &= (50.0)(0.800) = 40.0\ \text{m/s}^2 \end{aligned}

Magnitude of the resultant (the two components are perpendicular):

\begin{aligned} a &= \sqrt{a_{\text{rad}}^2 + a_{\text{tan}}^2} \\ &= \sqrt{(80.0)^2 + (40.0)^2} \approx 89.4\ \text{m/s}^2 \end{aligned}

Moment of inertia and energy of a three-mass system. Masses $m_1 = 0.100$ kg, $m_2 = 0.200$ kg, $m_3 = 0.300$ kg sit in a line; the spacings place $m_1$ at $0.500$ m and $m_2$ at $0.400$ m from an axis that passes through $m_3$ perpendicular to the line. For rotation at $\omega = 4.00\ \text{rad/s}$ , find $I$ and $K_{\text{rot}}$ .

\begin{aligned} I &= \sum_i m_i r_i^2 \\ &= (0.300)(0)^2 + (0.100)(0.500)^2 + (0.200)(0.400)^2 = 0.057\ \text{kg}\cdot\text{m}^2 \end{aligned}

\begin{aligned} K_{\text{rot}} &= \tfrac{1}{2}I\omega^2 \\ &= \tfrac{1}{2}(0.057)(4.00)^2 \approx 0.46\ \text{J} \end{aligned}

Summary

Rotational motion is described by angular position $\theta$ , angular velocity $\omega = d\theta/dt$ , and angular acceleration $\alpha = d\omega/dt$ , all in radians.
Each angular quantity maps to a linear one through the radius: $x = r\theta$ , $v = r\omega$ , $a_{\text{tan}} = r\alpha$ , with $a_{\text{rad}} = \omega^2 r = v^2/r$ pointing toward the center.
The constant- $\alpha$ kinematic equations mirror the linear ones exactly, but only apply when $\alpha$ is constant; otherwise differentiate directly.
The moment of inertia $I = \sum m_i r_i^2 = \int r^2\,dm$ is the rotational analog of mass and depends on the axis; distance from the axis matters as $r^2$ .
Rotational kinetic energy is $K_{\text{rot}} = \tfrac12 I\omega^2$ , the direct counterpart of $\tfrac12 mv^2$ .
Standard uniform-body results follow from integration: thin rod $\tfrac{1}{12}ML^2$ (center) or $\tfrac{1}{3}ML^2$ (end), solid cylinder/disk $\tfrac{1}{2}MR^2$ , hollow cylinder $\tfrac12 M(R_1^2+R_2^2)$ , solid sphere $\tfrac{2}{5}MR^2$ , and thin spherical shell $\tfrac{2}{3}MR^2$ .