At time t = 0 the body-fixed axes and inertial angular velocity of a rigid body are those given in Example 11.18, namely i0 = 0.40825I -0.40825J +0.81649K j0 = -0.10102I -0.90914J -0.40405K k0 = 0.90726I +0.082479J -0.41240K (a) and ωX = -3.1I +2.5J +1.7K (rad/s) (b) If the angular velocity is

Question

At time t = 0 the body-fixed axes and inertial angular velocity of a rigid body are those given in Example 11.18, namely

[latex]\hat{ i }_0=0.40825 \hat{ I }-0.40825 \hat{ J }+0.81649 \hat{ K }[/latex]

[latex]\hat{ j }_0=-0.10102 \hat{ I }-0.90914 \hat{ J }-0.40405 \hat{ K }[/latex] (a)

[latex]\hat{ k }_0=0.90726 \hat{ I }+0.082479 \hat{ J }-0.41240 \hat{ K }[/latex]

and

[latex]\omega _X=-3.1 \hat{ I }+2.5 \hat{ J }+1.7 \hat{ K } ( rad / s )[/latex] (b)

If the angular velocity is constant, find the time histories of the Euler angles and the quaternion.

Accepted Answer

Because the angular velocity is constant, the motion of the body will be pure rotation about the fixed axis of rotation defined by the angular velocity vector. Once we find the direction cosine matrix as a function of time, we can use Algorithm 4.3 to obtain the Euler angles at each time.
Step 1:
As in Example 11.18, we find that the direction cosine matrix at time t = 0 is

[latex]\left[ Q _0 ight]_{X x}=\left[\begin{array}{rrr}0.40825 & -0.40825 & 0.81649 \-0.10102 & -0.90914 & -0.40405 \0.90726 & 0.082479 & -0.41240\end{array} ight][/latex] (c)

Step 2:
As in Example 11.18, use [latex]\left[ Q _0 ight]_{X x}[/latex] to project the angular velocity [latex]\omega _X[/latex] onto the axes of the body-fixed frame

[latex]\{ \omega \}_X=\left[ Q _0 ight]_{X x}\{ \omega \}_X=\left[\begin{array}{rrr}0.40825 & -0.40825 & 0.81649 \-0.10102 & -0.90914 & -0.40405 \0.90726 & 0.082479 & -0.41240\end{array} ight]\left\{\begin{array}{r}-3.1 \2.5 \1.7\end{array} ight\}[/latex]

so that

[latex]\omega_x=-0.89817 rad / s \quad \omega_y=-2.6466 rad / s \quad \omega_z=-3.3074 rad / s[/latex] (d)

The magnitude of the constant angular velocity is

[latex]\omega=\sqrt{\omega_x^2+\omega_y^2+\omega_z^2}=4.3301 rad / s[/latex] (e)

The period of the rotation is T = 2π/ω = 1.451 s.
Step 3:
Use the angular velocities in (d) to form the matrix [Ω] in Eq. (11.168),

[latex][ \Omega ]=\left[\begin{array}{cccc}0 & -3.3074 & 2.6466 & -0.89817 \3.3074 & 0 & -0.89817 & -2.6466 \-2.6466 & 0.89817 & 0 & -3.3074 \0.89817 & 2.6466 & 3.3074 & 0\end{array} ight][/latex] (f)

[latex][ \Omega ]=\left[\begin{array}{ccc:c}0 & \omega_3 & -\omega_2 & \omega_1 \-\omega_3 & 0 & \omega_1 & \omega_2 \\omega_2 & -\omega_1 & 0 & \omega_3 \\hdashline-\omega_1 & -\omega_2 & -\omega_3 & 0\end{array} ight][/latex] (11.168)

[Ω] remains constant.
Step 4:
Use Algorithm 11.2 to obtain the quaternion at t = 0 from the direction cosine matrix in (c).

[latex]\widehat{ q }_0=\left\{\begin{array}{c}-0.82610 \0.15412 \-0.52165 \\hdashline0.14724\end{array} ight\}[/latex] (g)

Step 5
At each time through [latex]t_{ ext {final }}[/latex]:
Compute the quaternion [latex]\widehat{ q }(t)[/latex] from Eqs. (11.169) and (11.171).
Use [latex]\widehat{ q }(t)[/latex] to update the direction cosine matrix [latex][ Q (t)]_{X x}[/latex] using Algorithm 11.1.
Use [latex][ Q (t)]_{X x}[/latex] to calculate the Euler angles [latex]\phi(t), heta(t)[/latex], and [latex]\psi(t)[/latex] by means of Algorithm 4.3.

Fig. 11.30 shows the time variation of the four components of [latex]\hat{ q }[/latex] during one rotation of the body. The variation of the three Euler angles is shown in Fig. 11.31. Observe that their values at t = 0 agree with those found in Example 11.18.
Fig. 11.32 shows the initial orientation of the orthonormal body-fixed xyz axes given in Eq. (a). The dotted lines trace
out the subsequent motion of their end points as they rotate at 4.33 rad/s about the fixed angular velocity vector ω. Finally, Fig. 11.33 shows the initial position of the Euler axis [latex]\hat{ u }[/latex] and its subsequent motion during rotation of the body. The unit vector [latex]\hat{ u }[/latex] is obtained from the unit quaternion [latex]\widehat{ q }(t)[/latex] at any instant by means of Eq. (11.146),

[latex]\hat{ u }(t)=\frac{ q (t)}{\sin \left\{\cos ^{-1}\left[q_4(t) ight] ight\}}[/latex]

where q(t) is the vector part of [latex]\widehat{ q }(t)[/latex]. Fig. 11.33 amply illustrates the fact that the unit vectors [latex]\hat{ \omega }[/latex] and [latex]\hat{ u }[/latex] are not the same.

[latex]\{\widehat{ q }\}=\exp \left(\frac{[ \Omega ]}{2} t ight)\left\{\widehat{ q }_0 ight\}[/latex] (11.169)

[latex]\exp \left(\frac{[ \Omega ]}{2} t ight)=\left[\begin{array}{rrrr}\cos \frac{\omega t}{2} & \frac{\omega_z}{\omega} \sin \frac{\omega t}{2} & -\frac{\omega_y}{\omega} \sin \frac{\omega t}{2} & \frac{\omega_x}{\omega} \sin \frac{\omega t}{2} \-\frac{\omega_z}{\omega} \sin \frac{\omega t}{2} & \cos \frac{\omega t}{2} & \frac{\omega_x}{\omega} \sin \frac{\omega t}{2} & \frac{\omega_y}{\omega} \sin \frac{\omega t}{2} \\frac{\omega_y}{\omega} \sin \frac{\omega t}{2} & -\frac{\omega_x}{\omega} \sin \frac{\omega t}{2} & \cos \frac{\omega t}{2} & \frac{\omega_z}{\omega} \sin \frac{\omega t}{2} \-\frac{\omega_x}{\omega} \sin \frac{\omega t}{2} & -\frac{\omega_y}{\omega} \sin \frac{\omega t}{2} & -\frac{\omega_z}{\omega} \sin \frac{\omega t}{2} & \cos \frac{\omega t}{2}\end{array} ight][/latex] (11.171)

[latex]\widehat{ p } \otimes \widehat{ q }=\left\{\frac{p_4 q +q_4 p + p imes q }{p_4 q_4- p \cdot q } ight\}[/latex] (11.146)

Question 11.25: At time t = 0 the body-fixed axes and inertial angular veloc...

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