Definition. A rigid body is a system of particles in which the distance between every pair of particles is fixed.
A rigid body has 6 degrees of freedom: 3 translational (centre of mass position) and 3 rotational (orientation). The orientation is specified by three angles, most commonly the Euler angles ( ϕ , θ , ψ ) (\phi, \theta, \psi) ( ϕ , θ , ψ )
The angular velocity ω \boldsymbol{\omega} ω z z z x " x" x " z ′ ′ z'' z ′′
ω 1 = ϕ ˙ sin θ sin ψ + θ ˙ cos ψ \omega_1 = \dot{\phi}\sin\theta\sin\psi + \dot{\theta}\cos\psi ω 1 = ϕ ˙ sin θ sin ψ + θ ˙ cos ψ
ω 2 = ϕ ˙ sin θ cos ψ − θ ˙ sin ψ \omega_2 = \dot{\phi}\sin\theta\cos\psi - \dot{\theta}\sin\psi ω 2 = ϕ ˙ sin θ cos ψ − θ ˙ sin ψ
ω 3 = ϕ ˙ cos θ + ψ ˙ \omega_3 = \dot{\phi}\cos\theta + \dot{\psi} ω 3 = ϕ ˙ cos θ + ψ ˙
Definition. The moment of inertia tensor about the centre of mass is:
I j k = ∫ ρ ( r ) ( ∣ r ∣ 2 δ j k − r j r k ) d V I_{jk} = \int \rho(\mathbf{r})\left(\lvert\mathbf{r}\rvert^2 \delta_{jk} - r_j r_k\right) dV I j k = ∫ ρ ( r ) ( ∣ r ∣ 2 δ j k − r j r k ) d V
In matrix form, for a discrete system:
I = ( I x x I x y I x z I x y I y y I y z I x z I y z I z z ) \mathbf{I} = \begin{pmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{xy} & I_{yy} & I_{yz} \\ I_{xz} & I_{yz} & I_{zz} \end{pmatrix} I = I xx I x y I x z I x y I y y I y z I x z I y z I z z
Where:
I x x = ∑ i m i ( y i 2 + z i 2 ) , I x y = − ∑ i m i x i y i , e t c . I_{xx} = \sum_i m_i(y_i^2 + z_i^2), \quad I_{xy} = -\sum_i m_i x_i y_i, \quad \mathrm{etc}. I xx = ∑ i m i ( y i 2 + z i 2 ) , I x y = − ∑ i m i x i y i , etc .
Theorem 8.1 (Parallel Axis Theorem). The inertia tensor about a point displaced by a \mathbf{a} a
I j k ′ = I j k ( C M ) + M ( a 2 δ j k − a j a k ) I_{jk}' = I_{jk}^{(\mathrm{CM})} + M(a^2\delta_{jk} - a_j a_k) I j k ′ = I j k ( CM ) + M ( a 2 δ j k − a j a k )
Where M M M
Proof. Writing r i ′ = r i + a \mathbf{r}_i' = \mathbf{r}_i + \mathbf{a} r i ′ = r i + a r i \mathbf{r}_i r i
I j k ′ = ∑ i m i ( ∣ r i ′ ∣ 2 δ j k − r i ′ r i ′ , k ) = ∑ i m i ( ∣ r i + a ∣ 2 δ j k − ( r i , j + a j ) ( r i , k + a k ) ) I_{jk}' = \sum_i m_i\left(\lvert\mathbf{r}_i'\rvert^2\delta_{jk} - r_i' r_{i',k}\right) = \sum_i m_i\left(\lvert\mathbf{r}_i + \mathbf{a}\rvert^2\delta_{jk} - (r_{i,j} + a_j)(r_{i,k} + a_k)\right) I j k ′ = ∑ i m i ( ∣ r i ′ ∣ 2 δ j k − r i ′ r i ′ , k ) = ∑ i m i ( ∣ r i + a ∣ 2 δ j k − ( r i , j + a j ) ( r i , k + a k ) )
Expanding and using ∑ i m i r i = 0 \sum_i m_i \mathbf{r}_i = \mathbf{0} ∑ i m i r i = 0
I j k ′ = I j k ( C M ) + M ( a 2 δ j k − a j a k ) I_{jk}' = I_{jk}^{(\mathrm{CM})} + M(a^2\delta_{jk} - a_j a_k) I j k ′ = I j k ( CM ) + M ( a 2 δ j k − a j a k )
■ \blacksquare ■
Definition. The principal axes are the eigenvectors of I \mathbf{I} I principal moments of inertia I 1 , I 2 , I 3 I_1, I_2, I_3 I 1 , I 2 , I 3
In the principal axis frame, the inertia tensor is diagonal:
I = ( I 1 0 0 0 I 2 0 0 0 I 3 ) \mathbf{I} = \begin{pmatrix} I_1 & 0 & 0 \\ 0 & I_2 & 0 \\ 0 & 0 & I_3 \end{pmatrix} I = I 1 0 0 0 I 2 0 0 0 I 3
Since I \mathbf{I} I
Theorem 8.2 (Euler’s Equations). For torque-free rotation (τ = 0 \boldsymbol{\tau} = \mathbf{0} τ = 0
I 1 ω ˙ 1 = ( I 2 − I 3 ) ω 2 ω 3 I_1\dot{\omega}_1 = (I_2 - I_3)\omega_2\omega_3 I 1 ω ˙ 1 = ( I 2 − I 3 ) ω 2 ω 3
I 2 ω ˙ 2 = ( I 3 − I 1 ) ω 3 ω 1 I_2\dot{\omega}_2 = (I_3 - I_1)\omega_3\omega_1 I 2 ω ˙ 2 = ( I 3 − I 1 ) ω 3 ω 1
I 3 ω ˙ 3 = ( I 1 − I 2 ) ω 1 ω 2 I_3\dot{\omega}_3 = (I_1 - I_2)\omega_1\omega_2 I 3 ω ˙ 3 = ( I 1 − I 2 ) ω 1 ω 2
Proof. In the body frame, the angular momentum is L = I ω \mathbf{L} = \mathbf{I}\,\boldsymbol{\omega} L = I ω
( d L d t ) s p a c e = τ \left(\frac{d\mathbf{L}}{dt}\right)_{\mathrm{space} = \boldsymbol{\tau}} ( d t d L ) space = τ
Transforming to the body frame using ( d L d t ) s p a c e = ( d L d t ) b o d y + ω × L \left(\frac{d\mathbf{L}}{dt}\right)_{\mathrm{space} = \left(\frac{d\mathbf{L}}{dt}\right)_{\mathrm{body} + \boldsymbol{\omega} \times \mathbf{L}}} ( d t d L ) space = ( d t d L ) body + ω × L
I ω ˙ + ω × ( I ω ) = τ \mathbf{I}\,\dot{\boldsymbol{\omega}} + \boldsymbol{\omega} \times (\mathbf{I}\,\boldsymbol{\omega}) = \boldsymbol{\tau} I ω ˙ + ω × ( I ω ) = τ
In the principal axis frame with τ = 0 \boldsymbol{\tau} = \mathbf{0} τ = 0 ■ \blacksquare ■
Theorem 8.3. Torque-free rotation about a principal axis is stable if the axis corresponds to the largest or smallest principal moment of inertia, and unstable for the intermediate axis.
Proof. Consider rotation primarily about the 1-axis: ω = ( ω 1 , ϵ 2 , ϵ 3 ) \boldsymbol{\omega} = (\omega_1, \epsilon_2, \epsilon_3) ω = ( ω 1 , ϵ 2 , ϵ 3 ) ϵ 2 , ϵ 3 \epsilon_2, \epsilon_3 ϵ 2 , ϵ 3
I 2 ϵ ˙ 2 = ( I 3 − I 1 ) ω 1 ϵ 3 , I 3 ϵ ˙ 3 = ( I 1 − I 2 ) ω 1 ϵ 2 I_2\dot{\epsilon}_2 = (I_3 - I_1)\omega_1\epsilon_3, \quad I_3\dot{\epsilon}_3 = (I_1 - I_2)\omega_1\epsilon_2 I 2 ϵ ˙ 2 = ( I 3 − I 1 ) ω 1 ϵ 3 , I 3 ϵ ˙ 3 = ( I 1 − I 2 ) ω 1 ϵ 2
Combining: ϵ ¨ 2 = ( I 3 − I 1 ) ( I 1 − I 2 ) I 2 I 3 ω 1 2 ϵ 2 \ddot{\epsilon}_2 = \frac{(I_3 - I_1)(I_1 - I_2)}{I_2 I_3}\omega_1^2\,\epsilon_2 ϵ ¨ 2 = I 2 I 3 ( I 3 − I 1 ) ( I 1 − I 2 ) ω 1 2 ϵ 2
For stability, the coefficient must be negative. This requires ( I 1 − I 3 ) ( I 1 − I 2 ) > 0 (I_1 - I_3)(I_1 - I_2) \gt 0 ( I 1 − I 3 ) ( I 1 − I 2 ) > 0 I 1 I_1 I 1 I 1 I_1 I 1 ■ \blacksquare ■
is counterintuitive: a Rigid body spinning about its intermediate axis is unstable. This is not a violation of angular Momentum conservation --- the angular momentum vector remains fixed in space, but the body Tumbles relative to it.
Problem. A symmetric top (moments of inertia I 1 = I 2 ≠ I 3 I_1 = I_2 \neq I_3 I 1 = I 2 = I 3 M M M l l l
Solution Using Euler angles ( ϕ , θ , ψ ) (\phi, \theta, \psi) ( ϕ , θ , ψ ) θ \theta θ ϕ \phi ϕ ψ \psi ψ
The kinetic energy is:
T = 1 2 I 1 ( θ ˙ 2 + ϕ ˙ 2 sin 2 θ ) + 1 2 I 3 ( ψ ˙ + ϕ ˙ cos θ ) 2 T = \frac{1}{2}I_1(\dot{\theta}^2 + \dot{\phi}^2\sin^2\theta) + \frac{1}{2}I_3(\dot{\psi} + \dot{\phi}\cos\theta)^2 T = 2 1 I 1 ( θ ˙ 2 + ϕ ˙ 2 sin 2 θ ) + 2 1 I 3 ( ψ ˙ + ϕ ˙ cos θ ) 2
The potential energy is V = M g l cos θ V = Mgl\cos\theta V = M g l cos θ
The Lagrangian is L = T − V L = T - V L = T − V ϕ \phi ϕ ψ \psi ψ
p ϕ = I 1 ϕ ˙ sin 2 θ + I 3 ( ψ ˙ + ϕ ˙ cos θ ) cos θ = c o n s t p_\phi = I_1\dot{\phi}\sin^2\theta + I_3(\dot{\psi} + \dot{\phi}\cos\theta)\cos\theta = \mathrm{const} p ϕ = I 1 ϕ ˙ sin 2 θ + I 3 ( ψ ˙ + ϕ ˙ cos θ ) cos θ = const
p ψ = I 3 ( ψ ˙ + ϕ ˙ cos θ ) = c o n s t p_\psi = I_3(\dot{\psi} + \dot{\phi}\cos\theta) = \mathrm{const} p ψ = I 3 ( ψ ˙ + ϕ ˙ cos θ ) = const
The quantity p ψ = I 3 ω 3 p_\psi = I_3\omega_3 p ψ = I 3 ω 3 p ϕ p_\phi p ϕ
For steady precession at constant θ \theta θ ϕ ˙ = Ω \dot{\phi} = \Omega ϕ ˙ = Ω
θ ˙ = 0 , ϕ ˙ = Ω = c o n s t , ψ ˙ = ψ ˙ 0 = c o n s t \dot{\theta} = 0, \quad \dot{\phi} = \Omega = \mathrm{const}, \quad \dot{\psi} = \dot{\psi}_0 = \mathrm{const} θ ˙ = 0 , ϕ ˙ = Ω = const , ψ ˙ = ψ ˙ 0 = const
The equation of motion for θ \theta θ
I 1 Ω 2 sin θ cos θ − I 3 ( ψ ˙ 0 + Ω cos θ ) Ω sin θ + M g l sin θ = 0 I_1\Omega^2\sin\theta\cos\theta - I_3(\dot{\psi}_0 + \Omega\cos\theta)\Omega\sin\theta + Mgl\sin\theta = 0 I 1 Ω 2 sin θ cos θ − I 3 ( ψ ˙ 0 + Ω cos θ ) Ω sin θ + M g l sin θ = 0
Dividing by sin θ \sin\theta sin θ p ψ = I 3 n p_\psi = I_3 n p ψ = I 3 n n = ψ ˙ 0 + Ω cos θ n = \dot{\psi}_0 + \Omega\cos\theta n = ψ ˙ 0 + Ω cos θ
I 1 Ω 2 cos θ − I 3 n Ω + M g l = 0 I_1\Omega^2\cos\theta - I_3 n\Omega + Mgl = 0 I 1 Ω 2 cos θ − I 3 n Ω + M g l = 0
This is a quadratic in Ω \Omega Ω
Ω ± = I 3 n ± ( I 3 n ) 2 − 4 I 1 M g l cos θ 2 I 1 cos θ \Omega_\pm = \frac{I_3 n \pm \sqrt{(I_3 n)^2 - 4I_1 Mgl\cos\theta}}{2I_1\cos\theta} Ω ± = 2 I 1 c o s θ I 3 n ± ( I 3 n ) 2 − 4 I 1 M g l c o s θ
Real solutions exist when ( I 3 n ) 2 ≥ 4 I 1 M g l cos θ (I_3 n)^2 \geq 4I_1 Mgl\cos\theta ( I 3 n ) 2 ≥ 4 I 1 M g l cos θ n n n Ω ≈ ≈ I 3 n / ( I 1 cos θ ) \Omega_\approx \approx I_3 n / (I_1\cos\theta) Ω ≈ ≈ I 3 n / ( I 1 cos θ ) Ω ≈ ≈ M g l / ( I 3 n ) \Omega_\approx \approx Mgl / (I_3 n) Ω ≈ ≈ M g l / ( I 3 n )
■ \blacksquare ■
Problem. A bicycle wheel of mass m m m R R R ψ ˙ \dot{\psi} ψ ˙
Solution Model the wheel as a symmetric top with I 3 ≈ m R 2 I_3 \approx mR^2 I 3 ≈ m R 2 I 1 ≈ m R 2 / 2 I_1 \approx mR^2/2 I 1 ≈ m R 2 /2 l l l
For a horizontal axle (θ = π / 2 \theta = \pi/2 θ = π /2
I 1 Ω 2 ⋅ 0 − I 3 n Ω + M g l = 0 I_1\Omega^2 \cdot 0 - I_3 n\Omega + Mgl = 0 I 1 Ω 2 ⋅ 0 − I 3 n Ω + M g l = 0
Ω = M g l I 3 n = m g l m R 2 ψ ˙ = g l R 2 ψ ˙ \Omega = \frac{Mgl}{I_3 n} = \frac{mgl}{mR^2\dot{\psi}} = \frac{gl}{R^2\dot{\psi}} Ω = I 3 n M g l = m R 2 ψ ˙ m g l = R 2 ψ ˙ g l
This is the gyroscopic precession rate . Notice that it is inversely proportional to the spin rate --- the faster the wheel spins, the slower it precesses.
■ \blacksquare ■
A[8_Rigid Body Dynamics] --> B[Key Concepts]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
D --> G[Real-world usage]
Rigid body dynamics is the physics of how objects spin and tumble. The inertia tensor is like a map of how resistant an object is to rotation about different axes. A figure skater pulling in their arms speeds up because the same angular momentum must be conserved with a smaller moment of inertia. The truly counterintuitive result is the tennis racket theorem: a body spinning about its intermediate axis is unstable, while rotation about the largest or smallest axis is stable. Try flipping a phone end-over-end and then side-to-side and you will see this asymmetry. Gyroscopic precession is like a spinning top following the slope of a hill it cannot fall down: gravity tries to tilt the angular momentum vector, but the torque only rotates it around the vertical, creating that characteristic wobble.
Problem. Find the inertia tensor of a uniform thin rod of mass M M M L L L
Solution Place the rod along the x x x x = 0 x = 0 x = 0 x = L x = L x = L λ = M / L \lambda = M/L λ = M / L
I x x = ∫ 0 L λ ( y 2 + z 2 ) d x = 0 I_{xx} = \int_0^L \lambda(y^2 + z^2)\, dx = 0 I xx = ∫ 0 L λ ( y 2 + z 2 ) d x = 0
(since y = z = 0 y = z = 0 y = z = 0 x x x
I y y = ∫ 0 L λ ( x 2 + z 2 ) d x = λ ∫ 0 L x 2 d x = M L ⋅ L 3 3 = M L 2 3 I_{yy} = \int_0^L \lambda(x^2 + z^2)\, dx = \lambda \int_0^L x^2\, dx = \frac{M}{L} \cdot \frac{L^3}{3} = \frac{ML^2}{3} I y y = ∫ 0 L λ ( x 2 + z 2 ) d x = λ ∫ 0 L x 2 d x = L M ⋅ 3 L 3 = 3 M L 2
I z z = ∫ 0 L λ ( x 2 + y 2 ) d x = M L 2 3 I_{zz} = \int_0^L \lambda(x^2 + y^2)\, dx = \frac{ML^2}{3} I z z = ∫ 0 L λ ( x 2 + y 2 ) d x = 3 M L 2
The products of inertia all vanish by symmetry:
I x y = I x z = I y z = 0 I_{xy} = I_{xz} = I_{yz} = 0 I x y = I x z = I y z = 0
The inertia tensor is:
I = ( 0 0 0 0 M L 2 / 3 0 0 0 M L 2 / 3 ) \mathbf{I} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & ML^2/3 & 0 \\ 0 & 0 & ML^2/3 \end{pmatrix} I = 0 0 0 0 M L 2 /3 0 0 0 M L 2 /3
The principal moments about one end are 0 0 0 M L 2 / 3 ML^2/3 M L 2 /3 M L 2 / 3 ML^2/3 M L 2 /3 0 0 0 M L 2 / 12 ML^2/12 M L 2 /12 M L 2 / 12 ML^2/12 M L 2 /12
■ \blacksquare ■
Mistake 1: Treating the moment of inertia as a scalar instead of a tensor The moment of inertia is a rank-2 tensor I \mathbf{I} I I I I I ω \mathbf{I}\boldsymbol{\omega} I ω L = I ω \mathbf{L} = \mathbf{I}\boldsymbol{\omega} L = I ω
Mistake 2: Assuming all axes are stable for torque-free rotation Torque-free rotation about a principal axis is stable only if the axis corresponds to the largest or smallest principal moment of inertia. Rotation about the intermediate axis is unstable: small perturbations grow exponentially. This is the tennis racket theorem, and it explains why a spinning phone tumbles when flipped about its intermediate axis.
Mistake 3: Confusing the parallel axis theorem for tensors with the scalar version The parallel axis theorem for the inertia tensor is I j k ′ = I j k ( C M ) + M ( a 2 δ j k − a j a k ) I_{jk}' = I_{jk}^{(\mathrm{CM})} + M(a^2\delta_{jk} - a_j a_k) I j k ′ = I j k ( CM ) + M ( a 2 δ j k − a j a k ) I = I C M + M a 2 I = I_{\mathrm{CM}} + Ma^2 I = I CM + M a 2
