§9.6 — Rigid Body Dynamics

Now we know that:

But real objects (wheels, balls, planets) both translate and rotate. This section combines both types of motion, introduces moment of inertia, and shows how rigid bodies move under forces and torques.

Moment of inertia

Moment of inertia measures resistance of an object to changes in rotational motion — the rotational analog of mass.

For a system of particles:

\[ I = \sum_i m_i r_i^2 \]

where \( r_i \) is the distance of particle \( i \) from the rotation axis.

For a continuous body:

\[ I = \int r^2 \, dm = \int r^2 \rho(\mathbf{r}) \, dV \]

Unit: kilogram-meter-squared (kg·m²).

Table of moments of inertia for common shapes

For symmetric bodies about the axis of symmetry:

Object Axis Moment of inertia
Thin rod (length \( L \)) Perpendicular through center \( \frac{1}{12}ML^2 \)
Thin rod Through end \( \frac{1}{3}ML^2 \)
Cylinder (radius \( R \)) Central axis \( \frac{1}{2}MR^2 \)
Spherical shell (radius \( R \)) Through center \( \frac{2}{3}MR^2 \)
Solid sphere (radius \( R \)) Through center \( \frac{2}{5}MR^2 \)
Disk (radius \( R \)) Perpendicular through center \( \frac{1}{2}MR^2 \)

Note: Moment of inertia depends on the rotation axis. The same object rotating about different axes has different values.

Combined motion: Translation + Rotation

Most real objects have both translational motion (center of mass moves) and rotational motion (spinning about center of mass).

Total kinetic energy:

\[ KE_{\text{total}} = KE_{\text{trans}} + KE_{\text{rot}} = \frac{1}{2}M v_{\text{cm}}^2 + \frac{1}{2}I_{\text{cm}} \omega^2 \]

where:

Example 1: Rolling wheel

Solid cylinder (wheel) with radius \( R = 0.5 \) m, mass \( M = 20 \) kg rolls without slipping.

Question: Total kinetic energy?

Solution:

Moment of inertia of cylinder about central axis:

\[ I_{\text{cm}} = \frac{1}{2}MR^2 = \frac{1}{2}(20)(0.5)^2 = 2.5 \text{ kg·m}^2 \]

Translational energy:

\[ KE_{\text{trans}} = \frac{1}{2}M v_{\text{cm}}^2 = \frac{1}{2}(20)(4)^2 = 160 \text{ J} \]

Rotational energy:

\[ KE_{\text{rot}} = \frac{1}{2}I_{\text{cm}} \omega^2 = \frac{1}{2}(2.5)(8)^2 = 80 \text{ J} \]

Total energy:

\[ KE_{\text{total}} = 160 + 80 = 240 \text{ J} \]

Note: Approximately 67 percent is translational and 33 percent is rotational!

Example 2: Sliding ball

Solid sphere with radius \( R = 0.1 \) m, mass \( M = 2 \) kg on a rough floor (partial slipping).

Question: Total kinetic energy?

Solution:

Moment of inertia of solid sphere:

\[ I_{\text{cm}} = \frac{2}{5}MR^2 = \frac{2}{5}(2)(0.1)^2 = 0.008 \text{ kg·m}^2 \]

Translational energy:

\[ KE_{\text{trans}} = \frac{1}{2}(2)(3)^2 = 9 \text{ J} \]

Rotational energy:

\[ KE_{\text{rot}} = \frac{1}{2}(0.008)(10)^2 = 0.4 \text{ J} \]

Total energy:

\[ KE_{\text{total}} = 9 + 0.4 = 9.4 \text{ J} \]

Here mostly translational (96 percent).

Equations of motion for rigid bodies

For a rigid body under external forces and torques:

Translational motion:

\[ \mathbf{F}_{\text{ext}} = M\mathbf{a}_{\text{cm}} \]

Rotational motion (about fixed axis):

\[ \boldsymbol{\tau}_{\text{ext}} = I_{\text{cm}} \boldsymbol{\alpha} \]

where \( \boldsymbol{\alpha} = \frac{d\boldsymbol{\omega}}{dt} \) is angular acceleration.

These two equations are independent when you rotate about the center of mass.

Example 3: Cylinder on inclined plane

Solid cylinder on incline at angle \( \theta = 30° \) rolls without slipping.

Question: What is the acceleration of the center of mass?

Solution:

Forces:

Translational motion (down the incline):

\[ Mg\sin\theta - f = M a_{\text{cm}} \]

Rotational motion:

\[ \tau = fR = I_{\text{cm}} \alpha \]

For no slipping:

\[ a_{\text{cm}} = R\alpha \]

(kinematic constraint)

Substitute \( \alpha = a_{\text{cm}}/R \) into torque equation:

\[ fR = \frac{1}{2}MR^2 \cdot \frac{a_{\text{cm}}}{R} \]

\[ f = \frac{1}{2}M a_{\text{cm}} \]

Substitute into translational equation:

\[ Mg\sin\theta - \frac{1}{2}M a_{\text{cm}} = M a_{\text{cm}} \]

\[ Mg\sin\theta = \frac{3}{2}M a_{\text{cm}} \]

\[ a_{\text{cm}} = \frac{2}{3}g\sin\theta = \frac{2}{3}(9.8)\sin(30°) = \frac{2}{3}(9.8)(0.5) = 3.27 \text{ m/s}^2 \]

Note: This is less than a sliding object (\( a = g\sin\theta = 4.9 \) m/s²) because some energy goes into rotation!

Example 4: Two different cylinders

Two identical cylinders of equal mass and radius:

Both roll without slipping down the same incline.

Question: Which reaches the bottom faster?

Solution:

From formula \( a_{\text{cm}} = \frac{g\sin\theta}{1 + I/(MR^2)} \):

Solid cylinder: \[ a_1 = \frac{g\sin\theta}{1 + 1/2} = \frac{2}{3}g\sin\theta \]

Hollow cylinder: \[ a_2 = \frac{g\sin\theta}{1 + 1} = \frac{1}{2}g\sin\theta \]

Solid cylinder is faster! (\( a_1 > a_2 \))

Why? The hollow cylinder is "harder to spin" and more energy goes into rotation.

Key principles

  1. Moment of inertia depends on mass distribution; objects with mass far from the axis are harder to spin.

  2. Combined motion — translation and rotation about the center of mass — are independent.

  3. Rolling without slipping has simple kinematics: \( v_{\text{cm}} = R\omega \) and \( a_{\text{cm}} = R\alpha \).

  4. Energy is divided between translational and rotational — if friction acts!

What you should be able to do

Chapter 9 summary

Chapter 9 built the foundations of mechanics for systems of particles and rigid bodies:

These concepts are the foundation of classical mechanics and lead to advanced topics (fluid dynamics, astronomy, engineering).

📚 See also: Halliday Vol 1, Ch 9, §9.6.

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