Figure illustrates a Turbo Booster toy that launches a60-gram \"insect\' glider

Question

Figure illustrates a Turbo Booster toy that launches a60-gram "insect' glider (projectile) by compressing a helical spring and then releasing the spring when the trigger is pulled. When pointed upward, the glider should ascend approximately 8 m before falling. The launcher spring is made of carbon steel wire. with a diameter d = 1.1 mm. The coil diameter is D = 10 mm. Calculate the number of turns Nin the spring such that it would provide the necessary energy to the glider. The total spring working deflection is x = 150 mm with a clash allowance of 10%.

Explanation / Answer

KE Kinetic energy at the point of release of the toy = mv2/2 PE Potential energy of the spring =KX2/2 The velocity at the start must ensure 8 meters travel upward. We shall calculate the intital velocity u using Newtons laws of motion. For u=?, v= final velocity =0, height h=8m u = (2gh) = 2x 9.81 x 8 = 12.53 m/sec thus the kinetic energy (KE) in 60g toy at start must be = 60/1000 x 12.53^2/2 = 4.71 joules This KE must be equal to PE in the Spring . Therfore kx2/2 = k 0.150^2/2 = 4.71. Hence k= 4.71 x 2/0.1502 = 418N/m = 0.418N/mm D 0.01 mm d 0.0011 mm Deflection x 150 mm k = Spring stiffness =d4G/(8D3N) G = Shear Modulus = 79.3 Gpa We need to provide for 10% cushion in delfection. thus total deflection desired is 165mm. correspondigly we shall increase energy by 10% more = 4.71*1.1 = 5.2 joules After trial 32 active coils, Final spring force of 75N , deflection of 150mm results in sotred energy = 6.2 joules should be OK. the resulting shear stress is within acceptable range for Carbon Steel wire. Final spring stiffness = 0.453 N/mm.

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