Further Mathematics · AQA 7367 · Option M · Topic ME

Centres of mass and moments

Centres of mass of particles, laminas and solids, and when bodies slide or topple.

  • 6 sub-topics
  • Taught in Year 13
  • 93 practice questions
  • 6 animated lessons
Practice questions (93) Lessons Worksheets All AQA Further Mathematics topics

What the specification says

Every sub-topic AQA can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. ME1Centre of mass of a system of particlesY13
  2. ME2Centre of mass of a composite bodyY13
  3. ME3Centre of mass of a lamina by integrationY13
  4. ME4Centres of mass of solids of revolutionY13
  5. ME5Sliding and toppling, including suspended bodies and inclined planesY13
  6. ME6Rigid bodies in equilibrium: moments and couplesY13

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Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 93 questions in this topic, free, no account needed.

Centres of mass and moments

Not needed for Year 12Easy

Centre of mass of a cone · Year 13 · 1 mark

  1. How far is the centre of mass of a uniform solid cone of height hh from its base?
    (1)

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  1. It is 34h\tfrac{3}{4}h from the vertex, on the axis of symmetry.CC: 14h\tfrac{1}{4}h

Total1

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Centres of mass of plane figures

Not needed for Year 12Medium

Centres of mass of particles · Year 13 · 6 marks

Particles are placed in the x-y plane: 1 kg at (1, 4); 2 kg at (2, 3); 1 kg at (-2, 1).
  1. (a)
    Find the coordinates of the centre of mass of the system.
    (3)
  2. (b)
    A particle of mass 1 kg is added at the point (p,q)(p, q). The centre of mass of the new system is at (0,2)(0, 2). Find pp and qq.
    (3)

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Part (a)3 marks

  1. xˉ\bar{x}=∑mx∑m{}= \frac{\sum mx}{\sum m}=34{}= \frac{3}{4}
  2. xˉ=34\bar{x} = \frac{3}{4}
  3. yˉ=114\bar{y} = \frac{11}{4}

Part (b)3 marks

  1. 3+p5=0\frac{3 + p}{5} = 0
  2. p=−3p = -3
  3. q=−1q = -1

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Centres of mass of plane figures

Not needed for Year 12Hard

Centre of mass of a lamina · Year 13 · 7 marks

Figure 1 shows a uniform rectangular lamina OABC with OA=12OA = 12 cm and OC=8OC = 8 cm. A square of side 4 cm, with centre 8 cm from OCOC and 4 cm from OAOA, is removed.
OABC
Figure 1
  1. (a)
    Find the distance of the centre of mass of the remaining lamina from OC and from OA.
    (5)
  2. (b)
    The lamina is freely suspended from OO and hangs in equilibrium. Find the angle between OAOA and the vertical.
    (2)

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Part (a)5 marks

  1. Mass is proportional to area.areas: 96, 16, remaining 80\text{areas: } 96, \ 16, \ \text{remaining } 80
  2. Moments about OC; the removed square counts as negative.96(6)−16(8)=80xˉ96(6) - 16(8) = 80\bar{x}
  3. xˉ=285 cm\bar{x} = \frac{28}{5}\text{ cm}
  4. Moments about OA.yˉ=4 cm\bar{y} = 4\text{ cm}

Part (b)2 marks

  1. The centre of mass lies vertically below OO.tan⁡θ\tan\theta=yˉxˉ{}= \frac{\bar{y}}{\bar{x}}
  2. θ=35.5∘\theta = 35.5^\circ

Total7

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