Centre of mass — derivation for common shapes (rod, triangle, semicircle, hemisphere): Question bank

Prerequisites worth a glance first: Centre of mass — definition and system of particles, Integration as continuous summation, Symmetry arguments in physics, Centroid vs centre of mass vs centre of gravity.

True or false — justify

True/False items — every answer must give the reason, never a bare yes/no.

The COM of a uniform rod always lies inside the rod.

True — for a straight uniform rod the mass-weighted average of positions between

0

and

L

must land at

L /2

, which is a point of the rod itself.

For a uniform L-shaped bent wire, the COM can lie outside the material (in empty space).

True — the COM is an average position, not a physical point of the body; for concave/bent shapes the average can fall in a region with no mass, like the corner-gap of an "L".

For a uniform body, the centre of mass is identical to the geometric centroid.

True — when density is constant it cancels top and bottom in

\frac{\int r d m}{\int d m}

, leaving a purely geometric average, which is the definition of the centroid (Centroid vs centre of mass vs centre of gravity).

The semicircular wire and the semicircular disc of the same radius

R

have the same COM.

False — the wire has all mass on the rim (

\frac{2 R}{π} \approx 0.64 R

) while the disc piles area near the centre, dragging the average down to

\frac{4 R}{3 π} \approx 0.42 R

. Different

d m

distributions give different answers.

A triangular lamina and a triangular wire frame (three rods) have the same COM.

False — the lamina's COM is the vertex-average (medians' intersection), but the frame's COM is the mass-weighted average of the three edge midpoints, which is a generally different point unless the triangle is equilateral.

The COM of a solid hemisphere lies exactly halfway up its height.

False — it sits at

\frac{3 R}{8} = 0.375 R

, below the midpoint, because the wide slabs near the flat face carry more mass and pull the average down.

The hemispherical shell's COM (

R /2

) is higher than the solid hemisphere's (

3 R /8

True — the shell puts all its mass out on the curved surface, farther from the flat face on average, so its balance point sits higher than the mass-filled solid.

Doubling the mass of a uniform rod moves its COM.

False — the COM position is a mass-weighted average; scaling every

d m

by the same factor cancels in numerator and denominator, so the location is unchanged.

If a body has a plane of symmetry, its COM must lie on that plane.

True — for every mass element on one side there is a mirror-image element on the other; their contributions to the off-plane coordinate cancel in pairs, forcing the average onto the plane (Symmetry arguments in physics).

The COM of a triangle depends on which side you call the "base".

False — the point is the intersection of the three medians (vertex average

\frac{1}{3} (r_{1} + r_{2} + r_{3})

); the "

h /3

from base" description just reads out the same fixed point from a chosen edge.

Spot the error

Each item states a wrong line of reasoning; the reveal names the flaw and the fix.

"For the semicircular disc I used

d m = σ d r

over the shells."

Wrong — that copies the 1D linear pattern. A half-ring shell has area

d A = π r d r

(length

\times

thickness), so

d m = σ π r d r

; you must include the slice's area geometry.

"The triangle's COM is at

h /2

, like the rod was at

L /2

Wrong — the rod has constant width so its mass is evenly spread, but the triangle narrows toward the apex, making it bottom-heavy. The correct height is

h /3

"I integrated both

x

and

y

for the semicircular wire to be safe."

Wasteful, not wrong — the flat side lies on the

x

-axis making the shape symmetric about

x = 0

, so

x_{c m} = 0

instantly by symmetry; only

y_{c m}

needs integration.

"For the solid hemisphere I set slice mass

d m = ρ d z

Wrong — a slab at height

z

is a full disc of radius

R^{2} - z^{2}

, so

d V = π (R^{2} - z^{2}) d z

and

d m = ρ π (R^{2} - z^{2}) d z

. You dropped the disc's area factor.

"The semicircular arc element has length

d l = d θ

Wrong — arc length is radius times angle,

d l = R d θ

. Forgetting the

R

breaks the density

λ = M / (π R)

and the final

\frac{2 R}{π}

"To find the disc COM I weighted each half-ring by its own radius

r

Wrong — each half-ring's COM is not at

r

but at its own height

\frac{2 r}{π}

(result from the wire case). Weight each shell's mass by

\frac{2 r}{π}

, not

r

"Density

σ

affects the COM of a uniform lamina, so I kept it in the final answer."

Wrong — for a uniform body

σ

is constant and cancels between

\int x d m

and

\int d m

; the result is purely geometric with no

σ

in it.

"A strip at height

y

in the triangle has width

b

everywhere."

Wrong — by similar triangles the width shrinks linearly:

w (y) = b (1 - \frac{y}{h})

, zero at the apex. Using constant

b

would wrongly give

h /2

Why questions

Conceptual "why" prompts — the reveal is the mechanism, not just the fact.

Why does replacing the sum

\sum m_{i} x_{i} / \sum m_{i}

by an integral work for a continuous body?

A continuous body is the limit of infinitely many infinitesimal particles

d m

; the discrete average becomes

\frac{\int x d m}{\int d m}

exactly as summation becomes integration (Integration as continuous summation).

Why do we bother choosing the origin and axes cleverly before integrating?

A good choice lets symmetry zero out one or two coordinates immediately and makes the density function simple, so you integrate one clean 1D integral instead of a messy multi-variable one.

Why is the triangle's COM the average of its three vertices?

Splitting the area into strips parallel to any side and integrating gives

h /3

from that side; doing this for all three sides pins the point to the medians' intersection, which equals

\frac{1}{3} (r_{1} + r_{2} + r_{3})

Why does the filled disc COM (

\frac{4 R}{3 π}

) come out lower than the wire COM (

\frac{2 R}{π}

The disc has lots of area near the flat diameter (small

y

), and those low shells drag the mass-weighted average down, unlike the wire whose mass sits entirely on the high outer rim.

Why does momentum conservation only track the COM and ignore internal motion?

Internal forces come in action–reaction pairs that cancel, so only external forces move the COM; the system's total momentum is

M v_{c m}

regardless of how the parts jiggle (Momentum conservation).

Why can we build the semicircular disc from half-rings whose COM we already know?

The whole is the sum of its shells; treating each shell as a point mass at its own COM height

\frac{2 r}{π}

and averaging those points reproduces the exact disc COM — reuse instead of re-deriving.

Why does the

π

cancel in the solid-hemisphere calculation but appear in the semicircle answers?

In the hemisphere both numerator and denominator carry a

π

from the disc area, so it cancels; in the semicircle the

π

enters only through the total length/area

π R

, so it survives into

\frac{2 R}{π}

and

\frac{4 R}{3 π}

Why does a heavier particle "pull" the COM toward itself?

In the weighted average

\frac{\sum m _{i} x _{i}}{\sum m _{i}}

its large

m_{i}

multiplies its position more strongly, so the balance point shifts in its direction — exactly the see-saw effect.

Edge cases

Boundary and degenerate inputs — make sure no scenario surprises you.

Where is the COM of a rod whose density increases linearly from one end,

λ (x) = k x

Not at

L /2

— the mass is bottom-heavy toward the dense end, so

x_{c m} = \frac{\int _{0}^{L} x k x d x}{\int _{0}^{L} k x d x} = \frac{2 L}{3}

, shifted toward the heavier end.

What happens to the triangle COM formula

h /3

as the triangle degenerates to a thin sliver (apex angle

\to 0

It still holds —

h /3

is scale- and shape-independent for a triangle; even a needle-thin triangle has its COM one-third up from the base along the height.

As the semicircle "opens out" to a full circle, what should the COM

y

become?

Zero — a full ring or disc is symmetric top-and-bottom, so the upward pull that gave

\frac{2 R}{π}

is cancelled by an equal downward half, sending

y_{c m} \to 0

For a hemisphere in the limit

R \to 0

, where is the COM?

At the origin —

\frac{3 R}{8} \to 0

; a vanishingly small body collapses its COM onto its own point, as any average must.

If you carve a small hole exactly at the geometric centre of a symmetric plate, does the COM move?

No — the removed piece sits at the COM itself, so removing it does not shift the mass-weighted average (its "leverage" about the COM is zero).

Does the COM of a system of two separated balls lie inside either ball?

Not necessarily — it lies on the line joining them at the mass-weighted point, which is generally in the empty space between them, closer to the heavier ball.

Is the "centre of gravity" the same as the COM for a very tall object in a non-uniform gravitational field?

No — COM is purely a mass average, while centre of gravity weights by local

g

; they coincide only in a uniform field, and drift apart when

g

varies across the body (Centroid vs centre of mass vs centre of gravity).

Where is the COM of a uniform full circular disc vs a uniform full circular ring?

Both at the centre — full circular symmetry forces the COM to the geometric centre regardless of whether mass is on the rim or spread over the area.