Further Mathematics · AQA 7367 · Option M · Topic MD

Circular motion

Motion in horizontal and vertical circles, the conical pendulum and angular speed.

  • 6 sub-topics
  • 3 in Year 12, 3 in Year 13
  • 93 practice questions
  • 6 animated lessons

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. MD1Motion of a particle in a circle at constant speedY12
  2. MD2Angular speedY12
  3. MD3Relationships between speed, angular speed, radius and accelerationY12
  4. MD4Position, velocity and acceleration vectors in circular motionY13
  5. MD5Conical pendulum, with one or two stringsY13
  6. MD6Motion 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 93 questions in this topic, free, no account needed.

Circular motion

Easy

Angular speed · Year 12 · 7 marks

A coin of mass 0.05 kg rests on a turntable at a distance 0.3 m from the centre. The turntable rotates at 60 revolutions per minute and the coin does not slip.
  1. (a)
    Find the angular speed of the turntable in rad s⁻¹.
    (2)
  2. (b)
    Find the speed and the acceleration of the coin.
    (3)
  3. (c)
    Find the least possible coefficient of friction between the coin and the turntable.
    (2)
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Part (a)2 marks

  1. ω=60×2π60\omega = \frac{60 \times 2\pi}{60}
  2. =6.28 rad s−1= 6.28\text{ rad s}^{-1}

Part (b)3 marks

  1. v=rω=1.88 m s−1v = r\omega = 1.88\text{ m s}^{-1}
  2. Directed towards the centre.aa=rω2{}= r\omega^2=11.8 m s−2{}= 11.8\text{ m s}^{-2}

Part (c)2 marks

  1. Friction provides the central force.μmg≥mrω2\mu mg \geq mr\omega^2
  2. μ≥1.21\mu \geq 1.21

Total7

  • MMethod mark: for a correct method, even with a slip.
  • AAccuracy mark: for a correct answer from a correct method.
  • BIndependent mark: for a correct fact or result on its own, with no method needed.
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Circular motion

Medium

Angular speed · Year 12 · 7 marks

A coin of mass 0.05 kg rests on a turntable at a distance 0.15 m from the centre. The turntable rotates at 33 revolutions per minute and the coin does not slip.
  1. (a)
    Find the angular speed of the turntable in rad s⁻¹.
    (2)
  2. (b)
    Find the speed and the acceleration of the coin.
    (3)
  3. (c)
    Find the least possible coefficient of friction between the coin and the turntable.
    (2)
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Part (a)2 marks

  1. ω=33×2π60\omega = \frac{33 \times 2\pi}{60}
  2. =3.46 rad s−1= 3.46\text{ rad s}^{-1}

Part (b)3 marks

  1. vv=rω{}= r\omega=0.518 m s−1{}= 0.518\text{ m s}^{-1}
  2. Directed towards the centre.aa=rω2{}= r\omega^2=1.79 m s−2{}= 1.79\text{ m s}^{-2}

Part (c)2 marks

  1. Friction provides the central force.μmg≥mrω2\mu mg \geq mr\omega^2
  2. μ≥0.183\mu \geq 0.183

Total7

  • MMethod mark: for a correct method, even with a slip.
  • AAccuracy mark: for a correct answer from a correct method.
  • BIndependent mark: for a correct fact or result on its own, with no method needed.
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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}

Total8

  • MMethod mark: for a correct method, even with a slip.
  • AAccuracy mark: for a correct answer from a correct method.
  • BIndependent mark: for a correct fact or result on its own, with no method needed.
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