Applications — mass, centre of mass, moments of inertia: Question bank

True or false — justify

Each claim is a sentence someone believes. Decide true/false and say why — the justification is the point.

The centre of mass of a lamina always lies inside the region.

False. For a non-convex or ring-shaped region (e.g. an L-shape or annulus) the balance point can fall in empty space — it is the average position weighted by mass, not a physical material point.

ρ

is constant, the centre of mass equals the centroid.

True. A constant

ρ

factors out of both

\iint x ρ d A

and

\iint ρ d A

and cancels, leaving a purely geometric average — that is the definition of the centroid.

Moment of inertia can be negative if the region sits in the third quadrant.

False.

I_{x} = \iint y^{2} ρ d A

multiplies mass by

y^{2} \geq 0

and

ρ \geq 0

, so the integrand is never negative —

I

measures resistance to spinning and is always

\geq 0

regardless of quadrant.

The moment

M_{x}

can be negative.

True.

M_{x} = \iint y ρ d A

uses

y

to the first power; if the plate lies below the

x

-axis (

y < 0

) each tile contributes a negative lever arm, so

M_{x} < 0

— that is how the balance point ends up below the axis.

I_{0} = I_{x} + I_{y}

holds for any body.

False. This is the perpendicular-axis theorem, valid only for a flat lamina in the

x y

-plane, because it needs

r^{2} = x^{2} + y^{2}

with no

z

contribution. A 3-D solid has

z^{2}

terms that break it.

Doubling every density value doubles the centre of mass coordinates.

False.

\overset{x}{ˉ} = \frac{\iint x ρ d A}{\iint ρ d A}

— a global factor of

2

multiplies numerator and denominator and cancels. The balance point is unchanged; only the distribution of

ρ

moves it.

Doubling every density value doubles the mass and doubles

I_{0}

True. Both

m = \iint ρ d A

and

I_{0} = \iint r^{2} ρ d A

are linear in

ρ

, so a constant factor scales them directly — unlike the CoM, they do not divide the factor away.

For a uniform disc,

I_{0}

is proportional to

a^{2}

(radius squared).

False.

I_{0} = \iint r^{2} ρ d A

and

d A \sim (length)^{2}

, giving

I_{0} \propto a^{4}

. Only per unit mass,

I_{0} / m \propto a^{2}

, because

m \propto a^{2}

absorbs two powers.

The radius of gyration

R_{g} = I / m

is the distance of the centre of mass from the axis.

False.

R_{g}

is the single distance at which the whole mass concentrated would give the same

I

; it is generally larger than the CoM distance because

I

weights far mass by

r^{2}

Spot the error

Someone wrote the line shown. Find the flaw and state the correct idea.

"In polar, mass

= ρ \int_{0}^{2 π} \int_{0}^{a} d r d θ

The area element is wrong: a polar tile is a curved wedge of area

r d r d θ

, not

d r d θ

. The Jacobian factor

r

is mandatory — the correct integrand is

ρ r d r d θ

\overset{x}{ˉ} = \frac{M _{x}}{m}

Indices are swapped.

\overset{x}{ˉ}

balances about the

y

-axis whose lever arm is

x

, so

\overset{x}{ˉ} = M_{y} / m

. Using

M_{x}

(lever arm

y

) would give

\overset{y}{ˉ}

"About the

x

-axis:

I_{x} = \iint x^{2} ρ d A

Wrong lever arm. Distance to the $x$ -axis is the vertical gap

y

, so

I_{x} = \iint y^{2} ρ d A

. The

x^{2}

version is

I_{y}

(distance to the

y

-axis).

"Since it's about the

x

-axis,

M_{x} = \iint x ρ d A

The word "about the

x

-axis" fixes the lever arm as

y

, not

x

M_{x} = \iint y ρ d A

. The name refers to the axis you balance about, and the arm is the perpendicular distance to it.

"The plate is symmetric about the

y

-axis, so its mass is zero because negative

x

cancels positive

x

Cancellation happens for the moment

M_{y} = \iint x ρ d A

(odd integrand), giving

\overset{x}{ˉ} = 0

. Mass uses integrand

ρ \geq 0

with no

x

factor, so nothing cancels — mass is strictly positive.

"For a lamina in the

x y

-plane,

I_{z} = \iint z^{2} ρ d A

, which is

0

since

z = 0

The moment of inertia about the

z

-axis uses the distance from that axis,

x^{2} + y^{2}

, not the coordinate

z

. So

I_{z} = I_{0} = \iint (x^{2} + y^{2}) ρ d A \neq = 0

Why questions

Answer the "why" in one or two sentences of genuine reasoning.

Why does the moment of inertia use

r^{2}

but the moment (for CoM) use

r^{1}

Balance/leverage is linear in distance, so a moment needs

r^{1}

; rotational energy is

\frac{1}{2} (d m) v^{2}

with

v = ω r

, so energy carries

r^{2}

— different physics, different power.

Why do we divide the moment by the mass to get the centre of mass?

We demand that a single point mass

m

placed at

(\overset{x}{ˉ}, \overset{y}{ˉ})

reproduce the same moments:

m \overset{x}{ˉ} = M_{y}

. Solving forces

\overset{x}{ˉ} = M_{y} / m

— the division converts total leverage into a location.

Why must a varying-density plate keep

ρ

inside the CoM integral?

Position must be weighted by how much mass sits there; heavy regions pull the balance point toward them. Dropping

ρ

would treat all tiles equally and give the geometric centroid instead.

Why does the extra

r

appear only in polar coordinates and not in Cartesian?

A Cartesian tile

d x d y

is a true rectangle of area

d x d y

. A polar tile has sides

d r

and arc

r d θ

, so its area grows with radius — the change of variables records this stretch as the factor

r

Why is the moment of inertia relevant to spinning but the moment (CoM) is not?

Spinning stores kinetic energy

\frac{1}{2} I ω^{2}

and angular momentum

I ω

, both governed by

I = \iint r^{2} ρ d A

; see Rotational Kinetic Energy and Angular Momentum. The CoM only locates balance and translational motion.

Why can two plates with the same mass have very different

I_{0}

I_{0}

depends on where the mass sits, weighting far mass by

r^{2}

. Spreading the same mass outward raises

I_{0}

sharply while leaving

m

unchanged.

Why is the perpendicular-axis theorem restricted to flat plates?

It relies on

r^{2} = x^{2} + y^{2}

so that

I_{0} = \iint x^{2} + \iint y^{2} = I_{y} + I_{x}

. A solid adds

z^{2}

terms, so the clean split no longer holds.

Edge cases

Boundary and degenerate inputs — never let one surprise you.

What is the centre of mass of a plate whose density is zero everywhere except along a single point?

Undefined (a

0/0

form): both

\iint x ρ d A

and

\iint ρ d A

vanish over a set of zero area. A point carries no area, hence no mass, so no balance point exists.

For a region symmetric about the origin with even density

ρ (- x, - y) = ρ (x, y)

, what are

M_{x}

and

M_{y}

Both are

0

: the integrand

x ρ

(and

y ρ

) is odd under the symmetry, so contributions from opposite tiles cancel. Hence

\overset{x}{ˉ} = \overset{y}{ˉ} = 0

— the CoM sits at the origin.

If a plate lies entirely on the

x

-axis (a thin rod along

y = 0

), what is

I_{x}

I_{x} = \iint y^{2} ρ d A = 0

, since every tile has

y = 0

. A line has no thickness to fight rotation about the very axis it lies on.

A homogeneous plate is scaled up by a factor

k

in both dimensions (density fixed). How do

m

and

I_{0}

change?

Area scales as

k^{2}

m \to k^{2} m

; distances also scale by

k

, giving

I_{0} \to k^{4} I_{0}

(two powers from area, two from

r^{2}

). So

I_{0} / m \to k^{2} (I_{0} / m)

What happens to

\overset{x}{ˉ}

as the region shrinks to a single point

(x_{0}, y_{0})

It approaches

x_{0}

: both integrals concentrate at that point, and the weighted average of a single location is the location itself — the limit is well-behaved even though mass

\to 0

ρ

is negative somewhere (a modelling error), what breaks first?

Mass could become

0

or negative, so

\overset{x}{ˉ} = M_{y} / m

can blow up or flip sign, and

I_{0}

could go negative — all physically meaningless. Density must satisfy

ρ \geq 0

for the whole framework to make sense.

For a 3-D solid, does the centre-of-mass recipe still divide by total mass?

Yes — the pattern is identical with Triple Integrals:

\overset{x}{ˉ} = \frac{∭ x ρ d V}{∭ ρ d V}

. Only the dimension of the integral and the density (mass per volume) change.

Recall One-line self-test

Cover the right side and finish each sentence in your own words. " $M$ uses distance to the ___ power; $I$ uses the ___ power." ::: First; second. "In polar, every area integral wears an extra ___." ::: $r$ (from the Jacobian). "CoM divides by mass; $I$ ___ divide by mass." ::: does not.