A-Level Kinematics of uniform circular motion and centripetal acceleration: worked solution

Question

For the same aircraft (radius 12 km, period 250 s, lift L at angle θ to the vertical), determine the angle θ.

The aircraft flies in a horizontal circle at constant speed, so the resultant force on it is horizontal and points at the centre of the circle. First find the speed. In one period the aircraft covers one circumference: v = 2πr / T v = 2π × 12 000 / 250 v = 301.6 m/s (about 302 m/s) Now resolve the lift L, which acts at angle θ to the vertical. Vertically there is no acceleration, so the vertical component of L balances the weight: L cosθ = mg Horizontally the only force is the horizontal component of L, and it supplies the centripetal force: L sinθ = mv²/r Divide the horizontal equation by the vertical one. Both L and m cancel, which is why you do not need the mass of the aircraft: (L sinθ)/(L cosθ) = (mv²/r)/(mg) tanθ = v²/(rg) tanθ = 301.6²/(12 000 × 9.81) tanθ = 90 960/117 720 tanθ = 0.7727 θ = tan⁻¹(0.7727) θ ≈ 37.7°