Inertia tensor — principal axes, principal moments

2.1.22 · Physics › Analytical Mechanics

1. Inertia tensor KYA hai — scratch se derive kiya

Hum chahte hain $\vec{L}$ ek rigid body ke liye jo ek fixed point ke baare mein angular velocity $\vec{\omega }$ se rotate kar rahi hai.

Ek particle se shuru karo. Particle $a$ position ${\vec{r}}_a$ par, mass $m_a$. Ek rigid body mein uski velocity hai ${\vec{v}}_a=\vec{\omega }\times {\vec{r}}_a$. Uska angular momentum:

${\vec{L}}_a=m_a{\vec{r}}_a\times {\vec{v}}_a=m_a{\vec{r}}_a\times (\vec{\omega }\times {\vec{r}}_a)$

Yeh step kyun? Angular momentum by definition $\vec{r}\times m\vec{v}$ hai, aur rigidity force karta hai $\vec{v}=\vec{\omega }\times \vec{r}$.

BAC–CAB identity use karo $\vec{A}\times (\vec{B}\times \vec{C})=\vec{B}(\vec{A}\cdot \vec{C})-\vec{C}(\vec{A}\cdot \vec{B})$:

${\vec{L}}_a=m_a[\vec{\omega }(r_a^2)-{\vec{r}}_a({\vec{r}}_a\cdot \vec{\omega })]$

Yeh step kyun? Yeh $\vec{\omega }$ ko separate kar deta hai taaki hum us par act karne wala ek linear map padh sakein.

Particles pe sum karo (ya $\int dm$ integrate karo). Ek component dekho, maan lo $L_x$. $\vec{r}=(x,y,z)$ aur $\vec{\omega }=({\omega }_x,{\omega }_y,{\omega }_z)$ ke saath:

$L_x={\sum }_a{m}_a[{\omega }_x(x^2+y^2+z^2)-x_a(x_a{\omega }_x+y_a{\omega }_y+z_a{\omega }_z)]$ $={\omega }_x\underset{I_{xx}}{\underbrace{\sum m_a(y^2+z^2)}}-{\omega }_y\underset{-I_{xy}}{\underbrace{\sum m_{a}xy}}-{\omega }_z\underset{-I_{xz}}{\underbrace{\sum m_{a}xz}}$

Toh $\vec{L}=\mathbf{I}\vec{\omega }$ jahan:

2. $\vec{L}$, $\vec{\omega }$ ke parallel kyun nahi — aur fix

Kyunki $\mathbf{I}$ mein off-diagonal terms hain, $\mathbf{I}\vec{\omega }$ generally kahin aur point karta hai. Ek symmetric real matrix ko hamesha ek orthogonal (rotation) matrix se diagonalize kiya ja sakta hai. Iska matlab hai ek special rotated coordinate frame exist karta hai jahan:

${\mathbf{I}}^{\prime }=\left(\begin{array}{ccc}{I}_1& 0& 0\\ 0& I_2& 0\\ 0& 0& I_3\end{array}\right)$

Yeh exactly eigenvectors kyun hain: "$\vec{L}$, $\vec{\omega }$ ke parallel hai" ka matlab hai $\mathbf{I}\vec{\omega }=\lambda \vec{\omega }$. Yahi hai eigenvalue equation. Toh:

Inhe kaise find karein (recipe):

1. $\mathbf{I}$ ko kisi bhi convenient frame mein likho.
2. Characteristic cubic $\det (\mathbf{I}-\lambda 1)=0$ solve karo → teen roots $I_1,I_2,I_3$.
3. Har $\lambda =I_k$ ke liye, $(\mathbf{I}-I_{k}1){\hat{e}}_k=0$ solve karo → axis direction; normalize karo.
4. Shortcut: body ka koi bhi symmetry axis automatically ek principal axis hota hai.

3. Rotational KE aur ek key invariant

$T=\frac{1}{2}\sum m_a{v}_a^2=\frac{1}{2}\vec{\omega }\cdot \vec{L}=\frac{1}{2}{\vec{\omega }}^{\mathsf{T}}\mathbf{I}\vec{\omega }$

Principal axes mein yeh clean diagonal form ban jaata hai:

$\boxed{T=\frac{1}{2}(I_1{\omega }_1^2+I_2{\omega }_2^2+I_3{\omega }_3^2)}$

4. Worked examples

5. Common mistakes (Steel-manned)

6. Flashcards

Connections

• Angular momentum — woh quantity $\vec{L}=\mathbf{I}\vec{\omega }$ jo yahan derive ki gayi.
• Eigenvalues and eigenvectors — principal axes $\mathbf{I}$ ke eigenvectors hain.
• Euler's equations of rigid body motion — principal-axis frame mein likhi jaati hain.
• Parallel axis theorem — $\mathbf{I}$ ko centre of mass se dur axes par shift karta hai.
• Rigid body rotation — kinematics $\vec{v}=\vec{\omega }\times \vec{r}$ step 1 mein use ki gayi.
• Symmetric and orthogonal matrices — real eigenvalues aur orthogonal axes guarantee karte hain.

Concept Map