Moments of Inertia

The moment of inertia is the rotational analogue of mass. Mass measures a body's resistance to linear acceleration; moment of inertia measures its resistance to angular acceleration. But there is a crucial new ingredient: where mass is a single number intrinsic to a body, the moment of inertia depends on which axis you spin the body about and on how the mass is distributed relative to that axis. The same ring is hard to spin about a diameter but easy about its central axis. This lesson defines $I$I rigorously, derives the standard results from $I = \int r^2\,\mathrm{d}m$I=∫r2dm, proves the two great labour-saving theorems (parallel-axis and perpendicular-axis), and builds the technique for composite bodies and the radius of gyration. It is the foundation for the next two lessons on rotational dynamics and angular momentum.

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1 Where this sits in AQA 7367

This is Paper 3, the Mechanics option (7367/3M). Moments of inertia underpin the entire rotational-dynamics block of Further Mechanics, so this is the gateway lesson for $\tau = I\alpha$τ=Iα, rotational kinetic energy $\tfrac12 I\omega^2$21​Iω2 and angular momentum $I\omega$Iω. The derivations are pure AO1/AO2 (compute and prove standard results, deploy the two theorems), while composite-body and "set up the integral" parts reach into AO3 — and recall that Paper 3 is the more problem-solving-weighted paper (AO1 40% / AO2 25% / AO3 35%). The prerequisite is definite integration from A-Level Maths (you will integrate $x^2$x2, $r^3$r3 and trigonometric forms) and the centre-of-mass integration you met earlier in this course; the machinery $I = \int r^2\,\mathrm{d}m$I=∫r2dm is the centre-of-mass integral with the square of the distance.

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2 Core theory — definition and the standard results

For a system of $n$n particles, the moment of inertia about a chosen axis is the mass-weighted sum of squared perpendicular distances:

$$\boxed{\,I = \sum_{i=1}^{n} m_i r_i^2\,}$$I=i=1∑n​mi​ri2​​

where $r_i$ri​ is the perpendicular distance of the $i$i-th particle from the axis. For a continuous body the sum becomes an integral over the mass elements $\mathrm{d}m$dm:

$$I = \int r^2\,\mathrm{d}m .$$I=∫r2dm.

Property	Detail
SI unit	$\text{kg m}^2$kg m2
Always non-negative	since $r^2 \ge 0$r2≥0
Axis-dependent	different axes give different $I$I
Distribution-dependent	mass far from the axis contributes more (the $r^2$r2 weighting)
Additive	for parts about the same axis, $I = I_1 + I_2 + \cdots$I=I1​+I2​+⋯

The $r^2$r2 weighting is the whole story: doubling the distance of a mass element from the axis quadruples its contribution. This is why hollow shapes (mass at the rim) have larger $I$I than solid ones of the same mass and radius.

The standard results you must know

Body	Axis	$I$I
Thin rod mass $M$M, length $L$L	through centre, $\perp$⊥ rod	$\dfrac{1}{12}ML^2$121​ML2
Thin rod	through one end, $\perp$⊥ rod	$\dfrac{1}{3}ML^2$31​ML2
Uniform disc / solid cylinder mass $M$M, radius $R$R	through centre, $\perp$⊥ face	$\dfrac{1}{2}MR^2$21​MR2
Thin hoop / ring mass $M$M, radius $R$R	through centre, $\perp$⊥ plane	$MR^2$MR2
Solid sphere mass $M$M, radius $R$R	through centre (diameter)	$\dfrac{2}{5}MR^2$52​MR2
Thin spherical shell	through centre (diameter)	$\dfrac{2}{3}MR^2$32​MR2
Rectangular lamina $a \times b$a×b, mass $M$M	through centre, $\perp$⊥ lamina	$\dfrac{1}{12}M(a^2 + b^2)$121​M(a2+b2)

Notice the pattern across the "round" bodies: hoop $MR^2$MR2 > disc $\tfrac12 MR^2$21​MR2 > solid sphere $\tfrac25 MR^2$52​MR2. The more the mass huddles near the axis, the smaller $I$I. The spherical shell $\tfrac23 MR^2$32​MR2 exceeds the solid sphere $\tfrac25 MR^2$52​MR2 for exactly this reason.

Why $I = \int r^2\,\mathrm{d}m$I=∫r2dm — kinetic-energy motivation

The cleanest justification is energetic. A rigid body rotating at angular speed $\omega$ω has every element $\mathrm{d}m$dm at distance $r$r moving with speed $v = r\omega$v=rω, so its kinetic energy is

$$\text{KE} = \int \tfrac12 v^2\,\mathrm{d}m = \int \tfrac12 (r\omega)^2\,\mathrm{d}m = \tfrac12\omega^2\!\int r^2\,\mathrm{d}m = \tfrac12 I\omega^2 .$$KE=∫21​v2dm=∫21​(rω)2dm=21​ω2∫r2dm=21​Iω2.

Comparing with the linear $\tfrac12 mv^2$21​mv2, the quantity $\int r^2\,\mathrm{d}m$∫r2dm plays exactly the role of mass — which is what "moment of inertia" means.

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3 Derivations with mark scheme

Derivation 1 — uniform rod about its centre

Take a rod of length $L$L and mass $M$M, linear density $\rho = M/L$ρ=M/L, with its centre at the origin and the axis perpendicular at the origin. A slice at position $x$x of width $\mathrm{d}x$dx has mass $\mathrm{d}m = \rho\,\mathrm{d}x$dm=ρdx at distance $|x|$∣x∣ from the axis:

$$\begin{aligned} I &= \int_{-L/2}^{L/2} x^2 \rho\,\mathrm{d}x = \rho\left[\frac{x^3}{3}\right]_{-L/2}^{L/2} \quad (\textbf{M1 set up integral}) \\ &= \frac{M}{L}\cdot\frac{2}{3}\left(\frac{L}{2}\right)^3 = \frac{M}{L}\cdot\frac{2}{3}\cdot\frac{L^3}{8} = \frac{ML^2}{12}. \quad (\textbf{M1 A1}) \end{aligned}$$I​=∫−L/2L/2​x2ρdx=ρ[3x3​]−L/2L/2​(M1 set up integral)=LM​⋅32​(2L​)3=LM​⋅32​⋅8L3​=12ML2​.(M1 A1)​

(M1 correct integrand $x^2\rho$x2ρ with limits $\pm L/2$±L/2; M1 integrate and substitute; A1 $\tfrac{1}{12}ML^2$121​ML2. A frequent slip is integrating $0$0 to $L$L — that axis is the end, giving $\tfrac13 ML^2$31​ML2.)

Derivation 2 — uniform disc about its central axis

Treat the disc as nested rings. A ring of radius $r$r, width $\mathrm{d}r$dr, has area $2\pi r\,\mathrm{d}r$2πrdr; with surface density $\sigma = M/(\pi R^2)$σ=M/(πR2) its mass is $\mathrm{d}m = \sigma\,2\pi r\,\mathrm{d}r$dm=σ2πrdr. Every part of that ring is at distance $r$r from the central axis, so:

$$\begin{aligned} I &= \int_0^R r^2\,\mathrm{d}m = \int_0^R r^2\,\frac{M}{\pi R^2}\,2\pi r\,\mathrm{d}r \quad (\textbf{M1 ring element}) \\ &= \frac{2M}{R^2}\int_0^R r^3\,\mathrm{d}r = \frac{2M}{R^2}\cdot\frac{R^4}{4} = \frac{1}{2}MR^2. \quad (\textbf{M1 A1}) \end{aligned}$$I​=∫0R​r2dm=∫0R​r2πR2M​2πrdr(M1 ring element)=R22M​∫0R​r3dr=R22M​⋅4R4​=21​MR2.(M1 A1)​

(M1 ring mass $\sigma 2\pi r\,\mathrm{d}r$σ2πrdr; M1 integrate $r^3$r3; A1 $\tfrac12 MR^2$21​MR2. The same integral gives the solid cylinder, since every disc-slice contributes $\tfrac12 MR^2$21​MR2 and the masses add.)

Derivation 3 — solid sphere about a diameter

Slice the sphere into discs perpendicular to the axis. At height $x$x the disc has radius $y = \sqrt{R^2 - x^2}$y=R2−x2​, thickness $\mathrm{d}x$dx, volume density $\rho = \dfrac{M}{\frac{4}{3}\pi R^3}$ρ=34​πR3M​, and mass $\mathrm{d}m = \rho\,\pi y^2\,\mathrm{d}x$dm=ρπy2dx. Each disc contributes $\tfrac12\,(\mathrm{d}m)\,y^2$21​(dm)y2 (using Derivation 2):

$$\begin{aligned} I &= \int_{-R}^{R} \tfrac12 y^2\,\mathrm{d}m = \tfrac12\rho\pi\int_{-R}^{R} (R^2 - x^2)^2\,\mathrm{d}x \quad (\textbf{M1}) \\ &= \tfrac12\rho\pi\int_{-R}^{R}\!\bigl(R^4 - 2R^2x^2 + x^4\bigr)\mathrm{d}x = \tfrac12\rho\pi\left[R^4x - \tfrac{2}{3}R^2x^3 + \tfrac15 x^5\right]_{-R}^{R} \\ &= \tfrac12\rho\pi\cdot 2\left(R^5 - \tfrac23 R^5 + \tfrac15 R^5\right) = \rho\pi R^5\cdot\tfrac{8}{15}. \quad (\textbf{M1}) \end{aligned}$$I​=∫−RR​21​y2dm=21​ρπ∫−RR​(R2−x2)2dx(M1)=21​ρπ∫−RR​(R4−2R2x2+x4)dx=21​ρπ[R4x−32​R2x3+51​x5]−RR​=21​ρπ⋅2(R5−32​R5+51​R5)=ρπR5⋅158​.(M1)​

Now substitute $\rho = \dfrac{3M}{4\pi R^3}$ρ=4πR33M​:

$$I = \frac{3M}{4\pi R^3}\,\pi R^5\,\frac{8}{15} = \frac{2}{5}MR^2. \quad (\textbf{A1})$$I=4πR33M​πR5158​=52​MR2.(A1)

(M1 disc-element using the disc result; M1 expand and integrate the polynomial; A1 $\tfrac25 MR^2$52​MR2 after substituting $\rho$ρ. The bracket evaluates to $\tfrac{8}{15}R^5$158​R5: check $1 - \tfrac23 + \tfrac15 = \tfrac{15-10+3}{15} = \tfrac{8}{15}$1−32​+51​=1515−10+3​=158​.)

Derivation 4 — rectangular lamina about its central perpendicular axis

It is instructive to obtain the lamina result $\tfrac{1}{12}M(a^2+b^2)$121​M(a2+b2) from the perpendicular-axis theorem (proved in the next section) together with two one-dimensional integrals. Take a uniform rectangular lamina of side $a$a (in the $x$x-direction) and $b$b (in the $y$y-direction), surface density $\sigma = M/(ab)$σ=M/(ab), centred at the origin. The moment of inertia about the in-plane $x$x-axis treats each horizontal strip as a line of mass at height $y$y:

$$I_x = \int_{-b/2}^{b/2} y^2\,(\sigma a)\,\mathrm{d}y = \sigma a\left[\frac{y^3}{3}\right]_{-b/2}^{b/2} = \sigma a\cdot\frac{b^3}{12} = \frac{M}{ab}\cdot a\cdot\frac{b^3}{12} = \frac{1}{12}Mb^2.$$Ix​=∫−b/2b/2​y2(σa)dy=σa[3y3​]−b/2b/2​=σa⋅12b3​=abM​⋅a⋅12b3​=121​Mb2.

By the identical argument about the $y$y-axis, $I_y = \tfrac{1}{12}Ma^2$Iy​=121​Ma2. The perpendicular-axis theorem then gives the central perpendicular axis directly:

$$I_z = I_x + I_y = \tfrac{1}{12}Mb^2 + \tfrac{1}{12}Ma^2 = \frac{1}{12}M(a^2 + b^2). \quad \checkmark$$Iz​=Ix​+Iy​=121​Mb2+121​Ma2=121​M(a2+b2).✓

This single calculation simultaneously delivers all three central-axis results for the rectangle, and shows how the perpendicular-axis theorem turns two easy integrals into the harder perpendicular result for free.

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4 The two great theorems

The parallel-axis theorem

If $I_{\text{cm}}$Icm​ is the moment of inertia about an axis through the centre of mass, then about a parallel axis a distance $d$d away,

$$\boxed{\,I = I_{\text{cm}} + Md^2\,}.$$I=Icm​+Md2​.

Proof sketch. Place the centre of mass at the origin and the new axis through $(d,0)$(d,0). For an element at $(x,y)$(x,y), its squared distance from the new axis is $(x-d)^2 + y^2 = (x^2+y^2) - 2dx + d^2$(x−d)2+y2=(x2+y2)−2dx+d2. Integrating, $\int(x^2+y^2)\,\mathrm{d}m = I_{\text{cm}}$∫(x2+y2)dm=Icm​, $\int d^2\,\mathrm{d}m = Md^2$∫d2dm=Md2, and the cross term $-2d\int x\,\mathrm{d}m = 0$−2d∫xdm=0 because $\int x\,\mathrm{d}m = 0$∫xdm=0 at the centre of mass. Hence $I = I_{\text{cm}} + Md^2$I=Icm​+Md2. The theorem only works from the centre-of-mass axis — that is the entire point of the vanishing cross term.

Worked Example 1 — rod about its end, via the theorem

Find $I$I for a uniform rod (mass $M$M, length $L$L) about a perpendicular axis through one end.

$I_{\text{cm}} = \tfrac{1}{12}ML^2$Icm​=121​ML2, and the end is $d = L/2$d=L/2 from the centre. (M1)

$$I = \tfrac{1}{12}ML^2 + M\left(\tfrac{L}{2}\right)^2 = \tfrac{1}{12}ML^2 + \tfrac{1}{4}ML^2 = \tfrac{1}{3}ML^2. \quad (\textbf{M1 A1})\ \checkmark$$I=121​ML2+M(2L​)2=121​ML2+41​ML2=31​ML2.(M1 A1) ✓

This agrees with the direct integral — a perfect consistency check.

The perpendicular-axis theorem

For a flat lamina in the $xy$xy-plane, with $I_x,\,I_y$Ix​,Iy​ the moments about the two in-plane axes,

$$\boxed{\,I_z = I_x + I_y\,}\qquad(\text{lamina only}).$$Iz​=Ix​+Iy​​(lamina only).

Proof. For an element at $(x,y)$(x,y) in the plane, its distance from the $z$z-axis satisfies $r^2 = x^2 + y^2$r2=x2+y2. Integrate: $I_z = \int(x^2+y^2)\,\mathrm{d}m = \int y^2\,\mathrm{d}m + \int x^2\,\mathrm{d}m = I_x + I_y$Iz​=∫(x2+y2)dm=∫y2dm+∫x2dm=Ix​+Iy​. The step $r^2 = x^2 + y^2$r2=x2+y2 is only true for a planar body, which is why the theorem fails for solids.

Worked Example 2 — disc about a diameter

A uniform disc has $I_z = \tfrac12 MR^2$Iz​=21​MR2 about its central axis. Find $I$I about a diameter.

By symmetry the two in-plane diameters give $I_x = I_y$Ix​=Iy​. (M1) By the perpendicular-axis theorem,

$$I_z = I_x + I_y = 2I_x \ \Rightarrow\ I_x = \tfrac12 I_z = \tfrac14 MR^2. \quad (\textbf{M1 A1})$$Iz​=Ix​+Iy​=2Ix​ ⇒ Ix​=21​Iz​=41​MR2.(M1 A1)

So a disc spun about a diameter has half the moment of inertia of the same disc spun about its axis.

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5 Composite bodies and the radius of gyration

For a body built from parts, the moments of inertia about a common axis add: $I_{\text{total}} = I_1 + I_2 + \cdots$Itotal​=I1​+I2​+⋯. For a body with material removed (a hole), subtract the missing piece's moment of inertia. Combine freely with the parallel-axis theorem to shift each part to the common axis first.

Worked Example 3 — disc with a rim particle and a rod through its centre

A uniform disc (mass $4\,\text{kg}$4kg, radius $0.3\,\text{m}$0.3m) carries a $2\,\text{kg}$2kg particle on its rim and a thin rod (mass $1\,\text{kg}$1kg, length $0.6\,\text{m}$0.6m) lying along a diameter. Find $I$I about the central axis perpendicular to the disc.

$$\begin{aligned} I_{\text{disc}} &= \tfrac12(4)(0.3)^2 = \tfrac12(4)(0.09) = 0.18\ \text{kg m}^2, \quad (\textbf{M1})\\ I_{\text{particle}} &= 2(0.3)^2 = 0.18\ \text{kg m}^2,\\ I_{\text{rod}} &= \tfrac{1}{12}(1)(0.6)^2 = \tfrac{1}{12}(0.36) = 0.03\ \text{kg m}^2, \quad (\textbf{M1}) \end{aligned}$$Idisc​Iparticle​Irod​​=21​(4)(0.3)2=21​(4)(0.09)=0.18 kg m2,(M1)=2(0.3)2=0.18 kg m2,=121​(1)(0.6)2=121​(0.36)=0.03 kg m2,(M1)​

(the rod spins about a perpendicular axis through its own centre, so use $\tfrac{1}{12}ML^2$121​ML2).

$$I_{\text{total}} = 0.18 + 0.18 + 0.03 = 0.39\ \text{kg m}^2. \quad (\textbf{A1})$$Itotal​=0.18+0.18+0.03=0.39 kg m2.(A1)

(M1 each standard result correctly applied; M1 correct rod result for the central perpendicular axis; A1 sum. Common slip: using the rod's end result by mistake.)

The radius of gyration

The radius of gyration $k$k about an axis is defined by

$$I = Mk^2 \quad\Longleftrightarrow\quad k = \sqrt{\frac{I}{M}} .$$I=Mk2⟺k=MI​​.

It is the distance from the axis at which a point mass $M$M would have the same moment of inertia. For a disc, $k = \sqrt{\tfrac12 R^2} = R/\sqrt2 \approx 0.707R$k=21​R2​=R/2​≈0.707R; for a hoop, $k = R$k=R exactly (all the mass is at radius $R$R); for a solid sphere, $k = \sqrt{\tfrac25}\,R \approx 0.632R$k=52​​R≈0.632R. The radius of gyration is a compact way to quote a body's "spread" about an axis.

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6 Specimen-style exam question