Download Conservation of mechanical energy

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup

Relativistic mechanics wikipedia , lookup

Centripetal force wikipedia , lookup

Gibbs free energy wikipedia , lookup

Classical central-force problem wikipedia , lookup

Internal energy wikipedia , lookup

Eigenstate thermalization hypothesis wikipedia , lookup

Hunting oscillation wikipedia , lookup

Work (thermodynamics) wikipedia , lookup

Transcript
OpenStax-CNX module: m15102
1
Conservation of mechanical energy
∗
Sunil Kumar Singh
This work is produced by OpenStax-CNX and licensed under the
Creative Commons Attribution License 2.0†
Abstract
When only conservative force interacts within the system, the mechanical energy of an isolated system
can not change.
Conservation of mechanical energy is description of an ideal mechanical process. It is characterized by
absence of non-conservative force like friction. There is no external force either on the system. Therefore,
the elements of this ideal process are :
•
•
•
•
System : isolated
Energy : kinetic and potential
Forces : gravitational and elastic
Transfer of energy : No transfer of energy across the system.
1 Characteristics of an ideal mechanical process
Conservation of mechanical energy applies to a mechanical process in which external force and non-conservative
internal forces are absent.
There is no external force on the system. Hence, work by external force is zero. There is no exchange or
transfer of energy across the system. Therefore, we use an isolated system to apply conservation of energy.
We should, however, note that transfer of energy from one form to another takes place within the system,
resulting from work done by internal force.
There is no non-conservative force like friction in the system.
It means that there is no change in
thermal energy of the system. The internal forces are only conservative force. This ensures that transfer of
energy takes place only between kinetic and potential energy of the isolated system. Since potential energy
is regained during the process, there is no dissipation of energy.
As there is no dissipation of energy involved, the system represents the most energy ecient reference
for the particular process. One of the most striking feature of this system is that only force working in the
system is conservative force. This has great simplifying eect on the analysis. The work by conservative
force is independent of path and hence calculation of potential energy of the system is path independent as
well. The independence of path, in turn, allows analysis of motion along paths, which are not straight.
∗ Version
1.2: Sep 3, 2009 6:23 am -0500
† http://creativecommons.org/licenses/by/2.0/
http://cnx.org/content/m15102/1.2/
OpenStax-CNX module: m15102
2
2 Conservation of mechanical energy
The statement of conservation of energy for the ideal mechanical process is known as conservation of
mechanical energy". The equation for the mechanical process is :
WE + WF = ∆K + ∆U
Here,
WE = 0
WF = 0
Hence, for the isolated system,
⇒ ∆K + ∆U = 0
⇒ ∆Emech = 0
⇒ Emech = 0
This is what is known as conservation of mechanical energy.
We can interpret this equation in many
ways and in dierent words :
1: When only conservative forces interact within an isolated system, sum of the change in kinetic and
potential energy between two states is equal to zero.
∆K + ∆U = 0
2:
When only conservative forces interact within an isolated system, sum of the kinetic and potential
energy of an isolated system can not change.
∆K = −∆U
⇒ Kf − Ki = − (Uf − Ui )
⇒ Ki + Ui = Kf + Uf
3:
When only conservative forces interact within an isolated system, the change in mechanical energy
of an isolated system is zero.
∆Emech = 0
4:
When only conservative forces interact within an isolated system, the mechanical energy of an isolated
system can not change.
Emech = 0
We should be aware that there are two ways to apply conservation law.
We can apply it in terms of
energy for initial (subscripted with i) and nal (subscripted with f ) states or in terms of change in
energy. This point will be clear as we work with examples.
Here, we shall work an example that we had previously analyzed with the help of "work-kinetic energy"
theorem.
This will help us to understand (i) what are the elements of ideal mechanical process (ii) how
conservation principle works and (iii) how does it help simplify analysis.
http://cnx.org/content/m15102/1.2/
OpenStax-CNX module: m15102
3
2.1 Example
Problem :
A small block of 0.1 kg is released from a height 5 m as shown in the gure. The block following
a curved path transitions to a linear horizontal path and hits the spring xed to a wedge. If no friction is
involved and spring constant is 1000 N/m, nd the maximum compression of the spring.
Motion of a block
Figure 1
Solution :
Here block, curved incline and spring form isolated system. Earth is implicit element of the
system. We may argue why incline and why not horizontal surface? Curved incline (its shape) determines
the component of gravity driving block downward. Hence, it is included in the system. Horizontal surface
also applies force on block, but in normal direction to the motion.
Now, as friction is not involved, the
horizontal surface is not involved in the exchange of energy.
Two of the elements i.e. block and spring are subjected to motion. Application of conservation principle,
thus, involves more than one object unlike analysis based on laws of motion. As friction is not involved, we
can apply the conservation of mechanical energy. Now, initial energy of the system is :
⇒ Ei = K + U = 0 + mgh = mgh = 0.1 x 10 x 5 = 5 J
When the compression in the spring is maximum, kinetic energies of both block and spring are zero. Let
"x" be the maximum compression,
⇒ Ef = K + U = 0 +
1
2
2 kx
=
⇒ Ef = = 0.5 x 1000x2 = 500x2
According to conservation of mechanical energy :
⇒ 500x2 = 5
⇒ x = 0.1 m = 10 cm
http://cnx.org/content/m15102/1.2/
1
2
2 kx
OpenStax-CNX module: m15102
note:
1
4
We can compare this solution with the one given in the module titled " Conservative forces
".
3 Vertical circular motion
Study of this vertical motion was incomplete earlier as we were limited by constant force system and linear
motion in most of the cases. On the other hand, vertical circular motion involved variable force and nonlinear path. The analysis of the situation is suited to the energy analysis. Let us consider that a small object
of mass "m" is attached to a string and is whirled in a vertical plane.
Force analysis at the highest point gives the minimum speed required so that string does not slack. It
is important to understand that this is the most critical point, where string might slack.
y-direction :
Vertical circular motion
Figure 2: The object maintains a minimum speed.
P
Fy = mg + T =
mv 2
r
For the limiting case, when T = 0,
2
⇒ mg = mv
r
√
⇒ v =
( gr )
1 "Conservative force" <http://cnx.org/content/m14104/latest/#element-52>
http://cnx.org/content/m15102/1.2/
The forces in
OpenStax-CNX module: m15102
5
This expression gives the limiting value of the speed of particle in vertical circular motion. If the speed
of the particle is less than this limiting value, then the particle will fall. Once this value is known, we can
nd speed of the particle at any other point along the circular path by applying conservation of mechanical
energy. Here, we shall nd speed at two other positions - one at the bottom (C) and other at the middle
point (B or D).
Vertical circular motion
Figure 3: The object maintains a minimum speed.
In order to analyze mechanical energy, we need to have a reference zero gravitational potential energy. For
this instant case, we consider the horizontal level containing point "C" as the zero reference for gravitational
potential energy. Now, mechacnial energy at points "A", "C" and "D" are :
+ mg x 2r =
⇒ EC = K + U =
1
2
2 mvA
1
2
2 mvC
⇒ ED = K + U =
1
2
2 mvD
+ mgr
⇒ EA = K + U =
+ 0 =
1
2
2 mvC
From rst two equations, we have :
⇒ vC =
√
( 5gr )
From last two equations, we have :
⇒ vD =
http://cnx.org/content/m15102/1.2/
√
( 3gr )
1
2 mgr
+ 2mgr =
5
2 mgr
OpenStax-CNX module: m15102
6
This shows how energy analysis can be helpful in situations where force on the object is not constant
and path of motion is not straight.
However, this example also points out one weak point about energy
analysis. For example, the initial condition for minimum speed at the highest can not be obtained using
energy analysis and we have to resort to force analysis. What it means that if details of the motion are to
be ascertained, then we have to take recourse to force analysis.
3.1 Example
Problem 2:The ball is released from a height h along a smooth path shown in the gure. What should
be the height h so that ball goes around the loop without falling of the track ?
Vertical circular motion
Figure 4: The ball does not fall o the track.
Solution : The path here is smooth.
Hence, friction is absent.
No external force is working on the
Earth ball system. The only internal force is gravity, which is a conservative force. Hence, we can apply
conservation of mechanical energy.
The most critical point where ball can fall o the track is B. We have seen that the ball need to have a
minimum speed at "B" as given by :
vB =
http://cnx.org/content/m15102/1.2/
√
gr
OpenStax-CNX module: m15102
7
Corresponding to this requirement at B, the speed of the ball can be obtained at bottom point C by
applying conservation of mechanical energy. The speed of the ball as worked out earlier is given as :
vC =
p
5gr
In order to nd the height required to impart this speed to the ball at the bottom, we apply law of
conservation of energy between A and C.
Ki + Ui = Kf + Uf
Considering ground as zero reference gravitational potential, we have :
⇒ 0 + mgh =
1
mv 2 + 0
2 C
2mgh
= 2gh
m
p
⇒ vC = 2gh
2
⇒ vC
=
Equating two expressions of the speed at point C, we have :
⇒ 5gr = 2gh
⇒h=
http://cnx.org/content/m15102/1.2/
5r
2