Moment of inertia

At a Glance

• Frequency: 7 sub-parts across 7 of 14 years (2013, 2015, 2017, 2020, 2022, 2024, 2025)
• Priority tier: T2
• Marks (count): 10 (4), 15 (3)
• Average solve time: ~11 min
• Difficulty mix: easy 2, medium 4, hard 1
• Section: B | Dominant type: computation

Why This Chapter Matters

Moment of inertia appears in 7 of the last 14 years — consistently in the compulsory part (Q5e) or the optional Section B problems — making it one of the most reliable Paper 2 computation atoms. The range is broad: spheres about a diagonal of a square (2013), hemispheres by integration (2015), elliptic laminas via the pedal relation (2017), triangular laminas via area coordinates (2020), the hard cone-about-slant-side via the inertia tensor (2022), an elliptic disc with non-uniform density (2024), and a cylinder optimization (2025). The unifying skill is recognising which tool applies: coordinate integration for volumes with clean boundaries, the parallel-axis theorem for shifted axes, the perpendicular-axis theorem for flat laminas, and the inertia tensor for arbitrary axes through a symmetric body.

Minimum Theory

Definition. The moment of inertia of a body of mass $M$ about an axis $\ell$ is $I = \int r^2\,dm,$ where $r$ is the perpendicular distance from each mass element $dm$ to $\ell$.

Standard results (memorise these):

Parallel-axis theorem. If $I_G$ is the moment of inertia about an axis through the centre of mass $G$, then the moment of inertia about any parallel axis at distance $d$ is

$\boxed{I = I_G + Md^2.}$

Perpendicular-axis theorem (laminas only). For a flat lamina in the $xy$-plane: $I_z = I_x + I_y,$ where $I_x,I_y$ are moments about the $x$- and $y$-axes lying in the lamina’s plane, and $I_z$ is the moment about the $z$-axis perpendicular to the lamina.

Inertia tensor. For a body with a symmetry axis, the inertia tensor at a point on that axis is diagonal. For a cone with apex at the origin and symmetry axis $z$: $\mathbf{I}=\begin{pmatrix}I_x&0&0\\0&I_x&0\\0&0&I_z\end{pmatrix}.$ The moment of inertia about an axis with unit direction $\hat n=(n_1,n_2,n_3)$ through the apex is $I_{\hat n}=\hat n^T\mathbf{I}\hat n=n_1^2 I_x+n_2^2 I_x+n_3^2 I_z.$

Setting up the integral. For a uniform body of density $\rho$:

• Spherical: $dV=r^2\sin\phi\,dr\,d\phi\,d\theta$; $r^2_\perp=(x^2+y^2)=r^2\sin^2\phi$ for the $z$-axis.
• Cylindrical: $dV=s\,ds\,d\phi\,dz$; $r^2_\perp=s^2$ for the $z$-axis, $r^2_\perp=y^2+z^2=s^2\sin^2\phi+z^2$ for the $x$-axis.
• Elliptic: substitute $x=as\cos\phi$, $y=bs\sin\phi$ (Jacobian $abs$) for regions bounded by an ellipse.

Question Archetypes

All seven questions share the single archetype moment-of-inertia. The technique varies:

moment-of-inertia (7 question(s); 2013, 2015, 2017, 2020, 2022, 2024, 2025)

Recognition Cues

• “Find the moment of inertia of [body] about [axis].”
• “Prove that the MI is [expression] involving mass and geometric parameters.”
• “Find [ratio] for which the MI is minimum for a given mass.”

Solution Template

Parallel-axis theorem (2013 type — discrete system):

1. Identify the axis. Find the perpendicular distance $d_k$ from each component’s CM to the axis.
2. $I=\sum_k(I_{G_k}+m_k d_k^2)$.

Spherical integration (2015 type):

1. Write $\rho=M/V$. Set up $I_z=\rho\iiint r^2\sin^2\phi\cdot r^2\sin\phi\,dr\,d\phi\,d\theta$ in spherical coordinates.
2. Factor: $I_z=\rho\cdot(2\pi)\cdot(\int_0^{r_{\max}}r^4\,dr)\cdot(\int_0^{\phi_{\max}}\sin^3\phi\,d\phi)$.
3. Evaluate $\int\sin^3\phi\,d\phi$ via $u=\cos\phi$.

Elliptic lamina — general diameter (2017 type):

1. Write $I_\alpha=\sin^2\alpha\,I_{Oy}+\cos^2\alpha\,I_{Ox}=\tfrac{M}{4}(a^2\sin^2\alpha+b^2\cos^2\alpha)$ using the vanishing product of inertia.
2. Apply the ellipse pedal relation $1/r^2=\cos^2\alpha/a^2+\sin^2\alpha/b^2$ to convert $a^2\sin^2\alpha+b^2\cos^2\alpha=a^2b^2/r^2$.
3. For the tangent: use the parallel central diameter (at perpendicular distance $p$) plus the parallel-axis theorem.

Area-coordinate integration (2020 type):

1. Parametrise: $\vec{AP}=u\vec{AB}+v\vec{AC}$ with $u,v\ge0$, $u+v\le1$. Mass element $dm=2M\,du\,dv$.
2. Perpendicular distance from axis through $A$: $\delta=u\beta+v\gamma$ (linear in $u,v$).
3. $I=2M\iint_{u+v\le1}(u\beta+v\gamma)^2\,du\,dv$. Use $\iint u^2=\iint v^2=1/12$ and $\iint uv=1/24$.

Inertia tensor (2022 type):

1. Exploit symmetry: inertia tensor at the apex is diagonal with $I_x=I_y=I_{\text{apex,perp}}$, $I_z=I_{\text{axis}}$.
2. Slant direction $\hat n=(a,0,h)/\sqrt{a^2+h^2}$: apply $I_{\hat n}=n_1^2 I_x+n_3^2 I_z$.

Cylindrical integration (2025 type):

1. Set up $I_{zz}=\rho\iiint s^2\cdot s\,ds\,d\phi\,dz$ and $I_{xx}=\rho\iiint(s^2\sin^2\phi+z^2)\cdot s\,ds\,d\phi\,dz$.
2. For the optimisation: substitute the volume constraint $V=\pi R^2 L=\text{const}$ to eliminate $R$; differentiate $I_{xx}(L)$ and set to zero.

Worked Example(s)

2013 Paper 2, 2013-P2-Q5e (10 marks)

Four solid spheres $A,B,C,D$, each of mass $m$ and radius $a$, placed at the corners of a square of side $b$. Find the MI about a diagonal.

Place the square with corners $A=(b/2,b/2)$, $B=(-b/2,b/2)$, $C=(-b/2,-b/2)$, $D=(b/2,-b/2)$. Diagonal $AC$ lies along $y=x$.

Perpendicular distance from each corner to $y=x$:

• $A,C$: on the diagonal, $d=0$.
• $B,D$: distance $=|b/2-(-b/2)|/\sqrt2=b/\sqrt2$, so $d^2=b^2/2$.

MI of each solid sphere about any axis through its CM: $I_G=\tfrac{2}{5}ma^2$.

By the parallel-axis theorem: $I=\underbrace{2\cdot\frac{2}{5}ma^2}_{A,C}+\underbrace{2\cdot\left(\frac{2}{5}ma^2+m\cdot\frac{b^2}{2}\right)}_{B,D}=\frac{8ma^2}{5}+mb^2.$

$\boxed{\;I=m\!\left(\frac{8a^2}{5}+b^2\right).\;}$

Common Traps

• The off-diagonal spheres have perpendicular distance $b/\sqrt2$ to the diagonal, so $d^2=b^2/2$ — not $b^2$.
• A sphere’s MI through its CM is the same for any axis (spherical symmetry), so the parallel-axis theorem applies without choosing a specific axis direction.
• Don’t forget the two on-diagonal spheres — they contribute $\tfrac{2}{5}ma^2$ each, not zero.

2015 Paper 2, 2015-P2-Q5e (10 marks)

Find the MI of a solid uniform hemisphere $x^2+y^2+z^2=a^2$, $z\ge0$, mass $m$, about the $z$-axis.

Density $\rho=m/V=m/(\tfrac{2}{3}\pi a^3)=3m/(2\pi a^3)$.

In spherical coordinates, $x^2+y^2=r^2\sin^2\phi$: $I_z=\rho\int_0^{2\pi}\!d\theta\int_0^{\pi/2}\!\int_0^a r^2\sin^2\phi\cdot r^2\sin\phi\,dr\,d\phi.$

Factor: $I_z=\rho\cdot(2\pi)\cdot\frac{a^5}{5}\cdot\int_0^{\pi/2}\sin^3\phi\,d\phi.$

For the angular integral, substitute $u=\cos\phi$: $\int_0^{\pi/2}\sin^3\phi\,d\phi=\int_0^1(1-u^2)\,du=1-\tfrac{1}{3}=\tfrac{2}{3}.$

$I_z=\frac{3m}{2\pi a^3}\cdot2\pi\cdot\frac{a^5}{5}\cdot\frac{2}{3}=\frac{3m}{a^3}\cdot\frac{2a^5}{15}=\frac{2ma^2}{5}.$

$\boxed{\;I_z=\frac{2}{5}ma^2.\;}$

The result equals the MI of a full sphere of the same mass — expected by $z$-symmetry.

Common Traps

• In spherical coordinates, the square distance from the $z$-axis is $r^2\sin^2\phi$, not $r^2$.
• The upper hemisphere has $\phi\in[0,\pi/2]$, not $[0,\pi]$.
• $\int_0^{\pi/2}\sin^3\phi\,d\phi=2/3$ via $u=\cos\phi$: write $\sin^3\phi=\sin\phi(1-\cos^2\phi)$.

2017 Paper 2, 2017-P2-Q5e (10 marks)

Show the MI of an elliptic area, mass $M$, semi-axes $a,b$, about a semi-diameter of length $r$ is $\tfrac{1}{4}M\tfrac{a^2b^2}{r^2}$. Prove the MI about a tangent is $\tfrac{5M}{4}p^2$, where $p$ is the distance from the centre to the tangent.

Standard moments. For a uniform elliptic lamina: $I_{Ox}=\frac{Mb^2}{4},\quad I_{Oy}=\frac{Ma^2}{4},\quad I_{xy}=0.$

MI about a central diameter. The diameter in direction $\alpha$ has perpendicular distance $|x\sin\alpha-y\cos\alpha|$ from a general point. Since $I_{xy}=0$: $I_\alpha=\sin^2\alpha\cdot I_{Oy}+\cos^2\alpha\cdot I_{Ox}=\frac{M}{4}(a^2\sin^2\alpha+b^2\cos^2\alpha).$

The semi-diameter length $r$ in direction $\alpha$ satisfies the ellipse equation: $r^2\cos^2\alpha/a^2+r^2\sin^2\alpha/b^2=1$, giving $\frac{1}{r^2}=\frac{b^2\cos^2\alpha+a^2\sin^2\alpha}{a^2b^2}\quad\Longrightarrow\quad a^2\sin^2\alpha+b^2\cos^2\alpha=\frac{a^2b^2}{r^2}.$

Therefore: $\boxed{\;I_{\text{diameter}}=\frac{M}{4}\cdot\frac{a^2b^2}{r^2}.\;}$

MI about a tangent. Let the tangent be parallel to the central diameter with semi-length $r'$, at perpendicular distance $p$ from the centre. The MI about the parallel central diameter is $\tfrac{M}{4}a^2b^2/r'^2=\tfrac{M}{4}p^2$ (since $r'^2=a^2b^2/p^2$ by the same pedal relation). By the parallel-axis theorem: $I_{\text{tangent}}=\frac{M}{4}p^2+Mp^2=\boxed{\frac{5M}{4}p^2.}$

Common Traps

• $I_{Ox}=Mb^2/4$ (spread in $y$, semi-axis $b$) and $I_{Oy}=Ma^2/4$; swapping $a$ and $b$ is the most common error.
• The central diameter parallel to the tangent has direction equal to the tangent direction, not the conjugate diameter. Get the direction right before applying the pedal relation.
• The $\tfrac{5}{4}=\tfrac{1}{4}+1$ mnemonic: $\tfrac{1}{4}$ from the central-diameter MI, plus $1$ from the parallel-axis shift $Mp^2$.

2020 Paper 2, 2020-P2-Q5e (10 marks)

Prove the MI of a triangular lamina $ABC$ about any axis through $A$ in its plane is $\tfrac{M}{6}(\beta^2+\beta\gamma+\gamma^2)$, where $\beta,\gamma$ are the perpendicular distances from $B,C$ to the axis.

Area-coordinate setup. Write $\vec{AP}=u\vec{AB}+v\vec{AC}$ for a general point $P$, with $u,v\ge0$, $u+v\le1$. The mass element is $dm=2M\,du\,dv$ (since the Jacobian of the map to Cartesian is $2\Delta$, the triangle area, and $\sigma=M/\Delta$, giving $dm=\sigma\cdot2\Delta\,du\,dv=2M\,du\,dv$; check: $\int_0^1\int_0^{1-u}2M\,dv\,du=M$ ✓).

Perpendicular distance. The distance from $P$ to the axis through $A$ is linear in position and vanishes at $A$ (where $u=v=0$), equals $\beta$ at $B$ (where $u=1,v=0$), and equals $\gamma$ at $C$ (where $u=0,v=1$): $\delta(u,v)=u\beta+v\gamma.$

Integration. Standard moments over the unit simplex $\{u,v\ge0,\,u+v\le1\}$: $\iint u^2\,du\,dv=\frac{1}{12},\quad \iint v^2\,du\,dv=\frac{1}{12},\quad \iint uv\,du\,dv=\frac{1}{24}.$

$I=2M\iint(u\beta+v\gamma)^2\,du\,dv=2M\!\left(\beta^2\cdot\frac{1}{12}+2\beta\gamma\cdot\frac{1}{24}+\gamma^2\cdot\frac{1}{12}\right)=\frac{M}{6}(\beta^2+\beta\gamma+\gamma^2).$

$\boxed{\;I=\frac{M}{6}(\beta^2+\beta\gamma+\gamma^2).\;}$

Common Traps

• The mass element $dm=2M\,du\,dv$: the factor $2$ comes from $\sigma\cdot2\Delta$ (not $M\,du\,dv$).
• The cross term contributes $\beta\gamma$: the integral $\iint uv=1/24$, and $2\beta\gamma\cdot 2M\cdot1/24 = M\beta\gamma/6$ ✓.
• Verify the simplex integrals: $\int_0^1u^2(1-u)\,du=1/3-1/4=1/12$.

2022 Paper 2, 2022-P2-Q6c (15 marks)

Find the MI of a right circular solid cone (mass $M$, height $h$, base radius $a$) about one of its slant sides.

Place the cone with apex at the origin, axis along $z$, base at $z=h$.

Inertia tensor at the apex. By the cone’s rotational symmetry: $\mathbf{I}=\begin{pmatrix}I_x&0&0\\0&I_x&0\\0&0&I_z\end{pmatrix},\quad I_z=\frac{3Ma^2}{10},\quad I_x=I_y=\frac{3M(a^2+4h^2)}{20}.$

Slant direction. A slant side runs from the apex $(0,0,0)$ to the base point $(a,0,h)$; the unit direction is $\hat n=(a,0,h)/\sqrt{a^2+h^2}$.

MI formula. $I_{\hat n}=n_x^2 I_x+n_z^2 I_z$ (since $n_y=0$): $I_{\hat n}=\frac{a^2}{a^2+h^2}\cdot\frac{3M(a^2+4h^2)}{20}+\frac{h^2}{a^2+h^2}\cdot\frac{3Ma^2}{10}.$

Combine over the common denominator $20(a^2+h^2)$: $I_{\hat n}=\frac{3M}{20(a^2+h^2)}\!\left[a^2(a^2+4h^2)+2a^2h^2\right]=\frac{3Ma^2(a^2+6h^2)}{20(a^2+h^2)}.$

$\boxed{\;I_{\text{slant}}=\frac{3Ma^2(a^2+6h^2)}{20(a^2+h^2)}.\;}$

Common Traps

• $I_z=3Ma^2/10$ (axis) and $I_x=3M(a^2+4h^2)/20$ (transverse through apex) — do not confuse or swap.
• The slant direction from the apex is $(a,0,h)/\sqrt{a^2+h^2}$; choose coordinates so $n_y=0$.
• Final numerator: $a^2(a^2+4h^2)+2a^2h^2=a^2(a^2+4h^2+2h^2)=a^2(a^2+6h^2)$.

2024 Paper 2, 2024-P2-Q6c (15 marks)

Find the MI of a quadrant of the elliptic disc $x^2/a^2+y^2/b^2\le1$ (first quadrant, mass $M$, density $\propto xy$) about the axis through the centre perpendicular to the plane.

Elliptic-coordinate substitution. Set $x=ar\cos\theta$, $y=br\sin\theta$ for $r\in[0,1]$, $\theta\in[0,\pi/2]$. Jacobian $|J|=ab\cdot r$. Density $\rho(x,y)=kxy$ for some constant $k$.

Find $k$ from total mass: $M=\int_0^{\pi/2}\!\!\int_0^1 k(ar\cos\theta)(br\sin\theta)\cdot ab\,r\,dr\,d\theta=ka^2b^2\int_0^{\pi/2}\!\!\frac{\sin2\theta}{2}\,d\theta\int_0^1\!r^3\,dr=ka^2b^2\cdot\frac{1}{2}\cdot\frac{1}{4}=\frac{ka^2b^2}{8}.$ Hence $k=8M/(a^2b^2)$.

Compute $I$ ($r^2_\perp=x^2+y^2=r^2(a^2\cos^2\theta+b^2\sin^2\theta)$): $I=k\cdot\frac{a^2b^2}{2}\int_0^1 r^5\,dr\int_0^{\pi/2}(a^2\cos^2\theta+b^2\sin^2\theta)\sin2\theta\,d\theta.$

For the angular integral, substitute $u=\sin^2\theta$: $\int_0^{\pi/2}(a^2\cos^2\theta+b^2\sin^2\theta)\sin2\theta\,d\theta=\int_0^1(a^2(1-u)+b^2 u)\,du=a^2+\frac{b^2-a^2}{2}=\frac{a^2+b^2}{2}.$

$I=\frac{8M}{a^2b^2}\cdot\frac{a^2b^2}{2}\cdot\frac{1}{6}\cdot\frac{a^2+b^2}{2}=\frac{M(a^2+b^2)}{3}.$

$\boxed{\;I=\frac{M(a^2+b^2)}{3}.\;}$

Common Traps

• Use elliptic coordinates immediately — Cartesian integration over the elliptic region is very difficult.
• The Jacobian is $ab\cdot r$ (not $r$ alone); the factor $ab$ scales the area of the ellipse relative to the unit disc.
• Density is $kxy$, not $k(x^2+y^2)$; keep the density and MI distance-squared factors separate.

2025 Paper 2, 2025-P2-Q6c (15 marks)

Find the MI of a uniform solid cylinder (mass $M$, radius $R$, length $L$) about the cylinder axis ($z$) and a transverse axis through the centre ($x$). Find $L/R$ minimising the transverse MI for given mass.

Cylinder occupies $s\in[0,R]$, $\phi\in[0,2\pi]$, $z\in[-L/2,L/2]$. Density $\rho=M/(\pi R^2 L)$.

$I_{zz}$: distance from $z$-axis is $s$: $I_{zz}=\rho\cdot2\pi\cdot L\cdot\frac{R^4}{4}=\frac{\pi\rho LR^4}{2}=\boxed{\frac{1}{2}MR^2.}$

$I_{xx}$: distance from $x$-axis is $y^2+z^2=s^2\sin^2\phi+z^2$. Split: $\int s^2\sin^2\phi\,dV=\rho\cdot\frac{R^4}{4}\cdot\pi\cdot L=\frac{\pi\rho LR^4}{4},\quad \int z^2\,dV=\rho\cdot\frac{R^2}{2}\cdot2\pi\cdot\frac{L^3}{12}=\frac{\pi\rho R^2 L^3}{12}.$ Substituting $\rho$: $\boxed{I_{xx}=I_{yy}=\frac{M}{12}(3R^2+L^2).}$

Minimisation. “Given mass” with uniform density means given volume $V=\pi R^2 L$. Write $R^2=V/(\pi L)$: $I_{xx}(L)=\frac{M}{12}\!\left(\frac{3V}{\pi L}+L^2\right).$ Differentiate and set to zero: $\frac{dI_{xx}}{dL}=\frac{M}{12}\!\left(-\frac{3V}{\pi L^2}+2L\right)=0\;\Longrightarrow\;L^3=\frac{3V}{2\pi}=\frac{3R^2 L}{2}.$ Hence $L^2=\tfrac{3}{2}R^2$: $\boxed{\;\frac{L}{R}=\sqrt{\frac{3}{2}}=\frac{\sqrt6}{2}\approx1.22.\;}$

The second derivative is positive, confirming a minimum.

Common Traps

• $I_{xx}=$ MI about the $x$-axis uses $y^2+z^2$ (not $x^2+y^2$) as the square distance.
• The constraint is fixed volume (from fixed mass and constant density), not fixed radius or length separately.
• The optimization gives $L^2=3R^2/2$, not $L=R$ or $L=\sqrt3 R$.

Marks-Aware Writing

10-mark Section B questions (2013, 2015, 2017, 2020): These are compulsory (Q5e) short answers. One clear diagram or coordinate setup, the key formula, and the final boxed answer. For the parallel-axis theorem: table of distances is cleaner than prose. For integrals: display the setup once, factor the integrals, substitute the result.

15-mark Section B questions (2022, 2024, 2025): Structured in two or three parts. For the cone: state the inertia tensor components (with the standard formulas cited), write the slant direction, and compute $\hat n^T\mathbf{I}\hat n$ explicitly. For the cylinder: set up both integrals, then do the optimisation as a separate labelled step — this is a two-part question in disguise.

Practice Set