Question 9.20: Figure 9.24 shows a rotating platform on which is mounted a ......

The inertial XYZ frame is centered at O on the platform, and it is right handed (\hat{ I } \times \hat{ J }=\hat{ K }). The origin of the right handed comoving body frame xyz is at the shaft’s center of mass G, and it is aligned with the symmetry axes of the parallelepiped. The three Euler angles \phi, \theta, \text { and } \psi are shown in Figure 9.24. Since θ is constant, the nutation rate is zero \left(\omega_n=0\right) \text {. } Thus, Eqn (9.115) reduces to

\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)

\omega_x=\omega_p \sin \theta \sin \psi \quad \omega_y=\omega_p \sin \theta \cos \psi \quad \omega_z=\omega_p \cos \theta+\omega_s                          (a)

Since \omega_{p,} \omega_s, \text { and } \theta are constant, it follows (recalling Eqn (9.106)) that

\omega_p=\dot{\phi} \quad \omega_n=\dot{\theta} \quad \omega_s=\dot{\psi}                             (9.106)

\dot{\omega}_x=\omega_p \omega_s \sin \theta \cos \psi \quad \dot{\omega}_y=-\omega_p \omega_s \sin \theta \sin \psi \quad \dot{\omega}_z=0                           (b)

The principal moments of inertia of the parallelepiped are (Figure 9.10(c))

\begin{aligned}& A=\ell_x=\cfrac{1}{12}  m\left(h^2+\ell^2\right) \\& B=\ell_y=\cfrac{1}{12}  m\left(b^2+\ell^2\right) \\& C=\ell_z=\cfrac{1}{12}  m\left(b^2+h^2\right)\end{aligned}                        (c)

Figure 9.25 is a free-body diagram of the shaft. Let us assume that the bearings at D and E are such as to exert just the six body frame components of force shown. Thus, D is a thrust bearing to which the axial torque T_D is appliedfrom, say, a motor of some kind. At E, there is a simple journal bearing.

From Newton’s laws of motion, we have F _{ net }=m a _{ G }. But G is fixed in inertial space, so a_G=0. Thus,

It follows that

E_x=-D_x \quad E_y=-D_y \quad D_z=0                       (d)

Summing moments about G we get

where we made use of Eqn (d)_2. Thus,

M_{x_{\text {net }}}=D_y \ell \quad M_{y_{\text {net }}}=-D_x \ell \quad M_{z_{\text {net }}}=T_D                      (e)

We substitute Eqns (a)e(c) and (e) into Euler’s equations (Eqn (9.72)):

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

After making the substitutions and simplifying, the first Euler equation, Eqn (f)_1, becomes

D_x=\left\{\cfrac{1}{12} \cfrac{m}{\ell}\left[\left(\ell^2-h^2\right) \omega_p \cos \theta-2 h^2 \omega_s\right] \omega_p \sin \theta\right\} \cos \psi                            (g)

Likewise, from Eqn (f)_2 we obtain

D_y=\left\{\cfrac{1}{12} \cfrac{m}{\ell}\left[\left(\ell^2-b^2\right) \omega_p \cos \theta-2 b^2 \omega_S\right] \omega_p \sin \theta\right\} \sin \psi                        (h)

Finally, Eqn (f)_3 yields

T_D=\left[\cfrac{1}{24}  m\left(b^2-h^2\right) \omega_p^2 \sin ^2 \theta\right] \sin 2 \psi                       (i)

This completes the solution, since E_y=-D_y \text { and } E_z=-D_z. Note that the resultant transverse bearing load V at D (and E) is

V=\sqrt{D_x^2+D_y^2}                              (j)

As a numerical example, let

and

For these numbers, the variation of V and T_D \text { with } \psi are as shown in Figure 9.26.