For simplicity we demonstrate this important principle for one-dimensional motion. Suppose that a body undergoes displacement ∆x under the action of a conservative force F. Then from the WE theorem we have,
∆K = F(x)∆x
If the force is conservative, the potential energy function V(x) can be defined such that
−∆V = F(x)∆x
The above equations imply that
∆K + ∆V = 0
∆(K + V) = 0
which means that K + V, the sum of the kinetic and potential energies of the body is a constant. Over the whole path, xito xf , this means that
K$_i$ + V(x$_i$) = K$_f$ + V(x$_f$) (6.11)
The quantity K +V(x), is called the total mechanical energy of the system. Individually the kinetic energy K and the potential energy V(x) may vary from point to point, but the sum is a constant. The aptness of the term ‘conservative force’ is now clear.
Let us consider some of the definitions of a conservative force.
- A force F(x) is conservative if it can be derived from a scalar quantity V(x) by the relation given by Eq. (6.9). The three-dimensional generalisation requires the use of a vector derivative, which is outside the scope of this book.
- The work done by the conservative force depends only on the end points. This can be seen from the relatio,
which depends on the end points.
- A third definition states that the work done by this force in a closed path is zero. This is once again apparent from Eq. (6.11) since x$_i$ = x$_i$.
The total mechanical energy of a system is conserved if the forces, doing work on it, are conservative.
The above discussion can be made more concrete by considering the example of the gravitational force once again and that of the spring force in the next section. Fig. 6.5 depicts a ball of mass m being dropped from a cliff of height H.
The total mechanical energies E$_0$, E$_h$, and E$_H$ of the ball at the indicated heights zero (ground level), h and H, are
E$_H$ = mgH (6.11 a)
E$_h$ = mgh + $\frac{1}{2}mv_h^2$ (6.11 b)
E$_0$ = $\frac{1}{2}mv_f^2$ (6.11 c)
The constant force is a special case of a spatially dependent force F(x). Hence, the mechanical energy is conserved. Thus
E$_H$ = E$_0$
or, mgh = $\frac{1}{2}mv_f^2$
$v_f=\sqrt{2gH}$
a result that was obtained in section 3.7 for a freely falling body. Further,
E$_H$ = E$_h$
which implies,
$v_h^2=2g(H-h)$ (6.11 d)
and is a familiar result from kinematics.
At the height H, the energy is purely potential. It is partially converted to kinetic at height h and is fully kinetic at ground level. This illustrates the conservation of mechanical energy.
Example 1
A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity v$_0$ at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point, C. This is shown in Fig. 1. Obtain an expression for (i) v$_0$ ; (ii) the speeds at points B and C; (iii) the ratio of the kinetic energies (K$_B$/K$_C$) at B and C. Comment on the nature of the trajectory of the bob after it reaches the point C.
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| Fig.1 |
Answer
(i) There are two external forces on the bob: gravity and the tension (T) in the string. The latter does no work since the displacement of the bob is always normal to the string. The potential energy of the bob is thus associated with the gravitational force only. The total mechanical energy E of the system is conserved. We take the potential energy of the system to be zero at the lowest point A. Thus, at A:
$E=\frac{1}{2}mv_0^2$
$T_A-mg=\frac{mv_0^2}{L}$ [Newton’s Second Law]
where T$_A$ is the tension in the string at A. At the highest point C, the string slackens, as the tension in the string (T$_C$) becomes zero. Thus, at C
$E=\frac{1}{2}mv_C^2+2mgL$
$mg=\frac{mv_c^2}{L}$ [Newton’s Second Law
where v$_C$ is the speed at C. From Eqs. (6.13) and (6.14)
$E=\frac{5}{2}$mgL
Equating this to the energy at A
$\frac{5}{2}$mgL = $\frac{1}{2}mv_0^2$
Or, $v_0=\sqrt{5gL}$
(ii) It is clear from Eq. (6.14)
$v_C=\sqrt{gL}$
At B, the energy is
$E=\frac{1}{2}mv_B^2+mgL$
Equating this to the energy at A and employing the result from (i), namely $v_0^2$ = 5gL
$\frac{1}{2}mv_B^2+mgL=\frac{1}{2}mv_0^2=\frac{5}{2}mgL$
$v_B=\sqrt{3gL}$
(iii) The ratio of the kinetic energies at B and C is :
$\frac{K_B}{K_C}=\frac{\frac{1}{2}mv_B^2}{\frac{1}{2}mv_C^2}=\frac{3}{1}$
At point C, the string becomes slack and the velocity of the bob is horizontal and to the left. If the connecting string is cut at this instant, the bob will execute a projectile motion with horizontal projection akin to a rock kicked horizontally from the edge of a cliff. Otherwise the bob will continue on its circular path and complete the revolution.
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