Question bank — Parallel axis theorem — I = I_CM + Md² — proof

False. One axis must pass through the CM. The cross terms $-2a\sum m_i{x}_i$ only vanish when the origin is the CM; between two off-CM axes they survive and the clean formula breaks.

True. Since $I=I_{CM}+M{d}^2$ and $M{d}^2\ge 0$, any parallel axis adds a non-negative amount. The minimum occurs at $d=0$, i.e. the CM axis itself.

False. The correction is $M{d}^2$, so doubling $d$ quadruples it ($M(2d{)}^2=4M{d}^2$). Distance enters squared — moving out is punished sharply.

False. $I_{CM}$ is a fixed property of the body about its own CM axis. Only the added term $M{d}^2$ changes as you move the outer axis; $I_{CM}$ never does.

True. $I=I_{CM}+M{d}^2$ depends only on $I_{CM}$, $M$, and $d$ — nothing about the detailed shape once those three match.

False. The derivation used only $\sum m_i{x}_i=0$ at the CM, which holds for any rigid body of any shape. No symmetry is assumed.

True. The proof ignored the $z_i$ coordinate entirely (distance to a $z$-axis uses only $x,y$). The theorem is fully 3D; the axis just has to be parallel to the CM axis.

The error is $d$. The CM of a uniform rod is at its midpoint, so the end is only $d=L/2$ away. Correct: $I_{end}=\frac{1}{12}M{L}^2+M(L/2{)}^2=\frac{1}{3}M{L}^2$.

Wrong — you cannot connect two off-CM axes directly. Route through the CM: $I_A=I_{CM}+M{R}^2$, $I_B=I_{CM}+M(2R{)}^2$, so $I_B-I_A=3M{R}^2$, not $M{R}^2$.

$d$ is not measured from an arbitrary point. It is the perpendicular distance between the two parallel axes, equivalently CM-to-outer-axis. A corner is irrelevant unless the axis passes through it.

The theorem is only valid for parallel axes. If the axes are tilted relative to each other, the whole $r_i$ geometry changes and $I=I_{CM}+M{d}^2$ does not apply at all.

Impossible. $I_{CM}$ is the minimum, so no parallel axis gives less. If you got something smaller, you either mislabelled which term is $I_{CM}$ or used the theorem between two non-CM axes.

That $\frac{1}{2}M{R}^2$ is $I_{CM}$ (central axis). The rim axis is $d=R$ away, so $I_{rim}=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$. Don't forget the $M{d}^2$.

Because the origin was placed at the CM, and by the very definition of CM, $x_{CM}=\frac{\sum m_i{x}_i}{M}=0$, forcing $\sum m_i{x}_i=0$ (same for $y$). This is the sole reason the formula is clean.

You need one to make the cross terms vanish. Two parallel axes generally can't both pass through the CM (only one line through the CM has that fixed direction), and zero would leave the ugly cross terms in.

Because $M{d}^2$ is literally the moment of inertia of a single point of mass $M$ at perpendicular distance $d$ from the axis. The theorem splits inertia into "spread about CM" plus "whole mass as a point at the CM."

The added term is $M{d}^2$, and both $M>0$ and $d^2\ge 0$, so it can only add. The only way to reduce $I$ toward its minimum is to move the axis toward the CM.

The distance from a point to a $z$-axis depends only on its $x$ and $y$ (it's $\sqrt{x^2+y^2}$). Sliding a point along $z$ doesn't change its distance to a vertical axis, so $z_i$ never enters $I$.

Because $I_{CM}$ for standard shapes (rod, disc, sphere) is tabulated once, and the theorem lets you reach any parallel axis by adding one easy term $M{d}^2$ — no re-integration needed. This is why it underpins the Physical pendulum and Radius of gyration.

$I=I_{CM}+0=I_{CM}$. The "outer" axis coincides with the CM axis — the theorem gracefully reduces to the trivial identity, confirming the CM value is the floor.

A point mass has $I_{CM}=0$ about any axis through itself (all its mass is at distance zero). So $I=0+M{d}^2=M{d}^2$, recovering the point-mass formula directly — the theorem's building block.

No. $I_{CM}=\sum m_i{r}_i^2$ is a sum of masses times squares, so it is always $\ge 0$. Adding $M{d}^2\ge 0$ keeps $I$ non-negative for every axis.

The correction $M{d}^2\to 0$, so $I\to I_{CM}$, which is itself $0$ for zero mass. Everything degenerates to zero inertia — consistent, since there's no mass to resist rotation.

Yes. The CM is a geometric point defined by $\sum m_i{\vec{r}}_i=0$; it need not lie inside the material. The proof only used that mathematical property, so the theorem applies unchanged.

Formally $I=I_{CM}+M{d}^2\to \infty$ as $d\to \infty$. Physically sensible: swinging the whole mass at an enormous radius demands enormous rotational inertia — the $M{d}^2$ term dominates and $I_{CM}$ becomes negligible.