Normal-tangential acceleration; nozzle stream curvature

Normal and tangential components of acceleration

Let a particle move on a curved path with speed v. Its velocity is always tangent to the path: v = v e_t, where e_t is the unit tangent vector.

Differentiating, a = dv/dt = (dv/dt) e_t + v (de_t/dt).

Over a small time dt the tangent turns through dθ, and de_t = dθ e_n (e_n is the unit normal pointing to the centre of curvature). If ρ is the radius of curvature, ds = ρ dθ = v dt, so dθ/dt = v/ρ. Therefore de_t/dt = (v/ρ) e_n and

a = a_t e_t + a_n e_n, with a_t = dv/dt and a_n = v²/ρ

• a_t changes the magnitude of the velocity (tangential component).
• a_n changes its direction and always points to the centre of curvature (normal component).
• Magnitude: a = √(a_t² + a_n²).

Nozzle

The stream leaves A at v = 25 m/s along a slope 3 : 4, so θ = tan⁻¹(3/4) = 36.87° above the horizontal (cos θ = 0.8, sin θ = 0.6). The only acceleration is g = 9.81 m/s² downward.

(i) At the nozzle: the normal component of g is a_n = g cos θ = 9.81 × 0.8 = 7.848 m/s². ρ = v²/a_n = 25²/7.848 = 79.6 m

(ii) At the maximum height: the velocity is horizontal, v = v_x = 25 cos θ = 20 m/s, and g is entirely normal, a_n = 9.81 m/s². ρ = v²/a_n = 20²/9.81 = 40.8 m