MathsFurther Mechanics 2 › Centres of mass by integration

Centres of mass by integration

When the shape has no straight edges to split along, cut it into strips instead. Each strip is a rectangle whose centre you know, and the sum becomes an integral.

Year FMEDEXCEL 9FM0 FM2

Builds on Centres of mass of plane figures and frameworks and Areas and the limit of a sum.

IN THIS TOPIC

  • Find the centre of mass of a lamina bounded by a curve, by integration.
  • Find the centre of mass of a solid of revolution.
  • Handle a non-uniform body whose density varies with position.

WHAT YOU PROBABLY THINK

For a lamina under a curve, the y coordinate of the centre of mass is found by integrating y against the area, just as x is.

Strips for a lamina

Cut the region into vertical strips of width δx. Each strip is a rectangle to within a vanishing error of height y, so its area is y δx and its own centre is at height y/2. Summing the moments and letting the width tend to zero gives:

mean x = xy dx y dx,      mean y = 12y2 dx y dx

The two are not symmetric, and that is the point the opening claim misses. The x moment uses the whole strip at distance x, so the integrand is xy. The y moment uses the strip at its own halfway height, giving y × y/2, hence the half and the square.

The lamina under y = x² from 0 to 2, with its centre of mass at (1.5, 1.2)a thin stripGy = x²mean x = ∫xy dx / ∫y dxmean y = ∫½y² dx / ∫y dx
FIG. 1A lamina under a curve, cut into thin strips, with its centre of mass marked.

WORKED EXAMPLE

Under a parabola

Find the centre of mass of the uniform lamina bounded by y = x², the x-axis and x = 2.

Area = ∫x² dx from 0 to 2 = 8/3.

∫xy dx = ∫x³ dx = 4, so xG = 4 ÷ 8/3 = 1.5.

∫½y² dx = ½∫x⁴ dx = ½(32/5) = 3.2, so yG = 3.2 ÷ 8/3 = 1.2.

Both lie inside the region, and xG is well right of centre because the area is concentrated there.

Solids of revolution, and varying density

For a solid formed by rotating a curve about the x-axis, cut it into discs of radius y and thickness δx. Each disc has volume πy² δx and its centre on the axis at x, so by symmetry the centre of mass lies on the axis and only xG needs finding:

mean x = x y2 dx y2 dx

The π cancels. For a non-uniform body the density varies with position, so each element carries a mass ρ dV rather than dV, and ρ goes inside both integrals. That is the only change: the method is the same weighted average throughout.

The solid formed by rotating y = x²: thin discs of radius y, with the centre of mass on the axis at 5/3G at 5/3y = x²disc volume πy² δx, all on the axis
FIG. 2The solid of revolution cut into thin discs, with its centre of mass on the axis at five thirds.

YOUR TURN

A solid of revolution

The region under y = x² from x = 0 to x = 2 is rotated about the x-axis. Find the centre of mass of the solid formed.

Show the working

Volume ∝ ∫y² dx = ∫x⁴ dx = 32/5.

Moment ∝ ∫xy² dx = ∫x⁵ dx = 64/6 = 32/3.

xG = (32/3) ÷ (32/5) = 5/3 ≈ 1.67, on the axis by symmetry.

That is further right than the lamina's 1.5, because squaring the radius weights the wide end more heavily still.

THE EXAM BIT

  • Write down the element you are using, its mass and its own centre, before setting up any integral.
  • Remember the half and the square in the y integral for a lamina; the two coordinates are not symmetric.
  • For a solid of revolution, say that symmetry puts the centre on the axis rather than integrating for it.
  • Quote standard results from the formulae book where the question allows, and say that you are doing so.

CHECK YOURSELF

A uniform lamina lies under y = x from x = 0 to x = 3. Find xG.

Show a hint

Two integrals, then divide.

Show the answer

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Cut into strips: xG is ∫xy dx over ∫y dx, but yG is ∫½y² dx over ∫y dx, because each strip acts at its own halfway height.

For a solid of revolution use y² in place of y and take the centre on the axis by symmetry; for a non-uniform body put ρ inside both integrals.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

CHECK YOUR PROGRESS

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  • Find the centre of mass of a lamina bounded by a curve, by integration.
  • Find the centre of mass of a solid of revolution.
  • Handle a non-uniform body whose density varies with position.

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