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How do you find the centre of mass of a system of particles, a lamina or a composite body?

The centre of mass of a system of particles, of a uniform lamina (by symmetry or integration), and of a composite body, and the equilibrium of a suspended or tilting body.

A focused answer to the OCR A-Level Further Mathematics A Mechanics option content on centre of mass, covering the centre of mass of a system of particles, of a uniform lamina by symmetry or integration, and of a composite body by treating each part as a point mass, and applications to a suspended body hanging in equilibrium.

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  1. What this dot point is asking
  2. Centre of mass of a system of particles
  3. Uniform laminas by symmetry and integration
  4. Composite bodies
  5. Suspension and equilibrium
  6. Try this

What this dot point is asking

OCR's Mechanics option wants you to find the centre of mass of a system of point masses using the weighted-average formula, of a uniform lamina by symmetry or by integration, and of a composite body by treating each standard part as a point mass at its own centre of mass, and to apply this to a body hanging in equilibrium (where the centre of mass lies vertically below the suspension point) or about to tilt.

Centre of mass of a system of particles

The centre of mass is the mass-weighted average position, the point that behaves as if all the mass were concentrated there. Each coordinate is computed independently.

Uniform laminas by symmetry and integration

For a uniform lamina (a flat shape of constant density), the centre of mass lies on any axis of symmetry, which often locates it immediately. When there is no helpful symmetry, integrate: the centre of mass is found from the moments of thin strips.

Composite bodies

A composite body is built from standard shapes. Treat each shape as a point mass (proportional to its area for a lamina, or volume for a solid) at its own centre of mass, then apply the system formula. A hole or removed piece is included as a negative mass.

Suspension and equilibrium

When a body is freely suspended from a point, it settles so that its centre of mass is directly below the suspension point (otherwise gravity would exert a turning moment). This fixes the angle at which the body hangs, found from the position of the centre of mass relative to the suspension point. In practice you first find the centre of mass relative to convenient axes, then draw the line from the suspension point to the centre of mass; the angle this line makes with an edge of the body is the angle of tilt, obtained by simple trigonometry on the horizontal and vertical displacements between the two points. The same reasoning answers tilting questions: a body on the point of toppling has its centre of mass directly above the edge about which it would turn, so the toppling condition is found by comparing the horizontal position of the centre of mass with that pivoting edge.

Centre of mass completes the mechanics option, using the moment idea from statics and the integration skills from the core, and parallels the volume-of-revolution method for solids.

Try this

Q1. Two particles, 11 kg at x=0x = 0 and 33 kg at x=8x = 8, lie on a line. Find xˉ\bar{x}. [2 marks]

  • Cue. xˉ=1(0)+3(8)4=244=6\bar{x} = \dfrac{1(0) + 3(8)}{4} = \dfrac{24}{4} = 6.

Q2. State where the centre of mass of a uniform triangular lamina lies. [1 mark]

  • Cue. At the centroid, the intersection of the medians, one third of the way up from each side.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20185 marksParticles of mass 22 kg, 33 kg and 55 kg are placed at the points (0,0)(0, 0), (4,0)(4, 0) and (2,6)(2, 6). Find the coordinates of the centre of mass.
Show worked answer →

Total mass (M1): 2+3+5=102 + 3 + 5 = 10 kg.

xx-coordinate (M1): xˉ=∑mixi∑mi=2(0)+3(4)+5(2)10=0+12+1010=2210=2.2\bar{x} = \dfrac{\sum m_i x_i}{\sum m_i} = \dfrac{2(0) + 3(4) + 5(2)}{10} = \dfrac{0 + 12 + 10}{10} = \dfrac{22}{10} = 2.2 (A1).

yy-coordinate (M1): yˉ=2(0)+3(0)+5(6)10=3010=3\bar{y} = \dfrac{2(0) + 3(0) + 5(6)}{10} = \dfrac{30}{10} = 3 (A1).

So the centre of mass is at (2.2,3)(2.2, 3).

Markers reward the total mass, the weighted-sum formula for each coordinate, and the correct point.

OCR 20226 marksA uniform lamina is formed from a square of side 44 (mass proportional to area) with a square of side 22 removed from one corner. Taking the large square's corner as the origin with the removed square in the corner 0≤x≤20 \le x \le 2, 0≤y≤20 \le y \le 2, find the xx-coordinate of the centre of mass of the remaining shape.
Show worked answer →

Work with areas as masses. Full square: area 1616, centre at (2,2)(2, 2). Removed square: area 44, centre at (1,1)(1, 1) (M1).

The remaining lamina has area 16−4=1216 - 4 = 12 (M1). Use moments about the yy-axis: (area of full)×\times(its xˉ\bar{x}) −- (area removed)×\times(its xˉ\bar{x}) == (remaining area)×xˉ\times \bar{x} (M1).

So 16(2)−4(1)=12xˉ16(2) - 4(1) = 12\bar{x} (A1): 32−4=28=12xˉ32 - 4 = 28 = 12\bar{x}, giving xˉ=2812=73\bar{x} = \dfrac{28}{12} = \dfrac{7}{3} (A1, A1).

Markers reward treating areas as masses, subtracting the removed piece, taking moments about the axis, and the value 73\tfrac{7}{3}.

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