Question

For the sake of directional stability, the bullet fired from a rifle is given a spin angular velocity about its axis by means of spiral grooves ("rifling") cut into barrel. The bullet fired by a Lee-Enfield rifle is (approximately) a uniform cylinder of length equal to 3.18 cm, diameter equal to 0.790 cm, and mass equal to 13.9 g. The bullet emerges from the muzzle with a translational velocity of 622 m/s and a spin angular velocity of 2.78 103 rev/s.

(a) What is the translational kinetic energy of the bullet? J

(b) What is the rotational kinetic energy? J

(c) What fraction of the total kinetic energy is rotational?

Answer 1

Given;

Translational velocity of bullet; v = 622 m/s

Rotational velocity/ Spin angular velocity; w = 2.78103 rev/s = rad/s = 17.4754 rad/s

Mass of the bullet; m = 13.9 g = 0.0139 kg

Radius of the bullet; r = 0.395 cm = 0.00395 m

Moment of Inertia of the Bullet ; I = = kgm²

(a) Translational Kinetic energy ; K = = 2688.8438 J

(b) Rotational Kinetic Energy; K' = =

(c) Fraction of Rotational Kinetic energy = =