Further Mathematics · OCR H245 · Option M · Topic 6.05

Motion in a circle

Uniform motion in a circle, motion in a vertical circle.

  • 2 sub-topics
  • Taught in Year 13
  • 42 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. 6.05a-cUniform motion in a circleY13
  2. 6.05d-fMotion in a vertical circleY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 42 questions in this topic, free, no account needed.

Motion in a circle

Not needed for Year 12Easy

Acceleration in circular motion · Year 13 · 1 mark

  1. A particle moves in a horizontal circle of radius 0.8 m with angular speed 5 rad s⁻¹. What is the magnitude of its acceleration?
    (1)

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  1. a=rω2a = r\omega^2, directed towards the centre.DD: 20 m s⁻²

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Motion in a circle

Not needed for Year 12Medium

Friction on a level bend · Year 13 · 1 mark

  1. A car of mass 1200 kg travels at 15 m s⁻¹ round a level bend of radius 50 m. What friction force towards the centre is needed?
    (1)

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  1. Friction provides the centripetal force: F=mv2rF = \dfrac{mv^2}{r}.CC: 5400 N

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Motion in a circle

Not needed for Year 12Hard

Motion on a banked track · Year 13 · 8 marks

A car travels round a bend of radius 50 m on a track banked at 20° to the horizontal. The car is modelled as a particle moving in a horizontal circle.
  1. (a)
    Find the speed at which there is no tendency for the car to slip sideways.
    (3)
  2. (b)
    The coefficient of friction between the tyres and the road is 0.25. Find the maximum speed at which the car can go round the bend without slipping.
    (5)
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Part (a)3 marks

  1. Resolve vertically and horizontally (towards the centre).Rcos⁡20∘R\cos 20^\circ=mg, Rsin⁡20∘{}= mg, \ R\sin 20^\circ=mv250{}= \frac{mv^2}{50}
  2. vv=50×9.8tan⁡20∘{}= \sqrt{50 \times 9.8\tan 20^\circ}=13.4 m s−1{}= 13.4\text{ m s}^{-1}

Part (b)5 marks

  1. At maximum speed friction acts down the slope, and is limiting: F=μRF = \mu R.
  2. Vertical equilibrium.Rcos⁡20∘R\cos 20^\circ−0.25Rsin⁡20∘{}- 0.25R\sin 20^\circ=mg{}= mg
  3. Newton's second law towards the centre.Rsin⁡20∘R\sin 20^\circ+0.25Rcos⁡20∘{}+ 0.25R\cos 20^\circ=mv250{}= \frac{mv^2}{50}
  4. Divide the equations to eliminate RR and mm.v=18.2 m s−1v = 18.2\text{ m s}^{-1}

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