Questions on centres of mass by integration

Question 1

Write down the two integrals for the centre of mass of a lamina under a curve, and explain why they are not symmetric.

Accepted Answer

x(G) = ∫xy dx ÷ ∫y dx and y(G) = ∫½y² dx ÷ ∫y dx. A vertical strip of height y acts at horizontal distance x, giving the moment xy δx. But it acts at its own halfway height y/2, so the vertical moment is y × y/2 δx, hence the half and the square. B1 B1 for the two integrals, B1 for the reason the vertical one carries the half and the square. Both integrals are divided by the same area, so the asymmetry is entirely in the numerators.

Question 2

Find the centre of mass of the uniform lamina bounded by the curve y = 4 − x², the y-axis and the positive x-axis.

Accepted Answer

Area = ∫(4 − x²) dx = 8 − 8/3 = 16/3. ∫xy dx = ∫(4x − x³) dx = 8 − 4 = 4, so x(G) = 4 ÷ 16/3 = 0.75. ∫½y² dx = ½(32 − 64/3 + 32/5) = 8.533, so y(G) = 8.533 ÷ 16/3 = 1.6. B1 for the upper limit, M1 A1 for the area, M1 A1 for the x coordinate, A1 for the y coordinate. The curve meets the x-axis at x = 2, which is where the upper limit comes from.

Question 3

A uniform triangular lamina occupies the region under the line y = x from x = 0 to x = 3. Find the x coordinate of its centre of mass by integration, and confirm that it agrees with the centroid of the triangle.

Accepted Answer

Area = ∫x dx from 0 to 3 = 9/2. ∫xy dx = ∫x² dx = 9. So x(G) = 9 ÷ 9/2 = 2. The triangle has vertices (0, 0), (3, 0) and (3, 3), and its centroid is at the mean of the vertices, x = (0 + 3 + 3)/3 = 2, in agreement. M1 for the area and moment integrals, A1 for x(G) = 2, B1 for the centroid check.

Question 4

A uniform lamina occupies the region enclosed by the curve y² = x and the line x = 4. Explain why only one integral quotient is needed, and find the coordinates of its centre of mass.

Accepted Answer

The region is symmetric about the x-axis (the curve is y = √x above the axis and y = −√x below it), so y(G) = 0 with no integration. Area = 2∫√x dx from 0 to 4 = 2 × (2/3) × 8 = 32/3. Moment = 2∫x√x dx = 2 × (2/5) × 32 = 128/5. So x(G) = (128/5) ÷ (32/3) = 12/5 = 2.4, and the centre of mass is at (2.4, 0). B1 for the symmetry argument giving y(G) = 0, M1 for the area, M1 for the moment, A1 for (2.4, 0). The factor of 2 in both integrals cancels, so it is enough to work with the upper half alone. The answer sits three fifths of the way from the vertex to the chord x = 4, which is the general result for a parabolic segment.

Question 5

The region under y = √x from x = 0 to x = 4 is rotated about the x-axis. Find the centre of mass of the solid.

Accepted Answer

By symmetry the centre of mass lies on the axis, so only x(G) is needed. Volume ∝ ∫y² dx = ∫x dx = 8. Moment ∝ ∫xy² dx = ∫x² dx = 64/3. So x(G) = (64/3) ÷ 8 = 8/3 ≈ 2.67. B1 for the symmetry argument, M1 A1 for the volume integral, M1 for the moment integral, A1 for the position. It sits two thirds of the way along, since the solid widens towards the far end.

Question 6

A uniform solid hemisphere of radius r is formed by rotating y = √(r² − x²) about the x-axis for x from 0 to r. Show that its centre of mass is 3r/8 from the flat face.

Accepted Answer

Volume ∝ ∫(r² − x²) dx from 0 to r = r³ − r³/3 = 2r³/3. Moment ∝ ∫x(r² − x²) dx = r⁴/2 − r⁴/4 = r⁴/4. Dividing: x(G) = (r⁴/4) ÷ (2r³/3) = 3r/8. M1 A1 for the volume integral, M1 A1 for the moment integral, A1 for the printed result. It lies nearer the flat face than the pole, because the wide end is there.

Question 7

Explain how the method changes for a body whose density varies with position.

Accepted Answer

Each element carries a mass ρ dV rather than dV, so the density goes inside both integrals: the total mass is ∫ρ dV and the moment is ∫xρ dV. Everything else is unchanged, since the centre of mass is still the mass-weighted average of position. M1 for the element of mass, A1 A1 for the two integrals. A constant ρ cancels top and bottom, so uniform bodies never need it.

Question 8

A rod of length 2 m lying along the x-axis from the origin has density (1 + x) kg per metre. Find its centre of mass.

Accepted Answer

Mass = ∫(1 + x) dx from 0 to 2 = 2 + 2 = 4 kg. Moment = ∫x(1 + x) dx = 2 + 8/3 = 14/3. So x(G) = (14/3) ÷ 4 = 7/6 ≈ 1.17 m. M1 A1 for the mass, M1 for the moment, A1 for the position. It sits past the midpoint, towards the denser end, which is the check to make.

Question 9

A uniform lamina occupies the region between the curve y = sin x and the x-axis for 0 ≤ x ≤ π. Find the coordinates of its centre of mass.

Accepted Answer

The region is symmetric about the line x = π/2, so x(G) = π/2 without integrating. Area = ∫sin x dx from 0 to π = [−cos x] = 2. For y(G), ∫½sin²x dx = ∫¼(1 − cos 2x) dx = ¼[x − ½sin 2x] from 0 to π = π/4. So y(G) = (π/4) ÷ 2 = π/8 ≈ 0.393. B1 for x(G) = π/2 by symmetry, B1 for the area, M1 for the double-angle substitution in the y integral, A1 for π/4, A1 for π/8.

Question 10

A uniform solid consists of a cylinder of radius r and length 2r with a solid hemisphere of radius r fixed to one end, the flat face of the hemisphere coinciding with the end of the cylinder. Taking the centre of mass of a solid hemisphere to be 3r/8 from its flat face, find the distance of the centre of mass of the whole solid from the free end of the cylinder.

Accepted Answer

Measure x from the free end. The cylinder has volume πr² × 2r = 2πr³ with centre at x = r; the hemisphere has volume (2/3)πr³ with centre at x = 2r + 3r/8 = 19r/8. The volumes are in the ratio 3 : 1, so take masses 3 and 1. Moments about the free end: 3 × r + 1 × 19r/8 = 43r/8, and the total mass is 4, so x(G) = 43r/32 ≈ 1.34r. B1 for the volumes in the ratio 3 : 1, B1 for the hemisphere centre at 19r/8, M1 for the moments equation, A1 for 43r/32. The centre of mass lies inside the cylinder, past its midpoint towards the hemisphere, as it should for an attached mass on that side; the hemisphere result is quoted from the formulae booklet rather than derived, which the question allows.

Question 11

A lamina occupies the triangular region under the line y = x from x = 0 to x = 2. Its density per unit area at the point (x, y) is kx, where k is a positive constant. Find the coordinates of its centre of mass.

Accepted Answer

A strip at x has area x δx and density kx, so its mass is kx² δx. Mass = ∫kx² dx from 0 to 2 = 8k/3. x moment = ∫x × kx² dx = ∫kx³ dx = 4k, so x(G) = 4k ÷ 8k/3 = 1.5. The strip acts at height y/2 = x/2, so the y moment is ∫(x/2) × kx² dx = ∫½kx³ dx = 2k, giving y(G) = 2k ÷ 8k/3 = 0.75. Centre of mass (1.5, 0.75). M1 for the element of mass kx² δx, A1 for the mass 8k/3, M1 for a moment integral with the density inside, A1 for x(G) = 1.5, A1 for y(G) = 0.75. The uniform triangle would have its centroid at (4/3, 2/3); both coordinates have moved towards x = 2, where the material is densest, and k cancels in each quotient.

Question 12

A uniform lamina occupies the region enclosed by the curve y = x² and the line y = 2x. Find the coordinates of its centre of mass.

Accepted Answer

The curve and the line meet where x² = 2x, that is at x = 0 and x = 2, with the line above the curve between them.
Area = ∫(2x − x²) dx from 0 to 2 = 4 − 8/3 = 4/3.
For x(G), each vertical strip has height 2x − x² and acts at distance x: ∫x(2x − x²) dx = 16/3 − 4 = 4/3, so x(G) = (4/3) ÷ (4/3) = 1.
For y(G), a strip running from y = x² up to y = 2x acts at the average of its ends, so its moment is ½[(2x)² − (x²)²] δx. Then ∫½(4x² − x⁴) dx = ½(32/3 − 32/5) = 32/15, giving y(G) = (32/15) ÷ (4/3) = 1.6.
Centre of mass (1, 1.6). B1 for the upper limit, M1 A1 for the area, M1 A1 for the x coordinate, M1 A1 for the y coordinate.
The y moment is ½∫[(2x)² − (x²)²] dx: it is the difference of the squares that is needed, not the square of the difference ½∫(2x − x²)² dx.

Question 13

A uniform solid cone of height h and base radius r is formed by rotating the line y = rx/h, for 0 ≤ x ≤ h, about the x-axis. Show by integration that the centre of mass of the cone is a distance h/4 from its base.

Accepted Answer

The solid is symmetric about the x-axis, so the centre of mass lies on the axis and only x(G) is needed. Cut the cone into discs of radius y = rx/h and thickness δx.
Volume ∝ ∫y² dx = ∫(r²/h²)x² dx from 0 to h = (r²/h²) × h³/3 = r²h/3.
Moment ∝ ∫xy² dx = ∫(r²/h²)x³ dx = (r²/h²) × h⁴/4 = r²h²/4.
So x(G) = (r²h²/4) ÷ (r²h/3) = 3h/4 from the vertex, which is h − 3h/4 = h/4 from the base.
B1 for the symmetry argument, M1 A1 for the volume integral, M1 A1 for the moment integral, A1 for 3h/4 and the conversion to h/4 from the base. The r cancels, so every cone of height h has its centre of mass at the same height. The formulae booklet gives h/4 from the base, and the last line is where the vertex measurement is turned into that.

Question 14

The region under the curve y = e^x from x = 0 to x = 1 is rotated about the x-axis to form a uniform solid. Find the distance of its centre of mass from the plane face at x = 0, giving an exact expression and a value to 3 significant figures.

Accepted Answer

By symmetry the centre of mass is on the axis, so only x(G) is needed. Volume ∝ ∫e^2x dx from 0 to 1 = [½e^2x] = (e² − 1)/2 ≈ 3.19.
The moment ∝ ∫xe^2x dx needs integration by parts: [½xe^2x] − ∫½e^2x dx = ½e² − ¼(e² − 1) = (e² + 1)/4 ≈ 2.10.
So x(G) = [(e² + 1)/4] ÷ [(e² − 1)/2] = (e² + 1)/(2(e² − 1)) ≈ 0.657.
B1 for the symmetry argument, M1 A1 for the volume integral, M1 for integration by parts in the moment, A1 for (e² + 1)/4, A1 for the exact quotient and 0.657. The answer lies well past the midpoint 0.5, as the solid is wider at x = 1: the radius grows by a factor e over the length, so the volume is concentrated towards that end.

Question 15

The region under the curve y = 1/x from x = 1 to x = a, where a > 1, is rotated about the x-axis to form a uniform solid. Find the distance of the centre of mass from the face at x = 1 in terms of a. Show that as a increases the volume of the solid stays below a fixed bound while the centre of mass moves off without limit.

Accepted Answer

Volume = π∫x⁻² dx from 1 to a = π[−1/x] = π(1 − 1/a). Moment ∝ π∫x × x⁻² dx = π∫(1/x) dx = π ln a. Dividing, the centre of mass is at x = ln a ÷ (1 − 1/a), that is a ln a/(a − 1) from the origin, and so a ln a/(a − 1) − 1 from the face at x = 1.
As a → ∞, 1/a → 0 so the volume approaches π and never exceeds it. But ln a → ∞ while a/(a − 1) → 1, so a ln a/(a − 1) grows without bound: at a = 10 the centre of mass is at 2.56, at a = 1000 it is beyond 6.9, while the volume is still under π.
M1 A1 for the volume integral, M1 A1 for the moment integral ln a, A1 for a ln a/(a − 1), B1 for the bounded-volume, unbounded-position argument. The solid is a horn that gets ever thinner: the extra length adds almost no volume, yet what little it adds sits so far out that it keeps dragging the centre of mass along.