Two particles of mass m1 = m and m2 = 3m are moving with respective velocities v1 = (4v , -7v, 0) and v2 = (0, v, 4v) relative to frame Φ = {F; ik}. (a) Find the velocity of the center of mass in Φ. (b) At a certain instant the respective particles are at x1 = (0, -1 ,3) and x2 = (8, -1 ,3) from a

Question

Two particles of mass [latex]m_1 = m[/latex] and [latex]m_2 = 3m[/latex] are moving with respective velocities [latex]v_1 = (4v , -7v, 0)[/latex] and [latex]v_2 = (0, v, 4v)[/latex] relative to frame [latex]\phi = \{F; i_k\}[/latex]. (a) Find the velocity of the center of mass in Φ. (b) At a certain instant the respective particles are at [latex]x_1 = (0, -1 ,3)[/latex] and [latex]x_2 = (8, -1 ,3)[/latex] 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?

Accepted Answer

Solution of (a). To find v*, consider the momentum of the center of mass in [latex]\Phi: \mathbf{p}^*=m(\beta) \mathbf{v}^*[/latex], where [latex]m(\beta)=m_1 + m_2=4 m[/latex]. Then , by (5.7), the total momentum of the system is [latex]\mathbf{p}^*=m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2=m(4 v,-7 v, 0)+3 m(0, v, 4 v)[/latex], i.e.

[latex]\mathbf{p}^* \equiv m(\beta) \mathbf{v}^*=\sum_{k=1}^n m_k \mathbf{v}_k=\mathbf{p}(\beta, t)[/latex] (5.7)

[latex]\mathbf{p}^*=4 m \mathbf{v}^*=4 m v(1,-1,3)[/latex]; (8.28a)

and hence the center of mass has velocity

[latex]\mathbf{v}^*=v(\mathbf{i}-\mathbf{j} + 3 \mathbf{k})[/latex]. (8.28b)

Solution of (b). To determine [latex]\mathbf{h}_O^*=\mathbf{x}^* imes \mathbf{p}^*[/latex], 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

[latex]\mathbf{x}^*=6 \mathbf{i} - \mathbf{j} + 3 \mathbf{k}[/latex]; (8.28c)

and with (8.28a) the moment about 0 of the momentum of C in Φ is

[latex]\mathbf{h}_O^*=\mathbf{x}^* imes \mathbf{p}^*=4 m v\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \6 & -1 & 3 \ 1 & -1 & 3\end{array} ight|=-20 m v(3 \mathbf{j} + \mathbf{k})[/latex]. (8.28d)

Solution of (c). The moment about O of the momentum of C relative to the origin in Φ is given by [latex]\mathbf{h}_O^*=\mathbf{x}^* imes m(\beta) \mathbf{v}^*[/latex], in which [latex]\mathbf{v}^* \equiv \dot{\mathbf{X}}^*=\mathbf{v}_O + \dot{\mathbf{x}}^*[/latex] in Fig. 8.2. Because point O is fixed in [latex]\Phi, \mathbf{v}_O=\mathbf{0}[/latex] and hence [latex]\mathbf{p}^*=\mathbf{p}_O^*[/latex]. (In general, of course, [latex]\mathbf{p}^* eq \mathbf{p}_O^*[/latex].) Then, by (8.26) and (8.28d), [latex]\mathbf{h}_{r O}^*=\mathbf{h}_O^*=-20 m v(3 \mathbf{j} + \mathbf{k})[/latex].

More generally, (8.24) holds for a system consisting of only a single particle, hence for the center of mass particle alone,

[latex]\mathbf{h}_O(\beta, t)=\mathbf{h}_{r O}(\beta, t) + m(\beta) \mathbf{x}_O^* imes \mathbf{v}_O[/latex] (8.24)

[latex]\mathbf{h}_O^*=\mathbf{h}_{r O}^* + m(\beta) \mathbf{x}_O^* imes \mathbf{v}_O[/latex]. (8.28e)

Therefore, when [latex]v_O=0[/latex], we have the special result [latex]\mathbf{h}_O^*=\mathbf{h}_{r O}^*[/latex] found in the example. Characterization of other situations for which this holds is left for the reader.

Question 8.2: Two particles of mass m1 = m and m2 = 3m are moving with res...

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