Question 8.2: Two particles of mass m1 = m and m2 = 3m are moving with res...
Two particles of mass m_1 = m and m_2 = 3m are moving with respective velocities v_1 = (4v , -7v, 0) and v_2 = (0, v, 4v) relative to frame \phi = \{F; i_k\}. (a) Find the velocity of the center of mass in Φ. (b) At a certain instant the respective particles are at x_1 = (0, -1 ,3) and x_2 = (8, -1 ,3) from a fixed point O in Φ. What is the moment about O of the momentum of the center of mass of the system in Φ? (c) What is the moment of momentum of C relative to O?
Solution of (a). To find v*, consider the momentum of the center of mass in \Phi: \mathbf{p}^*=m(\beta) \mathbf{v}^*, where m(\beta)=m_1 + m_2=4 m. Then , by (5.7), the total momentum of the system is \mathbf{p}^*=m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2=m(4 v,-7 v, 0)+3 m(0, v, 4 v), i.e.
\mathbf{p}^* \equiv m(\beta) \mathbf{v}^*=\sum_{k=1}^n m_k \mathbf{v}_k=\mathbf{p}(\beta, t) (5.7)
\mathbf{p}^*=4 m \mathbf{v}^*=4 m v(1,-1,3); (8.28a)
and hence the center of mass has velocity
\mathbf{v}^*=v(\mathbf{i}-\mathbf{j} + 3 \mathbf{k}). (8.28b)
Solution of (b). To determine \mathbf{h}_O^*=\mathbf{x}^* \times \mathbf{p}^*, the moment about point O of the momentum of the center of mass in Φ, we need to determine the position vector x* of the center of mass from O at the instant of interest. From (5.5), the reader will find that the center of mass is at the place from O given by
\mathbf{x}^*=6 \mathbf{i} – \mathbf{j} + 3 \mathbf{k}; (8.28c)
and with (8.28a) the moment about 0 of the momentum of C in Φ is
\mathbf{h}_O^*=\mathbf{x}^* \times \mathbf{p}^*=4 m v\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \\6 & -1 & 3 \\ 1 & -1 & 3\end{array}\right|=-20 m v(3 \mathbf{j} + \mathbf{k}). (8.28d)
Solution of (c). The moment about O of the momentum of C relative to the origin in Φ is given by \mathbf{h}_O^*=\mathbf{x}^* \times m(\beta) \mathbf{v}^*, in which \mathbf{v}^* \equiv \dot{\mathbf{X}}^*=\mathbf{v}_O + \dot{\mathbf{x}}^* in Fig. 8.2. Because point O is fixed in \Phi, \mathbf{v}_O=\mathbf{0} and hence \mathbf{p}^*=\mathbf{p}_O^*. (In general, of course, \mathbf{p}^* \neq \mathbf{p}_O^*.) Then, by (8.26) and (8.28d), \mathbf{h}_{r O}^*=\mathbf{h}_O^*=-20 m v(3 \mathbf{j} + \mathbf{k}).
More generally, (8.24) holds for a system consisting of only a single particle, hence for the center of mass particle alone,
\mathbf{h}_O(\beta, t)=\mathbf{h}_{r O}(\beta, t) + m(\beta) \mathbf{x}_O^* \times \mathbf{v}_O (8.24)
\mathbf{h}_O^*=\mathbf{h}_{r O}^* + m(\beta) \mathbf{x}_O^* \times \mathbf{v}_O. (8.28e)
Therefore, when v_O=0, we have the special result \mathbf{h}_O^*=\mathbf{h}_{r O}^* found in the example. Characterization of other situations for which this holds is left for the reader.
