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Centre of mass of a disk with a hole

Solve a classic center-of-mass subtraction problem by treating the missing circular piece as negative mass.

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  • University Physics Year 1
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Why this matters + quick links

This page gives the UY1 working model/result for Pizza with a hole. You reuse it as part of the mechanics toolkit: choose coordinates, pick a method (Newton/energy/momentum), and check limits and signs.

At a glance

  • Treat the missing piece as a negative mass contribution.
  • Use the center-of-mass equation along one axis.
  • Result for this geometry: the remaining pizza center shifts by x_CM = R/20 away from the hole side.
Common traps (holes = negative mass)
  • Put the origin somewhere convenient (the full-disk center is best here) and keep signs.
  • The cut-out has negative mass but its center position xₕ keeps its physical sign (left/right).
  • Do a quick limit check: if the hole radius → 0, the shift must → 0.

Setup

A uniform circular pizza of radius R has a circular hole of radius R/4 removed from one side.

Scroll across the figure to read all labels.

The hole touches the rim from the inside, so its centre is R − R/4 = 3R/4 to the left of O.

Choose the full pizza center as origin. Let x be the center-of-mass position of the remaining piece.

Let areal mass density be σ.

Geometry reminder (where does 3R/4 come from?):

  • The hole radius is R/4 and it is tangent internally to the big pizza at the rim.
  • Along the diameter, the distance from the big center to the hole center is
    R-R/4 = 3R/4
    with the sign set by whether the hole is on the left (-) or right (+).

Core derivation/explanation

Use superposition:

  • Whole pizza mass: M_full = σπ R², centered at x = 0.
  • Removed circle mass: mₕ = σπ(R/4)² = (σπ R²)/16 with center at xₕ = -3R/4 (toward the removed side).

Write moment balance for the full pizza as M_full · 0 = (Mᵣₑₘ)x + mₕ xₕ where Mᵣₑₘ = M_full-mₕ = (15/16)σπ R²

Substitute: 0 = (15/16)σπ R² x + (1/16)σπ R²(-3R/4)

Solve: x = R/20

So the remaining pizza center is shifted by R/20 opposite to the hole.

Checks:

  • Units: x_CM is a length.
  • Limit R/4 → 0: hole mass mₕ → 0, so x_CM → 0 (full disk recovered).
  • Direction: removing mass on the left makes the right side relatively heavier, so x_CM > 0 for a left-side hole.

Worked examples

Worked example 1

A numerical check

Problem

For the pizza with a hole above, take R = 20 cm. Where is the centre of mass of the remaining piece?

Show full solution

  1. Substitute into the result

    Method

    Use x = R/20.

    Reason

    The superposition argument above gave x = R/20, on the side away from the hole.

    Working

    Answer

    x = 20/20 = 1.0 cm.

  2. State the answer

    Working

    Answer

    The centre of mass is 1.0 cm from the original centre, away from the hole.

Practice set

Check your understanding 1

The hole’s radius is changed to R/5, with its centre at x = -4R/5. Find mₕ/M_full, and the new centre of mass.

Show hint

The mass ratio depends only on the area ratio.

Show answer

For uniform density the mass ratio is the area ratio: (R/5)²/R² = 1/25. Then x = ((1/25)(4R/5))/(24/25) = R/30, again on the side away from the hole.

Check your understanding 2

In the original problem, why does the centre of mass shift away from the hole?

Show hint

Think about removing mass from one side.

Show answer

Removing mass from the left leaves the right side relatively heavier, so the balance point moves right, away from the hole.

Check your understanding 3

Is a shift of R/4 reasonable for the original pizza?

Show hint

Compare the hole’s mass fraction with the whole.

Show answer

No. The hole is only 1/16 of the full mass, so the shift should be much smaller than the hole’s distance 3R/4 from the centre: the exact value is R/20.

Summary + next steps

  • Composite-body COM problems become easy if you model removed parts as negative mass.
  • Write one clean moment equation about a convenient origin.
  • For this pizza-hole case, the center shifts by R/20.

Next step:

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Syllabus and review details

No official syllabus alignment is listed for this lesson.

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