Question 11.12: Calculate the net moment on the solar panel of Examples 11.2...
Calculate the net moment on the solar panel of Examples 11.2 and 11.8 (Fig. 11.17).
Since the comoving frame is rigidly attached to the panel, the Euler equation (Eq. 11.72a) applies to this problem. That is
\left.\left. M _G\right)_{\text {net }}=\dot{ H }_G\right)_{ 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 11.2, the angular velocity of the panel in the satellite’s x’y’z’ frame was found to be
\omega =-\dot{\theta} \hat{ j }^{\prime}+N \hat{ k }^{\prime} (d)
In Example 11.8, we showed that the direction cosine matrix for the transformation from the panel’s xyz frame to that of the satellite is
[ 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 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 11.2, N and \dot{\theta} were said to be constant. Therefore, the time derivatives of Eq. (f) are
\dot{\omega}_x=\frac{ d (N \cos \theta)}{ d t}=-N \dot{\theta} \sin \theta \quad \dot{\omega}_x=\frac{ d \theta}{ dt }=0 \quad \dot{\omega}_z=\frac{ d (N \sin \theta)}{ d t}=N \dot{\theta} \cos \theta (g)
In Example 11.8, the moments of inertia in the panel frame of reference were listed as
\left.\left.\left.A=\frac{1}{12} m\left(\ell^2+t^2\right) \quad B=\frac{1}{12} m\left(w^2+t^2\right) \quad C=\frac{1}{12} m\left(w^2+\ell^2\right) \quad I_G\right)_{x y}=I_G\right)_{x z}=I_G\right)_{y z}=0 (h)
Substituting Eqs. (b), (c), (f), (g), and (h) into Eq. (a) yields
\left. M _G\right)_{\text {net }}=\frac{1}{12} m\left(\ell^2+t^2\right)(-N \dot{\theta} \sin \theta) \hat{ i }+\frac{1}{12} m\left(w^2+t^2\right)(0) \hat{ j }+\frac{1}{12} m\left(w^2+\ell^2\right)(N \dot{\theta} \cos \theta) \hat{ k }
+\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\N \cos \theta & \dot{\theta} & N \sin \theta \\\frac{1}{12} m\left(\ell^2+t^2\right)(N \cos \theta) & \frac{1}{12} m\left(w^2+t^2\right) \dot{\theta} & \frac{1}{12} m\left(w^2+\ell^2\right)(N \sin \theta)\end{array}\right|
Upon expanding the cross product determinant and collecting terms, this reduces to
\left. M _G\right)_{\text {net }}=-\frac{1}{6} m t^2 N \dot{\theta} \sin \theta \hat{ i }+\frac{1}{24} m\left(t^2-w^2\right) N^2 \sin 2 \theta \hat{ j }+\frac{1}{6} m w^2 N \dot{\theta} \cos \theta \hat{ k }
Using the numerical data of Example 11.8 (m = 50 kg, N= 0.1 rad/s, θ = 40°, \dot{\theta} = 0.01 rad/s, \ell = 6 m, w = 2 m, and t = 0.025 m), we find
\left. M _G\right)_{\text {net }}=-3.348\left(10^{-6}\right) \hat{ i }-0.08205 \hat{ j }+0.02554 \hat{ k } ( N \cdot m )