Download 2000, W. E. Haisler Energy Principles for Kinetic Problems 1 ENGR

 2000, W. E. Haisler
Energy Principles for Kinetic Problems
ENGR 211
Principles of Engineering I
(Conservation Principles in Engineering Mechanics)
 Conservation of Energy
 Kinetic Energy
 Potential Energy
 Internal Energy
 Conservation Equation
 Energy Analysis
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 2000, W. E. Haisler
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Energy Principles for Kinetic Problems
The conservation of energy follows the same general format as
for mass, linear momentum and angular momentum:
 Accumulation of Energy
 within system

 during time period


 Energy entering
   system during



 time period


  Energy leaving 
 -  system during 
 

  time period

 

where
 Accumulation of Energy 
 Energy in system   Energy in system 
 within system
   at end of
 -  at beginning of 



 

 during dime period

 time period
  time period




 

In the above, we have not considered any generation or
consumption of energy, i.e., we have not considered certain
chemical or nuclear reactions although these may be important in
special cases.
 2000, W. E. Haisler
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Energy Principles for Kinetic Problems
Energy may take many forms including

Kinetic Energy (KE)
 Potential Energy (PE)
 Internal Energy (U)
 Heat Energy (Q)
 Work Energy (W)
Kinetic Energy (KE)
Kinetic energy is that energy that a body possesses as a result
of mass and its motion, specifically, its velocity. The Kinetic
Energy is defined by
and
KE  1 mv 2  1 m(vx2  v 2y  vz2 )
2
2
KE  1  (v  v )dm
2
concentrated mass
distributed mass
 2000, W. E. Haisler
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Energy Principles for Kinetic Problems
For a rigid body, it is convenient to think of the kinetic energy in
terms of its translational and rotational components. So we write


 Rotational





 Translational

 kinetic energy 








Kinetic
Energy
of
kinetic
energy




   with respect to 

  


 rigid body



 of center of mass 


 center of mass 




 of rigid body





 of rigid body




We need the position
vectors as before:
any particle
r  rG  G r
y
 v  vG  G v
ua
r
G
r
O
z
system
G
rG
x
r = rG + Gr
axis through the
center of mass
 2000, W. E. Haisler
Energy Principles for Kinetic Problems
The integrand of the KE becomes
v  v  (vG  G v )  (vG  G v )
 vG  vG  2vG  G v  G v  G v
and the KE becomes
KE  1  (v  v )dm
2
 1  (vG  vG )dm   (vG  G v )dm  1  ( G v  G v )dm
2
2
The velocity of the center of mass vG does not vary within the
body (mass) and may be taken outside of the integral. Thus,
KE  1 mvG2  vG 
2
  G v dm   12  ( G v  G v )dm
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 2000, W. E. Haisler
Energy Principles for Kinetic Problems
6
d Gr
The integrand in the second term can be written G v 
and
dt
d Gr
thus Gvdm 
dm  d G rdm  0 . The integral rdm 0


 dt
dt 
when r is measured from the center of mass.. Thus the KE
becomes
KE  1 mvG2  1  ( G v  G v )dm
2
2
For a rigid body we can write
Gv   G r
This says that with respect to the center of mass, Gv has only a
tangential component (no radial component). Then the integrand
in the last term of KE can be written (using a vector identity):
 2000, W. E. Haisler
G
Now
v  Gv  (  G r )(  G r )  (  )(G r  G r ) (  G r )(G r  )

   2
system
ua
2
r

r

r
G G
G
particle
G
rn
y
and  ua so that
ra
G
r
G
r
x
r = rG + Gr
z
( G r )( G r )  ( G ra)(G ra)  2G ra2
And finally:
G
rG
  G r  (ua) G r  G ra
Thus
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Energy Principles for Kinetic Problems
G v  G v  G r
2  2 r 2
G a
 2(G r 2  G ra2)
 2 G rn2
2
axis through
center of mass
 2000, W. E. Haisler
Energy Principles for Kinetic Problems
8
Substituting this result back into the previous KE equation gives
KE  1 mvG2  1  ( 2 G rn2 )dm
2
2
 1 mvG2  1  2  G rn2 dm
2
2
 1 mvG2  1  2 G I a
2
2
Thus, the KE for a rigid body translating with velocity vG and
rotating about its a-axis with angular velocity  is given by
KE  1 mvG2  1 G I a 2
2
2
It is sometimes useful to define the specific kinetic energy, or the
kinetic energy per unit mass as
I
KEˆ  KE / m  1 vG2  1 ( G a ) 2
2
2 m
 2000, W. E. Haisler
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Energy Principles for Kinetic Problems
Potential Energy (PE)
A body possesses potential energy as a result of its position in
a potential force field. For conservative forces, the change in
_
potential energy is defined by the negative
z
s ds F
of the work the force does to move from one
position to another in the direction of path s:
PE  PE  f c(s)ds
0 
y
PE = potential energy at the end of the
x
path followed
PE0 = potential energy at the beginning of the path
f c = force
s = path force follows
The dot product is required in order to obtain the magnitude of
the force in the direction of the path followed.
 2000, W. E. Haisler
Energy Principles for Kinetic Problems
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Gravity is an example of an important conservative force field
where the force is given by f c  mg . The the potential energy
definition becomes PE  PE  mg ds .
0 
Suppose that the gravity vector is directed in the -z coordinate
direction with magnitude g. Then g gk .
The path can be described in a Cartesian coordinate system as
ds  dxi  dyj  dzk .
The potential energy equation becomes
z
PE  PE m ( g )dz  mg (z  z0)
0
0
The above assumes that m and g do not change in moving from 0
to z.
 2000, W. E. Haisler
Energy Principles for Kinetic Problems
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Another way to write this is to say
PE
 PE
 mg(z
z
)
end
beginning
end beginning
For simplicity, PEbeginning is usually taken to be zero at zbeg so that
PE  mgz  mgh
end
end
where h is distance in the +z direction.
 2000, W. E. Haisler
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Energy Principles for Kinetic Problems
Notes on Sign Convention and Sign of PE
The expression for potentional energy of a mass, PE  mgh , has a
sign convention associated with with it.
 When the direction of h is opposite the
direction of gravity, g, PE is assumed positive:
PE  mgh1
 Thus, when the direction of h is in the same
direction of gravity, g, PE is negative:
mg
h1
h2
PE  mgh2
mg
 2000, W. E. Haisler
Energy Principles for Kinetic Problems
13
Internal Energy (U)
A deformable body also possesses internal potential energy as
a result of atomic bonds being stretched. These bonds create
forces between particles. When the particles are moved relative
to one another, these forces do work and result in energy being
stored in the body. We use the symbol U to define this internal
potential energy.
An example is a spring. When we apply an external force to
the end of a spring, the spring is stretched (or compressed) and
internal potential energy is stored in the spring. If the force is
removed, the internal energy within the spring is converted to
translational energy and the spring returns to its original shape.
We can also define specific internal energy or internal energy per
unit mass as Uˆ U .
m
 2000, W. E. Haisler
Energy Principles for Kinetic Problems
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k
F
Internal Energy, U, of a Linear, Elastic
Spring

F
For a linear, elastic spring, we observe
experimentally that the stretching of the
spring,  , is proportional to the amount of
force, F, applied to the spring. The
proportionality constant is called the spring
constant, k, and we thus have F  k .
F
The amount of internal energy U stored in k
the spring is equal to the area under the
F vs.  curve. Hence,
2
2
1
1
U   Fd   k d 

1 k 2 2
2
1
F  k
k

F  k
U  12 k 2
U

 12 k ( 22  12 )
 2000, W. E. Haisler
Energy Principles for Kinetic Problems
15
Work (W)
Work done on a body by an external force is defined to be the
(product of the force component in the direction of the resultant
motion of the body) times (the displacement it produces). The
work that we will be interested in is that due to forces that act on
the external boundary of the system.
W

f ext ds
Work is considered positive when the external force is in the
direction of the displacement.
We can also define the work rate (or power) as
W  dW
dt
 2000, W. E. Haisler
Energy Principles for Kinetic Problems
16
Work is defined to be positive when energy is added to the
system and negative when energy leaves the system.
Note: Work is also produced then a torque causes a rotation.
Note: Gravity forces also do work on the volume of the body but
they are included in the potential energy (PE) term.
 2000, W. E. Haisler
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Energy Principles for Kinetic Problems
Examples of Work, W, Done on
System
Work is positive if force and
motion are in same direction.
S
A
C
A
10 lb
 When block B moves through
the distance S B , the force F will
move through the same
distance. Hence the work done
by F is W  FS B
 When block A moves through the
distance S A (note, down the plane), the
work done by the frictional force Fk is
W   Fk S A . The negative is because the
frictional force and motion are always in
opposite direction.
SB
20o
B
6 lb
F  100 lbf
10 lb
y
x
20o
Fk
F
N
 2000, W. E. Haisler
Energy Principles for Kinetic Problems
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Heat Energy (Q)
Heat energy (defined by the symbol Q) may enter or leave the
system through the system boundary at a rate Q  dQ . We may
dt
also consider internal heat sources or sinks within the body (to be
discussed in ENGR 214). Heat energy has units of Joules and is
defined to be positive when it enters the system.
 2000, W. E. Haisler
Energy Principles for Kinetic Problems
Conservation of Energy Equation
The total energy of the system is defined to be Esys and
consists of the internal, kinetic and potential energy terms:
Esys U  KE  PE
For a complete system, we can write the conservation of energy
equation in the same manner as was done for conservation of
linear and angular momentum. We write
dEsys
  (mUˆ  mKEˆ  mPEˆ )
  Q   WTOT
in / out
dt
The first term on the right represents energy entering or leaving
the system due to mass flow across the system boundary. The
second term is the net rate of heat entering the system. The last
term is the net rate of work applied to the system boundary.
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 2000, W. E. Haisler
Energy Principles for Kinetic Problems
20
Integral Form of Conservation of Energy
The conservation of energy equation may be integrated over a
time period (tbeg to tend) to obtain:
tend dEsys
tend
(
)dt 
( (mUˆ  mKEˆ  mPEˆ )
)dt
tbeg dt
tbeg
in / out
tend
tend

tbeg ( Q)dt  tbeg ( WTOT )dt
If the energy flow terms are constant over the time period, then
(Esys )
 (Esys)
  (U  KE  PE)
end
beg
in / out
  Q   WTOT
Note:

ˆ  dmUdt
ˆ  mUˆ U
mUdt
 dt
 2000, W. E. Haisler
Energy Principles for Kinetic Problems
21
Closed System
For a closed system with no mass crossing the system
boundary, the only work is non-flow work. In other words, when
m 0 , there is no work associated with the mass flow terms
( mUˆ ,mKEˆ ,mPEˆ ). Thus,
dEsys
  Q   Wnon  flow
dt
W
means the work done by forces, pressures, etc. that
non flow
act on the system boundary (not due to mass flow across a
boundary).
Steady State
For steady state, the system properties do no change with time
and the rate equation becomes
0   (U  KE  PE)
  Q   Wnon - flow
in / out
 2000, W. E. Haisler
Energy Principles for Kinetic Problems
Energy Analysis of Dynamic Systems
We now return to the consideration of a rigid body
undergoing translational and rotational motion. For a closed
system, we can write the conservation of energy equation as
dEsys
dt
  Q   Wnon  flow
where
Esys U  KE  PE
The kinetic energy (KE) is given by
KE  1 mvG2  1 G I a 2
2
2
The potential energy (PE) due to gravitational forces is given by
PE  mgh
G
where hG=position of center of mass relative to the position
where PE=0.
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 2000, W. E. Haisler
Energy Principles for Kinetic Problems
The work energy due to boundary forces is given by
W  f ext ds

For a rigid body, the internal energy (U) is zero. We take the
special case when there is no heat flow through the boundary
( Q 0 ).
The system energy becomes
Esys  1 mvG2  1 G I a 2  mghG
2
2
The right side of COE becomes
W
d( f
ds)   ( f
 ds )
non flow dt  surface
surface dt
 ( f
v
)
surface surface
23
 2000, W. E. Haisler
24
Energy Principles for Kinetic Problems
The Conservation of Energy equation (for a rigid body and no
heat flow) becomes
d ( 1 mv 2  1 I  2  mgh )   ( f
 vsurface )
G
G
a
G
surface
dt 2
2
RIGID BODY
The term on the right side will be non-zero only if the surface
boundary is moving where the surface forces are applied. For
friction without slipping, the frictional forces does no work since
it does not move relative to the point of contact.
Conservation of Energy is more conveniently written as:
( 1 mvG2  1 G I a 2  mghG )end  ( 1 mvG2  1 G I a 2  mghG )beg  WTOT
2
2
2
2
OR
FOR RIGID BODY ONLY
( KE  PE )end  ( KE  PE )beg  WTOT
 2000, W. E. Haisler
Energy Principles for Kinetic Problems
Remember that the Conservation of Energy equation for a
general system is given by
d (KE  PE U )  (mKEˆ  mPEˆ  mUˆ )
  Q   WTOT

in / out
dt
where
KE  1  (v  v )dm
2
; for rigid body: KE  1 mvG2  1 G I a 2
2
2
PE  f c(s)ds

; for gravity, PE  mghG
U = internal energy (a further topic for ENGR 214)
Q = heat flow rate into body
 WTOT   ( f surface vsurface)
25
 2000, W. E. Haisler
Energy Principles for Kinetic Problems
When the system has no mass flow in/out and no heat input,
above reduces to:
d (KE  PE U )  W
 TOT
dt
Note that the term on the right is the work rate of the external
forces, or power.
Integration of the above (between the "end" and "beginning"
states) gives:
( KE  PE  U )end  ( KE  PE  U )beg  W
26
 2000, W. E. Haisler
Energy Principles for Kinetic Problems
27
E
For many problems, we
k D
B
must simply "count up" all
A
the energy contributions of
the system. For the
problem shown, we have:
 KE of mass E
 KE of disk AB
C
 KE of mass C
 PE of mass C
F
 W done by force F
 W done be frictional force on E
 U for spring k
There could also be:
 W due to frictional force on pin at disk support
 W due to a torque applied to disk
 If rope has mass, then KE and PE of rope, etc., etc., …