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

$I$ depends on the chosen axis: the same rod is $\frac{1}{12}M{L}^2$ about its centre but $\frac{1}{3}M{L}^2$ about an end. Mass is one number; $I$ is one number per axis.

$I$ depends on where the mass sits, not just how much. A ring ($M{R}^2$) beats a disk of the same $M,R$ ($\frac{1}{2}M{R}^2$) because its mass is pushed out to the rim.

$I=\int r^{2}dm$ sums squared distances times positive mass elements, so every term is $\ge 0$. It is zero only if all mass lies exactly on the axis.

$r^2$, doubling $r$ quadruples that element's contribution. This is why "far mass costs so much."

$M$ and $R$ have the same $I$ about the long axis. ::: True. Both are $\frac{1}{2}M{R}^2$. A cylinder is just a stack of disks, and length cancels out entirely.

$I=\frac{2}{3}M{R}^2$ about their symmetry axis. ::: False. The shell cylinder has all mass at radius $R$, so it is $M{R}^2$ (like a ring). Only the hollow sphere is $\frac{2}{3}M{R}^2$, because its mass ranges from $r=0$ (poles) to $r=R$ (equator).

$M,R$), the ring has the largest $I$. ::: True. The mean-square distance $⟨r^2⟩=I/(M{R}^2)$ equals $1$ (ring), $\frac{2}{3}$ (hollow sphere), $\frac{1}{2}$ (disk), $\frac{2}{5}$ (solid sphere). Since no mass can exceed distance $R$, the largest possible $⟨r^2⟩$ is $1$ — the ring, which puts all mass exactly at $R$. Figure 2 shows these mass-distributions side by side.

$M{R}^2$, achieved when all mass is at the farthest allowed distance $R$ — that is exactly the ring/thin-shell-cylinder case.

$r$ is the distance from each point to the disk's centre." ::: Wrong. For a diameter axis, $r$ is the perpendicular drop to that line, so points on the diameter have $r=0$. Distance-to-centre only equals $r$ when the axis passes through the centre perpendicular to the plane.

$I_{\text{end}}=I_{\text{cm}}+M{d}^2$ with $d=L$ gives the rod-about-end value." ::: Wrong $d$. Parallel-axis uses $d$ = the gap between the two parallel axes. The centre-of-mass axis sits at the rod's midpoint, and the end axis at the tip; a uniform rod's midpoint is halfway along its length, so the gap is $d=L/2$. That gives $\frac{1}{12}M{L}^2+M(L/2{)}^2=\frac{1}{3}M{L}^2$.

$I$ about its long axis." ::: Wrong. About the long axis every slice is an identical disk of $\frac{1}{2}dm{R}^2$; when you sum, only total mass and $R$ survive. Length is irrelevant: $\frac{1}{2}M{R}^2$.

$I$; using rings is a mistake." ::: Both are valid. Slicing into flat rings and integrating $\int r^{2}dm$ gives the correct $\frac{2}{5}M{R}^2$; it just needs a two-variable (radius and height) geometry setup. The spherical-shell route is quicker because each shell already contributes the known $\frac{2}{3}{r}^{2}dm$ — a shortcut, not a correctness issue.

$I$ needed a hard integral because the mass is spread around a circle." ::: Wrong feeling. Every point of a ring is at the same distance $R$, so $R^2$ pulls straight out of $\int dm$, giving $M{R}^2$ with no real integration.

$I_{\text{diameter}}=2I_z$ for a disk." ::: Backwards. For a flat body, $I_z=I_x+I_y=2I_{\text{diameter}}$, so $I_{\text{diameter}}=\frac{1}{2}{I}_z=\frac{1}{4}M{R}^2$, not double. See Perpendicular Axis Theorem.

$r$ appear squared and not just to the first power? ::: A particle's speed is $v=\omega r$, and its kinetic energy is $\frac{1}{2}m{v}^2=\frac{1}{2}m{\omega }^2{r}^2$. Defining $I=\sum m{r}^2$ makes rotational KE read $\frac{1}{2}I{\omega }^2$, mirroring $\frac{1}{2}m{v}^2$. See Rotational Kinetic Energy.

$\frac{2}{5}$ smaller than the hollow sphere's $\frac{2}{3}$? ::: In a solid sphere much of the mass sits close to the centre ($r<R$), lowering the average $r^2$. A hollow sphere puts all mass on the surface, farther out on average.

half of the ring? ::: A ring has all mass at $R$; a disk spreads the same mass inward across $0\to R$, so the average $r^2$ drops. The exact integral ${\int }_0^R{r}^{3}dr$ produces the factor $\frac{1}{2}$.

$I$ (per unit $M{R}^2$) means more energy goes into spinning rather than sliding, so the object accelerates slower — a ring loses to a solid sphere. See Rolling Motion.

$I$ called the "rotational analogue of mass"? ::: In $\tau =I\alpha$ (see Torque and Angular Acceleration), $I$ plays the exact role $m$ plays in $F=ma$: it measures resistance to being angularly accelerated.

$I$ of a point mass sitting on the rotation axis? ::: Zero. Its perpendicular distance is $r=0$, so $r^{2}dm=0$. A mass on the axis contributes nothing.

$I$ about the axis running along its own length? ::: Exactly zero. Every point of a one-dimensional rod lies on the axis, so $r=0$ everywhere and $\int r^{2}dm=0$. The familiar $\frac{1}{12}M{L}^2$ applies only to an axis perpendicular to the rod.

$I$ approach? ::: It approaches $M{R}^2$, identical to a ring — because a thin shell has all its mass at the single radius $R$.

$M$ fixed), what happens to $I$? ::: $I=M{R}^2\to 0$. All mass collapses onto the axis, so there is nothing left resisting rotation — it becomes a point on the axis.

$\varphi$ with the rod. How does $I$ change as $\varphi$ goes from $0$ to $90^{\circ }$? ::: Only the perpendicular part of each element counts, so an element at distance $x$ along the rod sits at $r=x\sin \varphi$, giving $I(\varphi )=\frac{1}{12}M{L}^2{\sin }^2\varphi$. At $\varphi =0$ (axis along rod) $I=0$; at $\varphi =90^{\circ }$ (perpendicular) $I=\frac{1}{12}M{L}^2$. Figure 3 shows the same centre-point axis tilted through three angles — same $M$, same centre, different $I$.

$I$? ::: Not necessarily. Parallel-axis (see Parallel Axis Theorem) only compares parallel axes; different directions can slice the mass differently and give different $I$ (see the tilted-rod case above). The radius of gyration (see Radius of Gyration) captures this per axis.

less than about its centre-of-mass axis? ::: No, for parallel axes. Parallel-axis says $I=I_{\text{cm}}+M{d}^2\ge I_{\text{cm}}$ since $M{d}^2\ge 0$. The centre-of-mass axis always gives the minimum among axes parallel to it.