--- title: "Nozzle stream curvature; dependent motion" subject: "Applied Mechanics" chapter: 8 key: "semester_1/applied_mechanics:nozzle-curvature-dependent-motion" source: https://www.bctengineer.com/q/semester_1/applied_mechanics/nozzle-curvature-dependent-motion site: BCT Engineering --- # Nozzle stream curvature; dependent motion Applied Mechanics · Chapter 8: Kinematics of Particles and Rigid Body - Asked in: 2073 Chaitra, 8+2 marks - Video: [a nozzle releasing stream of water (applied mechanics) @ioecrack2935](https://www.youtube.com/watch?v=XbYK2rbZdaU) — ioe crack section architecture - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/nozzle-curvature-dependent-motion - Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-8 ## Question A nozzle discharges a stream of water in the direction shown below with an initial velocity of 25 m/sec. Determine the radius of curvature of the stream (a) as it leaves the nozzle, (b) at the maximum height of the stream. What do you mean by dependent motion of particle? Explain with suitable example. [Figure: nozzle at A, stream leaves at 25 m/s along a slope of 3 vertical to 4 horizontal, reaching maximum point B.] ## Short answer Dependent motion occurs when particles are connected by an inextensible cord, e.g., block A and hanging block B with constraint x_A + 2y_B = L, giving v_A = -2v_B. For the nozzle: at exit, ρ = v²/(g cosθ) = 79.6 m; at maximum height, ρ = v_x²/g = 40.8 m. ## Answer #### Nozzle stream The stream leaves A at v = 25 m/s along a slope 3 : 4, so θ = tan⁻¹(3/4) = 36.87° (cos θ = 0.8, sin θ = 0.6). The only acceleration is g = 9.81 m/s² downward. Radius of curvature ρ = v²/a_n, where a_n is the component of g normal to the path. **(a) As it leaves the nozzle:** a_n = g cos θ = 9.81 × 0.8 = 7.848 m/s² ρ_A = 25²/7.848 = **79.6 m** **(b) At the maximum height B:** velocity is horizontal: v_B = v_x = 25 × 0.8 = 20 m/s; g is entirely normal, so a_n = 9.81 m/s². ρ_B = 20²/9.81 = **40.8 m** (For reference, the maximum height is (25 × 0.6)²/(2 × 9.81) = 11.47 m above the nozzle and the time to reach it is 15/9.81 = 1.53 s.) #### Dependent motion of particles When the position of one particle depends on the position of another because they are joined by an inextensible cord (or a rigid link), their motions are called **dependent**. The constraint gives a relation between the coordinates; differentiating it once gives the velocity relation and twice the acceleration relation. Example: block A moves on a horizontal table and is joined by a cord over a pulley to block B, which hangs from a movable pulley supported by the same cord. If x_A is the distance of A from the fixed pulley and y_B the drop of the movable pulley, the cord length is constant: x_A + 2y_B = L so v_A = −2v_B and a_A = −2a_B. If B moves down 1 m, A moves 2 m toward the pulley. Only one coordinate is independent, so the system has one degree of freedom. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is a_n = g cos θ at the nozzle? A. Only the part of g perpendicular to the path (tangent at angle θ) is normal B. g is always fully tangential C. g is multiplied by sin θ D. Normal acceleration equals v^2/g Answer: A. The tangent is inclined at θ to the horizontal, so g splits into g sin θ tangential and g cos θ normal. (Key point: Normal acceleration at A) 2. What is the radius of curvature at the maximum height? A. 79.6 m B. 20.4 m C. 61.2 m D. 40.8 m Answer: D. At B v = 20 m/s and a_n = g, so ρ = 400/9.81 = 40.8 m. (Key point: Radius at B) 3. When are the motions of two particles dependent? A. When they have equal masses B. When they are on the same table C. When a constraint such as an inextensible cord relates their coordinates D. When they have equal velocities Answer: C. A constant cord length gives a constraint equation relating the coordinates. (Key point: Definition of dependent motion) 4. For x_A + 2y_B = L, how do velocities relate? A. v_A = -v_B B. v_A = -2v_B C. v_A = -v_B/2 D. v_A = 2v_B Answer: B. Differentiating gives v_A + 2v_B = 0, so A moves twice as fast as B, in the opposite sense of the coordinate. 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