Download Figure 58: Gravitational potential-energy experiment. • Betty: fixes

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 yrock = +1 m.
– Using (8.4), gravitational potential energy of rock is
UgA = m g yrock = 1 × 9.80 × 1 = 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 yrock = 0 m.
– Using (8.4), gravitational potential energy of rock is
UgA = m g yrock = 1 × 9.80 × 0 = 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
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, yrock = +0.5 m.
– Gravitational potential energy of the rock is now
UgB = m g yrock = 1 × 9.80 × 0.5 = 4.9 J .
• Bill: in his co-ordinate system, yrock = −0.5 m.
– Gravitational potential energy of rock is now
UgB = m g yrock = 1 × 9.80 × (−0.5) = −4.9 J .
So both observers continue to disagree over the value of Ug 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 yrock = −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: ∆Ug = UgB − UgA = 4.90 − 9.80 = −4.90 J.
• Bill: ∆Ug = UgB − UgA = −4.90 − 0 = −4.90 J.
So both observers agree on the change in the gravitational potential energy
of the rock.
Therefore, we find that
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• Ug depends upon the choice of the co-ordinate system.
• ∆Ug is independent of the choice of co-ordinate system.
8.2
Conservation of Mechanical Energy
Recalling (8.2), i.e.
1
1 2
mvf + mgyf = mvi2 + mgyi ,
2
2
what this equation really says is
Ki + Ugi = Kf + Ugf ,
(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, EM , to be
EM = K + Ug ,
(8.6)
equation (8.5) may be rewritten in terms of the change in mechanical energy,
(EM )i = (EM )f
−→ ∆EM = (Kf + Ugf ) − (Ki + Ugi ) = 0 ,
(8.7)
or, defining ∆K = (Kf − Ki ) and ∆Ug = (Ugf − Ugi ),
∆EM = ∆K + ∆Ug = 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 Ug 6= 0, i.e. EM = Ug .
106
• As the object falls, K increases (as v increases) while Ug decreases (as y
decreases), such that EM still remains constant.
As the object falls, its potential energy is transferred/converted into
kinetic energy.
Example:
49
a rock is released from rest, as in figure 59.
50
. 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 vi = 0, then
1
∆K = Kf − Ki = mvf2 .
2
49
50
Knight, Example 10.2, page 275
Knight, Figure 10.9, page 275
107
(8.9)
• Potential energy: recall that, although both observers measure different
values for Ug , they measure the same value for ∆Ug .
– In Betty’s frame, where yi = 1 m and yf = 0 m,
∆Ug = Ugf − Ugi = 0 − m g yi .
(8.10)
Using (8.9) and (8.10) in the law of conservation of mechanical energy (8.8),
in Betty’s frame,
∆K + ∆Ug = 0
−→
1 2
mv − m g yi = 0 ,
2 f
i.e.
vf =
8.3
r
2 m g yi √
= 2 × 9.80 × 1 = 4.43 m s−1 .
m
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, L0 , to a length L, and released, experiences a restoring
force returning it to its equilibrium position (e.g. a rubber band, ruler).
8.3.1
Hooke’s Law (1676)
Consider a spring stretched or compressed from its equilibrium length (see
figure 60
51
). 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 281
108
• Let the position co-ordinate of the stretched/compressed length be s.
• Let us define the displacement from equilibrium, ∆s, as
∆s = s − se .
(8.11)
Hooke’s Law states that provided the elastic limit of the spring is not
exceeded, the spring force (Fsp )s is given by
(Fsp )s = −k ∆s ,
(8.12)
where k is the spring constant of the spring.
• 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 (Fsp )s versus ∆s graph.
• As figure 60 shows, the sign of (Fsp )s is always opposite to the sign of
∆s:
– For a stretched spring: ∆s > 0, (Fsp )s < 0.
– For a compressed spring, ∆s < 0, (Fsp )s > 0.
Note: Hooke’s law is written as
(Fsp )s = −ks
if and only if the origin of the co-ordinate system is at the equilibrium value,
i.e. se = 0 in (8.11).
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Figure 60: Stretched/compressed spring.
110
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 61
52
.
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 283
111
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
1
1 2 1
mvf + k (∆sf )2 = mvi2 + k (∆si )2 ,
2
2
2
2
(8.13)
where ∆si and ∆sf 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, Us :
1
Us = k (∆s)2 .
2
(8.14)
Us is the energy stored in a spring.
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
EM = K + Ug + Us .
8.5
(8.15)
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 62
54
.
• 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,
http://www.nanotech.uwaterloo.ca/∼ne131/
54
Knight, Figure 9.1, page 240
112
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:
8.5.1
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.
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• The colliding pair approach each other (both particles make up the isolated system).
• The pair collide and move away (figure 63
55
).
• Linear momentum is conserved:
m1 (vix )1 + m2 (vix )2 = m1 (vf x )1 + m2 (vf x )2 ,
(8.16)
where (vix )2 (vf x )2 = 0 in figure 63.
• Mechanical energy is conserved:
1
1
1
1
m1 (vix )21 + m2 (vix )22 = m1 (vf x )21 + m2 (vf x )22 ,
2
2
2
2
where (vix )2 = (vf x )2 = 0 in figure 63.
Figure 63: A perfectly elastic collision.
55
Knight, Figure 10.24, page 287
114
(8.17)
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 64
56
)..
• 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:
m1 (vix )1 + m2 (vix )2 = m1 (vf x )1 + m2 (vf x )2 ,
(8.18)
where (vix )2 (vf x )2 = 0 in figure 63.
• Mechanical energy is not conserved (but the total energy is).
8.5.3
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 255
115