A 3.084-kg steel projectile is fired from a gun. At a particular point in its tr

Question

A 3.084-kg steel projectile is fired from a gun. At a particular point in its trajectory it has a velocity of 300 m/s, an altitude of 700 m, and a temperature of 350 K. The projectile is spinning about its longitudinal axis at an angular speed of 100 rad/s. Treating the projectile as a thermodynamic system, calculate the total system energy relative to a reference state of zero velocity, zero altitude, and a temperature of 298 K. The rotational moment of inertia for the projectile is 9.64x10^-4 kg*m^2 and the specific heat is 460.5 J/kg*K.

Explanation / Answer

the translational KE = .5 * mv^2
Translational KE = 0.5 * 3.084 * 300^2 = 138780 J ;
rotational kinetic Energy = 0.5 * I^2 = 0.5 * ( 9.64 e -4 ) ( 100 ) ^2 = 4.82 J

its Potential energy = mgh = 3.084 * 9.8 * 700 = 21156.24 J ;

its energy gained due to heat = m c T = 3.084 * 460.5 * ( 350 - 298 ) = 73849.464 J

to get the total energy just sum all the above values in red = 138780 + 4.82 + 21156.24 + 73849.464 J

ans = 233790.524 J = 233.790524 kJ

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