Question

For a rigid axisymmetric satellite, the mass moment of inertia about its long axis is 1000 kg$m2 , and the moment of inertia about transverse axes through the center of mass is 5000 kg$m2 . It is spinning about the minor principal body axis in torque-free motion at 6 rad/s with the angular velocity lined up with the angular momentum vector H. Over time, the energy degrades due to internal effects and the satellite is eventually spinning about a major principal body axis with the angular velocity lined up with the angular momentum vector H. Calculate the change in rotational kinetic energy between the two states. {Ans.: 14.4 kJ}

Answer 1

Answer: Delta T = - 14.4 kJ

Step-by-step explanation:

Firstly the initial momentum of the satellite is;

H0 = L0*W0

L0 is the momentum of inertia of the satellite about its longer axis (1000 kg.m^2) and Wo is the initial angular velocity of the satellite ( 6 rad/s)

now we substitute

H0 = 1000 kg.m^2 * 6 rad/s

H0 = 6000 kg.m^2

Now the initial rotational kinetic energy of the satellite is;

0o = 1/2L0W0^2

we substitute

T0 = 1/2 * (1000 kg.m^2) * (6 rad/s)^2

T0 = 18000

Next is the final angular momentum of the satellite;

H = IW

I is the moment of inertia of the satellite about its transverse axes through center of mass (5000 kgm^2)

we know that the law of conservation angular momentum, the total initial angular momentum of the satellite is equal to the total final angular momentum of the satellite.

simply put H = H0

we know that our H0 = 6000 kgm^2

so we substitute

H0 = IW

6000 kg.m^2 = 5000 kgm^2 * W

W = 1.2 rad/s

The final rotational kinetic energy of the satellite is;

T = 1/2IW^2

T = 1/2 * 5000 kgm^2 * 1.2^2

T = 3600 J

so the change in rotational kinetic energy of the satellite is;

Delta T = T - T0

Delta T = 3600 - 18000

DeltaT = - 14400 J

Delta T = - 14.4 kJ