MG9, MG10 - Lamina centres of mass and equilibrium | OCR A-Level Further Mathematics B (MEI)

MG9 finds a uniform plane lamina's centre of mass by adding the moments of thin strips, including regions between curves. MG10 then uses the calculated centre in rigid-body equilibrium, particularly suspension and the competition between sliding and toppling.

A thin strip has its own midpoint

A uniform solid hemisphere has its centre $3a/8$ above the flat face. Does a uniform semicircular plate with the same outline have its centre at the same fractional height? The outline agrees, but the mass distribution does not: a plate is weighted by strip area, whereas the solid was weighted by disc volume.

You already know

For a uniform lamina, mass is proportional to area. The centre satisfies $A\bar x=\sum A_i x_i$ and $A\bar y=\sum A_i y_i$ for component areas. The weight acts through that centre when moments are calculated.

In the lamina below $y=x^2$ , a strip near $x=1$ has height $1$ and centre height $1/2$ . The material is spread through the whole strip, not concentrated along the curve.

For a region between $y=f(x)\ge0$ , the $x$ -axis and $x=a,b$ , a thin vertical strip has area $y\,dx$ . Its centre is approximately $(x,y/2)$ . It therefore contributes $xy\,dx$ to the horizontal area moment and $y^2\,dx/2$ to the vertical area moment.

[DIAGRAM: asset_name: A vertical strip and its centre; asset_slug: h645_b077_strip; caption: A strip of height y has centre halfway up, so its vertical moment is one half y squared dx; description: The uniform lamina under y equals x squared from zero to two is shaded. A thin vertical strip is highlighted at a labelled x, with height y and width dx. Its midpoint is marked at x comma y over two, and the whole lamina centre is marked at three halves comma six fifths. A side panel distinguishes strip area y dx from the moment integrands x y dx and one half y squared dx.]
A strip of height y has centre halfway up, so its vertical moment is one half y squared dx

Lamina below a non-negative curve

For $A>0$ and uniform mass per unit area,

A=\int_a^b y\,dx,\qquad A\bar x=\int_a^b xy\,dx,\qquad A\bar y=\frac12\int_a^b y^2\,dx

Here the coordinate weights come from each strip's centre, not from the curve's upper boundary.

Not in the formula booklet: learn it.

There is no factor $\pi$ : this is a plane lamina, not a solid formed by rotation. The area moments have units of length cubed; dividing by area gives a length.

Worked example 1: a parabolic boundary

Question. A uniform lamina occupies $0\le x\le2$ , $0\le y\le x^2$ . Find its centre of mass.

Plan. Integrate the strip areas and their two coordinate moments, then divide by area.

\begin{aligned} A&=\int_0^2x^2\,dx=\frac83 && \text{strip area} \\ A\bar x&=\int_0^2x^3\,dx=4 && \text{horizontal moment} \\ A\bar y&=\frac12\int_0^2x^4\,dx=\frac{16}{5} && \text{strip midpoint height} \\ \bar x&=\frac4{8/3}=\frac32,\qquad \bar y=\frac{16/5}{8/3}=\frac65 && \text{divide moments by area} \end{aligned}

Why this step? The factor $1/2$ comes from the strip centre at height $y/2$ . Weighting its area at height $y$ would put all its mass on the upper boundary.

Answer. $G=(3/2,6/5)$ .

Check. The centre is inside the lamina: at $x=3/2$ , its upper boundary is $9/4>6/5$ . Its horizontal coordinate is towards the wider right end.

Change only the right endpoint from $2$ to $3$ . This stretches the lamina horizontally by factor $3/2$ and vertically by factor $(3/2)^2$ .

Worked example 2: extending the same curve

Question. Find the centre for $0\le x\le3$ , $0\le y\le x^2$ .

Plan. Use the same three integrands with the new endpoint.

\begin{aligned} A&=\int_0^3x^2\,dx=9 && \text{new strip-area total} \\ A\bar x&=\int_0^3x^3\,dx=\frac{81}{4} && \text{horizontal moment} \\ A\bar y&=\frac12\int_0^3x^4\,dx=\frac{243}{10} && \text{vertical moment} \\ G&=\left(\frac94,\frac{27}{10}\right) && \text{divide both moments by }9 \end{aligned}

Why this step? Changing the endpoint changes the entire weighted distribution. It is not enough to move the original centre by one unit.

Answer. $G=(9/4,27/10)$ .

Check. The coordinates have scaled by $3/2$ and $9/4$ , respectively, matching the geometric stretches described above.

For a curve supplied as $x=g(y)$ , horizontal strips may be simpler; the corresponding formulae are developed later in the lesson. Always choose the direction that makes each strip a single clear interval.

Watch out

Wrong: $A\bar y=\int y^2\,dx$ .

Right: $A\bar y=\frac12\int y^2\,dx$ . The missing half is a misplaced strip centre, not an integration error.

Use another endpoint and give both coordinates.

Use the midpoint between both boundaries

If a vertical strip lies between $y=g(x)$ and $y=f(x)$ , where $f\ge g$ , its height is $f-g$ , but its midpoint height is $(f+g)/2$ . A difference of heights alone does not locate that midpoint.

Multiplying area by midpoint gives

\frac{f+g}{2}(f-g)\,dx=\frac12(f^2-g^2)\,dx

for the vertical moment. This works even when a boundary is below the axis, provided the upper and lower curves are identified correctly.

A uniform lamina between curves

For $a\le x\le b$ , $g(x)\le y\le f(x)$ , and positive area,

\begin{aligned} A&=\int_a^b(f-g)\,dx \\ A\bar x&=\int_a^b x(f-g)\,dx \\ A\bar y&=\frac12\int_a^b(f^2-g^2)\,dx \end{aligned}

Not in the formula booklet: learn it.

You can also obtain these equations by subtracting lower-region area moments from upper-region moments when that partition is appropriate. The strip derivation shows directly why the squared boundaries are subtracted.

Worked example 3: a curved region between two graphs

Question. A uniform lamina lies between $y=x$ and $y=x^2$ , for $0\le x\le1$ . Find its centre.

Plan. Use upper curve $f=x$ , lower curve $g=x^2$ , then integrate their area and moment differences.

\begin{aligned} A&=\int_0^1(x-x^2)\,dx=\frac16 && \text{strip-height difference} \\ A\bar x&=\int_0^1(x^2-x^3)\,dx=\frac1{12} && \text{weight by }x \\ A\bar y&=\frac12\int_0^1(x^2-x^4)\,dx=\frac1{15} && \text{strip midpoint height} \\ G&=\left(\frac12,\frac25\right) && \text{divide moments by area} \end{aligned}

Why this step? The strip's centre is at $(x+x^2)/2$ , not at half the gap $(x-x^2)/2$ , which would be measured from the wrong baseline.

Answer. $G=(1/2,2/5)$ .

Check. At $x=1/2$ , the bounds are $1/4$ and $1/2$ , containing $2/5$ . Also the strip-area distribution $x(1-x)$ is symmetric about $x=1/2$ , agreeing with $\bar x=1/2$ .

Change only the vertical scale, multiplying both boundary heights by $2$ . The horizontal centre remains fixed, while the vertical centre doubles.

Worked example 4: stretching both boundaries

Question. Find the centre of the uniform lamina between $y=2x$ and $y=2x^2$ , for $0\le x\le1$ .

Plan. Update the strip height and squared-boundary difference.

\begin{aligned} A&=\int_0^1(2x-2x^2)\,dx=\frac13 && \text{new strip heights} \\ A\bar x&=\int_0^1(2x^2-2x^3)\,dx=\frac16 && \text{horizontal moment} \\ A\bar y&=\frac12\int_0^1(4x^2-4x^4)\,dx=\frac4{15} && \text{squared boundaries} \\ G&=\left(\frac12,\frac45\right) && \text{divide moments by area} \end{aligned}

Why this step? Area scales by $2$ , while the vertical area moment scales by $4$ : it contains both a height in the strip area and a height in the lever arm.

Answer. $G=(1/2,4/5)$ .

Check. Dividing the scaled vertical moment by the scaled area gives twice the original $\bar y$ , as a vertical stretch requires.

[DIAGRAM: asset_name: The centre of a strip between two curves; asset_slug: h645_b077_between; caption: The strip midpoint lies halfway between the actual upper and lower heights; description: The region between y equals x and y equals x squared on zero to one is shaded. A vertical strip at x equals one half runs from one quarter to one half, with midpoint three eighths marked. The whole lamina centre one half comma two fifths is marked separately, showing that a single strip midpoint is not automatically the whole centre.]
The strip midpoint lies halfway between the actual upper and lower heights

Regions that change which curve is upper must be split at their intersections. For a hole, subtract its area and both area moments, just as in the Minor composite-body method.

Watch out

Wrong: use $\frac12(f-g)^2$ as the vertical-moment integrand.

Right: use $\frac12(f^2-g^2)$ . The first expression would shift every strip down to the axis before calculating its moment.

Stretch the same region once more.

Horizontal strips can make a curved lamina simple

Horizontal strips reverse the coordinate roles. For boundaries $x=l(y)$ and $x=r(y)$ , the strip has area $(r-l)\,dy$ and centre $((r+l)/2,y)$ . A symmetry line may make one moment zero before any integration.

Horizontal-strip moments

For $c\le y\le d$ , $l(y)\le x\le r(y)$ , and positive area,

\begin{aligned} A&=\int_c^d(r-l)\,dy \\ A\bar x&=\frac12\int_c^d(r^2-l^2)\,dy \\ A\bar y&=\int_c^d y(r-l)\,dy \end{aligned}

For symmetry about the $y$ -axis, $l=-r$ , so $\bar x=0$ .

Not in the formula booklet: learn it.

For an upper semicircle, a horizontal strip has width $2\sqrt{a^2-y^2}$ . Its first moment about the diameter contains $y\sqrt{a^2-y^2}$ , which is straightforward with $u=a^2-y^2$ .

Worked example 5: a uniform semicircular plate

Question. Find the centre of the uniform lamina $x^2+y^2\le4$ , $y\ge0$ .

Plan. Use symmetry for $\bar x$ , then horizontal-strip moments for $\bar y$ .

\begin{aligned} A&=2\pi,\qquad \bar x=0 && \text{semicircle area and symmetry} \\ A\bar y&=\int_0^2 2y\sqrt{4-y^2}\,dy \\ &=\left[-\frac23(4-y^2)^{3/2}\right]_0^2=\frac{16}{3} && u=4-y^2 \\ \bar y&=\frac{16/3}{2\pi}=\frac8{3\pi} && \text{divide by area} \end{aligned}

Why this step? The full width is twice the positive $x$ -boundary. Forgetting the left half would halve both area and moment if done consistently, but mixing half-width and full area gives a wrong centre.

Answer. $G=(0,8/(3\pi))$ .

Check. The height is approximately $0.849$ , between $0$ and the semicircle's height $2$ .

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MG9, MG10 - Lamina centres of mass and equilibrium

A thin strip has its own midpoint

Use the midpoint between both boundaries

Horizontal strips can make a curved lamina simple

Continue this lesson.

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