Question 9.19: The mass moments of inertia of a body about the principal bo......

(a) We must use Euler’s equations (Eqns (9.72)) to calculate the net moment, which means we must first obtain \omega_x, \omega_y, \omega_z, \dot{\omega}_x, \dot{\omega}_y, \text { and } \dot{\omega}_z . Since we are given the Euler angles as a function of time, we can compute their time derivatives and then use Eqn (9.115) to find the body frame angular velocity components and their derivatives. Starting with Eqn (b), we get

\boxed{\left. M _{\text {net }}=\dot{ H }\right)_{\text {rel }}+ \omega \times H}                            (9.72a)

\begin{aligned}& M_{x_{ net }}=A \dot{\omega}_x+(C-B) \omega_y \omega_z \\& M_{y_{ net }}=B \dot{\omega}_y+(A-C) \omega_z \omega_x \\& M_{z_{ net }}=C \dot{\omega}_z+(B-A) \omega_x \omega_y\end{aligned}                      (9.72b)

\begin{aligned}& \omega_x=\omega_p \sin \theta \sin \psi+\omega_n \cos \psi \\\\& \omega_y=\omega_p \sin \theta \cos \psi-\omega_n \sin \psi \\\\& \omega_z=\omega_s+\omega_p \cos \theta\end{aligned}                                   (9.115)

Proceeding to the remaining two Euler angles leads to

Evaluating all these quantities, including those in Eqn (b), at t = 10 s yields

Equation (9.115) relates the Euler angle rates to the angular velocity components,

\begin{aligned}& \omega_x=\omega_p \sin \theta \sin \psi+\omega_n \cos \psi \\& \omega_y=\omega_p \sin \theta \cos \psi-\omega_n \sin \psi \\& \omega_z=\omega_s+\omega_p \cos \theta\end{aligned}                      (d)

Taking the time derivative of each of these equations in turn leads to the following three equations:

Substituting the data in Eqn (c) into Eqns (d) and (e) yields

With Eqns (a) and (f) we have everything we need for Euler’s equations, namely,

from which we find

(b) Since the comoving xyz frame is a body frame, rigidly attached to the solid, we know from Eqn (9.74) that

\left\{\begin{array}{c}\dot{\omega}_x \\\dot{\omega}_y \\\dot{\omega}_z\end{array}\right\}=[ Q ]_{X \text{x}}\left\{\begin{array}{l} d \omega_X / d t \\d \omega_Y / d t \\d \omega_Z / d t\end{array}\right\}                                      (9.74)

\left\{\begin{array}{c}\alpha_X \\\alpha_Y \\\alpha_Z\end{array}\right\}=[ Q ]_{\text{x} X}\left\{\begin{array}{c}\dot{\omega}_X \\\dot{\omega}_y \\\dot{\omega}_z\end{array}\right\}                                 (g)

In other words, the absolute angular acceleration and the relative angular acceleration of the body are thesame. All we have to do is project the components of relative acceleration in Eqn (f) onto the axes of the inertial frame. The required orthogonal transformation matrix is given in Eqn (9.105),

Upon substituting the numerical values of the Euler angles from Eqn (c), this becomes

Substituting this and the relative angular velocity rates from Eqn (f) into Eqn (g) yields