Question 8.8: Two small blocks of weight W1 and W2 are connected by a perf...
Two small blocks of weight W_1 and W_2 are connected by a perfectly flexible and inextensible cable of negligible mass, as shown in Fig. 8.9. The weight W_1 rests on a smooth horizontal surface and is attached to a linear spring of stiffness k fastened to a rigid wall. The cable is free to slide over a smooth pulley at P and suspends the weight W_2 . The system is at rest initially when W_2 is displaced vertically and released. (i) Find the total energy of the system. (ii) Derive the equation of motion and determine the frequency of the vibration. (iii) Describe alternative formulations of these issues .
Solution of (i). The physical system is modeled as a system of two particles of masses m_1 and m_2 for each of which the free body diagram is shown in Fig. 8.9a. The weight W_1 and the normal, smooth surface reactions at W_1 and at P do no work in the motion . The cable has negligible mass, so its motion around the smooth pulley may be ignored , and hence the oppositely directed, internal cable tensions T_1 and T_2 have equal magnitude T, say. The cable is inextensible and perfectly flexible, so the total internal potential energy of the system is zero: B = 0, and W_1 and W_2 share the same displacement so that x = y. In the equilibrium state, x_o=y_o is the static displacement of W_1 and W_2 so that kx_o=m_2g. The relevant external forces W_2 and the elastic spring force are conservative. The system , therefore, is conservative with total potential energy (8.84) equal to the total external potential energy \Phi(\beta)=\phi_1+\phi_2 from (8.72), namely,
V(\beta) \equiv \Phi(\beta) + B(\beta) (8.84)
\Phi(\beta) \equiv \sum_{k=1}^n \phi_k\left(\mathbf{x}_k\right) (8.72)
V(\beta)=\frac{1}{2} k\left(x + x_0\right)^2 – m_2 g\left(y + y_0\right), (8.96a)
obtained directly by inspection of the system diagram . The total kinetic energy of the system, with the inextensibility constraint x = y in mind, is
K(\beta, t)=K_1 + K_2=\frac{1}{2} m_1 \dot{x}^2 + \frac{1}{2} m_2 \dot{y}^2=\frac{1}{2} m(\beta) \dot{x}^2, (8.96b)
wherein m(\beta) \equiv m_1 + m_2 Hence , the principle of conservation of energy gives the constant total energy of the system :
E=K+V=\frac{1}{2} m(\beta) \dot{x}^2 + \frac{1}{2} k\left(x + x_0\right)^2 – m_2 g\left(x + x_0\right) . (8.96c)
Solution of (ii). Because the system has only one degree of freedom, the equation of motion may be found by differentiation of (8.96c) with respect to the path variable x (or the time t) to obtain
\ddot{x} + \frac{k}{m(\beta)} x=\frac{m_2 g – k x_0}{m(\beta)} \text {. } (8.96d)
The system is in its static equilibrium state at x = 0 where m_2 g – k x_0=0, and hence the equation of motion for the system is
\ddot{x} + p^2 x=0, \quad p=\sqrt{\frac{k}{m_1 + m_2}}, (8.96e)
in which p is the circular frequency .
Solution of (iii). Other methods will lead to the same results. Because the spring is linear, one alternative approach is to consider the motion from the static equilibrium position directly . The kinetic energy is unchanged in (8.96b), while the total potential energy may be written as
V(\beta)=\frac{1}{2} k x^2. (8.96f)
This procedure, however, cannot be used when the spring is nonlinear, whereas the earlier method leading to (8.96a) can. The principle of conservation of energy (8.86) yields
\frac{1}{2} m(\beta) \dot{x}^2 + \frac{1}{2} k x^2=E \text {. } (8.96g)
Differentiation of this equation with respect to x (or t) returns (8.96e) .
Another approach starts with the separate equations of motion for each particle. With reference to the free body diagrams in Fig. 8.9a, we find easily
m_1 \ddot{x}=T – k\left(x + x_0\right), \quad m_2 \ddot{y}=m_2 g – T, \quad W_1=N . (8.96h)
Eliminating T and introducing the inextensibility constraint y = x and the equilibrium condition m_2 g=k x_0, we recover (8.96e). The equations of motion also may be formulated relative to the static state : m_1 \ddot{x}=T-k x, m_2 \ddot{y}=-T, which again lead to (8.96e).