Question 8.5: Find the kinetic energy of the antenna system in Example 8.3...

The center of mass C of the two coil system has velocity v* = v_o , and m(β) =2m; so, by (8.51), the kinetic energy of the center of mass is

K^*(\beta, t) \equiv \frac{1}{2} m(\beta) \mathbf{v}^* \cdot \mathbf{v}^*                     (8.51)

K^*(\beta, t)=m v_O^2.                  (8.54a)

The velocity of each coil relative to C is given in (8.48b); therefore, by (8.52), the kinetic energy of the system relative to C is

\dot{\rho}_1=-\dot{\rho}_2=-\mathbf{v}  +  \omega_f \times \mathbf{d}=-v \mathbf{i}  +  \omega d \mathbf{j}                      (8.48b)

K_{r C}(\beta, t) \equiv \sum_{k=1}^n \frac{1}{2} m_k \dot{\boldsymbol{\rho}}_k \cdot \dot{\boldsymbol{\rho}}_k                          (8.52)

K_{r C}=\frac{1}{2} m\left(\dot{\rho}_1 \cdot \dot{\rho}_1  +  \dot{\rho}_2 \cdot \dot{\rho}_2\right)=m\left(v^2  +  \omega^2 d^2\right)               (8.54b)

Finally, (8.53) yields the kinetic energy of the antenna coil system:

K(\beta, t)=K^*(\beta, t)  +  K_{r C}(\beta, t) .                        (8.53)

K(\beta, t)=m\left(v_O^2  +  v^2  +  \omega^2 d^2\right) .                (8.54c)

The reader will find the same result on starting from (8.50).

K(\beta, t) \equiv \sum_{k=1}^n K_k(t)=\sum_{k=1}^n \frac{1}{2} m_k \mathbf{v}_k \cdot \mathbf{v}_k,                  (8.50)