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Centres of mass by integration questions
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.
7 original questions · 33 marks · the centres of mass by integration notes · Further Mechanics 2
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Write down the two integrals for the centre of mass of a lamina under a curve, and explain why they are not symmetric.
Worked 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.Find the centre of mass of the uniform lamina bounded by the curve y = 4 − x², the y-axis and the positive x-axis.
Worked 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.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.
Worked 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.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.
Worked 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 well inside the flat face, because the wide end is there.Explain how the method changes for a body whose density varies with position.
Worked 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.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.
Worked 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.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.
Worked 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 commonest loss of marks here is writing ∫½(2x − x²)² dx for the y moment. It is the difference of the squares that is needed, not the square of the difference.
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