Edexcel International A Level Maths · M2 · Mechanics 2
Centres of mass
28 free, high-quality past-paper-style questions on Centres of mass for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 239 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-M2-3-B-01M2 · A2Easy5 marks
A light rod PQ has length 3.0 m. Particles of mass 4 kg, 2 kg and 6 kg are attached to the rod at P, at the point 1.2 m from P, and at Q respectively. The particles are modelled as point masses.
Figure: the light rod PQ with the three particles attached; not to scale.
(a)
Find the distance of the centre of mass of the system from P.
(2)
(b)
The 2 kg particle is moved to a new position on the rod so that the centre of mass of the system is 1.6 m from P. Find the distance of the new position from P.
(3)
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Answer
Mark
Notes
Part (a)
moments about P: 12x̄ = 4(0) + 2(1.2) + 6(3.0)
M1
x̄ = 20.4 ÷ 12 = 1.7 m
A1
Part (b)
new moment equation: 12 × 1.6 = 2d + 18
M1
correct equation with 19.2
A1
d = 0.6 m from P
A1
Exam tip: Write the moment equation as (total mass) × (position of centre of mass) = sum of the individual moments. It works the same way whether the unknown is a mass or a distance.
Three particles have masses 3 kg, 5 kg and 2 kg. Their position vectors are (2i + j) m, (−i + 4j) m and (6i − 3j) m respectively.
(a)
Find the position vector of the centre of mass of the three particles.
(3)
(b)
A fourth particle, of mass m kg, is placed at the point with position vector (4i + kj) m. The centre of mass of the four particles is at the point with position vector (2.5i + 2.5j) m. Find the value of m and the value of k.
(4)
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Answer
Mark
Notes
Part (a)
Σmr = 3(2i + j) + 5(−i + 4j) + 2(6i − 3j)
M1
= 13i + 17j
A1
r̄ = (13i + 17j) ÷ 10 = (1.3i + 1.7j) m
A1
Part (b)
i: 13 + 4m = 2.5(10 + m)
M1
m = 8
A1
j: 17 + 8k = 2.5 × 18
M1
k = 3.5
A1
Exam tip: Use the i equation first when it is the only one with a single unknown, then substitute into the j equation.
A piece of uniform wire has length 0.4π m. It is bent into a semicircular arc with centre O.
(a)
Find, in terms of π, the distance of the centre of mass of the arc from O.
(2)
(b)
The same wire is straightened and then bent into a circular arc which subtends an angle of 120° at its centre. Find the distance of the centre of mass of this arc from the centre of the circle, giving your answer to 3 significant figures.
(3)
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Answer
Mark
Notes
Part (a)
radius = 0.4 m
B1
distance = 2r/π = 0.8/π m
A1
Part (b)
radius: 0.4π = r × 2π/3, so r = 0.6 m
B1
use of r sin α/α with α = π/3: 0.6 sin(π/3) ÷ (π/3)
M1
= 0.9√3/π = 0.496 m
A1
Common error: Putting the whole angle (2π/3) into r sin α/α. The formula uses half the angle at the centre.
Figure 1 shows four particles A, B, C and D, of masses 3 kg, 2 kg, 4 kg and 1 kg respectively, at points of a coordinate grid in a horizontal plane. The grid squares have side 1 m and the particles lie at grid intersections.
(a)
Use Figure 1 to find the coordinates of the centre of mass of the four particles.
(3)
(b)
Particle C is removed. Find the distance through which the centre of mass moves.
(3)
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Answer
Mark
Notes
Part (a)
positions read from the grid: A(1, 4), B(5, 2), C(−2, 1), D(3, −2); total mass 10 kg
Common error: Sign slips with the negative coordinates. Write each particle's coordinates next to its mass before forming Σmx and Σmy; C contributes 4 × (−2) = −8 to Σmx.
Figure 1 shows a uniform lamina in the shape of a letter T. It consists of a rectangular bar, 12 cm by 3 cm, whose top edge is AB, joined to a rectangular stem, 4 cm by 9 cm, whose lower edge is PQ. The stem is placed centrally below the bar.
(a)
Explain why the centre of mass of the lamina lies on the line through the midpoint of PQ that is perpendicular to PQ.
(1)
(b)
Show that the centre of mass of the lamina is 7.5 cm from PQ.
(3)
(c)
The lamina is freely suspended from A and hangs in equilibrium. Find the angle between AB and the vertical.
(3)
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Answer
Mark
Notes
Part (a)
the lamina is symmetrical about this line, so its centre of mass lies on it
B1
Part (b)
areas: bar 36 cm2, stem 36 cm2
B1
centres 10.5 cm and 4.5 cm from PQ: 72ȳ = 36(10.5) + 36(4.5)
M1
ȳ = 540 ÷ 72 = 7.5 cm
A1
Part (c)
G is 6 cm from A measured along AB and 12 − 7.5 = 4.5 cm below AB
M1
tan θ = 4.5/6
A1
θ = 36.9°
A1
Exam tip: Measure from the edge you are working with: the 7.5 cm is measured from PQ, so G is 12 − 7.5 = 4.5 cm from AB.
A piece of uniform wire is bent into the shape ABCD shown in Figure 1, where AB = 24 cm, BC = 18 cm and CD = 12 cm. AB and CD are both perpendicular to BC and lie on the same side of BC.
(a)
Find the distance of the centre of mass of the wire from AB and from BC.
(4)
(b)
The wire is freely suspended from A and hangs in equilibrium. Find the angle between AB and the vertical.
(3)
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Answer
Mark
Notes
Part (a)
lengths 24, 18 and 12, total 54
B1
54x̄ = 24(0) + 18(9) + 12(18) = 378
M1
x̄ = 7 cm from AB
A1
54ȳ = 24(12) + 18(0) + 12(6) = 360, so ȳ = 20/3 = 6.67 cm from BC
A1
Part (b)
G is 24 − 20/3 = 52/3 cm along AB from A and 7 cm from AB
M1
tan θ = 7 ÷ 52/3 = 21/52
A1
θ = 22.0°
A1
Common error: Treating the wire as a lamina. For a framework each piece is weighted by its length and acts at its midpoint; the region enclosed by the wire plays no part.
Figure 1 shows a uniform lamina in the shape of a sector OAB of a circle with centre O and radius 15 cm. The angle AOB is 1.2 radians.
(a)
Find the distance of the centre of mass G of the lamina from O.
(2)
(b)
The lamina is freely suspended from A and hangs in equilibrium. Find the angle between OA and the vertical.
(4)
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Answer
Mark
Notes
Part (a)
2r sin α/(3α) with α = 0.6
M1
OG = 30 sin 0.6 ÷ 1.8 = 9.41 cm
A1
Part (b)
G lies on the bisector of angle AOB, so angle AOG = 0.6 rad
B1
G is 9.411 cos 0.6 = 7.767 cm from O along OA and 9.411 sin 0.6 = 5.314 cm from OA
M1
tan θ = 5.314 ÷ (15 − 7.767)
M1
θ = 36.3°
A1
Common error: Using α = 1.2. In 2r sin α/(3α) the angle α is half the angle of the sector, so here it is 0.6, and the calculator must be in radian mode.
A uniform lamina is made by joining a rectangle PQRS and a semicircle. PQ = 16 cm and QR = 12 cm. The semicircle has diameter QR and lies outside the rectangle. Take PQ as the x-axis and PS as the y-axis.
Figure: the lamina formed from rectangle PQRS and the semicircle on diameter QR; not to scale.
(a)
Explain why the centre of mass of the lamina is 6 cm from PQ.
(1)
(b)
Show that the distance of the centre of mass of the lamina from PS is (1680 + 288π)/(192 + 18π) cm.
(4)
(c)
The lamina is freely suspended from P and hangs in equilibrium. Find the angle between PQ and the vertical.
(3)
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Answer
Mark
Notes
Part (a)
the lamina is symmetrical about the line joining the midpoints of PS and QR, which is 6 cm from PQ
B1
Part (b)
areas 192 and 18π
B1
semicircle's centre of mass at 16 + 4(6)/3π = 16 + 8/π from PS
= 1536 + 288π + 144 = 1680 + 288π, giving the result
A1
Part (c)
x̄ = 10.40… cm
B1
tan θ = 6/10.40
M1
θ = 30.0°
A1
Exam tip: In a “show that”, write the moment of the semicircle before simplifying: 18π × (16 + 8/π). The π in 8/π cancels, which is how the given form appears.
A uniform circular disc has centre O and radius 6a. A circular hole of radius 2a is cut from the disc. The centre of the hole, X, is at a distance 3a from O. The remaining lamina is called L.
Figure: the disc of radius 6a with the hole of radius 2a removed (dashed); not to scale.
(a)
Show that the centre of mass of L is at a distance 3a/8 from O.
(4)
(b)
The point B is on the circumference of L and OB is perpendicular to OX. L is freely suspended from B and hangs in equilibrium. Find the angle between OB and the vertical.
(3)
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Answer
Mark
Notes
Part (a)
areas: 36πa2, 4πa2, remainder 32πa2
B1
moments about O along OX: 32πa2d = 36πa2 × 0 − 4πa2 × 3a
M1
correct equation
A1
d = −3a/8, so 3a/8 from O (on the opposite side of O from X)
A1
Part (b)
G is 6a from B along BO and 3a/8 from the line BO
M1
tan θ = (3a/8) ÷ 6a = 1/16
A1
θ = 3.58°
A1
Exam tip: With a hole, the table has three rows: the whole disc, the hole (negative area), and the remainder. The factor π cancels, so leave it in until it does.
A uniform lamina OAB is a sector of a circle with centre O, radius 10 cm and angle AOB = 90°. The triangle OAB is removed, leaving a uniform lamina in the shape of the segment bounded by the chord AB and the arc AB.
Figure: the sector OAB with triangle OAB removed (dashed), leaving the shaded segment; not to scale.
(a)
Show that the centre of mass of the sector OAB is at a distance 40√2/(3π) cm from O.
(2)
(b)
Find the distance of the centre of mass of the segment from O, giving your answer to 3 significant figures.
centre of mass of triangle OAB on the line of symmetry, ⅔ × 10 cos 45° = 10√2/3 from O
B1
(25π − 50)d = 25π × 40√2/(3π) − 50 × 10√2/3
M1
d = 500√2/3 ÷ (25π − 50) = 8.26 cm
A1
Exam tip: The median of triangle OAB from O lies along the line of symmetry and has length 10 cos 45°. Two-thirds of it, measured from O, gives the triangle's centre of mass.
A uniform rectangular sign ABCD has mass 3 kg, with AB = 0.6 m and BC = 0.4 m. It is free to rotate in a vertical plane about a fixed smooth horizontal axis which passes through the point P on AB, where AP = 0.2 m, and which is perpendicular to the sign. A particle of mass m kg is attached to the sign at A. The sign rests in equilibrium with AB horizontal and C and D below AB.
Figure: the sign ABCD on the axis through P, with the particle at A; not to scale.
(a)
Find the value of m.
(3)
(b)
Find the magnitude of the force exerted on the sign by the axis.
(2)
(c)
The particle falls off. The sign comes to rest in a new equilibrium position. Find the angle between AB and the vertical in this position.
(3)
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Answer
Mark
Notes
Part (a)
the combined centre of mass is vertically below P, or moments about P
M1
3g × 0.1 = mg × 0.2
A1
m = 1.5
A1
Part (b)
resolving vertically: R = (3 + 1.5)g
M1
R = 44.1 N (accept 44 N) ft on m
A1
Part (c)
centre of mass of the sign is 0.1 m from P along AB and 0.2 m from AB
B1
PG vertical: tan θ = 0.2/0.1
M1
θ = 63.4°
A1
Exam tip: For a body on a smooth axis, moments about the axis avoid the unknown force from the axis. Then resolve to find that force if you need it.
A uniform lamina ABCD is in the shape of a trapezium with AB = 12 cm, DC = 6 cm and AD = 8 cm. AB is parallel to DC, and angle DAB = angle ADC = 90°.
Figure: the trapezium lamina ABCD; not to scale.
(a)
Show that the centre of mass of the lamina is 14/3 cm from AD and 32/9 cm from AB.
(4)
(b)
The lamina stands in a vertical plane with AB in contact with a rough plane, along a line of greatest slope, with A lower than B. The plane is inclined at angle α to the horizontal. Assuming the lamina does not slide, find the value of α at which the lamina is about to topple.
(3)
(c)
The coefficient of friction between the lamina and the plane is 0.8. The angle of the plane is slowly increased from zero. Determine whether the lamina slides or topples first.
(2)
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Answer
Mark
Notes
Part (a)
split into rectangle (area 48, centre (3, 4)) and triangle (area 24, centre (8, 8/3))
B1
72x̄ = 48(3) + 24(8) = 336 → x̄ = 14/3
A1
72ȳ = 48(4) + 24(8/3)
M1
= 256 → ȳ = 32/9
A1
Part (b)
about to topple when G is vertically above A
M1
tan α = (14/3) ÷ (32/9) = 21/16
A1
α = 52.7°
A1
Part (c)
slides when tan α > 0.8, i.e. at α = 38.7°
M1
38.7° < 52.7°, so the lamina slides first
A1
Common error: Measuring the distance along AB from B instead of A. The lamina tips about the lower end of its base, so measure from A.
A uniform rectangular tray PQRS of mass 1.2 kg has PQ = 50 cm and QR = 30 cm. Three small objects, modelled as particles, rest on the tray: 0.3 kg at the point 10 cm from PS and 20 cm from PQ, 0.5 kg at the point 40 cm from PS and 5 cm from PQ, and 0.4 kg at the point 35 cm from PS and 25 cm from PQ. The thickness of the tray is ignored.
Figure: the tray PQRS (50 cm by 30 cm) with the three objects, positions in cm from PS and PQ.
(a)
Find the distance of the centre of mass of the tray and objects from PS and from PQ.
(4)
(b)
A fourth object, of mass 0.6 kg, is placed on the tray so that the centre of mass of the whole system is at the centre of the tray. Find the distances of this object from PS and from PQ.
(4)
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Answer
Mark
Notes
Part (a)
tray's centre of mass at (25, 15); total mass 2.4 kg
Figure 1 shows a uniform lamina shaped like an arrow. It consists of a rectangle of length 20 cm and width 6 cm, with one end EF, joined to an isosceles triangle PQR. The base PQ of the triangle is 12 cm long and passes through the other end of the rectangle, and the height of the triangle is 12 cm. The lamina is symmetrical about the line through R perpendicular to PQ.
(a)
Show that the centre of mass of the lamina is 15.25 cm from EF.
(4)
(b)
The lamina is freely suspended from Q and hangs in equilibrium. Find the angle between the axis of symmetry and the vertical.
(3)
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Answer
Mark
Notes
Part (a)
areas: rectangle 120 cm2, triangle 72 cm2
B1
triangle's centre of mass is ⅓ × 12 = 4 cm from PQ, so 24 cm from EF
B1
192x̄ = 120(10) + 72(24)
M1
x̄ = 2928 ÷ 192 = 15.25 cm
A1
Part (b)
G is 20 − 15.25 = 4.75 cm from PQ and 6 cm from Q measured parallel to PQ
M1
tan θ = 6/4.75
A1
θ = 51.6°
A1
Exam tip: The centre of mass of a triangle is one third of its height from the base, which is the same as two thirds of the way along the median from the opposite vertex.
Figure 1 shows a uniform lamina L of mass M in the shape of half of a ring. It is the region between two semicircles with common centre O and radii 10 cm and 6 cm. The points A, C, O, D and B lie in that order on its straight edge, with OA = OB = 10 cm and OC = OD = 6 cm.
(a)
Show that the centre of mass of L is at a distance 49/(3π) cm from O.
(4)
(b)
L is freely suspended from C and hangs in equilibrium. Find the angle between AB and the vertical.
(3)
(c)
A particle of mass kM is attached to L at A. The lamina is again suspended from C, and now hangs in equilibrium with AB horizontal. Find the value of k.
(3)
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Answer
Mark
Notes
Part (a)
areas: 50π and 18π, so L has area 32π
B1
centres of mass of the semicircles: 4(10)/(3π) = 40/(3π) and 4(6)/(3π) = 8/π from O
B1
32πd = 50π × 40/(3π) − 18π × 8/π
M1
32πd = 2000/3 − 144 = 1568/3, so d = 1568/(96π) = 49/(3π)
A1
Part (b)
G is 6 cm from C measured along AB and 49/(3π) = 5.199 cm from AB
M1
tan θ = 5.199/6
A1
θ = 40.9°
A1
Part (c)
the combined centre of mass is vertically below C, so it is 6 cm from O towards A
M1
moments along AB about O: kM × 10 = (M + kM) × 6
A1
k = 1.5
A1
Exam tip: In (c) only distances measured along AB matter: the combined centre of mass must be directly below C, so its distance from the axis of symmetry must equal OC.
A uniform triangular lamina ABC has mass 4 kg. It is free to rotate in a vertical plane about a fixed smooth horizontal axis which passes through A and is perpendicular to the lamina. A particle of mass m kg is attached to the lamina at B. Figure 1 shows the lamina resting in equilibrium with AC horizontal. AC = 15 cm, and B is 6 cm horizontally from the vertical through A, on the opposite side from C, and 12 cm below the level of AC.
(a)
Find the value of m.
(4)
(b)
Find the magnitude of the force exerted on the lamina by the axis.
(2)
(c)
The particle falls off, and the lamina comes to rest hanging freely from the axis. Find the angle between AC and the vertical in this position.
(3)
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Answer
Mark
Notes
Part (a)
with A as origin, the lamina's centre of mass is at ((0 + 15 − 6)/3, (0 + 0 − 12)/3) = (3, −4)
B1
the combined centre of mass is on the vertical through A: moments about A
M1
4g × 3 = mg × 6
A1
m = 2
A1
Part (b)
resolving vertically: R = (4 + 2)g
M1
R = 58.8 N ft on m
A1
Part (c)
AG is vertical; G is 3 cm from A measured along AC and 4 cm from AC
M1
tan θ = 4/3
A1
θ = 53.1°
A1
Exam tip: Moments about the axis leave out the unknown force at A, and only horizontal distances from the vertical through A are needed.
Figure 1 shows a uniform lamina in the shape of an arch. It consists of a rectangle with base AB = 8 cm and height 10 cm, together with a semicircle of diameter 8 cm whose diameter is the top edge of the rectangle. The lamina stands in a vertical plane with AB along a line of greatest slope of a rough plane inclined at angle α to the horizontal, with A lower than B.
(a)
Show that the centre of mass of the lamina is at a distance (166 + 30π)/(30 + 3π) cm from AB.
(4)
(b)
The lamina does not slide. Find the value of α at which the lamina is about to topple.
(3)
(c)
The coefficient of friction between the lamina and the plane is μ. The angle α is slowly increased from zero and the lamina topples before it slides. Find the range of possible values of μ.
(2)
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Answer
Mark
Notes
Part (a)
areas: rectangle 80 cm2, semicircle 8π cm2
B1
semicircle's centre of mass is 10 + 4(4)/(3π) = 10 + 16/(3π) from AB
B1
(80 + 8π)ȳ = 80 × 5 + 8π(10 + 16/(3π))
M1
= 400 + 80π + 128/3; multiplying numerator and denominator by 3/8 gives ȳ = (166 + 30π)/(30 + 3π)
A1
Part (b)
about to topple when G is vertically above A
M1
tan α = 4 ÷ 6.601
A1
α = 31.2°
A1
Part (c)
the lamina must not slip at the toppling angle, so μ > tan 31.2° = 4/6.601
M1
μ> 0.606
A1
Common error: Measuring from B. On the slope the lamina turns about its lower corner A, so the distance that matters is from A to the axis of symmetry, 4 cm.
A uniform wire is bent to form the framework shown in Figure 1. It consists of two straight pieces OA and OB, each of length 20 cm, with angle AOB = 90°, and the circular arc AB with centre O.
(a)
Show that the centre of mass of the framework is at a distance 60√2/(4 + π) cm from O.
(5)
(b)
The framework is freely suspended from A and hangs in equilibrium. Find the angle between OA and the vertical.
(3)
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Answer
Mark
Notes
Part (a)
lengths: OA = OB = 20 and arc 10π
B1
arc's centre of mass on the bisector of angle AOB, 20 sin(π/4) ÷ (π/4) = 40√2/π from O
B1
each straight piece has its midpoint 10 cm from O, which is 10 cos 45° = 5√2 along the bisector
B1
(40 + 10π)d = 10π × 40√2/π + 40 × 5√2
M1
d = 600√2/(40 + 10π) = 60√2/(4 + π)
A1
Part (b)
G is 60/(4 + π) = 8.401 cm from OA, and 20 − 8.401 = 11.60 cm from A measured along AO
M1
tan θ = 8.401/11.60
A1
θ = 35.9°
A1
Exam tip: By symmetry G lies on the bisector of angle AOB, so its distances from OA and from OB are equal, each d cos 45°. Using these distances makes (b) quick.
Figure 1 shows a uniform lamina L. It is formed by removing the sector AOB, where angle AOB = 60°, from a circular disc with centre O and radius 9 cm. The dashed line is the line of symmetry of L.
(a)
Show that the centre of mass of L is at a distance 18/(5π) cm from O.
(4)
(b)
L is freely suspended from A and hangs in equilibrium. Find the angle between the line of symmetry of L and the vertical.
(4)
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Answer
Mark
Notes
Part (a)
areas: disc 81π, sector 13.5π, so L has area 67.5π
B1
sector's centre of mass on the line of symmetry, 2(9) sin(π/6) ÷ (3 × π/6) = 18/π from O
B1
67.5πd = 81π × 0 − 13.5π × 18/π
M1
d = −243/(67.5π), so 18/(5π) from O, on the opposite side of O from the gap
A1
Part (b)
A is 9 cos 30° = 7.794 cm from O along the line of symmetry and 9 sin 30° = 4.5 cm from it
B1
measured along the line of symmetry, A and G are 7.794 + 1.146 = 8.940 cm apart
M1
tan θ = 4.5/8.940
M1
θ = 26.7°
A1
Common error: Placing G on the same side of O as the gap. Removing material moves the centre of mass away from the part removed.
A uniform wire, of total mass M, is bent to form a framework consisting of a straight piece AB of length 24 cm and a semicircular arc with diameter AB. O is the midpoint of AB.
Figure: the framework made from the straight wire AB and the semicircular arc on AB; not to scale.
(a)
Show that the centre of mass of the framework is at a distance 24/(π + 2) cm from AB.
(4)
(b)
The framework is freely suspended from A and hangs in equilibrium. Find the angle between AB and the vertical.
(3)
(c)
A particle of mass kM is now attached to the framework at B. The framework is again freely suspended from A and hangs in equilibrium with AB at 10° to the vertical. Find the value of k.
(4)
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Answer
Mark
Notes
Part (a)
lengths: arc 12π, diameter 24
B1
arc's centre of mass 2(12)/π = 24/π from O
B1
(12π + 24)d = 12π × 24/π + 24 × 0
M1
d = 288/(12π + 24) = 24/(π + 2)
A1
Part (b)
G is 12 cm along AB from A and 24/(π + 2) from AB
M1
tan θ = (24/(π + 2)) ÷ 12 = 2/(π + 2)
A1
θ = 21.3°
A1
Part (c)
combined centre of mass: along AB from A, (12M + 24kM)/(M + kM); from AB, (24/(π + 2))M/(M + kM)
M1
tan 10° = (24/(π + 2)) ÷ (12 + 24k)
M1
correct equation
A1
k = 0.603
A1
Exam tip: When a particle is added, the factor (M + kM) appears in both coordinates of the combined centre of mass and cancels in the tangent.
A uniform square lamina OABC has side 6a, with O(0, 0), A(6a, 0), B(6a, 6a) and C(0, 6a). A quarter-circle with centre O and radius 3a is removed. The remaining lamina is L.
Figure: the square OABC with the quarter-circle of radius 3a, centre O, removed; not to scale.
(a)
Show that the centre of mass of L is at the point (ka, ka), where k = 44/(16 − π).
(5)
(b)
L is freely suspended from A and hangs in equilibrium. Find the angle between AB and the vertical.
(3)
(c)
L is placed in a vertical plane with the edge AB in contact with a rough plane inclined at angle β to the horizontal, AB along a line of greatest slope and B lower than A. The coefficient of friction is 1.2. The angle β is slowly increased from zero. Show that L topples before it slides, and state the value of β at which it topples.
(4)
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Answer
Mark
Notes
Part (a)
by symmetry about OB, x̄ = ȳ
B1
areas: square 36a2, quarter-circle 9πa2/4
B1
quarter-circle's centre of mass at x = 4(3a)/3π = 4a/π
A uniform triangular lamina ABC has mass 4m, with angle ABC = 90°, AB = 9 cm and BC = 12 cm. A particle of mass m is attached at A and a particle of mass 2m is attached at C. The system is modelled as a lamina with two point masses.
Figure: the triangular lamina ABC (mass 4m) with particles of mass m at A and 2m at C; not to scale.
(a)
Find the distance of the centre of mass of the system from AB and from BC.
(5)
(b)
The system is freely suspended from A and hangs in equilibrium. Find the angle between AB and the vertical.
(3)
(c)
The particle at C is replaced by a particle of mass km; the particle at A is unchanged. The system is freely suspended from B and hangs in equilibrium with BC at 45° to the vertical. Find the value of k.
(4)
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Answer
Mark
Notes
Part (a)
lamina's centre of mass is 4 cm from AB and 3 cm from BC
B1
7mx̄ = 4m(4) + m(0) + 2m(12)
M1
x̄ = 40/7 = 5.71 cm from AB
A1
7mȳ = 4m(3) + m(9) + 2m(0)
M1
ȳ = 3 cm from BC
A1
Part (b)
G is 40/7 from AB and 9 − 3 = 6 cm along AB from A
M1
tan θ = (40/7) ÷ 6
A1
θ = 43.6°
A1
Part (c)
BG at 45° to BC, so G is equidistant from AB and BC
M1
distance from AB: (16 + 12k)/(5 + k); from BC: 21/(5 + k)
A1
16 + 12k = 21
M1
k = 5/12
A1
Common error: Forgetting to include the mass of the lamina in the total. Here the total is 4m + m + 2m = 7m, not 3m.
A uniform semicircular lamina has mass 2 kg, centre O and diameter AB, where AB = 0.6 m. The lamina is free to rotate in a vertical plane about a fixed smooth horizontal axis through A, perpendicular to the lamina.
Figure: the semicircular lamina with diameter AB, free to rotate about the axis through A; not to scale.
(a)
Find, in terms of π, the distance of the centre of mass G of the lamina from O.
(2)
(b)
The lamina hangs freely in equilibrium. Find the angle between AB and the vertical.
(3)
(c)
The lamina is held in equilibrium with AB vertical and B below A by a horizontal force of magnitude P N acting at B in the plane of the lamina. Find P.
(3)
(d)
Find the magnitude of the force exerted on the lamina by the axis in this position.
(3)
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Answer
Mark
Notes
Part (a)
4r/3π with r = 0.3
M1
OG = 0.4/π m
A1
Part (b)
AG vertical; G is 0.3 m along AB from A and 0.4/π m from AB
M1
tan θ = (0.4/π) ÷ 0.3 = 4/(3π)
A1
θ = 23.0°
A1
Part (c)
moments about A
M1
P × 0.6 = 2g × 0.4/π
A1
P = 4.16
A1
Part (d)
horizontal component P, vertical component 2g
M1
√(4.162 + 19.62) ft on P
M1
= 20.0 N
A1
Exam tip: Take moments about the axis so the unknown force at the axis does not appear. Then resolve horizontally and vertically to find that force.
Figure 1 shows a uniform lamina PQTUS of mass 3 kg. It is formed from the rectangle PQRS, where PQ = 24 cm and PS = 18 cm, by cutting off the triangle UTR, where U is the point on SR with SU = 12 cm and T is the point on QR with QT = 6 cm.
(a)
Find the distance of the centre of mass of the lamina from PS and from PQ.
(5)
(b)
The lamina is free to rotate in a vertical plane about a fixed smooth horizontal axis through U, perpendicular to the lamina. A particle of mass m kg is attached at T, and the lamina rests in equilibrium with PQ horizontal, as shown in Figure 1. Find the value of m.
(4)
(c)
The particle is removed and the lamina comes to rest hanging freely from the axis. Find the angle between PQ and the vertical in this position.
G is 12 − 10.4 = 1.6 cm horizontally from U and 18 − 8 = 10 cm below SR
M1
UG is vertical and PQ is parallel to SR: tan θ = 10/1.6
A1
θ = 80.9°
A1
Exam tip: A table with columns for area, distance from PS and distance from PQ, with a negative row for the triangle cut off, keeps (a) organised and gives both coordinates at once.
Figure 1 shows a uniform lamina ABCD. It is a rectangle with AB = 12 cm and BC = h cm, where h > 4, from which a semicircle of radius 4 cm has been removed. The diameter of the semicircle lies along AB and its centre is the midpoint of AB.
(a)
Show that the centre of mass of the lamina is at a distance (9h2 − 64)/(18h − 12π) cm from AB.
(5)
(b)
The lamina stands in a vertical plane with AB along a line of greatest slope of a rough plane inclined at 30° to the horizontal, with A lower than B, as shown in Figure 1 with α = 30°. The plane is rough enough to prevent sliding. Find the greatest value of h for which the lamina does not topple.
(5)
(c)
In this part h = 15 and the coefficient of friction between the lamina and the plane is 0.5. The angle α of the plane is slowly increased from zero. Determine whether the lamina slides or topples first.
(3)
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Answer
Mark
Notes
Part (a)
areas: rectangle 12h, semicircle 8π
B1
semicircle's centre of mass is 4(4)/(3π) = 16/(3π) from AB
B1
(12h − 8π)ȳ = 12h × h/2 − 8π × 16/(3π)
M1
(12h − 8π)ȳ = 6h2 − 128/3
A1
multiplying numerator and denominator by 3/2: ȳ = (9h2 − 64)/(18h − 12π)
A1
Part (b)
on the point of toppling, G is vertically above A
M1
tan 30° = 6/ȳ, so ȳ = 6√3
A1
9h2 − 64 = 6√3(18h − 12π)
M1
9h2 − 187.06h + 327.78 = 0, solved by formula
M1
h = 18.85 (18.853…; accept 18.8 or 18.85, not 18.9, which would topple; the other root, 1.93, is less than 4)
A1
Part (c)
ȳ = (2025 − 64)/(270 − 12π) = 8.442
B1
topples when tan α = 6/8.442, α = 35.4°; slides when tan α = 0.5, α = 26.6°
M1
26.6° < 35.4°, so the lamina slides first
A1
Common error: Thinking the gap under the arch moves the pivot. The lamina still turns about its lowest point of contact, the corner A, so the horizontal half-width that matters is 6 cm.
A framework is made from three uniform rods AB, BC and AC, rigidly joined to form a right-angled triangle with angle ABC = 90°. The rods are not all made of the same material. Figure 1 shows the length and mass of each rod.
(a)
Find the distance of the centre of mass of the framework from AB and from BC.
(5)
(b)
The framework is freely suspended from A and hangs in equilibrium. Find the angle between AB and the vertical.
(3)
(c)
A particle of mass m kg is attached to the framework at C. The framework is again suspended from A, and now hangs in equilibrium with AB at 30° to the vertical. Find the value of m.
(4)
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Answer
Mark
Notes
Part (a)
with A as origin and AB along the x-axis, the midpoints are (8, 0), (16, 6) and (8, 6)
B1
10x̄ = 2(8) + 3(16) + 5(8) = 104
M1
x̄ = 10.4, so the distance from BC is 16 − 10.4 = 5.6 cm
A1
10ȳ = 2(0) + 3(6) + 5(6) = 48
M1
ȳ = 4.8 cm from AB
A1
Part (b)
G is 10.4 cm from A measured along AB and 4.8 cm from AB
M1
tan θ = 4.8/10.4
A1
θ = 24.8°
A1
Part (c)
new centre of mass: ((104 + 16m)/(10 + m), (48 + 12m)/(10 + m))
M1
tan 30° = (48 + 12m)/(104 + 16m)
M1
correct equation, e.g. 12m − 16m tan 30° = 104 tan 30° − 48
A1
m = 4.36
A1
Exam tip: The rods have different masses, so weight each midpoint by the mass of its rod, not by its length.
A uniform square lamina ABCD has side 12 cm. The points E on DC and F on DA are such that DE = DF = 6 cm. The corner triangle DEF is folded along EF so that it lies flat on the rest of the lamina, with D at the point D′, as shown in Figure 1.
(a)
Show that the centre of mass of the folded lamina is 6.25 cm from AD and 5.75 cm from AB.
(5)
(b)
The folded lamina is freely suspended from A and hangs in equilibrium. Find the angle between AB and the vertical.
(3)
(c)
The folded lamina stands in a vertical plane with AB along a line of greatest slope of a rough plane inclined at angle β to the horizontal, with B lower than A. The lamina does not slide. Find the value of β at which it is about to topple.
(3)
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Answer
Mark
Notes
Part (a)
with A as origin: square, area 144, centre of mass (6, 6)
B1
triangle DEF, area 18, has centre of mass (2, 10) before folding and (4, 8) after
B1
144x̄ = 144(6) − 18(2) + 18(4)
M1
x̄ = 900 ÷ 144 = 6.25
A1
144ȳ = 864 − 18(10) + 18(8) = 828, so ȳ = 5.75
A1
Part (b)
G is 6.25 cm from A measured along AB and 5.75 cm from AB
M1
tan θ = 5.75/6.25
A1
θ = 42.6°
A1
Part (c)
about to topple when G is vertically above B
M1
tan β = (12 − 6.25)/5.75 = 1
A1
β = 45°
A1
Exam tip: Folding does not change the mass. Treat it as removing the triangle from its old position and adding it at its new position: one negative row and one positive row in the table.
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