Question 7.12: (i) Apply the principle of conservation of energy to find th...
(i) Apply the principle of conservation of energy to find the velocity v_B at which the particle in Example 7.10, page 249, is projected from point B shown in Fig. 7.9, and show that an arbitrary constant reference potential energy does not alter the conclusion. (ii) Derive from the energy equation the equivalent Newton-Euler scalar equation of motion for the mass on the circular path AB.
(i), First, we need to confirm that the energy principle (7.73) may be applied. The force s that act on the mass m on the circular path AB are shown in Fig. 7.9. Since the weight W is a conservative force and the normal surface reaction force does no work on AB , the total energy is conserved.
K+V=E \text {, a constant. } (7.73)
The point A is clearly a convenient datum for zero gravitational potential energy. However, we recall that only differences in the potential energy are relevant. Moreover, an arbitrary reference potential energy V_0 does not alter the energy balance in (7.73), for the same constant potential energy will appear in both sides of the equation. To demonstrate this, let us choose an arbitrary value V_0 for the reference potential energy at A. Since the particle is at rest at A, the kinetic energy at A is zero. Thus , initially the total energy is E=K_A + V_A=V_0. With A as the reference state, the potential energy of the mass m at point B is V_B=V_0-m g R, and its kinetic energy is K_B=\frac{1}{2} m v_B^2, where \mathbf{v}_B=v_B \mathbf{i} is the velocity of m when it projects from B. The energy principle (7.73) now yields
K_B+V_B=\frac{1}{2} m v_B^2 + V_0 – m g R=E=V_0, (7.77a )
that is,
v_B=\sqrt{2 g R} \text {, } (7.77b)
so \mathbf{v}_B=\sqrt{2 g R} \mathbf{i}, the same result found in Example 7.10 by application of the work-energy principle. Notice that the arbitrary reference potential energy V_o cancels from (7.77a). Hence , the particular value assigned to the datum potential energy V_o has no effect whatsoever on the solution .
Solution of (ii). The scalar equation of motion equivalent to the energy principle is readily derived from the energy equation. At an arbitrary point on the path A B, K=\frac{1}{2} m R^2 \dot{\phi}^2 and V=-m g R \sin \phi, where we now fix V_0=0 at A. Since E = 0 at A, (7.73) yields the energy equation on the path AB,
\frac{1}{2} m R^2 \dot{\phi}^2 – m g R \sin \phi=0 (7.77c)
Differentiation of (7.77c) with respect to the path variable Φ (or with respect to t) yields the equivalent tangential component of the Newton–Euler equation of motion, namely,
m R \ddot{\phi} – W \cos \phi=0 . (7.77d)
Notice, in agreement with (7.76), that R \ddot{\phi}=\ddot{s} is the tangential component of the acceleration, and W \cos \phi=F_t is the conservative tangential component of the total force F = W +N acting on m in Fig. 7.9, whose workless normal component is F_n=N – W \sin \phi.