Physics · OCR H556 · Topic 5.2

Circular motion

Kinematics of circular motion, centripetal force.

  • 2 sub-topics
  • Taught in Year 13
  • 31 practice questions
  • 2 animated lessons
Practice questions (31) Lessons Worksheets coming soon All OCR Physics topics

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. 5.2.1Kinematics of circular motionY13
  2. 5.2.2Centripetal forceY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 31 questions in this topic, free, no account needed.

Further mechanics

Not needed for Year 12Easy

Centripetal force · Year 13 · 1 mark

  1. An object of mass 800 kg moves at 6 m s⁻¹ in a circle of radius 0.5 m. What is the centripetal force on it?
    (1)

Show worked solutionHide worked solution
  1. F=mv2rF = \dfrac{mv^2}{r}, directed towards the centre.C: 57600 N57600\ \mathrm{N}

Total1

  • M
  • A
  • B
How did it go?

Report a problem with this question

Further mechanics

Not needed for Year 12Medium

Angular velocity · Year 13 · 3 marks

A disc rotates at 33 revolutions per minute.
  1. (a)
    Calculate its angular velocity.
    (2)
  2. (b)
    Calculate the speed of a point 0.15 m from the axis.
    (1)

Check your answers before you look at the solution

Show worked solutionHide worked solution

Part (a)2 marks

  1. ω=2π×3360\omega = \frac{2\pi \times 33}{60}
  2. ω=3.46 rad s−1\omega = 3.46\ \mathrm{rad\,s^{-1}}

Part (b)1 mark

  1. v=0.518 m s−1v = 0.518\ \mathrm{m\,s^{-1}}

Total3

  • M
  • A
  • B
How did it go?

Report a problem with this question

Further mechanics

Not needed for Year 12Hard

Centripetal force · Year 13 · 5 marks

A car of mass 1500 kg goes round a flat bend of radius 80 m. The maximum frictional force between the tyres and the road is 0.6 times the car's weight.
  1. (a)
    Identify the force that provides the centripetal force.
    (1)
  2. (b)
    Calculate the maximum speed at which the car can take the bend.
    (3)
  3. (c)
    Explain why the mass of the car does not affect your answer.
    (1)

Check your answer before you look at the solution

Show worked solutionHide worked solution

Part (a)1 mark

  1. Friction between the tyres and the road, towards the centre.

Part (b)3 marks

  1. mv2r=0.6mg\frac{mv^2}{r} = 0.6mg
  2. vv=0.6×9.81×80{}= \sqrt{0.6 \times 9.81 \times 80}
  3. v=21.7 m s−1v = 21.7\ \mathrm{m\,s^{-1}}

Part (c)1 mark

  1. Both the friction and the required centripetal force are proportional to m.

Total5

  • M
  • A
  • B
How did it go?

Report a problem with this question