Circular Motion: Horizontal, Vertical & Banked

1. Foundations of Uniform Circular Motion

An object in uniform circular motion (UCM) moves at constant speed but constantly changing direction. Because velocity is a vector, any change in direction — even at constant speed — constitutes acceleration. This acceleration always points toward the centre of the circle and is called centripetal acceleration.

The two core kinematic equations for UCM are:

• Centripetal acceleration: a = v² / r, where v is the linear (tangential) speed in m s⁻¹ and r is the radius in m. Units of a: m s⁻².
• Period and speed: v = 2πr / T, where T is the period (time for one complete revolution) in seconds. Frequency f = 1/T in Hz.
• Combining: a = 4π²r / T² — useful when period rather than speed is given.

By Newton's second law, the net centripetal force required is:

F_net = ma = mv² / r = 4π²mr / T²

This is not a new type of force — it is always provided by one or more real forces already present in the problem (tension, normal force, friction, gravity, or combinations thereof). In free-body diagrams, label the real forces only; centripetal force is the net result pointing inward.

Key sign convention: take the inward (centripetal) direction as positive when applying Newton's second law along the radial direction.

Worked Example 1. A 0.50 kg ball on a 1.2 m horizontal string makes 2.0 revolutions per second. Find (a) the period, (b) the linear speed, (c) the centripetal acceleration, and (d) the tension in the string (assume horizontal plane, ignore gravity for part d).

• (a) Period: T = 1/f = 1/2.0 = 0.50 s
• (b) Speed: v = 2π × 1.2 / 0.50 = 2π × 2.4 = 15.08... ≈ 15 m s⁻¹
• (c) Centripetal acceleration: a = v²/r = (15.08)² / 1.2 = 227.4 / 1.2 = 190 m s⁻² (3 s.f.)
  Check via alternate formula: a = 4π² × 1.2 / (0.50)² = 4 × 9.870 × 1.2 / 0.25 = 47.37 / 0.25 = 189.5 ≈ 190 m s⁻² ✓
• (d) Tension: T_string = ma = 0.50 × 189.5 = 95 N

2. Horizontal Circular Motion — Conical Pendulum

A conical pendulum is a classic horizontal UCM problem. A mass m hangs on a string of length L that makes angle θ with the vertical, tracing a horizontal circle of radius r = L sin θ.

Two forces act on the mass: tension T along the string, and weight mg downward. Because the mass has no vertical acceleration, the vertical component of tension balances weight. The horizontal component of tension provides centripetal force.

• Vertical: T cos θ = mg
• Horizontal (radial, centripetal): T sin θ = mv² / r

Dividing the horizontal equation by the vertical:

tan θ = v² / (r × g)

Since r = L sin θ, substituting and solving for period:

T² = 4π² L cos θ / g → T = 2π √(L cos θ / g)

Note: as θ increases (faster spin), cos θ decreases, so the period decreases and the string sweeps higher — physically sensible.

Worked Example 2. A conical pendulum has a string of length L = 0.80 m making θ = 30° with the vertical. Mass = 0.20 kg, g = 9.8 m s⁻².

• Radius: r = 0.80 × sin 30° = 0.80 × 0.500 = 0.40 m
• Tension: From T cos 30° = mg: T = (0.20 × 9.8) / cos 30° = 1.96 / 0.8660 = 2.26 N
• Speed: T_string sin 30° = mv²/r → 2.26 × 0.500 = 0.20 × v² / 0.40 → 1.131 = 0.50 v² → v² = 2.262 → v = 1.50 m s⁻¹
• Period: T = 2π√(0.80 × cos 30° / 9.8) = 2π√(0.80 × 0.8660 / 9.8) = 2π√(0.6928 / 9.8) = 2π√(0.07069) = 2π × 0.2659 = 1.67 s
  Verify via speed: T = 2πr/v = 2π × 0.40 / 1.504 = 2.513 / 1.504 = 1.67 s ✓