Question 9.12: Calculate the net moment on the solar panel of Examples 9.2 ......
Calculate the net moment on the solar panel of Examples 9.2 and 9.8 (Figure 9.17).
Since the comoving frame is rigidly attached to the panel, Euler’s equation (Eqn (9.72a)) applies to this problem, that is
\boxed{\left. M _{\text {net }}=\dot{ H }\right)_{\text {rel }}+ \omega \times H} (9.72a)
\left. M _{G_{\text {net }}}=\dot{ H }_G\right)_{\text {rel }}+\omega \times H _G (a)
where
H _G=A \omega_x \hat{ i }+B \omega_y \hat{ j }+C \omega_z \hat{ k } (b)
and
\left.\dot{ H }_G\right)_{ rel }=A \dot{\omega}_x \hat{ i }+B \dot{\omega}_y \hat{ j }+C \dot{\omega}_z \hat{ k } (c)
In Example 9.2, the angular velocity of the panel in the satellite’s x^{\prime} y^{\prime} z^{\prime} frame was found to be
\omega=-\dot{\theta} \hat{ j }^{\prime}+N \hat{ k }^{\prime} (d)
In Example 9.8, we showed that the transformation from the panel’s xyz frame to that of the satellite is represented by the matrix
[ Q ]=\left[\begin{array}{ccc}-\sin \theta & 0 & \cos \theta \\0 & -1 & 0 \\\cos \theta & 0 & \sin \theta\end{array}\right] (e)
We use the transpose of [Q] to transform the components of ω into the panel frame of reference,
\{ \omega \}_{x y z}=[ Q ]^T\{ \omega \}_{x^{\prime} y^{\prime} z^{\prime}}=\left[\begin{array}{ccc}-\sin \theta & 0 & \cos \theta \\0 & -1 & 0 \\\cos \theta & 0 & \sin \theta\end{array}\right]\left\{\begin{array}{c}0 \\-\dot{\theta} \\N\end{array}\right\}=\left\{\begin{array}{c}N \cos \theta \\\dot{\theta} \\N \sin \theta\end{array}\right\}or
\omega_x=N \cos \theta \quad \omega_y=\dot{\theta} \quad \omega_z=N \sin \theta (f)
In Example 9.2, N and \dot{\theta} were said to be constant. Therefore, the time derivatives of Eqn (f) are
\begin{aligned}& \dot{\omega}_x=\cfrac{ d (N \cos \theta)}{ d t}=-N \dot{\theta} \sin \theta \\\\& \dot{\omega}_y=\cfrac{ d \dot{\theta}}{ d t}=0 \\\\& \dot{\omega}_z=\cfrac{ d (N \sin \theta)}{ d t}=N \dot{\theta} \cos \theta\end{aligned} (g)
In Example 9.8, the moments of inertia in the panel frame of reference were listed as
Substituting Eqns (b), (c), (f), (g) and (h) into Eqn (a) yields
Upon expanding the cross product and collecting terms, this reduces to
M _{G_{\text {net }}}=-\cfrac{1}{6} m t^2 N \dot{\theta} \sin \theta \hat{ i }+\cfrac{1}{24} m\left(t^2- \text{w}^2\right) N^2 \sin 2 \hat{\theta} \hat{ j }+\cfrac{1}{6} m \text{w}^2 N \dot{\theta} \cos \theta \hat{ k }
Using the numerical data of Example 9.8 \left(m=50 kg , N=0.1 rad / s , \theta=40^{\circ}, \dot{\theta}=0.01 rad / s , \ell=6 m , \text{w}=2 m ,\right. and t = 0.025 m), we find\boxed{M _{G_{\text {net }}}=-3.348 \times 10^{-6} \hat{ i }-0.08205 \hat{ j }+0.02554 \hat{ k } ( N \cdot m )}