Question 8.3: A communications van has an antenna system modeled in Fig. 8...
A communications van has an antenna system modeled in Fig. 8.4 as two coils of equal mass m that move radially along a rigid control shaft that rotates with angular speed ω about the vertical antenna axis. At an instant of interest, each coil is at a distance d from the center C and is moving with center directed variable speed v relative to the shaft frame \varphi=\left\{C ; \mathbf{i}_k\right\}. The van moves with speed v_o in the ground frame \Phi=\left\{F ; \mathbf{I}_k\right\}. (i) What is the total momentum of the system? (ii) What is the moment of momentum of the system relative to the center of mass? (iii) Apply (8.19) to determine the moment of momentum about the center of mass . Ignore the mass of the shaft and model the coils as particles.
Solution of (i). The total momentum in Φ of the system of coils P_1 and P_2 is equal to the momentum of its center of mass whose velocity in Φ is \mathbf{v}^*=\mathbf{v}_O=v_O \mathbf{I}, the velocity of the van. Hence , \mathbf{p}^*=m(\beta) \mathbf{v}^*=2 m v_O \mathbf{I}.
Solution of (ii). The moment of momentum of the system relative to the center of mass is determined by (8.22) . The respective relative position vectors of P_1 and P_2 from C are
\mathbf{h}_C(\beta, t)=\sum_{k=1}^n \rho_k \times m_k \dot{\rho}_k=\mathbf{h}_r(\beta, t) (8.22)
\rho_1=-\rho_2=d \mathbf{i}. (8.48a)
The angular velocity of the shaft frame φ is \boldsymbol{\omega}_f=\omega \mathbf{k}, and hence their velocity vectors relative to the center of mass at the instant of interest are
\dot{\rho}_1=-\dot{\rho}_2=-\mathbf{v} + \omega_f \times \mathbf{d}=-v \mathbf{i} + \omega d \mathbf{j}. (8.48b)
Use of (8.48a) and (8.48b) in (8.22) yields
\mathbf{h}_{r C}=\rho_1 \times m_1 \dot{\rho}_1 + \rho_2 \times m_2 \dot{\rho}_2=2 m \rho_1 \times \dot{\rho}_1=2 m d \mathbf{i} \times(\omega d \mathbf{j} – v \mathbf{i}).
That is,
\mathbf{h}_{r C}=2 m d^2 \omega \mathbf{k}. (8.48c)
Solution of (iii). To find the moment about C of the momentum relative to the origin in Φ, we shall need the total velocity of each particle in the inertial frame Φ , namely, \dot{\mathbf{X}}_k=\mathbf{v}_O + \dot{\rho}_k. Then, with (8.48a) and (8.48b), (8.19) gives
\mathbf{h}_C(\beta, t)=\sum_{k=1}^n \rho_k \times m_k \dot{\mathbf{X}}_k (8.19)
\mathbf{h}_C=\rho_1 \times m_1\left(\mathbf{v}_O + \dot{\rho}_1\right) + \rho_2 \times m_2\left(\mathbf{v}_O + \dot{\rho}_2\right)=2 \rho_1 \times m_1 \dot{\rho}_1=\mathbf{h}_{r C},
in agreement with the general rule (8.22): \mathbf{h}_C=\mathbf{h}_r C=2 m d^2 \omega \mathbf{k}. Notice in passing that the velocity of the center of mass \mathbf{v}^*=\mathbf{v}_O does not affect the final result. In view of (5.6), the term \left(m_1 \rho_1 + m_2 \rho_2\right) \times \mathbf{v}_o \equiv \mathbf{0}.
\sum_{k=1}^n m_k \rho_k=\mathbf{0} (5.6)