Circular motion

Further Maths · Further Mechanics. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

A particle moving in a circle is always changing direction, so it is accelerating even when its speed is constant. This acceleration points towards the centre of the circle. By Newton's second law the resultant force must also point to the centre. It is not an extra force: it is supplied by tension, friction, a normal contact force or a component of weight.

Angular speed ω measures how fast the radius turns, in radians per second.

In a horizontal circle the speed is constant: resolve vertically (forces balance) and towards the centre (F = ma). In a vertical circle the speed changes, so use conservation of energy to find the speed at a point, then F = ma towards the centre at that point.

Rules to remember

Angular speed ω = 2π ÷ (time for one revolution); speed = radius × angular speed (v = rω)
Acceleration towards the centre = v²/r = rω²; resultant force towards the centre = mv²/r = mrω²
Conical pendulum, string of length L at angle θ to the vertical: T cos θ = mg, T sin θ = mrω², with r = L sin θ
Vertical circle, energy: v² = u² − 2gh, where h is the height gained since the speed was u
String of length r: at the lowest point T − mg = mv²/r; at the highest point T + mg = mv²/r
Complete circles on a string (or on the inside of a surface): speed at the top must satisfy v² ≥ gr, so speed at the bottom must satisfy u² ≥ 5gr
Complete circles on a light rod, or a bead on a wire or in a tube: the particle only has to reach the top, so u² > 4gr at the bottom

Common mistake

Students say a particle on a string completes the circle if it reaches the top with any speed above zero. A string goes slack when the tension reaches zero, so the condition is v² ≥ gr at the top. Speed zero at the top is the condition for a rod or a bead on a wire only.

Worked example

A particle of mass 0.5 kg is attached to a fixed point by a light inextensible string of length 0.4 m and moves in a vertical circle. Its speed at the lowest point is 6 m s⁻¹. Take g = 10 m s⁻². Find the tension in the string at the highest point.

The highest point is 0.8 m above the lowest, so v² = 6² − 2 × 10 × 0.8 = 36 − 16 = 20.
At the top, tension and weight both act towards the centre: T + mg = mv²/r.
T = (0.5 × 20) ÷ 0.4 − 0.5 × 10 = 25 − 5 = 20 N.

Answer: 20 N

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