Questions on centres of mass of plane figures and frameworks

Question 1

State the centre of mass of a uniform triangular lamina and of a uniform rectangular one, and say how to use symmetry to save work.

Accepted Answer

A rectangle's centre of mass is at its centre, where the diagonals cross. A triangle's is at the centroid, a third of the way from each side towards the opposite vertex. Any axis of symmetry must contain the centre of mass, so a symmetric shape needs only one coordinate calculating. B1 for the two standard positions, B1 for the symmetry argument.

Question 2

A uniform lamina is made of a 6 by 2 rectangle with corners (0, 0) and (6, 2), and a 2 by 4 rectangle with corners (0, 2) and (2, 6). Find its centre of mass.

Accepted Answer

Split the shape into the two rectangles and treat each as a particle at its own centre. Areas 12 and 8, total 20, with centres (3, 1) and (1, 4).
x(G) = (12 × 3 + 8 × 1)/20 = 44/20 = 2.2.
y(G) = (12 × 1 + 8 × 4)/20 = 44/20 = 2.2.
M1 for the areas acting at their own centres, A1 for the x coordinate, A1 for the y coordinate.
The L is symmetric about the line y = x, so the two coordinates were bound to agree. That makes a quick check on the arithmetic.

Question 3

A uniform lamina consists of the rectangle with corners (0, 0), (4, 0), (4, 2) and (0, 2), with the triangle with vertices (0, 2), (4, 2) and (2, 5) attached along its top edge. Find the centre of mass.

Accepted Answer

Both pieces are symmetric about x = 2, so x(G) = 2. Areas: rectangle 8 with centre at height 1; triangle ½ × 4 × 3 = 6 with centroid at height (2 + 2 + 5)/3 = 3. y(G) = (8 × 1 + 6 × 3)/14 = 26/14 = 13/7 ≈ 1.86. Centre of mass (2, 1.86). B1 for x(G) = 2 by symmetry, M1 for the areas acting at their own centres, A1 for 13/7. The triangle's centre is a third of the way up from its base, at height 3, not at the midpoint of its height.

Question 4

A uniform semicircular lamina of radius 6 has its diameter along the x-axis with its centre at the origin, and lies above the axis. Use the formulae booklet to write down the position of its centre of mass, and state why one coordinate needs no calculation.

Accepted Answer

The lamina is symmetric about the y-axis, so x(G) = 0. The booklet gives the centre of mass of a semicircular lamina at 4r/(3π) from the centre of the diameter, so y(G) = 4 × 6/(3π) = 8/π ≈ 2.55. Centre of mass (0, 2.55). B1 for x(G) = 0 by symmetry, M1 for the booklet result with r = 6, A1 for 8/π. This is the sector formula 2r sin α/(3α) with α = π/2, since a semicircle is a sector of angle π; the value lies below half the radius, because the wide part of the semicircle is near the diameter.

Question 5

A uniform disc of radius 4 centred at the origin has a disc of radius 2 centred at (2, 0) removed. Find the centre of mass of what is left.

Accepted Answer

Areas 16π and 4π, leaving 12π. By symmetry the centre of mass lies on the x-axis. Treating the hole as a negative area: x(G) = (16π × 0 − 4π × 2)/(12π) = −8/12 = −2/3. B1 for the symmetry, M1 for treating the hole as a negative area, A1 for the moment equation, A1 for the position. It has moved away from the hole, as it must, and the π has cancelled throughout.

Question 6

A uniform wire framework consists of three rods forming three sides of a rectangle: from (0, 0) to (6, 0), from (6, 0) to (6, 4), and from (0, 0) to (0, 4). Find its centre of mass.

Accepted Answer

Lengths 6, 4 and 4, total 14, acting at (3, 0), (6, 2) and (0, 2). x(G) = (18 + 24 + 0)/14 = 42/14 = 3, which symmetry about x = 3 confirms. y(G) = (0 + 8 + 8)/14 = 16/14 = 8/7 ≈ 1.14. M1 for the lengths acting at their midpoints, A1 for the x coordinate, M1 A1 for the y coordinate. It sits below the middle because the long rod lies along the bottom.

Question 7

Explain why a triangular lamina and a triangular wire framework of the same shape have different centres of mass.

Accepted Answer

The lamina has its mass spread over the area, so each element is weighted by area and the answer is the centroid of the region. The framework has its mass along the perimeter, so each rod is weighted by its length and acts at its own midpoint. B1 for the lamina weighted by area, B1 for the framework weighted by rod length, B1 for each rod acting at its own midpoint.

Question 8

A uniform lamina is the quarter-circle sector of radius 6 bounded by the positive x axis, the positive y axis and the arc between them, with its centre at the origin. Use the formula in the booklet to find the coordinates of its centre of mass.

Accepted Answer

Symmetry first: the sector is symmetric in the line y = x, so the centre of mass lies on that line and only its distance from the origin needs calculating. B1 for the symmetry argument. The booklet gives 2r sin α/(3α) from the centre for a sector of angle 2α at the centre, so a quarter circle has α = π/4, not π/2. M1 for the formula with α = π/4. The distance is 2 × 6 × sin(π/4)/(3π/4) = 8√2/π ≈ 3.60. A1. Splitting that along y = x, each coordinate is (8√2/π) × cos(π/4) = 8/π, so the centre of mass is at (8/π, 8/π), about (2.55, 2.55). A1. Read α as the half-angle of the sector, not the whole angle: with α = π/2 the formula returns 2.55, which by an unhappy coincidence matches the coordinate and not the distance, so the check that 3.60 × cos 45° = 2.55 is worth writing down.

Question 9

A lamina is made of two rectangular plates joined along an edge: a 6 by 2 plate with corners (0, 0) and (6, 2), and a 2 by 4 plate with corners (0, 2) and (2, 6). The upright plate is made of material twice as heavy per unit area as the horizontal one. Find the centre of mass.

Accepted Answer

Mass is proportional to area × density, so the masses are in the ratio 12 × 1 : 8 × 2 = 12 : 16, acting at the centres (3, 1) and (1, 4). x(G) = (12 × 3 + 16 × 1)/28 = 52/28 = 13/7 ≈ 1.86. y(G) = (12 × 1 + 16 × 4)/28 = 76/28 = 19/7 ≈ 2.71. M1 for weighting each plate by area times density, A1 for the masses 12 : 16, A1 for x(G), A1 for y(G). Doubling the density of the upright drags the centre of mass left and up, towards that plate, which is the check to make: the heavier piece always pulls the point towards its own centre (1, 4).

Question 10

A uniform wire is bent into a semicircular arc of radius 3 and its ends are joined by a straight piece of the same wire along the diameter, so the framework is a closed D shape. Taking the diameter along the x-axis with its midpoint at the origin, use the formulae booklet to find the centre of mass of the framework.

Accepted Answer

By symmetry x(G) = 0. The arc has length 3π and, from the booklet, its own centre of mass is at r sin α/α with α = π/2, that is 2r/π = 6/π ≈ 1.91 from the centre. The diameter has length 6 and acts at the origin. Weighting by length: y(G) = (3π × 6/π + 6 × 0)/(3π + 6) = 18/(3π + 6) = 1.17. Centre of mass (0, 1.17). B1 for the symmetry, B1 for the arc's centre at 6/π from the booklet, M1 for the lengths acting at their own centres, A1 for 1.17. The arc result is for a wire, not the semicircular lamina result 4r/(3π); the two have different formulae because the arc carries all its mass at radius r.

Question 11

A uniform lamina is formed from the rectangle with corners (0, 0) and (9, 6) by removing the triangle with vertices (9, 0), (9, 6) and (3, 6). Find the centre of mass of the lamina that remains.

Accepted Answer

Take the removed triangle as a negative area acting at its own centroid. The rectangle has area 54 with centre (4.5, 3). The triangle has area ½ × 6 × 6 = 18, and its centroid is the mean of its vertices, ((9 + 9 + 3)/3, (0 + 6 + 6)/3) = (7, 4). The area left is 54 − 18 = 36.
x(G) = (54 × 4.5 − 18 × 7)/36 = 117/36 = 3.25.
y(G) = (54 × 3 − 18 × 4)/36 = 90/36 = 2.5.
M1 for the removed piece entering with a negative area at its own centroid, A1 for the areas and centres, A1 for x(G), A1 for y(G). Both coordinates fall short of the rectangle's own centre (4.5, 3), so the centre of mass has moved left and down, away from the corner that was cut off, which is the check to make. The same point comes from splitting the remaining trapezium into positive pieces instead: the 3 by 6 rectangle with corners (0, 0) and (3, 6), area 18 at (1.5, 3), and the triangle with vertices (3, 0), (9, 0) and (3, 6), area 18 at (5, 2).

Question 12

A uniform wire of total length 12 cm is bent to form the triangle with vertices (0, 0), (4, 0) and (0, 3). Find the centre of mass of the framework, and compare it with the centre of mass of the uniform triangular lamina having the same three vertices.

Accepted Answer

For a framework the mass of each rod is proportional to its length, and each rod acts at its own midpoint. Set the work out as a table of length against midpoint.
The three rods have lengths 4, 3 and 5, the last by Pythagoras, giving a total of 12. Their midpoints are (2, 0), (0, 1.5) and (2, 1.5).
x(G) = [4(2) + 3(0) + 5(2)]/12 = 18/12 = 1.5.
y(G) = [4(0) + 3(1.5) + 5(1.5)]/12 = 12/12 = 1.
For the lamina the mass is spread over the area, and the centre of mass is the centroid, the average of the three vertices: x(G) = (0 + 4 + 0)/3 = 4/3 and y(G) = (0 + 0 + 3)/3 = 1.
The two heights agree at y = 1, which is a coincidence of these particular numbers rather than a rule. The x coordinates differ, 1.5 against 4/3, because the framework's heaviest rod is the hypotenuse, whose midpoint is at x = 2, and that pulls the framework to the right.
M1 for the lengths and their midpoints, A1 for the x coordinate, A1 for the y coordinate, M1 A1 for the centroid of the lamina, B1 for the comparison.

Question 13

The booklet places the centre of mass of a solid hemisphere of radius r at 3r/8 from the centre of its flat face, and that of a hemispherical shell at r/2. Explain why the shell's centre of mass lies further from the flat face than the solid's.

Accepted Answer

Both centres lie on the axis of symmetry through the centre of the flat face, so the comparison is one distance against another. B1 for the shared axis. The solid carries its mass throughout its interior, and the wide slices near the flat face hold far more material than the narrowing dome above, which drags the average height down to 3r/8. B1 for where the solid's mass sits. The shell carries none of its mass in the middle: everything lies on the curved surface, which sits at distance r from the centre and rises well above the flat face, so its average height is the greater, at r/2. B1 for the shell's mass being all at radius r. The same pattern repeats for the cone against the conical shell, h/4 against h/3.

Question 14

A uniform rectangular lamina has corners (0, 0), (12, 0), (12, 8) and (0, 8). A circular hole of radius 3 is cut from it, the centre of the hole being at the point (a, b). The centre of mass of the lamina that remains is at (5, 4). Explain why b = 4, and find a, giving an exact value and a value to 3 significant figures.

Accepted Answer

The rectangle has area 96 with centre (6, 4); the hole has area 9π at (a, b) and enters every moment with a minus sign, leaving an area of 96 − 9π.
For b, the given centre of mass lies on y = 4, which is an axis of symmetry of the rectangle. The pierced lamina can have its centre of mass on that line only if the hole is placed symmetrically about it too, so b = 4. The moment equation says the same thing: (96 × 4 − 9πb)/(96 − 9π) = 4 gives 384 − 9πb = 384 − 36π, so b = 4.
For a, take moments about the y-axis: (96 × 6 − 9πa)/(96 − 9π) = 5, so 576 − 9πa = 480 − 45π and 9πa = 96 + 45π. Hence a = 32/(3π) + 5, which is 8.40 to 3 significant figures.
B1 for b = 4 with the symmetry reason, M1 for a moment equation with the hole as a negative area, A1 for 9πa = 96 + 45π, A1 for the exact value, A1 for 8.40.
Check the answer against the picture. The hole then runs from x = 5.40 to x = 11.40 and from y = 1 to y = 7, so it lies wholly inside the rectangle; and it sits to the right of the centre (6, 4), so the centre of mass has moved left of it, which removing material on that side must do. Multiply through by 96 − 9π before touching the π: the working then stays in whole numbers and multiples of π until the division on the last line.

Question 15

A uniform circular disc of mass M and radius 6 has its centre at the origin. A circular hole of radius 2 is cut from it with its centre at (3, 0). State where the pierced disc would balance on a point support, with a reason. A particle of mass m is then attached at the point (6, 0) on the rim so that the centre of mass of the disc and particle together is at the origin. Find m in terms of M.

Accepted Answer

The hole has area π × 2² against π × 6², so its mass is M/9 and the pierced disc has mass 8M/9. By symmetry the centre of mass is on the x-axis. Treating the hole as a negative mass: x(G) = (M × 0 − (M/9) × 3)/(8M/9) = −3/8. The disc balances on a point support placed at (−3/8, 0), since the moments of its weight about the support cancel only when the support is under the centre of mass.
With the particle at (6, 0), the combined centre of mass is at the origin when the total moment about O is zero: (8M/9)(−3/8) + 6m = 0, so −M/3 + 6m = 0 and m = M/18.
B1 for the hole as M/9, M1 for the negative-mass moment, A1 for −3/8, B1 for the balance point with a reason, M1 for the moment equation with the particle, A1 for M/18. The particle is small because it acts at the rim, far from the origin, while the missing mass acted at only 3 units; a lever ratio of 6 to 3/8 is what leaves m at a sixteenth of the pierced disc's mass.