Inertia tensor — principal axes, principal moments: Exercises

Quick reminder of the tools we lean on (all defined in the parent):

Intuition Three facts we lean on — sketched here, not just cited

(A) Vanishing off-diagonals $\Leftrightarrow$ principal axes. The entry $I_{x y}$ measures how much a spin about $x$ leaks angular momentum into the $y$ -direction: $L_{x} = I_{xx} ω_{x} + I_{x y} ω_{y} + \dots$ . If a spin purely about $\overset{x}{^}$ (so $ω = (ω, 0, 0)$ ) is to give $L$ purely along $\overset{x}{^}$ (i.e. $L ∥ ω$ ), the leaked components $L_{y} = I_{x y} ω$ and $L_{z} = I_{x z} ω$ must be zero for all $ω$ — forcing $I_{x y} = I_{x z} = 0$ . Demand this for spins about $\overset{y}{^}$ and $\overset{z}{^}$ too and every off-diagonal must vanish. So "all three axes are principal" and "all off-diagonals are zero" are literally the same statement.

(B) The factor of 2 in the trace. Add the diagonal entries: $I_{xx} + I_{y y} + I_{z z} = \sum m [(y^{2} + z^{2}) + (x^{2} + z^{2}) + (x^{2} + y^{2})] = \sum m [2 x^{2} + 2 y^{2} + 2 z^{2}] = 2 \sum m r^{2},$ where $r^{2} = x^{2} + y^{2} + z^{2}$ . Each coordinate appears in two of the three moments (e.g. $x^{2}$ shows up in $I_{y y}$ and $I_{z z}$ but not $I_{xx}$ ) — that double-counting is exactly where the $2$ comes from. Rotating axes never changes $\sum m r^{2}$ (distance to the origin is frame-independent), so the trace is invariant.

(C) The clean kinetic-energy formula. From the parent, $T = \frac{1}{2} ω^{T} I ω$ . In the principal frame $I$ is diagonal, so with $ω = (ω_{1}, ω_{2}, ω_{3})$ : $ω^{T} I ω = (ω_{1}, ω_{2}, ω_{3}) I_{1} ω_{1} I_{2} ω_{2} I_{3} ω_{3} = I_{1} ω_{1}^{2} + I_{2} ω_{2}^{2} + I_{3} ω_{3}^{2},$ giving $T = \frac{1}{2} (I_{1} ω_{1}^{2} + I_{2} ω_{2}^{2} + I_{3} ω_{3}^{2})$ . The cross terms $I_{ij} ω_{i} ω_{j}$ that clutter the general formula all vanish because the off-diagonals are zero.

Level 1 — Recognition

Recall Solution L1·Q1

(a) The off-diagonal entries are all exactly zero, so yes, the lab axes are already principal axes. (By fact (A) above: zero off-diagonals $\Leftrightarrow$ principal.) (b) Principal moments are just the diagonal entries: $I_{1} = 5, I_{2} = 8, I_{3} = 8 kg\cdotm^{2}$ . (c) $L = I ω = (5 \cdot 3, 0, 0) = (15, 0, 0)$ . This is parallel to $ω = (3, 0, 0)$ — yes. WHY: $\overset{x}{^}$ is an eigenvector, so $L = I_{1} ω$ .

Recall Solution L1·Q2

The trace invariant (fact (B)) says $I_{1} + I_{2} + I_{3} = 2 \sum m r^{2}$ . Left side: $2 + 3 + 4 = 9$ . Right side: $2 \times 5 = 10$ . Since $9 \neq = 10$ , the data is inconsistent. Somewhere a moment or the mass sum is wrong.

Level 2 — Application

Recall Solution L2·Q1

Both masses have $x = 0, z = 0, y = \pm b$ .

$I_{xx} = \sum m (y^{2} + z^{2}) = m b^{2} + m b^{2} = 2 m b^{2}$ .
$I_{y y} = \sum m (x^{2} + z^{2}) = 0$ (both lie on the $y$ -axis → zero distance from it).
$I_{z z} = \sum m (x^{2} + y^{2}) = 2 m b^{2}$ .
Products: every one contains a factor of $x$ or $z$ , all zero here, so $I_{x y} = I_{x z} = I_{y z} = 0$ . $I = 2 m b^{2} 00 000 00 2 m b^{2} .$ Already diagonal → principal moments $2 m b^{2}, 0, 2 m b^{2}$ ; principal axes are $x, y, z$ . The $0$ belongs to the $y$ -axis because the masses lie on it (spinning about it sweeps nothing).

Recall Solution L2·Q2

$L = I ω = 2 m a^{2} 1 - 1 0 - 1 10 002 ω 00 = 2 m a^{2} (ω, - ω, 0) .$ So $L$ points along $(1, - 1, 0)$ while $ω$ points along $(1, 0, 0)$ — not parallel. Angle, done with the full norms (no shortcuts). Write $L = 2 m a^{2} ω (1, - 1, 0)$ and $ω = ω (1, 0, 0)$ . $L \cdot ω = (2 m a^{2} ω) (ω) [(1) (1) + (- 1) (0) + (0) (0)] = 2 m a^{2} ω^{2} .$ $∣ L ∣ = 2 m a^{2} ω 1^{2} + (- 1)^{2} + 0^{2} = 2 m a^{2} ω 2, ∣ ω ∣ = ω 1^{2} = ω .$ $cos θ = \frac{2 m a ^{2} ω ^{2}}{( 2 m a ^{2} ω 2 ) ( ω )} = \frac{2 m a ^{2} ω ^{2}}{2 m a ^{2} ω ^{2} 2} = \frac{1}{2} .$ The common factor $2 m a^{2} ω^{2}$ cancels top and bottom — that is why we could have ignored it — so $θ = 4 5^{\circ}$ .

Figure below (s01): the orange arrow is $ω$ (along $x$ ), the magenta arrow is $L$ (pointing into the fourth quadrant along $y = - x$ ). The dashed violet line is the mass line $y = x$ ; notice $L$ is reflected away from $ω$ across it, and the navy arc marks the $4 5^{\circ}$ tilt — the geometric signature of a non-principal spin.

Figure — Inertia tensor — principal axes, principal moments

Level 3 — Analysis

Recall Solution L3·Q1

The $z$ -axis is decoupled (its column/row are zero off the diagonal), so $I_{3} = 6$ with axis $\overset{e}{^}_{3} = (0, 0, 1)$ . For the top-left block solve $det (4 - λ 1 1 4 - λ) = 0$ : $(4 - λ)^{2} - 1 = 0 \Rightarrow 4 - λ = \pm 1 \Rightarrow λ = 3 or 5.$ Eigenvectors: for $λ = 5$ , $(4 - 5) x + y = 0 \Rightarrow y = x$ , axis $\frac{1}{2} (1, 1, 0)$ . For $λ = 3$ , $y = - x$ , axis $\frac{1}{2} (1, - 1, 0)$ . Principal moments: $3, 5, 6$ . Check: trace $4 + 4 + 6 = 14 = 3 + 5 + 6$ . ✓

Recall Solution L3·Q2

Directly from the definitions with $x = d, y = z = 0$ :

$I_{xx} = m (y^{2} + z^{2}) = 0$ .
$I_{y y} = m (x^{2} + z^{2}) = m d^{2}$ .
$I_{z z} = m (x^{2} + y^{2}) = m d^{2}$ .
All products vanish (each needs a zero coordinate). $I = 000 0 m d^{2} 0 00 m d^{2} .$ So $I_{z z} = m d^{2}$ . WHY it matches parallel axis: about its own centre a point mass has $0$ ; shifting by $d$ perpendicular to the $z$ -axis adds $m d^{2}$ , exactly the theorem's $I = I_{cm} + M d^{2}$ .

Level 4 — Synthesis

Recall Solution L4·Q1

The $x$ -axis decouples: $I_{1} = 2$ with $\overset{e}{^}_{1} = (1, 0, 0)$ . Block $(3113)$ : $(3 - λ)^{2} - 1 = 0 \Rightarrow λ = 2$ or $4$ .

$λ = 4$ : $y = z$ , axis $\overset{e}{^}_{2} = \frac{1}{2} (0, 1, 1)$ .
$λ = 2$ : $y = - z$ , axis $\overset{e}{^}_{3} = \frac{1}{2} (0, 1, - 1)$ . Principal moments: $I_{1} = 2, I_{2} = 4, I_{3} = 2$ . Trace check: $2 + 3 + 3 = 8 = 2 + 4 + 2$ . ✓

Kinetic energy. First in the lab frame, $T = \frac{1}{2} ω^{T} I ω$ : $I ω = (2 \cdot 1, 3 \cdot 2 + 1 \cdot 2, 1 \cdot 2 + 3 \cdot 2) = (2, 8, 8) .$ $T = \frac{1}{2} (1, 2, 2) \cdot (2, 8, 8) = \frac{1}{2} (2 + 16 + 16) = 17 J .$ Cross-check in principal frame using fact (C). Components of $ω = (1, 2, 2)$ along each principal axis (project via dot product): $ω_{1} = (1, 2, 2) \cdot (1, 0, 0) = 1$ ; $ω_{2} = (1, 2, 2) \cdot \frac{1}{2} (0, 1, 1) = \frac{2 + 2}{2} = 22$ ; $ω_{3} = (1, 2, 2) \cdot \frac{1}{2} (0, 1, - 1) = \frac{2 - 2}{2} = 0$ . Pair each component with its own moment: $ω_{2}$ (along $\overset{e}{^}_{2}$ ) carries $I_{2} = 4$ , and $ω_{3}$ carries $I_{3} = 2$ : $T = \frac{1}{2} (I_{1} ω_{1}^{2} + I_{2} ω_{2}^{2} + I_{3} ω_{3}^{2}) = \frac{1}{2} (2 \cdot 1^{2} + 4 \cdot (22)^{2} + 2 \cdot 0^{2}) = \frac{1}{2} (2 + 32 + 0) = 17 J . ✓$ Both frames agree — energy is a scalar invariant.

Level 5 — Mastery

Recall Solution L5·Q1

By the cube's symmetry ( $x \to - x$ etc.), each product of inertia integral $\int x y d m$ pairs $+$ and $-$ contributions and vanishes, so $I$ is diagonal in these axes. One-line derivation of $\int x^{2} d m = \frac{1}{12} M L^{2}$ . The cube has uniform density $ρ = M / L^{3}$ . Slice it into slabs perpendicular to $x$ ; a slab at position $x$ (with $- L /2 \leq x \leq L /2$ ) has mass $d m = ρ L^{2} d x$ . Then $\int x^{2} d m = ρ L^{2} \int_{- L /2}^{L /2} x^{2} d x = ρ L^{2} [\frac{x ^{3}}{3}]_{- L /2}^{L /2} = ρ L^{2} \cdot \frac{1}{3} \cdot \frac{L ^{3}}{4} = ρ \frac{L ^{5}}{12} = \frac{M}{L ^{3}} \frac{L ^{5}}{12} = \frac{1}{12} M L^{2} .$ By symmetry $\int y^{2} d m = \int z^{2} d m = \frac{1}{12} M L^{2}$ too. Hence $I_{z z} = \int (x^{2} + y^{2}) d m = \frac{1}{12} M L^{2} + \frac{1}{12} M L^{2} = \frac{1}{6} M L^{2},$ identically for $I_{xx}, I_{y y}$ . So $I = \frac{1}{6} M L^{2} 1$ . Because $I = I 1$ (a scalar times identity), for any $ω$ , $I ω = I ω$ . Every direction is an eigenvector → every axis is principal ("spherical top"). Spinning about the diagonal: $L = \frac{1}{6} M L^{2} ω$ , perfectly parallel to $ω$ — no wobble, no tilt.

Figure below (s02): the violet cube is drawn with edges along the axes; the orange arrow is $ω$ pointing along the body diagonal $(1, 1, 1)$ , and the magenta arrow is $L$ lying exactly on top of it (drawn shorter only so both are visible). The geometric message: for a spherical top there is no special direction — $L$ never leaves $ω$ , unlike the tilted case in s01.

Recall Solution L5·Q2

Since $x, y, z$ are principal axes, $I$ is diagonal: $I = 6000600010 .$ $L = I ω = (6 \cdot 2, 6 \cdot 0, 10 \cdot 5) = (12, 0, 50) .$ Not parallel to $ω = (2, 0, 5)$ because $ω$ has components along axes with different moments ( $6$ vs $10$ ). Angle, with full norms: $L \cdot ω = 12 \cdot 2 + 0 \cdot 0 + 50 \cdot 5 = 24 + 250 = 274,$ $∣ L ∣ = 1 2^{2} + 0^{2} + 5 0^{2} = 2644, ∣ ω ∣ = 2^{2} + 0^{2} + 5^{2} = 29 .$ $cos θ = \frac{274}{2644 29} \approx \frac{274}{51.42 \cdot 5.385} \approx \frac{274}{276.9} \approx 0.9895,$ giving $θ \approx 8. 3^{\circ}$ . A small but nonzero tilt — the signature of a symmetric top spun off its symmetry axis. This tilt is exactly what drives precession in Euler's equations of rigid body motion.

Recap

Recall Self-test: can you answer these cold?

Off-diagonals of $I$ must be exactly what for principal axes? ::: Exactly zero. Trace invariant relation (and where the 2 comes from)? ::: $I_{xx} + I_{y y} + I_{z z} = I_{1} + I_{2} + I_{3} = 2 \sum m r^{2}$ ; each coordinate square appears in two of the three moments. When is $L ∥ ω$ ? ::: Only when $ω$ is along a principal axis, or when all principal moments are equal (spherical top). Formula for $T$ that only works in the principal frame? ::: $T = \frac{1}{2} (I_{1} ω_{1}^{2} + I_{2} ω_{2}^{2} + I_{3} ω_{3}^{2})$ . Sign convention for products of inertia? ::: Negative: $I_{x y} = - \sum m x y$ .