Exercises — Moment of inertia of - rod (about end, centre), disk, ring, sphere (solid, hollow), cylinder

WHAT body, WHAT axis? A ring about its central perpendicular axis. Every scrap of mass sits at the same perpendicular distance $R$ from the axis, so $r=R$ for all of it — that is exactly the case $I_{\text{ring}}=M{R}^2$. $I=M{R}^2=(2)(0.5{)}^2=2\times 0.25=0.5\mathrm{kg}{\mathrm{m}}^2$

Which of the two spheres? Solid, so "Full Five" → $I=\frac{2}{5}M{R}^2$. $I=\frac{2}{5}(3)(0.2{)}^2=0.4\times 3\times 0.04=0.048\mathrm{kg}{\mathrm{m}}^2$

End axis → $I=\frac{1}{3}M{L}^2$ (mass sits, on average, farther from an end than from the centre). $I=\frac{1}{3}(1.2)(1{)}^2=0.4\mathrm{kg}{\mathrm{m}}^2$

Step 1 — centre value. $I_{\text{cm}}=\frac{1}{12}M{L}^2=\frac{1}{12}(1.2)(1{)}^2=0.1\mathrm{kg}{\mathrm{m}}^2.$ Step 2 — shift by the theorem. The parallel-axis theorem says $I=I_{\text{cm}}+M{d}^2$, where $d$ is the gap between the two parallel axes. Centre-to-end is $d=L/2=0.5\mathrm{m}$. $I_{\text{end}}=0.1+(1.2)(0.5{)}^2=0.1+0.3=0.4\mathrm{kg}{\mathrm{m}}^2$ This matches Solution 1.3 exactly. ✓ The theorem adds $M{d}^2$ because moving the axis away from the centre of mass always makes the body harder to spin.

Step 1 — $I$. Disk about central axis: $I=\frac{1}{2}M{R}^2=\frac{1}{2}(0.8)(0.3{)}^2=0.036\mathrm{kg}{\mathrm{m}}^2.$ Step 2 — energy. Rotational KE mirrors $\frac{1}{2}m{v}^2$ but with $I$ for $m$ and $\omega$ for $v$: $KE=\frac{1}{2}I{\omega }^2=\frac{1}{2}(0.036)(10{)}^2=0.5\times 0.036\times 100=1.8\mathrm{J}$

The theorem. For any flat (planar) body, $I_z=I_x+I_y$, where $z$ is the perpendicular axis and $x,y$ are two in-plane axes through the same point. Look at the figure: the two diameters $x$ and $y$ are interchangeable by symmetry, so $I_x=I_y\equiv I_d$.

Solve. We know $I_z=\frac{1}{2}M{R}^2$ (central perpendicular axis). Then $I_z=I_x+I_y=2I_d\Rightarrow I_d=\frac{1}{2}{I}_z=\frac{1}{4}M{R}^2$ So a disk about its diameter is $\boxed{\frac{1}{4}M{R}^2}$ — half of the perpendicular-axis value.

Additivity. Moment of inertia adds: total $I$ = rod's $I$ + each point mass's $I$, all about the same axis.

• Rod: $I_{\text{rod}}=\frac{1}{12}M{L}^2=\frac{1}{12}(1)(1.2{)}^2=\frac{1.44}{12}=0.12\mathrm{kg}{\mathrm{m}}^2.$
• Each point mass sits at $r=L/2=0.6\mathrm{m}$: contribution $m{r}^2=0.5(0.6{)}^2=0.18\mathrm{kg}{\mathrm{m}}^2$. Two of them: $0.36$. $I_{\text{total}}=0.12+0.36=0.48\mathrm{kg}{\mathrm{m}}^2$

Compare the formulas. Hollow $=\frac{2}{3}M{R}^2$, solid $=\frac{2}{5}M{R}^2$. Since $\frac{2}{3}>\frac{2}{5}$, the hollow shell is harder to spin — its mass all sits at the outer radius $R$, giving the largest possible average $r^2$; the solid sphere hides much of its mass near the centre where $r<R$. Percentage. $\frac{I_{\text{hollow}}}{I_{\text{solid}}}=\frac{2/3}{2/5}=\frac{5}{3}\approx 1.667$ So the hollow sphere's $I$ is larger by $\frac{5}{3}-1=\frac{2}{3}\approx 66.7\%$.

Definition. The radius of gyration $k$ is the single distance at which you could put all the mass to reproduce the real $I$: $I=M{k}^2\Rightarrow k=\sqrt{I/M}$.

• Cylinder: $M{k}^2=\frac{1}{2}M{R}^2\Rightarrow k^2=\frac{1}{2}{R}^2\Rightarrow k/R=\frac{1}{\sqrt{2}}\approx \mathrm{0.707.}$
• Ring: $M{k}^2=M{R}^2\Rightarrow k/R=1.$ The ring's mass is effectively all at the rim ($k=R$); the cylinder's is bunched inward ($k\approx 0.71R$), so it wins the roll-down race.

Direct integration. Put the ring in the $xy$-plane, radius $R$, and choose the $x$-axis as our rotation axis. A mass element at angle $\varphi$ sits at position $(R\cos \varphi ,R\sin \varphi )$. Its perpendicular distance to the $x$-axis is the $y$-coordinate: $r=R\sin \varphi$.

With $dm=\frac{M}{2\pi }d\varphi$ (mass evenly spread around the circle): $I_d={\int }_0^{2\pi }(R\sin \varphi {)}^2\frac{M}{2\pi }d\varphi =\frac{M{R}^2}{2\pi }{\int }_0^{2\pi }{\sin }^2\varphi d\varphi$ Using ${\int }_0^{2\pi }{\sin }^2\varphi d\varphi =\pi$: $I_d=\frac{M{R}^2}{2\pi }\cdot \pi =\frac{1}{2}M{R}^2$ Check with perpendicular-axis theorem. For the flat ring, $I_z=I_x+I_y=2I_d$. Since $I_z=M{R}^2$, we get $I_d=\frac{1}{2}M{R}^2$. ✓ Both agree.

Step 1 — central value. Disk, central perpendicular axis: $I_{\text{cm}}=\frac{1}{2}M{R}^2=\frac{1}{2}(1)(0.2{)}^2=0.02\mathrm{kg}{\mathrm{m}}^2.$ Step 2 — shift to the rim. The tangent axis is parallel to the central axis, offset by $d=R=0.2\mathrm{m}$. Parallel-axis theorem: $I=I_{\text{cm}}+M{d}^2=0.02+(1)(0.2{)}^2=0.02+0.04=0.06\mathrm{kg}{\mathrm{m}}^2$ Equivalently $I=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$ — a neat closed form worth remembering.

Insight. Energy $=\frac{1}{2}I{\omega }^2$, and with $M,R,\omega$ fixed, the winner is simply whichever has the larger $I$. The ring puts every gram at $R$ (maximum $r^2$), so it must win.

• Disk: $I=\frac{1}{2}M{R}^2=\frac{1}{2}(4)(0.25{)}^2=0.125\mathrm{kg}{\mathrm{m}}^2.$ $E_{\text{disk}}=\frac{1}{2}(0.125)(50{)}^2=\frac{1}{2}(0.125)(2500)=156.25\mathrm{J}$
• Ring: $I=M{R}^2=(4)(0.25{)}^2=0.25\mathrm{kg}{\mathrm{m}}^2.$ $E_{\text{ring}}=\frac{1}{2}(0.25)(2500)=312.5\mathrm{J}$ The ring stores exactly twice the energy. Design note: real flywheels do push mass to the rim, but material strength limits how much rim stress you can survive at high $\omega$ — a genuine engineering trade-off beyond pure $I$.

Rod part (about its own end axis): $I_{\text{rod}}=\frac{1}{3}{M}_r{L}^2=\frac{1}{3}(1)(0.6{)}^2=\frac{0.36}{3}=0.12\mathrm{kg}{\mathrm{m}}^2.$ Sphere part. The sphere spins about a far axis, so use parallel-axis: its own $I_{\text{cm}}=\frac{2}{5}{M}_s{R}^2$ plus $M_s{d}^2$ with $d=L=0.6\mathrm{m}$. $I_{\text{cm}}=\frac{2}{5}(2)(0.1{)}^2=0.4\times 2\times 0.01=0.008\mathrm{kg}{\mathrm{m}}^2$ $M_s{d}^2=(2)(0.6{)}^2=0.72\mathrm{kg}{\mathrm{m}}^2\Rightarrow I_{\text{sphere}}=0.008+0.72=0.728$ Total. $I_{\text{total}}=0.12+0.728=0.848\mathrm{kg}{\mathrm{m}}^2$ Notice the sphere's own spin term ($0.008$) is tiny next to the $M_s{d}^2$ transport term ($0.72$) — when a compact body sits far from the axis, it behaves almost like a point mass.

Parallel-axis always increases $I$ ::: because $M{d}^2\ge 0$; moving off the CM never helps. Perpendicular-axis works only for ::: thin flat (planar) bodies, since it needs $z=0$. Ring vs disk flywheel energy ratio ::: $2:1$ (ring stores twice, all mass at $R$). Radius of gyration $k$ ::: the distance where all mass could sit to give the same $I$: $I=M{k}^2$.