Moment of inertia I = Σmᵢrᵢ² — concept: Question bank

True or false — justify

State true/false AND the one-sentence reason. A right answer with a wrong reason is a wrong answer.

Moment of inertia is a fixed property of an object, just like its mass.

False. Mass is fixed, but

I

depends on the chosen axis — the same object has different

I

about different axes because each particle's perpendicular distance

r_{i}

changes.

I

can be negative if you choose a strange axis.

False.

I = \sum m_{i} r_{i}^{2}

is a sum of positive masses times squared distances, and a square is never negative, so

I \geq 0

always.

A particle sitting exactly on the axis of rotation still adds a little to

I

because it has mass.

False. Its perpendicular distance is

r = 0

, so its term

m r^{2} = 0

— mass on the axis contributes nothing to the moment of inertia.

Two objects of equal mass always have equal moment of inertia about their centres.

False.

I

depends on where the mass sits: a hoop (

M R^{2}

) beats a disk (

\frac{1}{2} M R^{2}

) of the same mass because the hoop keeps all its mass at maximum radius.

ω = 0

(the body is at rest) then its moment of inertia is also zero.

False.

I = \sum m_{i} r_{i}^{2}

has no

ω

in it at all —

I

is a geometry-and-mass property that exists whether or not the body is spinning.

Doubling every particle's distance from the axis doubles

I

False. Since distance is squared, doubling

r

multiplies each term by

2^{2} = 4

, so

I

becomes 4×, not 2×.

Adding mass to an object can only increase its moment of inertia, never decrease it.

True. Every added term

m r^{2} \geq 0

, so any mass you glue on can only add to the sum — you can never lower

I

by adding mass.

Moving mass closer to the axis while keeping total mass fixed lowers

I

True. Smaller

r

means smaller

r^{2}

in each term, so

I

drops — this is exactly the skater pulling arms in.

Spot the error

Each line has a mistake baked in. Name it and correct it.

"Use each particle's distance from the origin to compute

I

The error is origin — you must use the perpendicular distance to the axis line. A mass sitting far up the axis has a big position vector but

r = 0

I = \sum m_{i} r_{i}

, mass times distance, summed up."

Missing the square. That linear-in-

r

formula is the first moment (centre of mass), not inertia;

I

needs

r^{2}

because it descends from

v^{2}

in kinetic energy.

"Since

K E_{rot} = \frac{1}{2} I ω^{2}

, the moment of inertia only exists while the object spins."

I

is defined by geometry alone (

\sum m_{i} r_{i}^{2}

); the formula for kinetic energy uses

I

but does not create it. A stationary object still has a well-defined

I

"The skater speeds up because pulling her arms in gives her more energy."

Energy isn't conserved here in the naive sense — she does work pulling in. The speed-up follows from angular momentum

L = I ω

staying constant: smaller

I

forces larger

ω

I

has units of

kg m

because it's mass times a distance."

The distance is squared, so the units are ==

kg m^{2}

==, matching the

r^{2}

in the definition.

"Because

I

is a scalar, it must be the same no matter how you turn the object."

Scalar (for a fixed axis) means "a single number, not a direction," but that number still changes with the axis you pick — scalar does not mean axis-independent.

"A ring and a solid disk of the same mass and radius store the same rotational KE at the same

ω

Different

I

means different KE. The ring has

I = M R^{2}

, the disk

I = \frac{1}{2} M R^{2}

, so at equal

ω

the ring stores twice the rotational energy.

Why questions

Answer the "why," not just the "what."

Why is the distance squared rather than appearing linearly?

It is inherited: kinetic energy is

\frac{1}{2} m v^{2}

v^{2}

), and for rotation

v = ω r

, so

v^{2} = ω^{2} r^{2}

— the square on

r

comes straight from the square on

v

Why must

r_{i}

be the perpendicular distance to the axis, not the straight-line distance to a point?

Each particle travels a circle whose radius is its perpendicular reach to the axis; that circle's radius sets its speed

v = ω r

, so only the perpendicular distance controls the kinetic energy.

Why does far-away mass "count double-extra"?

A particle at distance

r

moves at speed

ω r

, and its KE goes as

(ω r)^{2}

— so being twice as far means moving twice as fast and squaring that, giving four times the resistance to spinning.

Why is

I

called the "rotational analogue of mass"?

K E = \frac{1}{2} I ω^{2}

it sits exactly where

m

sits in

K E = \frac{1}{2} m v^{2}

; it measures how hard it is to change the spin, just as mass measures how hard it is to change straight-line motion.

Why can the same object have many different moments of inertia?

Because

I

is defined relative to an axis, and moving the axis changes every

r_{i}

; the mass distribution stays but its spread relative to the new axis differs.

Why does a rigid body let us pull

\frac{1}{2} ω^{2}

out of the sum in the derivation?

"Rigid" means every particle shares the same

ω

, so

ω

is a common factor across all terms and can be factored out, leaving the pure geometric sum

\sum m_{i} r_{i}^{2}

Edge cases

Push the definition to its limits.

All the mass lies on the axis. What is

I

Every

r_{i} = 0

, so

I = \sum m_{i} \cdot 0 = 0

— there is nothing off-axis to swing, so the body offers no resistance to that particular rotation.

A single point mass, and the axis passes right through it. What is

I

I = 0

: with

r = 0

the lone term vanishes, matching intuition that a point on the axis simply spins in place.

Two axes are parallel but different; can the same mass distribution give the same

I

Only if the two axes are equidistant from the mass in the right way — generally no. The Parallel axis theorem shows

I

is smallest about the axis through the centre of mass.

As you push the axis infinitely far from a fixed lump of mass, what happens to

I

Every

r_{i} \to \infty

, so

I \to \infty

— the far-off axis makes even a tiny mass sweep an enormous circle, so it becomes infinitely hard to spin.

A massless rigid rod connects two point masses. Does the rod contribute to

I

No — with zero mass, each rod element gives

0 \cdot r^{2} = 0

; only the two point masses (

m_{1} r_{1}^{2} + m_{2} r_{2}^{2}

) count.

Does

ω

appear anywhere in the value of

I

No.

I = \sum m_{i} r_{i}^{2}

contains only masses and distances;

ω

enters kinetic energy and angular momentum, but

I

itself is spin-independent.

Can two very different-looking objects have identical

I

about some axis?

Yes —

I

collapses the whole mass distribution into one number, so a compact heavy object and a spread-out lighter one can produce the same

\sum m_{i} r_{i}^{2}

Connections

Kinetic energy ½mv² — the $v^{2}$ that forces the $r^{2}$
Rotational kinetic energy ½Iω² — where $I$ plays mass's role
Angular velocity ω — the common factor in a rigid body
Centre of mass (first moment Σmr) — the linear-in- $r$ impostor
Parallel axis theorem — how $I$ shifts with the axis
Angular momentum L = Iω — why the skater speeds up
Torque τ = Iα — Newton's 2nd law for rotation