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How do you analyse motion in a horizontal or vertical circle, and what provides the centripetal force?

Angular speed, the centripetal acceleration and force, motion in a horizontal circle (including the conical pendulum and banked tracks), and motion in a vertical circle with the conditions for maintaining contact or tension.

A focused answer to the OCR A-Level Further Mathematics A Mechanics option content on circular motion, covering angular speed, the centripetal acceleration and force directed to the centre, motion in a horizontal circle such as the conical pendulum and banked tracks, and motion in a vertical circle including the conditions for a string to stay taut or a particle to maintain contact.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. Angular speed and centripetal acceleration
  3. The centripetal force
  4. Motion in a horizontal circle
  5. Motion in a vertical circle
  6. Try this

What this dot point is asking

OCR's Mechanics option wants you to use angular speed and the centripetal acceleration directed to the centre, find the centripetal force and identify what provides it, analyse horizontal-circle problems (such as the conical pendulum and a car on a banked track), and analyse vertical-circle problems, including the conditions for a string to stay taut or a particle to maintain contact with a surface.

Angular speed and centripetal acceleration

Even at constant speed, circular motion is accelerated motion, because the direction of the velocity is always changing. That acceleration points towards the centre.

The centripetal force

By Newton's second law, the centre-directed acceleration requires a net force towards the centre. The skill is identifying which real force (or component) supplies it.

Motion in a horizontal circle

For horizontal-circle problems the particle has no vertical acceleration, so resolving vertically balances the vertical forces, while resolving horizontally gives the centripetal equation. The conical pendulum (a mass on a string sweeping a horizontal circle) and a car on a banked track are the standard cases.

Motion in a vertical circle

In a vertical circle the speed changes with height, so gravity does work. Combine conservation of energy (to relate speeds at different heights) with the centripetal equation at the point in question.

Circular motion combines the energy methods (for varying speed) with Newton's second law and feeds into the link between uniform circular motion and simple harmonic motion.

Try this

Q1. A particle moves in a circle of radius 22 m at ω=4\omega = 4 rad s−1^{-1}. Find its speed. [1 mark]

  • Cue. v=rω=2×4=8v = r\omega = 2 \times 4 = 8 m s−1^{-1}.

Q2. Find the centripetal force on a 0.50.5 kg mass moving at 66 m s−1^{-1} in a circle of radius 33 m. [2 marks]

  • Cue. mv2r=0.5×363=6\dfrac{mv^2}{r} = \dfrac{0.5 \times 36}{3} = 6 N.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20196 marksA particle of mass 0.40.4 kg moves in a horizontal circle of radius 0.50.5 m at a constant angular speed of 33 rad s−1^{-1} on a smooth table, attached to the centre by a string. Find the speed of the particle and the tension in the string.
Show worked answer →

Speed from angular speed (M1): v=rω=0.5×3=1.5v = r\omega = 0.5 \times 3 = 1.5 m s−1^{-1} (A1).

The tension provides the centripetal force (M1): T=mrω2T = m r\omega^2 (or mv2r\dfrac{mv^2}{r}).

Compute (M1, A1): T=0.4×0.5×32=0.4×0.5×9=1.8T = 0.4 \times 0.5 \times 3^2 = 0.4 \times 0.5 \times 9 = 1.8 N (A1).

Markers reward v=rωv = r\omega, identifying the tension as the centripetal force, the formula mrω2mr\omega^2, and the value 1.81.8 N.

OCR 20226 marksA particle of mass 0.20.2 kg is attached to one end of a light string of length 0.50.5 m and swung in a vertical circle. Find the minimum speed at the top of the circle for the string to remain taut, and the corresponding tension at the bottom. (Take g=9.8g = 9.8.)
Show worked answer →

At the top, for minimum speed the tension is zero and gravity alone provides the centripetal force (M1): mg=mvtop2rmg = \dfrac{mv_{\text{top}}^2}{r}, so vtop2=gr=9.8×0.5=4.9v_{\text{top}}^2 = gr = 9.8 \times 0.5 = 4.9 (A1).

Use energy conservation between top and bottom (height difference 2r=12r = 1 m) (M1): 12vbottom2=12vtop2+g(2r)\tfrac{1}{2}v_{\text{bottom}}^2 = \tfrac{1}{2}v_{\text{top}}^2 + g(2r), so vbottom2=4.9+2(9.8)(0.5)⋅2=4.9+19.6=24.5v_{\text{bottom}}^2 = 4.9 + 2(9.8)(0.5)\cdot 2 = 4.9 + 19.6 = 24.5 (A1).

At the bottom, tension minus weight provides the centripetal force (M1): T−mg=mvbottom2rT - mg = \dfrac{mv_{\text{bottom}}^2}{r}, so T=m(g+vbottom2r)=0.2(9.8+24.50.5)=0.2(9.8+49)=11.76T = m\left(g + \dfrac{v_{\text{bottom}}^2}{r}\right) = 0.2\left(9.8 + \dfrac{24.5}{0.5}\right) = 0.2(9.8 + 49) = 11.76 N (A1).

Markers reward the zero-tension condition at the top, energy conservation to the bottom, the bottom force equation, and the tension.

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