Figure 58: Gravitational potential-energy experiment. Betty: fixes her co-ordinate system to the ground, with the origin at ground level.

Transcription

1 Figure 58: Gravitational potential-energy experiment. Betty: fixes her co-ordinate system to the ground, with the origin at ground level. Initially, rock is at y rock = +1 m. Using (8.4), gravitational potential energy of rock is U ga = m g y rock = = 9.80 J. Bill: fixes his co-ordinate system such that the origin coincides with the initial position of the rock. Initially, rock is at y rock = 0 m. Using (8.4), gravitational potential energy of rock is U ga = m g y rock = = 0 J. So both observers disagree about the gravitational potential energy of the rock! This disagreement is obviously due to the differing choices of co-ordinate system. 104

2 Suppose the experiment is repeated, but with the rock held initially at 0.5 m above the ground (let this be position B). Betty: in her co-ordinate system, y rock = +0.5 m. Gravitational potential energy of the rock is now U gb = m g y rock = = 4.9 J. Bill: in his co-ordinate system, y rock = 0.5 m. Gravitational potential energy of rock is now U gb = m g y rock = ( 0.5) = 4.9 J. So both observers continue to disagree over the value of U g for the rock. Note: A U < 0 value is perfectly okay, since its value depends upon the co-ordinate system. e.g. in the above example, all we are saying is that the rock has less gravitational potential energy at y rock = 0.5 m than it has at y = 0 m. However, since it depends upon a velocity, K 0. Considering the change in gravitational potential energy, U g, we find that Betty: U g = U gb U ga = = 4.90 J. Bill: U g = U gb U ga = = 4.90 J. So both observers agree on the change in the gravitational potential energy of the rock. Therefore, we find that 105

3 U g depends upon the choice of the co-ordinate system. U g is independent of the choice of co-ordinate system. 8.2 Conservation of Mechanical Energy Recalling (8.2), i.e. what this equation really says is 1 2 mv2 f + mgy f = 1 2 mv2 i + mgy i, K i + U gi = K f + U gf, (8.5) that is, the sum of the kinetic and gravitational potential energies at an initial time is identical to that at a later time. Defining the mechanical energy, E M, to be E M = K + U g, (8.6) equation (8.5) may be rewritten in terms of the change in mechanical energy, (E M ) i = (E M ) f E M = (K f + U gf ) (K i + U gi ) = 0, (8.7) or, defining K = (K f K i ) and U g = (U gf U gi ), E M = K + U g = 0. (8.8) Equation (8.8) is the Law of Conservation of Mechanical Energy. What does this law actually tell us for an object in free fall, for example? With respect to equation (8.5), we see that Initially, the object held at rest has K = 0 but U g 0, i.e. E M = U g. 106

4 As the object falls, K increases (as v increases) while U g decreases (as y decreases), such that E M still remains constant. As the object falls, its kinetic energy. potential energy is transferred/converted into Example: 49 a rock is released from rest, as in figure Use both Bill and Betty s perspectives to calculate its speed just before it hits the ground. Figure 59: Motion of rock in free fall. Solution: considering the kinetic and potential energies of the m = 1 kg rock in turn: Kinetic energy: since K depends only upon the rock s speed, and not position, both observers agree on the initial and final values of K. Since v i = 0, then 49 Knight, Example 10.2, page Knight, Figure 10.9, page 275 K = K f K i = 1 2 mv2 f. (8.9) 107

5 Potential energy: recall that, although both observers measure different values for U g, they measure the same value for U g. In Betty s frame, where y i = 1 m and y f = 0 m, U g = U gf U gi = 0 m g y i. (8.10) Using (8.9) and (8.10) in the law of conservation of mechanical energy (8.8), in Betty s frame, i.e. K + U g = mv2 f m g y i = 0, v f = 2 m g yi m = = 4.43 m s Restoring Forces and Hooke s Law Recall that any material which returns to its equilibrium position/configuration after the removal of an applied force can be considered to possess spring-like properties. elastic material: a material such that, when displaced from its equilibrium length, L 0, to a length L, and released, experiences a restoring force returning it to its equilibrium position (e.g. a rubber band, ruler) Hooke s Law (1676) Consider a spring stretched or compressed from its equilibrium length (see figure ). We assume that its elastic limit is not exceeded i.e. that the material will return to its equilibrium length when the spring force is removed. Let the position co-ordinate of the equilibrium length be s e. 51 Knight, Figure 10.17, page

6 Let the position co-ordinate of the stretched/compressed length be s. Let us define the displacement from equilibrium, s, as s = s s e. (8.11) Hooke s Law states that provided the elastic limit of the spring is not exceeded, the spring force (F sp ) s is given by where k is the spring constant of the spring. (F sp ) s = k s, (8.12) k is an intrinsic property of the spring. Large k spring is stiff. Small k spring is soft. SI units of k are N m 1. k is the gradient of a (F sp ) s versus s graph. As figure 60 shows, the sign of (F sp ) s is always opposite to the sign of s: For a stretched spring: s > 0, (F sp ) s < 0. For a compressed spring, s < 0, (F sp ) s > 0. Note: Hooke s law is written as (F sp ) s = ks if and only if the origin of the co-ordinate system is at the equilibrium value, i.e. s e = 0 in (8.11). 109

7 Figure 60: Stretched/compressed spring. 110

8 8.4 Elastic Potential Energy Intuitively, we are aware that an elastic material, displaced from its equilibrium position, stores energy, e.g. a stretched and loaded catapult! Suppose a ball of mass m, moving without friction on an ideal spring, is launched as in figure Figure 61: Motion of a spring-ball system. Before the ball s release: The kinetic energy of the ball is zero. The total energy of the system is stored as some sort of potential energy. After the ball s release: The stored potential energy is converted into kinetic energy, as the speed of the ball increases. 52 Knight, Figure 10.19, page

9 Performing a similar type of calculation as for the generalised motion of an object on a frictionless surface (see Additional Materials 53 ), we may derive the expression 1 2 mv2 f k ( s f) 2 = 1 2 mv2 i k ( s i) 2, (8.13) where s i and s f are the equilibrium displacements before and after release. Equation (8.13) has a similar form to equation (8.2), where in addition to the kinetic energy mv 2 /2, we define the elastic potential energy, U s : U s is the energy stored in a spring. U s = 1 2 k ( s)2. (8.14) Equation (8.13) is the conservation of mechanical energy for our idealised ball-spring system. Note: a spring moving vertically has both gravitational and elastic potential energy, and a total mechanical energy E M = K + U g + U s. (8.15) 8.5 Elastic and Inelastic Collisions Let us recall the concept of a collision. An example of a microscopic model of a typical collision is shown in figure Before collision: objects approach each other with a kinetic energy due to their movement. 53 Derivation of the Kinetic and Potential Energies of Some Simple Systems, ne131/ 54 Knight, Figure 9.1, page

10 Figure 62: Microscopic model of a collision. During collision: objects interact a very large number of molecular bonds are compressed. The kinetic energy is transformed into elastic potential energy stored in the spring-like molecular bonds. After collision: the stored elastic potential energy is converted into kinetic energy as the objects move away from each other. Although in reality, real collisions lie somewhere in between, we may consider two limiting cases of a collision: Perfectly Elastic Collisions The situation discussed above in which all of the initial kinetic energy is stored as elastic potential energy, and then all of the stored elastic potential energy is transformed into kinetic energy characterises a perfectly elastic collision. 113

11 The colliding pair approach each other (both particles make up the isolated system). The pair collide and move away (figure ). Linear momentum is conserved: m 1 (v ix ) 1 + m 2 (v ix ) 2 = m 1 (v fx ) 1 + m 2 (v fx ) 2, (8.16) where (v ix ) 2 (v fx ) 2 = 0 in figure 63. Mechanical energy is conserved: 1 2 m 1 (v ix ) m 2 (v ix ) 2 2 = 1 2 m 1 (v fx ) m 2 (v fx ) 2 2, (8.17) where (v ix ) 2 = (v fx ) 2 = 0 in figure 63. Figure 63: A perfectly elastic collision. 55 Knight, Figure 10.24, page

12 8.5.2 Perfectly Inelastic Collisions A collision in which the two objects stick together and move away with a common final velocity is called a perfectly inelastic collision. The colliding pair approach each other and interact (both particles make up the isolated system). Instead of moving away from each other, the two objects stick together (figure ).. Some of the kinetic energy is dissipated inside the object not all of the kinetic energy is recovered. The form of such dissipation may be, for example, A change in the molecular structure of either or both masses (e.g. bonds broken, etc.). Sound. Heat. Linear momentum is conserved: m 1 (v ix ) 1 + m 2 (v ix ) 2 = m 1 (v fx ) 1 + m 2 (v fx ) 2, (8.18) where (v ix ) 2 (v fx ) 2 = 0 in figure 63. Mechanical energy is not conserved (but the total energy is) Special Cases Let us consider elastic collisions only. In obeying both linear-momentum and mechanical-energy conservation laws, given by (8.16) and (8.17), respectively, they can be solved simultaneously, yielding the results 56 Knight, Figure 9.20, page