Further Mathematics · OCR H245 · Option M · Topic 6.04

Centre of mass

Centre of mass, rigid bodies.

  • 2 sub-topics
  • Taught in Year 13
  • 81 practice questions
  • 2 animated lessons
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What the specification says

Every sub-topic OCR 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. 6.04a-dCentre of massY13
  2. 6.04eRigid bodiesY13

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Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 81 questions in this topic, free, no account needed.

Centres of mass of plane figures

Not needed for Year 12Easy

The centre of mass of particles on a line · Year 13 · 1 mark

  1. Particles of mass 2 kg, 3 kg and 5 kg are at the points x=1x = 1, x=4x = 4 and x=6x = 6 on the xx-axis. Where is their centre of mass?
    (1)

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  1. xˉ=∑mx∑m\bar{x} = \dfrac{\sum mx}{\sum m}.A: x=4.4x = 4.4

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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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