2022 May/June · Paper 32 · Q4 (a, b)8 marksCentre of mass of composite bodies · Suspension, toppling & sliding
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Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
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Q1Centre of mass of composite bodies · Core · Paper 3 style · 4 marks
(a)A uniform L-shaped lamina is a sign for a shop in Peshawar. Taking axes along two edges, its vertices are (0, 0), (6, 0), (6, 2), (2, 2), (2, 5) and (0, 5), with units in decimetres. Find the coordinates of the centre of mass of the lamina.[4]
Q2Centre of mass of composite bodies · Core · Paper 3 style · 5 marks
(a)A uniform square lamina OABC has side 8 cm. A circular hole of radius 2 cm is cut from it, with its centre at the point 2 cm from both OA and OC. Find the distance of the centre of mass of the remaining lamina from OC, giving your answer correct to 3 significant figures.[4]
(b)State the distance of the centre of mass from OA, giving a reason.[1]
Show answer
(a) 4.49 cm
B1 areas 64 and 4π; M1 moments: 64 × 4 − 4π × 2 = (64 − 4π)x̄; A1 x̄ = (256 − 8π)/(64 − 4π); A1 4.49 cm.
(b) 4.49 cm, by symmetry about the diagonal OB
B1
Q3Centre of mass of composite bodies · Core · Paper 3 style · 4 marks
(a)A toy is made from a uniform solid cylinder of radius 3 cm and height 10 cm joined to a uniform solid hemisphere of radius 3 cm made of the same material, so that the plane face of the hemisphere coincides with one end of the cylinder. Find the distance of the centre of mass of the toy from the free plane end of the cylinder.[4]
Show answer
(a) 289/48 cm ≈ 6.02 cm
B1 volumes 90π and 18π; B1 hemisphere centre of mass 3 × 3/8 = 9/8 cm from its plane face, i.e. 10 + 9/8 from the end; M1 (90π × 5 + 18π × 89/8) = 108π x̄; A1 x̄ = 6.02 cm.
Q4Centre of mass of composite bodies · Core · Paper 3 style · 4 marks
(a)A uniform rectangular lamina ABCD of mass 2 kg has AB = 0.4 m and BC = 0.3 m. A particle of mass 0.5 kg is attached at C. Find the distances of the centre of mass of the combined body from AB and from AD.[4]
Show answer
(a) 0.18 m from AB, 0.24 m from AD
M1 moments about AD: 2 × 0.2 + 0.5 × 0.4 = 2.5x̄; A1 x̄ = 0.24 m; M1 moments about AB: 2 × 0.15 + 0.5 × 0.3 = 2.5ȳ; A1 ȳ = 0.18 m.
Q5Centre of mass of composite bodies · Standard · Paper 3 style · 6 marks
(a)A uniform lamina is formed from a rectangle ABCD, with AB = 12 cm and BC = 4 cm, and a semicircle with diameter DC drawn outside the rectangle. Show that the distance of the centre of mass of the lamina from AB is (40 + 12π)/(8 + 3π) cm.[5]
(b)Find this distance correct to 3 significant figures.[1]
Show answer
(a) Shown
B1 areas 48 and 18π; B1 semicircle centre of mass 4r/(3π) = 8/π from DC; M1 moments about AB: 48 × 2 + 18π(4 + 8/π); M1 divide by 48 + 18π; A1 (240 + 72π)/(48 + 18π) = (40 + 12π)/(8 + 3π).
(b) 4.46 cm
B1
Q6Centre of mass of composite bodies · Standard · Paper 3 style · 5 marks
(a)A uniform solid cone has base radius 4 cm and height 12 cm. The top part, a cone of height 6 cm with the same vertex, is removed to leave a frustum. Show that the centre of mass of the frustum is 33/14 cm from its larger plane face.[5]
Show answer
(a) Shown
B1 volumes in ratio 16 × 12 : 4 × 6 = 192 : 24 (small cone radius 2); B1 centres of mass 3 cm and 6 + 1.5 = 7.5 cm from the base; M1 192 × 3 − 24 × 7.5 = 168x̄; M1 x̄ = 396/168; A1 33/14.
Q7Centre of mass of composite bodies · Standard · Paper 3 style · 4 marks
(a)A framework is made from three uniform rods of the same material: AB = 6 m, BC = 8 m and AC = 10 m, forming a right-angled triangle with the right angle at B. Find the distances of the centre of mass of the framework from BC and from AB.[4]
Show answer
(a) 2 m from BC, 3 m from AB
B1 masses proportional to 6, 8, 10 with midpoints (0, 3), (4, 0), (4, 3) taking B as origin, BC along the x-axis; M1 x̄ = (6 × 0 + 8 × 4 + 10 × 4)/24 = 3; M1 ȳ = (6 × 3 + 0 + 10 × 3)/24; A1 2 m from BC and 3 m from AB.
Q8Centre of mass of composite bodies · Standard · Paper 3 style · 6 marks
(a)A uniform wire is bent into the shape of a semicircular arc of radius a together with its diameter, forming a closed 'D' shape. Show that the centre of mass is at a distance 2a/(π + 2) from the diameter.[4]
(b)The shape is freely suspended from one end of the diameter. Find the angle the diameter makes with the vertical.[2]
Show answer
(a) Shown
B1 arc centre of mass 2a/π from the centre (from a sin α/α with α = π/2); B1 masses ∝ πa and 2a; M1 πa × 2a/π + 2a × 0 = (πa + 2a)ȳ; A1 2a/(π + 2).
(b) tan⁻¹(2/(π + 2)) ≈ 21.3°
M1 tan θ = (2a/(π + 2))/a; A1 21.3°.
Q9Centre of mass of composite bodies · Standard · Paper 3 style · 4 marks
(a)A lampshade is modelled as a closed hollow cone made from thin uniform sheet metal: a conical curved surface of base radius 3 cm and height 4 cm, together with a circular base of radius 3 cm made from the same sheet. Find the distance of the centre of mass of the lampshade from its base.[4]
Show answer
(a) 5/6 cm
B1 curved surface area πrl = 15π (slant height 5); B1 its centre of mass is h/3 = 4/3 cm from the base; M1 15π × 4/3 + 9π × 0 = 24π ȳ; A1 5/6 cm.
Q10Centre of mass of composite bodies · Challenge · Paper 3 style · 6 marks
(a)A uniform solid hemisphere of radius 2a has a concentric hemisphere of radius a removed, the two plane faces being in the same plane. Show that the centre of mass of the remaining solid is at a distance 45a/56 from the plane face.[5]
(b)This solid rests in equilibrium with its curved surface on a smooth horizontal plane. Explain why the plane face is horizontal.[1]
Show answer
(a) Shown
B1 volumes in ratio 8 : 1; B1 centres of mass 3(2a)/8 = 3a/4 and 3a/8; M1 8 × 3a/4 − 1 × 3a/8 = 7ȳ; M1 6a − 3a/8 = 45a/8; A1 ȳ = 45a/56.
(b) The normal reaction passes through the centre O (normal to a sphere), so O and the centre of mass must be in a vertical line; this line is the axis, so the plane face is horizontal
B1
Q11Centre of mass of composite bodies · Challenge · Paper 3 style · 7 marks
(a)A uniform square lamina OABC has vertices O(0, 0), A(6a, 0), B(6a, 6a) and C(0, 6a). The triangle OPC, where P is the point (ka, 3a) with 0 < k < 6, is removed. Show that the x-coordinate of the centre of mass of the remaining lamina is (108 − k²)a/(36 − 3k).[4]
M1 3(108 − k²) = 11(36 − 3k); A1 k² − 11k + 24 = 0; A1 k = 3 (k = 8 rejected since k < 6).
Q12Centre of mass of composite bodies · Challenge · Paper 3 style · 7 marks
(a)A child's spinning top is made from a uniform solid cone of height h and base radius r joined at its base to a uniform solid hemisphere of radius r of the same material. Show that the centre of mass of the top is on the axis at a distance (3r² − h²)/(4(h + 2r)) from the common plane face, on the hemisphere side.[5]
(b)The top is placed with its hemispherical surface on a horizontal table and released from a tilted position. Find the set of values of h, in terms of r, for which it returns to the upright position.[2]
Show answer
(a) Shown
B1 volumes (1/3)πr²h and (2/3)πr³; B1 centres of mass h/4 (cone side) and 3r/8 (hemisphere side); M1 moments about the plane face with a consistent sign convention; A1 πr²(3r² − h²)/12; A1 divide by πr²(h + 2r)/3.
(b) h < √3 r
M1 needs the centre of mass inside the hemisphere (below the centre of the hemisphere's plane face), i.e. 3r² − h² > 0; A1 h < √3 r.
Q13Suspension, toppling & sliding · Core · Paper 3 style · 5 marks
(a)A uniform rectangular lamina ABCD has AB = 8 cm, BC = 4 cm and mass 3 kg. A particle of mass 1 kg is fixed at C. Find the distances of the centre of mass of the combined body from AD and from AB.[3]
(b)The body is freely suspended from A. Find the angle that AB makes with the vertical.[2]
Q14Suspension, toppling & sliding · Core · Paper 3 style · 3 marks
(a)A uniform solid hemisphere is freely suspended from a point A on the rim of its plane face. Find the angle between the plane face and the vertical.[3]
Show answer
(a) 20.6°
B1 centre of mass 3r/8 from O along the axis; M1 tan θ = (3r/8)/r; A1 20.6°.
Q15Suspension, toppling & sliding · Core · Paper 3 style · 4 marks
(a)A uniform solid cylinder of diameter 8 cm and height 20 cm stands on a rough plane. The coefficient of friction is 0.3. The angle of inclination of the plane is slowly increased. Show that the cylinder slides before it topples, and state the angle at which it starts to slide.[4]
Show answer
(a) Slides at 16.7°; toppling would need 21.8°
M1 sliding when tan α = μ = 0.3, α = 16.7°; M1 toppling when the centre of mass is vertically above the lowest edge: tan α = 4/10; A1 α = 21.8°; A1 16.7° < 21.8°, so it slides first, at 16.7°.
Q16Suspension, toppling & sliding · Standard · Paper 3 style · 5 marks
(a)A uniform solid cylinder of radius 3 cm and height 10 cm stands on a plane inclined at α to the horizontal. The plane is rough enough to prevent sliding. Find the greatest value of α for which the cylinder does not topple.[3]
(b)A second cylinder of the same height is made from the same material and stands on the same plane at α = 40°. Find the least radius it must have so that it does not topple.[2]
Show answer
(a) 31.0°
M1 centre of mass 5 cm above the base on the axis; M1 tan α = 3/5; A1 31.0°.
(b) 4.20 cm
M1 r = 5 tan 40°; A1 4.20 cm.
Q17Suspension, toppling & sliding · Standard · Paper 3 style · 5 marks
(a)A uniform solid frustum has a larger base of radius 4 cm and its centre of mass is 33/14 cm from that base. It stands on its larger base on a rough plane inclined at α to the horizontal. Assuming it does not slide, find the greatest value of α for which it does not topple.[3]
(b)The coefficient of friction between the frustum and the plane is 0.8. The angle α is gradually increased from zero. Determine whether the frustum slides or topples first.[2]
Show answer
(a) 59.5°
M1 tan α = 4/(33/14); A1 tan α = 56/33; A1 59.5°.
(b) It slides first, at 38.7°
M1 sliding when tan α = 0.8, α = 38.7°; A1 38.7° < 59.5°, slides first.
Q18Suspension, toppling & sliding · Standard · Paper 3 style · 5 marks
(a)A uniform solid cone of base radius 4 cm and height 12 cm is freely suspended from a point on the rim of its base. Find the angle between the axis of the cone and the vertical.[3]
(b)Instead the cone is placed with its base on a rough plane inclined at β, and it is on the point of toppling. Find β.[2]
Show answer
(a) 53.1°
B1 centre of mass 3 cm from the base on the axis; M1 tan θ = 4/3; A1 53.1°.
(b) 53.1°
M1 tan β = 4/3 (same geometry: centre of mass vertically above the lowest point of the rim); A1 53.1°.
Q19Suspension, toppling & sliding · Standard · Paper 3 style · 5 marks
(a)A uniform square lamina ABCD of side 2a has mass M. A particle of mass m is attached at B. The lamina is freely suspended from A and hangs in equilibrium with AB making an angle tan⁻¹(1/2) with the vertical. Find m in terms of M.[5]
Show answer
(a) m = M/2
M1 take A as origin, AB along x-axis: x̄ = (Ma + 2am)/(M + m); A1 ȳ = Ma/(M + m); M1 tan θ = ȳ/x̄ = M/(M + 2m); M1 M/(M + 2m) = 1/2; A1 m = M/2.
Q20Suspension, toppling & sliding · Challenge · Paper 3 style · 9 marks
(a)A uniform cuboid packing crate stands on rough horizontal ground. Its weight is 200 N, its height is 1.2 m and its square base has side 0.6 m. A horizontal force of magnitude P N is applied at the midpoint of a top edge, perpendicular to that edge. Find the value of P at which the crate would be on the point of tipping, assuming it does not slide.[3]
(b)The coefficient of friction between the crate and the ground is μ. Find the set of values of μ for which the crate tips over before it slides, as P is gradually increased.[3]
(c)If instead the force is applied at height h m above the ground (0 < h ≤ 1.2), find the set of values of h for which the crate tips before it slides when μ = 0.4.[3]
Show answer
(a) P = 50
M1 moments about the lower edge on the far side: P × 1.2 = 200 × 0.3; A1 equation; A1 P = 50.
(b) μ > 0.25
M1 sliding needs P > 200μ; M1 tipping first requires 50 < 200μ; A1 μ > 0.25.
(c) 0.75 < h ≤ 1.2 (h > 0.75 m)
M1 tipping when P × h = 200 × 0.3, so P = 60/h; sliding when P = 0.4 × 200 = 80; M1 tips first if 60/h < 80; A1 h > 0.75 m (up to the top of the crate, 1.2 m).
Q21Suspension, toppling & sliding · Challenge · Paper 3 style · 8 marks
(a)A uniform solid hemisphere of radius r rests in equilibrium with its curved surface in contact with a rough plane inclined at α to the horizontal. Show that the plane face of the hemisphere makes an angle θ with the horizontal, where sin θ = (8/3) sin α.[4]
(b)Find θ when α = 15°, and find the greatest value of α for which such an equilibrium is possible (assuming friction is sufficient).[3]
(c)State the least possible coefficient of friction when α = 15°.[1]
Show answer
(a) Shown
M1 the reaction (normal and friction) acts at the contact point; for equilibrium the centre of mass G is vertically above the contact point; M1 the normal at the contact point passes through the centre O, so the horizontal distance from O to G equals r sin α; A1 OG sin θ = r sin α with OG = 3r/8; A1 sin θ = (8/3) sin α.
(b) θ = 43.6°; greatest α = 22.0°
A1 sin θ = (8/3) sin 15° = 0.690, θ = 43.6°; M1 need (8/3) sin α ≤ 1; A1 α ≤ 22.0°.
(c) tan 15° ≈ 0.268
B1 friction resolves along the plane: F = W sin α, R = W cos α, μ ≥ tan α.
Q22Suspension, toppling & sliding · Challenge · Paper 3 style · 7 marks
(a)A uniform rectangular lamina OABC has vertices O(0, 0), A(8a, 0), B(8a, 4a) and C(0, 4a). The triangle with vertices A, B and P(6a, 2a) is removed. Find the coordinates of the centre of mass of the remaining lamina.[3]
(b)The remaining lamina is freely suspended from O. Find the angle between OA and the vertical.[2]
(c)The lamina is instead held in a vertical plane with the edge OA in contact with a rough horizontal surface, and is slowly tilted in its own plane about A. Find the angle through which it can be turned before it topples.[2]
Show answer
(a) (74a/21, 2a) ≈ (3.52a, 2a)
B1 triangle area 4a², centroid (22a/3, 2a); M1 x̄ = (32a² × 4a − 4a² × 22a/3)/28a²; A1 x̄ = 74a/21, ȳ = 2a by symmetry about y = 2a.
(b) 29.6°
M1 tan θ = 2a/(74a/21) = 21/37; A1 θ = 29.6°.
(c) tan⁻¹(21/47) ≈ 24.1°
M1 horizontal distance of G from A is 8a − 74a/21 = 94a/21, height 2a; A1 tan φ = 2/(94/21) = 21/47, φ = 24.1°.
Q23Friction & moments · Core · Paper 3 style · 6 marks
(a)A uniform ladder of length 5 m and mass 20 kg rests with its foot on rough horizontal ground and its top against a smooth vertical wall. The ladder makes an angle of 60° with the ground. Take g = 10 m s⁻². Find the magnitude of the reaction at the wall.[3]
(b)Find the least possible value of the coefficient of friction between the ladder and the ground.[3]
Show answer
(a) 57.7 N (100/√3 N)
M1 moments about the foot: S × 5 sin 60° = 200 × 2.5 cos 60°; A1 equation; A1 S = 57.7 N.
(b) μ = 1/(2√3) ≈ 0.289
B1 R = 200 N; B1 F = S = 57.7 N; M1 μ ≥ F/R = 0.289.
Q24Friction & moments · Core · Paper 3 style · 7 marks
(a)A uniform plank AB of length 4 m and weight 300 N rests with end A on rough horizontal ground and is supported by a smooth peg at C, where AC = 3 m. The plank is inclined at 30° to the horizontal. Find the magnitude of the force exerted by the peg on the plank.[3]
(b)Find the normal reaction and the friction force at A, and hence find the least possible coefficient of friction.[4]
Show answer
(a) 173 N (100√3 N)
B1 the peg force is perpendicular to the plank; M1 moments about A: R × 3 = 300 × 2 cos 30°; A1 R = 173 N.
(b) Normal reaction 150 N, friction 86.6 N, μ ≥ 1/√3 ≈ 0.577
M1 resolve vertically: N + R cos 30° = 300, N = 150; M1 resolve horizontally: F = R sin 30° = 86.6; M1 μ ≥ F/N; A1 0.577.
Q25Friction & moments · Standard · Paper 3 style · 6 marks
(a)A uniform ladder of length 6 m and weight 150 N rests on rough horizontal ground, coefficient of friction 0.4, against a smooth vertical wall. The ladder makes an angle of 65° with the ground. A painter of weight 700 N climbs the ladder. Find how far up the ladder he can climb before the ladder slips, giving your answer correct to 3 significant figures.[6]
Show answer
(a) 5.61 m
B1 R = 850 N; B1 F_max = 0.4 × 850 = 340 N, so wall reaction S = 340 N at limiting equilibrium; M1 moments about the foot: S × 6 sin 65° = 150 × 3 cos 65° + 700d cos 65°; M1 rearrange for d; A1 d = (2040 tan 65° − 450)/700; A1 5.61 m.
Q26Friction & moments · Standard · Paper 3 style · 6 marks
(a)A uniform rod AB of length 1.2 m and mass 3 kg has end A in contact with a rough vertical wall. The rod is held horizontal, perpendicular to the wall, by a light string attached to B and to a point C on the wall 0.9 m vertically above A. Take g = 10 m s⁻². Find the tension in the string.[3]
(b)Find the least possible coefficient of friction between the rod and the wall.[3]
Show answer
(a) 25 N
B1 string length 1.5 m, sin(angle ABC) = 0.6; M1 moments about A: T × 0.6 × 1.2 = 30 × 0.6; A1 T = 25 N.
(b) 0.75
M1 horizontal: normal reaction = T cos = 25 × 0.8 = 20 N; M1 vertical: F = 30 − 25 × 0.6 = 15 N; A1 μ ≥ 15/20 = 0.75.
Q27Friction & moments · Standard · Paper 3 style · 7 marks
(a)A uniform ladder rests in limiting equilibrium with its foot on rough horizontal ground (coefficient of friction μ₁) and its top against a rough vertical wall (coefficient of friction μ₂). The ladder makes an angle θ with the ground. Show that tan θ = (1 − μ₁μ₂)/(2μ₁).[5]
(b)Find θ when μ₁ = 0.5 and μ₂ = 0.25.[2]
Show answer
(a) Shown
M1 at both contacts friction is limiting: F₁ = μ₁R, F₂ = μ₂S; M1 horizontally S = μ₁R, vertically R + μ₂S = W, giving R = W/(1 + μ₁μ₂); M1 moments about the foot: S × 2a sin θ + μ₂S × 2a cos θ = W a cos θ; A1 substitute S; A1 tan θ = (1 − μ₁μ₂)/(2μ₁).
(b) 41.2°
M1 tan θ = 0.875; A1 41.2°.
Q28Friction & moments · Challenge · Paper 3 style · 7 marks
(a)A uniform rod AB of weight W has end A on rough horizontal ground. The rod is held at 40° to the horizontal by a light string attached at B, the string being perpendicular to the rod and in the same vertical plane. Find the tension in the string in terms of W.[3]
(b)Find the least possible coefficient of friction at A, correct to 3 significant figures.[4]
Show answer
(a) T = (W cos 40°)/2 ≈ 0.383W
M1 moments about A, rod length 2a: T × 2a = W a cos 40°; A1 equation; A1 T = 0.383W.
(b) 0.348
M1 the string pulls at 40° to the vertical, towards A's side horizontally: horizontal component T sin 40°, vertical T cos 40°; M1 N = W − T cos 40° = W(1 − cos²40°/2); M1 F = T sin 40°; A1 μ ≥ sin 40° cos 40°/(2 − cos²40°) = 0.348.
Q29Friction & moments · Challenge · Paper 3 style · 7 marks
(a)A uniform rod AB of length 2a and weight W hangs below a fixed rough horizontal wire. End A is attached to a small light ring threaded on the wire, and a horizontal force P, in the vertical plane containing the wire, is applied at B. The rod rests in equilibrium at an angle θ to the vertical. Show that P = (W/2) tan θ.[3]
(b)The coefficient of friction between the ring and the wire is 0.35. Find the greatest value of θ for which equilibrium is possible.[4]
Show answer
(a) Shown
M1 moments about A; A1 P × 2a cos θ = W × a sin θ; A1 P = (W/2) tan θ.
(b) 35.0°
M1 resolving vertically for the whole system: normal reaction at the ring = W; M1 horizontally: friction = P; M1 limiting: (W/2) tan θ = 0.35W, tan θ = 0.7; A1 θ = 35.0°.
Q30Friction & moments · Challenge · Paper 3 style · 6 marks
(a)A uniform ladder rests with its foot on rough horizontal ground and its top against a rough vertical wall. The coefficient of friction is μ at both contacts. The ladder is on the point of slipping when it makes an angle of 60° with the ground. Find the exact value of μ.[6]
Show answer
(a) μ = 2 − √3 ≈ 0.268
M1 resolve: S = μR and R + μS = W; M1 moments about the foot: S × 2a sin 60° + μS × 2a cos 60° = Wa cos 60°; M1 obtain tan 60° = (1 − μ²)/(2μ); A1 μ² + 2√3μ − 1 = 0; M1 quadratic formula; A1 μ = 2 − √3 (positive root).
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Centre of mass of composite bodies6 questions · 6/9 sessions
