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Topic 5 of 11

Rigid Body Dynamics

A rigid body has six degrees of freedom — three translational and three rotational. The rotational dynamics are governed by Euler's equations for the body-frame angular velocity, and the moment of inertia tensor replaces the scalar mass. Gyroscopic effects and precession emerge naturally.

Key Concepts

Moment of Inertia Tensor

I_{ij}=\sum_k m_k(r_k^2\delta_{ij}-r_{ki}r_{kj})

. For a principal-axis frame,

I

is diagonal with principal moments

I_1,I_2,I_3

. Angular momentum:

\vec L=\mathbf{I}\vec\omega

Euler's Equations

I_1\dot\omega_1-(I_2-I_3)\omega_2\omega_3=\tau_1

and cyclic permutations. Written in the body frame where

I

is diagonal. For a torque-free symmetric top (

I_1=I_2

), these give steady precession of

\vec\omega

around the symmetry axis.

Parallel Axis Theorem

I=I_{\rm cm}+Md^2

, where

I_{\rm cm}

is the moment about the center of mass and

d

is the distance to the new axis. Allows computation of

I

about any parallel axis.

Gyroscopic Precession

A spinning top with angular momentum

L

subject to torque

\tau

(from gravity) precesses at

\Omega_{\rm prec}=\tau/L=Mgd/(I\omega_s)

, where

d

is the distance from pivot to CM.

Key Equations

Rotational KE (principal axes)

T_{\rm rot} = \frac{1}{2}(I_1\omega_1^2 + I_2\omega_2^2 + I_3\omega_3^2)

Kinetic energy of rotation in the principal-axis frame.

Parallel axis theorem

I = I_{\rm cm} + Md^2

Moment of inertia about an axis parallel to the CM axis, distance d away.

Precession rate

\Omega_{\rm prec} = \frac{Mgd}{I_3\omega_s}

Gyroscopic precession rate of a symmetric top spinning at ωₛ, CM distance d from pivot.

Worked Example

Moment of Inertia of a Rod

Problem

Find the moment of inertia of a uniform rod of mass $M=2.0$ kg and length $L=1.2$ m about: (a) its center, (b) one end.

Solution

(a) About center: $I_{\rm cm}=\frac{1}{12}ML^2=\frac{1}{12}(2.0)(1.44)=0.240$ kg·m².

(b) About one end (distance $d=L/2=0.6$ m from CM):

I_{\rm end} = I_{\rm cm} + Md^2 = 0.240 + (2.0)(0.36) = 0.240 + 0.720 = 0.960\text{ kg·m}^2

Check: $I_{\rm end}=\frac{1}{3}ML^2=\frac{1}{3}(2.0)(1.44)=0.960$ kg·m² ✓

Answer

I_{\rm cm}=0.240

kg·m²,

I_{\rm end}=0.960

kg·m².

Practice

Exercises

7 problems

1 of 7

A uniform rod has $M=3.0$ kg and $L=2.0$ m. Find its moment of inertia (in kg·m²) about its center.

Your answer

kg·m²

2 of 7

Using the parallel axis theorem, find $I$ (in kg·m²) for the same rod ( $M=3.0$ kg, $L=2.0$ m) about one end.

Your answer

kg·m²

3 of 7

A solid disk has $M=5.0$ kg and $R=0.40$ m. Find its moment of inertia (in kg·m²) about its symmetry axis.

Unlock Exercise 3