Calculate the net moment on the solar panel of Examples 11.2 and 11.8 (Fig. 11.17).

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Accepted Answer

Since the comoving frame is rigidly attached to the panel, the Euler equation (Eq. 11.72a) applies to this problem. That is

[latex]\left.\left. M _G ight)_{ ext {net }}=\dot{ H }_G ight)_{ rel }+ \omega imes H _G[/latex] (a)

where

[latex]H _G=A \omega_x \hat{ i }+B \omega_{ y } \hat{ j }+C \omega_z \hat{ k }[/latex] (b)

and

[latex]\left.\dot{ H }_G ight)_{ rel }=A \dot{\omega}_x \hat{ i }+B \dot{\omega}_y \hat{ j }+C \dot{\omega}_z \hat{ k }[/latex] (c)

In Example 11.2, the angular velocity of the panel in the satellite’s x'y'z' frame was found to be

[latex]\omega =-\dot{ heta} \hat{ j }^{\prime}+N \hat{ k }^{\prime}[/latex] (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

[latex][ Q ]=\left[\begin{array}{ccc}-\sin heta & 0 & \cos heta \0 & -1 & 0 \\cos heta & 0 & \sin heta\end{array} ight][/latex] (e)

We use the transpose of [Q] to transform the components of ω into the panel frame of reference,

[latex]\{ \omega \}_{x y z}=[ Q ]^T\{ \omega \}_{x^{\prime} y z^{\prime}}=\left[\begin{array}{ccc}-\sin heta & 0 & \cos heta \0 & -1 & 0 \\cos heta & 0 & \sin heta\end{array} ight]\left\{\begin{array}{c}0 \-\dot{ heta} \N\end{array} ight\}=\left\{\begin{array}{c}N \cos heta \\dot{ heta} \N \sin heta\end{array} ight\}[/latex]

or

[latex]\omega_x=N \cos heta \quad \omega_y=\dot{ heta} \quad \omega_z=N \sin heta[/latex] (f)

In Example 11.2, N and [latex]\dot{ heta}[/latex] were said to be constant. Therefore, the time derivatives of Eq. (f) are

[latex]\dot{\omega}_x=\frac{ d (N \cos heta)}{ d t}=-N \dot{ heta} \sin heta \quad \dot{\omega}_x=\frac{ d heta}{ dt }=0 \quad \dot{\omega}_z=\frac{ d (N \sin heta)}{ d t}=N \dot{ heta} \cos heta[/latex] (g)

In Example 11.8, the moments of inertia in the panel frame of reference were listed as

[latex]\left.\left.\left.A=\frac{1}{12} m\left(\ell^2+t^2 ight) \quad B=\frac{1}{12} m\left(w^2+t^2 ight) \quad C=\frac{1}{12} m\left(w^2+\ell^2 ight) \quad I_G ight)_{x y}=I_G ight)_{x z}=I_G ight)_{y z}=0[/latex] (h)

Substituting Eqs. (b), (c), (f), (g), and (h) into Eq. (a) yields

[latex]\left. M _G ight)_{ ext {net }}=\frac{1}{12} m\left(\ell^2+t^2 ight)(-N \dot{ heta} \sin heta) \hat{ i }+\frac{1}{12} m\left(w^2+t^2 ight)(0) \hat{ j }+\frac{1}{12} m\left(w^2+\ell^2 ight)(N \dot{ heta} \cos heta) \hat{ k }[/latex]

[latex]+\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \N \cos heta & \dot{ heta} & N \sin heta \\frac{1}{12} m\left(\ell^2+t^2 ight)(N \cos heta) & \frac{1}{12} m\left(w^2+t^2 ight) \dot{ heta} & \frac{1}{12} m\left(w^2+\ell^2 ight)(N \sin heta)\end{array} ight|[/latex]

Upon expanding the cross product determinant and collecting terms, this reduces to

[latex]\left. M _G ight)_{ ext {net }}=-\frac{1}{6} m t^2 N \dot{ heta} \sin heta \hat{ i }+\frac{1}{24} m\left(t^2-w^2 ight) N^2 \sin 2 heta \hat{ j }+\frac{1}{6} m w^2 N \dot{ heta} \cos heta \hat{ k }[/latex]

Using the numerical data of Example 11.8 (m = 50 kg, N= 0.1 rad/s, θ = 40°, [latex]\dot{ heta}[/latex] = 0.01 rad/s, [latex]\ell[/latex] = 6 m, w = 2 m, and t = 0.025 m), we find

[latex]\left. M _G ight)_{ ext {net }}=-3.348\left(10^{-6} ight) \hat{ i }-0.08205 \hat{ j }+0.02554 \hat{ k } ( N \cdot m )[/latex]

Question 11.12: Calculate the net moment on the solar panel of Examples 11.2...

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