Parallel axis theorem — I = I_CM + Md² — proof: Worked examples

This page is a workout. The parent proof built the single line $I = I_{C M} + M d^{2}$ . Here we run that one line through every kind of situation it can meet — so that when an exam or a real machine throws a case at you, you have already seen its twin.

Before we start, one reminder in plain words. $I$ (the moment of inertia, see Moment of inertia — definition) is a number that tells you how hard it is to start or stop spinning a body about a chosen line in space (the "axis"). $I_{C M}$ is that number when the line passes through the balance point — the center of mass (CM). $M$ is the total mass. And $d$ is the perpendicular distance — the shortest straight-line gap, measured at a right angle — between the CM-axis and whatever parallel axis you actually care about.

The scenario matrix

Every problem this theorem can produce falls into one of these cells. The examples below are labelled with the cell(s) they cover.

Cell	Case class	What makes it tricky	Covered by
C1	Direct: known $I_{C M}$ → parallel axis at distance $d$	just add $M d^{2}$	Ex 1, Ex 2
C2	Degenerate: $d = 0$ (axis IS the CM-axis)	correction vanishes	Ex 3
C3	Reverse: known $I$ at a point → find $I_{C M}$ (subtract)	you must go backwards	Ex 4
C4	Between two non-CM axes	you may NOT connect them directly	Ex 5
C5	$d$ built from two components $(a, b)$	$d^{2} = a^{2} + b^{2}$ , not $a + b$	Ex 6
C6	Limiting / minimisation: which axis gives least $I$ ?	$M d^{2} \geq 0$ ⇒ CM wins	Ex 7
C7	Real-world word problem (physical pendulum)	translate words → $d$	Ex 8
C8	Exam twist: combined with Perpendicular axis theorem or Radius of gyration	two theorems chained	Ex 9, Ex 10

Worked example Example 1 — Rod about its end (Cell C1)

Thin uniform rod, mass $M$ , length $L$ . Given $I_{C M} = \frac{1}{12} M L^{2}$ about a perpendicular axis through its middle. Find $I$ about a perpendicular axis through one end.

Forecast: guess before reading — will the end-value be bigger or smaller than the middle-value, and by roughly how much?

Locate the CM. For a uniform rod the balance point is the midpoint. Why this step? $d$ is measured from the CM, so we must know where the CM sits.
Find $d$ . From midpoint to end is $d = \frac{L}{2}$ . Why this step? $d$ is the ⟂ distance between the (parallel) middle-axis and end-axis — for these two axes that gap is just half the rod.
Add $M d^{2}$ . $I_{e n d} = \frac{1}{12} M L^{2} + M (\frac{L}{2})^{2} = \frac{1}{12} M L^{2} + \frac{1}{4} M L^{2} = \frac{4}{12} M L^{2} = \frac{1}{3} M L^{2}$ Why this step? One axis (the middle one) passes through the CM, so the theorem applies with no cross terms.

Verify: $\frac{1}{3} M L^{2}$ is the standard textbook value for a rod about its end (from direct integration). The end-value is bigger — spinning a rod from its tip is harder than from its middle. ✓ Units: $[kg] [m^{2}]$ throughout. ✓

Figure — Parallel axis theorem — I = I_CM + Md² — proof

Worked example Example 2 — Solid disc about its rim (Cell C1, geometric)

Solid disc, mass $M$ , radius $R$ . Perpendicular central axis gives $I_{C M} = \frac{1}{2} M R^{2}$ . Find $I$ about a perpendicular axis piercing a point on the rim.

Forecast: the rim is distance $R$ from the centre. Guess the multiplier of $M R^{2}$ .

Confirm axes are parallel. Both point straight out of the disc's face (perpendicular to the disc), so they are parallel. Why this step? The theorem only relates parallel axes; if they weren't parallel we couldn't use it.
Read off $d$ . Centre to rim $= R$ , so $d = R$ . Why this step? $d$ is the ⟂ gap between the two parallel lines, which here is simply the radius.
Add the correction. $I_{r im} = \frac{1}{2} M R^{2} + M R^{2} = \frac{3}{2} M R^{2}$

Verify: $\frac{3}{2} M R^{2} = 1.5 M R^{2}$ . Sanity: it must exceed $I_{C M} = 0.5 M R^{2}$ , and it does — moving the axis out always adds inertia. ✓

Worked example Example 5 — Two non-CM axes (Cell C4, the classic trap)

A solid sphere, mass $M$ , radius $R$ , $I_{C M} = \frac{2}{5} M R^{2}$ . Axis A is $d_{A} = R$ from the CM; axis B is $d_{B} = 2 R$ from the CM; both parallel to each other and to the CM-axis. Find $I_{B} - I_{A}$ .

Forecast: the tempting shortcut says $I_{B} = I_{A} + M (d_{B} - d_{A})^{2} = I_{A} + M R^{2}$ . Is that right?

Refuse the direct link. You cannot write $I_{B} = I_{A} + M (gap)^{2}$ , because neither A nor B passes through the CM, so the cross terms would NOT vanish. Why this step? The theorem's clean form needs one axis on the CM.
Route each through the CM. $I_{A} = \frac{2}{5} M R^{2} + M R^{2}, I_{B} = \frac{2}{5} M R^{2} + M (2 R)^{2} = \frac{2}{5} M R^{2} + 4 M R^{2}$ Why this step? The CM is the shared reference both axes can legally connect to.
Subtract. $I_{B} - I_{A} = (4 M R^{2}) - (M R^{2}) = 3 M R^{2}$

Verify: the naive shortcut gave $M R^{2}$ ; the correct answer is $3 M R^{2}$ . They differ, which proves the shortcut is wrong. The $\frac{2}{5} M R^{2}$ terms cancel in the difference, as they must (same body, same $I_{C M}$ ). ✓

Worked example Example 6 —

$d$ from two components (Cell C5) A thin plate spins about a perpendicular axis through its CM with $I_{C M} = 0.040 kg\cdotm^{2}$ , mass $M = 0.50 kg$ . The new (parallel) axis pierces the plane at the point $(a, b) = (0.30 m, 0.40 m)$ measured from the CM. Find $I$ .

Forecast: careful — is $d = a + b = 0.70$ , or something else?

Build $d$ from the geometry. The ⟂ distance from the CM to the point $(a, b)$ is the hypotenuse: $d^{2} = a^{2} + b^{2}$ . Why this step? Distance in a plane is Pythagoras, NOT the sum of the two legs — this is exactly the $a^{2} + b^{2} = d^{2}$ line from the proof. $d^{2} = (0.30)^{2} + (0.40)^{2} = 0.09 + 0.16 = 0.25 m^{2} \Rightarrow d = 0.50 m$
Add $M d^{2}$ . $I = 0.040 + 0.50 \times 0.25 = 0.040 + 0.125 = 0.165 kg\cdotm^{2}$ Why this step? We need $M d^{2}$ , and we already have $d^{2} = 0.25$ directly, so no need to even take the square root.

Verify: the trap answer using $d = 0.70$ would give $M d^{2} = 0.50 \times 0.49 = 0.245$ , way too big. The (3,4,5) triangle gives $d = 0.5$ cleanly. Final $0.165 kg\cdotm^{2} > I_{C M}$ ✓.

Worked example Example 7 — Which axis is cheapest? (Cell C6, limiting behaviour)

Among all axes parallel to a fixed direction, one gives the minimum moment of inertia. For a rod ( $M$ , $L$ , $I_{C M} = \frac{1}{12} M L^{2}$ ), plot $I (d) = \frac{1}{12} M L^{2} + M d^{2}$ as the axis slides a distance $d$ from the CM, and find where $I$ is smallest.

Forecast: guess the shape of the graph and the location of its lowest point.

Read the function. $I (d) = I_{C M} + M d^{2}$ is a parabola in $d$ , opening upward, since $M d^{2} \geq 0$ . Why this step? Recognising the shape tells us the minimum without calculus.
Find the minimum. $M d^{2}$ is smallest when $d = 0$ , giving $I_{min} = I_{C M} = \frac{1}{12} M L^{2}$ . Why this step? A squared term can never be negative, so the lowest point is where it is zero — at the CM.
Check symmetry. $I (+ d) = I (- d)$ : moving the axis left or right of the CM by the same $d$ costs the same $M d^{2}$ . Why this step? The parabola is symmetric about the CM, confirming there is exactly one cheapest axis.

Verify: at $d = 0$ , $I = \frac{1}{12} M L^{2} \approx 0.0833 M L^{2}$ ; at $d = L /2$ , $I = \frac{1}{3} M L^{2} \approx 0.333 M L^{2}$ (matching Ex 1). The min really is at the CM. ✓

Worked example Example 8 — Real-world: a swinging sign (Cell C7, physical pendulum)

A uniform rectangular shop sign, mass $M = 8.0 kg$ , height $h = 1.2 m$ , hangs from a horizontal pivot along its top edge and swings like a Physical pendulum. Its CM is at the geometric centre. For a thin flat plate the moment of inertia about a horizontal axis through the CM in the plane of the sign is $I_{C M} = \frac{1}{12} M h^{2}$ . Find $I$ about the top-edge pivot.

Forecast: the pivot is at the top edge; how far is that from the centre?

Translate words → $d$ . The CM is at the centre, the pivot at the top edge, so $d = \frac{h}{2} = 0.60 m$ . Why this step? The phrase "top edge" is a geometry clue we must convert into a number for $d$ .
Compute $I_{C M}$ . $I_{C M} = \frac{1}{12} (8.0) (1.2)^{2} = \frac{1}{12} (8.0) (1.44) = 0.96 kg\cdotm^{2}$ . Why this step? We need the CM value before adding the correction.
Add $M d^{2}$ . $M d^{2} = 8.0 \times (0.60)^{2} = 8.0 \times 0.36 = 2.88 kg\cdotm^{2}$ . $I_{p i v o t} = 0.96 + 2.88 = 3.84 kg\cdotm^{2}$ Why this step? The pivot axis is parallel to the CM-axis (both horizontal, along the top), so the theorem applies.

Verify: notice $I_{p i v o t} = \frac{1}{12} M h^{2} + M (h /2)^{2} = \frac{1}{3} M h^{2} = \frac{1}{3} (8.0) (1.44) = 3.84$ ✓ — same $\frac{1}{3} M h^{2}$ form as the rod-about-end, because along its height a thin plate behaves like a rod. Units kg·m² ✓.

Worked example Example 9 — Exam twist: chain with the perpendicular axis theorem (Cell C8)

A thin disc, mass $M$ , radius $R$ . The Perpendicular axis theorem says for a flat plate $I_{z} = I_{x} + I_{y}$ , where $z$ is perpendicular to the plate and $x, y$ lie in it through the CM. By symmetry $I_{x} = I_{y}$ , and $I_{z} = \frac{1}{2} M R^{2}$ . Find $I$ about a diameter that lies along the rim's tangent (a line in the plane of the disc, tangent to the edge).

Forecast: two theorems in a row. First get the in-plane CM value, then shift it out to the tangent.

Perpendicular axis theorem → in-plane $I_{C M}$ . $I_{z} = I_{x} + I_{y} = 2 I_{x}$ , so $I_{x} = \frac{1}{2} I_{z} = \frac{1}{2} \cdot \frac{1}{2} M R^{2} = \frac{1}{4} M R^{2}$ . Why this step? We need the moment of inertia about a diameter (in-plane, through CM) before we can slide it.
Identify $d$ to the tangent line. A tangent line at the rim is parallel to a central diameter and $d = R$ away. Why this step? Parallel axis theorem needs the ⟂ gap between the parallel lines; centre to tangent is one radius.
Apply parallel axis theorem. $I_{t an g e n t} = \frac{1}{4} M R^{2} + M R^{2} = \frac{5}{4} M R^{2}$

Verify: $\frac{5}{4} M R^{2} = 1.25 M R^{2}$ . Sanity: it must exceed the diameter value $\frac{1}{4} M R^{2}$ , and it does. Both theorems used legally (one axis through CM at each stage). ✓

Worked example Example 10 — Exam twist: radius of gyration (Cell C8)

The Radius of gyration $k$ is defined by $I = M k^{2}$ — the distance at which the whole mass, as a point, would give the same $I$ . A solid sphere, $M = 3.0 kg$ , $R = 0.10 m$ , $I_{C M} = \frac{2}{5} M R^{2}$ , spins about a tangent axis (distance $d = R$ from CM). Find its radius of gyration $k$ about that tangent.

Forecast: will $k$ be larger or smaller than $R$ ?

Parallel axis theorem for the tangent $I$ . $I = \frac{2}{5} M R^{2} + M R^{2} = \frac{7}{5} M R^{2}$ . Why this step? Tangent axis is parallel to the CM-axis, $d = R$ away.
Convert to $k$ via the definition. $I = M k^{2} \Rightarrow k^{2} = \frac{I}{M} = \frac{7}{5} R^{2}$ . Why this step? The radius of gyration just repackages $I$ as an equivalent distance. $k = R \frac{7}{5} = 0.10 1.4 \approx 0.1183 m$

Verify: $k \approx 0.118 m > R = 0.10 m$ ✓ — sensible, since the tangent axis is farther out than the sphere's own radius, so the "equivalent point-mass distance" exceeds $R$ . Numerically $1.4 \approx 1.1832$ , times $0.10 = 0.11832$ . ✓

Recall Quick self-test across all cells

Direct add ( $C 1$ ) means... ::: $I = I_{C M} + M d^{2}$ , just add the correction. If $d = 0$ ( $C 2$ ) then $I$ equals... ::: exactly $I_{C M}$ ; no correction. To find $I_{C M}$ from a known far-axis $I$ ( $C 3$ ) you... ::: subtract, $I_{C M} = I - M d^{2}$ . Two non-CM axes ( $C 4$ ) — the safe route is... ::: through the CM separately, then subtract. Given $(a, b)$ , the distance is ( $C 5$ )... ::: $d = a^{2} + b^{2}$ , never $a + b$ . The cheapest axis ( $C 6$ ) is... ::: the one through the CM, since $M d^{2} \geq 0$ .

Connections

Parent proof
Moment of inertia — definition
Center of mass — definition and computation
Perpendicular axis theorem
Radius of gyration
Rotational kinetic energy
Physical pendulum

Worked examples — Parallel axis theorem — I = I_CM + Md² — proof

The scenario matrix

Connections

Scenario Map