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ScotlandMathematics of MechanicsSyllabus dot point

How do you analyse motion in a circle, and how do circular-motion ideas explain the conical pendulum, banked tracks, vertical circles and orbits under gravity?

Analyse circular motion using angular velocity and centripetal acceleration; apply Newton's second law radially to the conical pendulum, banked tracks and motion in a vertical circle; and model gravitation with the inverse-square law.

A focused answer to the SQA Advanced Higher Mathematics of Mechanics content on circular motion, covering angular velocity, centripetal acceleration and force, the conical pendulum, banked tracks, motion in a vertical circle, and Newton's inverse-square law of gravitation with orbital motion.

Generated by Claude Opus 4.814 min answer

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  1. What this dot point is asking
  2. Angular velocity and centripetal acceleration
  3. The conical pendulum and banked tracks
  4. The vertical circle and gravitation
  5. Try this

What this dot point is asking

When a body travels in a circle at, say, constant speed, its direction is always changing, so it is accelerating even though its speed is steady. The SQA wants you to quantify that acceleration, identify the resultant force that causes it, and apply Newton's second law in the radial direction to the standard models: the conical pendulum, a car on a banked track, a particle in a vertical circle, and a body orbiting under gravity.

Angular velocity and centripetal acceleration

Circular motion is most naturally described with the angular velocity ω\omega, the rate at which the radius sweeps angle. Linear and angular speed are linked by the radius.

The centripetal force is not a new kind of force. It is the name for whatever real force (tension, friction, gravity, a normal reaction, or a combination) supplies the net inward pull. The whole method is: identify the real forces, resolve toward the centre, and set their resultant equal to mv2r\dfrac{mv^2}{r}.

The conical pendulum and banked tracks

In a conical pendulum a mass on a string moves in a horizontal circle while the string traces a cone. The two governing equations come from resolving vertically (equilibrium, since there is no vertical motion) and horizontally (the radial equation).

For a rough banked track the friction force is added along the slope, and you get a range of safe speeds rather than a single design speed.

The vertical circle and gravitation

In a vertical circle the speed varies because gravity does work as the body rises and falls, so energy conservation is combined with the radial equation. The critical point is usually the top, where the minimum speed to stay in contact is v=grv = \sqrt{gr}, found by setting the normal reaction to zero.

For gravitation, Newton's law of universal gravitation gives the inward force between two masses, and for a body in a circular orbit this force is exactly the centripetal force.

Try this

Q1. A stone on a string 0.80.8 m long is whirled in a horizontal circle at 55 rad s−1^{-1}. Find its speed and centripetal acceleration. [3 marks]

  • Cue. v=rω=0.8×5=4v = r\omega = 0.8\times 5 = 4 m s−1^{-1}; a=rω2=0.8×25=20a = r\omega^2 = 0.8\times 25 = 20 m s−2^{-2}.

Q2. A satellite orbits at radius rr where GMr=4.0×107\dfrac{GM}{r} = 4.0\times 10^7 m2^2 s−2^{-2}. Find its orbital speed. [2 marks]

  • Cue. v=GM/r=4.0×107≈6.3×103v = \sqrt{GM/r} = \sqrt{4.0\times 10^7} \approx 6.3\times 10^3 m s−1^{-1}.

Exam-style practice questions

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

AH style: conical pendulum6 marksA particle of mass 0.50.5 kg is attached to a string of length 1.21.2 m and moves in a horizontal circle as a conical pendulum, the string making an angle of 40∘40^\circ with the vertical. Find the tension in the string and the angular speed. Take g=9.8g = 9.8 m s−2^{-2}.
Show worked answer →

Vertical equilibrium: Tcos⁡40∘=mgT\cos 40^\circ = mg, so T=0.5×9.8cos⁡40∘=4.90.766=6.40T = \dfrac{0.5\times 9.8}{\cos 40^\circ} = \dfrac{4.9}{0.766} = 6.40 N (2 marks).

The radius of the circle is r=ℓsin⁡40∘=1.2×0.643=0.771r = \ell\sin 40^\circ = 1.2\times 0.643 = 0.771 m (1 mark).

Radial (horizontal) equation: Tsin⁡40∘=mω2rT\sin 40^\circ = m\omega^2 r, so 6.40×0.643=0.5 ω2(0.771)6.40\times 0.643 = 0.5\,\omega^2 (0.771) (2 marks).

4.11=0.386 ω24.11 = 0.386\,\omega^2, giving ω2=10.7\omega^2 = 10.7, so ω=3.27\omega = 3.27 rad s−1^{-1} (1 mark). Markers reward resolving vertically for the tension and applying the radial equation Tsin⁡θ=mω2rT\sin\theta = m\omega^2 r.

AH style: vertical circle5 marksA particle moves in a vertical circle of radius rr on the inside of a smooth track. Show that the least speed at the top for the particle to maintain contact is v=grv = \sqrt{gr}.
Show worked answer →

At the top, both the weight mgmg and the normal reaction NN act downward toward the centre, so the radial equation is mg+N=mv2rmg + N = \dfrac{mv^2}{r} (2 marks).

Contact is maintained while N≥0N \ge 0. The limiting case is N=0N = 0 (just in contact) (1 mark).

Setting N=0N = 0: mg=mv2rmg = \dfrac{mv^2}{r}, so v2=grv^2 = gr (1 mark).

Hence the least speed at the top is v=grv = \sqrt{gr} (1 mark). Markers reward the radial equation at the top with weight and reaction both toward the centre, and the condition N=0N = 0 for the minimum.

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