Question 8.4: The antenna coil system in the previous example has an addit...
The antenna coil system in the previous example has an additional angular velocity Ω normal to the plane of \mathbf{i}^{\prime} and k in Fig. 8.5, and relative to its initially oriented shaft frame 1=\varphi=\left\{C ; \mathbf{i}_k\right\}, which is turning with angular velocity ω relative to the ground frame 0=\Phi=\left\{F ; \mathbf{I}_k\right\} at the instant shown. Find the applied torque about the center of mass required to sustain the motion of the system referred to the shaft frame 2=\varphi^{\prime}=\left\{C ; \mathbf{i}_k^{\prime}\right\} . Ignore the mass of the control shaft.
The torque about C is given by (8.45), so we must first find h_C in (8.22) referred to the moving frame . The total angular velocity of the moving frame 2=\varphi^{\prime}=\left\{C ; \mathbf{i}_k^{\prime}\right\} fixed in the control shaft is \boldsymbol{\omega}_f \equiv \boldsymbol{\omega}_{20}=\boldsymbol{\omega}_{21} + \boldsymbol{\omega}_{10}=\boldsymbol{\Omega} + \boldsymbol{\omega}. Hence, with reference to Fig. 8.5, referred to φ′,
\mathbf{M}_C(\beta, t)=\frac{d \mathbf{h}_C(\beta, t)}{d t}=\frac{d \mathbf{h}_r c(\beta, t)}{d t} . (8.45)
\mathbf{h}_C(\beta, t)=\sum_{k=1}^n \rho_k \times m_k \dot{\rho}_k=\mathbf{h}_r(\beta, t) (8.22)
\omega_f=-\Omega \mathbf{j}^{\prime} + \omega\left(\sin \theta \mathbf{i}^{\prime} + \cos \theta \mathbf{k}^{\prime}\right). (8.49a)
The velocity of each coil relative to C is \dot{\rho}_k=(-1)^k \mathbf{v} + \omega_f \times \rho_k, where we recall (8.48a) in which \mathbf{i} \rightarrow \mathbf{i}^{\prime}. Specifically,
\dot\rho_1=-\dot\rho_2=vi^\prime+\omega d \cos \theta j^\prime+\Omega d k^\prime. (8.49b)
Then (8.22) yields
\mathbf{h}_C=2 \rho_1 \times m \dot{\rho}_1=2 m d^2\left(-\Omega \mathbf{j}^{\prime} + \omega \cos \theta \mathbf{k}^{\prime}\right). (8.49c)
When θ = 0 and Ω = 0, we recover (8.48c) .
The total torque about C required to sustain the motion of the system is determined by (8.45) for h_C given in (8.49c). But h_C is a vector referred to a moving reference frame, so we shall need to apply (8.47). With the aid of (8.49a), (8.49c), and noting that \dot d(t) = –v(t) and \dot{\theta}=\Omega, we determine
\frac{\delta \mathbf{h}_C}{\delta t}=2 m d\left[(2 v \Omega – d \dot{\Omega}) \mathbf{j}^{\prime} + (d \dot{\omega} \cos \theta – d \omega \Omega \sin \theta – 2 v \omega \cos \theta) \mathbf{k}^{\prime}\right],
\begin{aligned}\boldsymbol{\omega}_f \times \mathbf{h}_C &=2 m d^2\left|\begin{array}{ccc}\mathbf{i}^{\prime} & \mathbf{j}^{\prime} & \mathbf{k}^{\prime} \\\omega \sin \theta & -\Omega & \omega \cos \theta \\0 & -\Omega & \omega \cos \theta\end{array}\right| \\&=2 m d^2\left(-\omega^2 \sin \theta \cos \theta \mathbf{j}^{\prime}-\Omega \omega \sin \theta \mathbf{k}^{\prime}\right) .\end{aligned}
Thus, by (8.47), the total moment about C of all external forces exerted on the coil system by the control shaft, by gravity, and by the drive mechanism is
\mathbf{M}_C(\beta, t)=\frac{\delta \mathbf{h}_C(\beta, t)}{\delta t}+\omega_f \times \mathbf{h}_C(\beta, t) (8.47)
\begin{aligned}\mathbf{M}_C=& 2 m d\left[\left(2 v \Omega-d \dot{\Omega} – d \omega^2 \sin \theta \cos \theta\right) \mathbf{j}^{\prime}\right.\\&\left. + (d \dot{\omega} \cos \theta – 2 d \Omega \omega \sin \theta – 2 v \omega \cos \theta) \mathbf{k}^{\prime}\right]\end{aligned} (8.49d)
When Ω = 0 and θ = 0, the applied torque required to sustain the motion of the system considered initially is \mathbf{M}_C=2 m d(d \dot{\omega} – 2 v \omega) \mathbf{k}, which also follows easily from (8.45) and (8.48c) wherein now \mathbf{k}^{\prime}= \mathbf{k} is fixed in Φ.