Question 8.7: A system shown in Fig. 8.8a consists of two blocks of mass m...

Solution of (i). We model the physical system in Fig. 8.8a as a system of two particles (center of mass objects) with mass m_1 \text{ and }m_2. respectively. Their separate free body diagrams are shown in Fig. 8.8b. To find the total energy of the system   \beta=\left\{m_1, m_2\right\},  we first note that the weights  W_1  and  W_2,  and the normal, smooth surface reaction forces  N_1  and  N_2  are external forces that do no work in the motion , and the elastic spring forces are conservative. As usual , the infinitesimal internal, mutual gravitational force between the particles is ignored. The system  \beta=\left\{m_1, m_2\right\},  therefore, is conservative.

Let   x_1  and  x_2  denote the respective displacements of  m_1  and  m_2  from the natural state of the springs  S_1  and  S_2 in Fig . 8.8a. There are no constraints relating these variables, so the system has two degrees of freedom. The spring  S_1  exerts an external force on  m_1  with external potential energy  \phi_1;  but no relevant external forces act on  m_2, \mathrm{so}   \phi_2=0.  The total external potential energy (8.72) is thus given by

\Phi(\beta) \equiv \sum_{k=1}^n \phi_k\left(\mathbf{x}_k\right)                (8.72)

\Phi=\phi_1=\frac{1}{2} k_1 x_1^2 .                 (8.95a)

The mutual internal force on  m_1  and  m_2  is due to the spring  S_2.  These forces, shown in Fig. 8.8b, are equal but oppositely directed, so the total internal force is zero; but the total internal potential energy is not. The internal potential energy arising from the elastic force exerted on  m_1  by  m_2  is   \beta_{12}=\frac{1}{2} k_2\left(x_2  –  x_1\right)^2,  and  the internal potential energy arising from the equal but oppositely directed elastic force exerted on  m_2  by  m_1  is  \beta_{21}=\frac{1}{2} k_2\left(x_2  –  x_1\right)^2.  Clearly, the mutual internal potential energy  \beta_{12}=\beta_{21}  is a symmetric function of the change in distance, hence , also the current distance between the particles, as indicated in (8.75) and (8.76). The total internal potential energy (8.79) is thus determined by  B=\frac{1}{2}\left(\beta_{12}  +  \beta_{21}\right),  that is,

\beta_{j k}=\psi\left(\left|\mathbf{r}_{j k}\right|\right)                         (8.75)

\beta_{j k}=\beta_{k j}                  (8.76)

B=\beta_{12}=\frac{1}{2} k_2\left(x_2  –  x_1\right)^2.                 (8.95b)

Consequently, the total internal potential energy of the system in Fig. 8.8a is simply the elastic potential energy due to the spring  S_2.  Therefore, with (8.95a) and (8.95b), the total potential energy (8.84) of the system is

V(\beta) \equiv \Phi(\beta)  +  B(\beta)             (8.84)

V(\beta)=\Phi  +  B=\frac{1}{2} k_1 x_1^2  +  \frac{1}{2} k_2\left(x_2  –  x_1\right)^2                   (8.95c)

It is evident that the total potential energy function   V(\beta)  may be written down immediately by inspection of the system in Fig. 8.8a. In fact, this is the usual procedure to follow.

The total kinetic energy (8.50) of the system in Fig. 8.8a is

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)

K(\beta, t)=\frac{1}{2} m_1 \dot{x}_1^2  +  \frac{1}{2} m_2 \dot{x}_2^2 .                   (8.95d)

The total energy of the conservative system   \beta=\left\{m_1, m_2\right\}  now follows from (8.86):

K  +  V=E \text {, a constant. }             (8.86)

K  +  V=\frac{1}{2} m_1 \dot{x}_1^2  +  \frac{1}{2} m_2 \dot{x}_2^2  +  \frac{1}{2} k_1 x_1^2  +  \frac{1}{2} k_2\left(x_2  –  x_1\right)^2=E, \text { a constant. }                     (8.95e)

Solution of (ii). Application of the law of motion to each particle shown separately in the free body diagrams of Fig. 8.8b yields the following equations of motion for the system :

m_1 \ddot{x}_1=-k_1 x_1  +  k_2\left(x_2  –  x_1\right), \quad m_2 \ddot{x}_2=-k_2\left(x_2  –  x_1\right)                       (8.95f)

Notice that the sum of these equations is  m_1 \ddot{x}_1  +  m_2 \ddot{x}_2=-k_1 x_1,  or  m(\beta) \ddot{x}^*= F(\beta),   which is the equation of motion (8.5) of the system with total external force  F(\beta)=-k_1 x_1.  This system equation is not useful. The motions  x_1  and  x_2  of the particles are coupled, but independent-the motion of one particle influences but does not determine the motion of the other. As a consequence, these equations cannot be separately integrated. The solution of coupled equations of this kind is considered in Chapter 11.

Solution of (iii). We next determine the horizontal forces and their totals from their potential functions. The total external force acting on each particle is determined by use of (8.95a) in (8.70):

\mathbf{f}_k=-\boldsymbol{\nabla}_k \phi_k\left(\mathbf{x}_k\right), \quad k=1,2, \ldots, n                       (8.70)

\mathbf{f}_1=-\nabla_1 \phi_1=-\frac{\partial \phi_1}{\partial x_1} \mathbf{i}=-k_1 x_1 \mathbf{i}, \quad \mathbf{f}_2=-\nabla_2 \phi_2=-\frac{\partial \phi_2}{\partial x_2} \mathbf{i}=\mathbf{0}                               (8.95g)

Hence, the total external force on the system, by (8.4), is   \mathbf{F}(\beta, t)=\mathbf{f}_1  +  \mathbf{f}_2=-k_1 x_1 \mathbf{i}  , noted earlier. All other forces that act must be the equal, oppositely directed external and internal contributions. Specifically, the equal and oppositely directed external forces are  N_1 = -W_1, N_2 = -W_2;  these have no influence on the motion. The total internal force acting on each particle is obtained by use of (8.95b) in (8.80):

\mathbf{b}_j=\sum_{\substack{k=1 \\ k \neq j}}^n \mathbf{b}_{j k}=-\sum_{\substack{k=1 \\ k \neq j}}^n \nabla_j \beta_{j k}\left(\left|\mathbf{r}_{j k}\right|\right)                             (8.80)

\begin{aligned}&\mathbf{b}_1=-\nabla_1 \beta_{12}=-\frac{\partial \beta_{12}}{\partial x_1} \mathbf{i}=k_2\left(x_2   –  x_1\right) \mathbf{i}, \\&\mathbf{b}_2=-\nabla_2 \beta_{21}=-\frac{\partial \beta_{21}}{\partial x_2} \mathbf{i}=-k_2\left(x_2  –  x_1\right) \mathbf{i} .\end{aligned}                     (8.95h)

Of course, the total internal force acting on the system is  b_1  +  b_2 = 0.  From  (8.95g) and (8.95h), the total force  F_k = f_k  +  b_k  acting on each particle separately is

\mathbf{F}_1=\left[-k_1 x_1  +  k_2\left(x_2  –  x_1\right)\right] \mathbf{i}, \quad \mathbf{F}_2=-k_2\left(x_2  –  x_1\right) \mathbf{i}                       (8.95i)

These are the totals of the forces on the right-hand side of the equations in  (8.95f) and shown in the free body diagrams of Fig. 8.8b.

Finally, we apply equations (8.87), (8.89) , and (8.90). Using (8.95c) in the first relation of (8.87), we have

\mathbf{F}_k=-\nabla_k V\left(\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_n\right)=-\nabla_k \Phi  –  \nabla_k B                  (8.87)

\nabla_k B=\sum_{\substack{j=1 \\ k \neq j}}^n \nabla_k \beta_{k j}=-\mathbf{b}_k                     (8.89)

\mathbf{F}(\beta, t)=\sum_{k=1}^n \mathbf{F}_k=-\sum_{k=1}^n \nabla_k V(\beta)=-\sum_{k=1}^n \nabla_k \Phi=\sum_{k=1}^n \mathbf{f}_k                        (8.90)

\mathbf{F}_1=-\boldsymbol{\nabla}_1 V=\left[-k_1 x_1  +  k_2\left(x_2  –  x_1\right)\right] \mathbf{i}, \quad \mathbf{F}_2=-\boldsymbol{\nabla}_2 V=-k_2\left(x_2  –  x_1\right) \mathbf{i},                      (8.95j)

in agreement with (8.95i). It is seen immediately that use of these relations in
(8.91) returns the separate equations on motion (8.95f) . With (8.95b) in (8.89), we find

m_k \ddot{\mathbf{x}}_k=-\nabla_k V(\beta), \quad k=1,2, \ldots, n .                        (8.91)

\mathbf{b}_1=-\nabla_1 B=k_2\left(x_2  –  x_1\right) \mathbf{i}, \quad \mathbf{b}_2=-\nabla_2 B=-k_2\left(x_2  –  x_1\right) \mathbf{i}                 (8.95k)

in accord with (8.95h). And, finally, substitution of (8.95j) into (8.90) yields

\mathbf{F}(\beta, t)=-\nabla_1 V  –  \nabla_2 V=-k_1 x_1 \mathbf{i}                     (8.95l)

the total external force on the system .