Seely F.B. Analytical Mechanics for Engineers
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POTENTIAL AND
KINETIC ENERGY
441
That
is,
in
any
displacement
of
a
conservative
mass-system
(or
one
on
which
the
external
forces do no
work),
the
gain
(or
loss)
in the
kinetic
energy
of the
system
is
equal
to
the loss
(or
gain)
in
the
potential
energy
of the
system.
If
the
system
is not
conservative,
the
transformation of
the
energy
within
the
system
from
potential
to
kinetic
energy
(or
vice
versa)
is
accompanied by
an increase
in
other
forms of
energy
such
as
heat
energy.
And,
although
the total
energy
remains
con-
stant,
the
mechanical
(potential
and
kinetic)
energy
does
not
remain
constant.
The relations
between
potential
and kinetic
energy
for
conservative
systems,
only,
are
considered herein.
ILLUSTRATIVE PROBLEMS
471. A
body
which
weighs
50 Ib. falls
from
a
height, h,
of
50 in.
upon
a
helical
spring
(Fig. 434),
the modulus of which is
800 Ib.
per
inch. What
is
the maximum
shortening,
s,
of
the
spring?
Solution.
Little error will
be
introduced
by assuming
the
spring
and
body
to
be
perfectly
elastic;
by neglecting
the
air resistance on
the
falling body;
and
by
neglecting
the inertia
of the
spring.
If
this is
done,
the
earth,
the
body,
and
the
spring, may
be
considered
to be
a
conservative
system
in which
the
earth
is assumed
to be fixed.
Thus the
mechanical
energy
of the
system
re-
mains constant
and
hence,
But
AEk=Q,
since
the
kinetic
energy
of the
system
is zero
at the
beginning
and
at the end of the
displacement.
And
the
change
in the
potential energy
of the
system
is,
Or,
800s2
in which
h
and
s
are
expressed
in inches. This
equation
states
that
the loss
of
potential
energy
of
the
body
is
equal
to
the
gain
of
potential
energy
of the
spring.
Solving
the
equation,
we
have,
50(50
+s)=
400s
2
.
Or,
8s
2
-s
-50=0.
Whence,
s
=
2.56 in.
FIG
434.
472.
A
simple
pendulum
(Fig.
435)
consists of
a cord
having
a
length,
r,
of
4
ft. and a
bob
(assumed
to be a
particle)
having
a
weight, W,
of 6
Ib.
The
442
WORK
AND ENERGY
pendulum
is
displaced
an
angle, 0,
of
60.
What
will
be
the
velocity,
v,
of
the
bob when
in
its
lowest
position,
if
air
resistance is
neglected?
Solution. The
tension
T
in
the
string
does no
work.
Thus,
the
earth
and
bob
may
be
treated as
a
conservative
system.
Hence,
AE
V
=-
w
6(4-4
cos
60)
=>\
whence,
v
=
11.36
ft./sec.
473.
Solve
Prob.
462
by
use of
the
principle expressed
by
the
equation,
f
FIG. 435.
Solution. As shown in
Prob.
442,
the
only
force which does work
is the
earth-pull. Hence,
the
earth and the
cylinder
form
a
conservative
system
Thus,
AEp=-AE
k
,
This
is the same
equation
that was obtained and
solved in
Prob. 462.
Substituting
the
given
values
in
the above
equation
as
was done in
Prob.
462,
the
value of
s
is found to be
36
ft.
NOTE.
All three of
the above
problems may
be
solved
by
the
principle
of
work and
kinetic
energy.
It is
important, however,
to
note
the relation
between
potential
and kinetic
energy
for conservative
systems.
4.
EFFICIENCY. DISSIPATION
OF
ENERGY
197.
Efficiency
Defined. The
efficiency
of a
machine,
as
for
example,
a steam
engine,
an electric
motor,
a
chain
hoist,
a
jack
screw, etc.,
is
the ratio
of the
energy output
of
the
machine
in
a
given period
of
time to
the
energy
input
in
the same
period,
pro-
vided
that no
energy
is stored
in
the
machine which becomes avail-
able at a
later
period. By
input
is
meant the amount
of
energy
received
by
the
machine,
a
portion
of which is
transformed or
transmitted into the work
or
energy
for
which
the
machine
is
designed.
The work
done
or
energy
delivered
by
the machine is
called
the
output.
Thus, denoting efficiency
by
e,
we
have,
_
energy output
energy
input
or e
=
power
output
.
power
input
DISSIPATION
OF
ENERGY
443
As
already
noted,
in
the transformation and
transference
of
energy
(which
is the main function of
many machines),
some
of the
energy always
takes the form of
a
lower
grade
of
energy
(heat
energy)
and
thereby
becomes unavailable for
the
particular
process
for which
the machine is used. The
amount of
energy
which thus
miscarries or leaks
out
in
the
process
is
spoken
of
by
various
names,
such as
lost
energy
(or
lost
work),
energy leak, dissipated
energy,
etc.
Since
dissipation
of
energy
occurs
with
every
physical
process,
the
output
is
always
less
than
the
input and,
therefore,
the
efficiency
is
always
less than
unity.
The
efficiency
as
defined
above is the
over-all
efficiency
of
the
machine. Certain
parts
of
the
machine,
however,
may
have their
individual
efficiencies;
and the
over-all
efficiency
is
the
product
of
the efficiencies of
the several elements
of the
machine.
198.
Dissipation
of
Energy.
The
work
done
against
friction
is
the
most
frequent
cause of
dissipation
of
energy
in
machines.
Energy dissipated
in
doing
work
against
frictional forces
is trans-
formed into heat
energy.
Electrical
resistance
in
connection with
electrical
machinery
also causes
a loss of
available
energy by
devel-
oping
heat.
The
work done
against
friction
is,
in
some
machines,
a
necessary
evil
to be
reduced to a minimum
as
in
the case of
prime
movers,
bearings,
teeth of
gears,
etc.
;
whereas,
in
other machines or
machine
elements,
the main
object
of the
machine is to
dissipate
all
the
energy
received
by
the
machine,
as in
the
case
of friction
brakes
and
absorption dynamometers.
Two
typical
examples
involving dissipation
of
energy
in
machines are
here
considered:
(1)
a
bearing
in
which
the
loss of
energy
is
made as small as
practicable
by
lubrication
and
(2)
a
simple
Prony
brake or
absorption
dynamometer,
the
object
of
which is not
only
to
absorb all the
energy
delivered to it but
also
to
measure that
energy (or
power).
Another class of
dynamom-
eters measures
the
power
delivered
by
a
machine
without
absorbing
(dissipating)
the
energy. Dynamometers
of
this
class are called
transmission
dynamometers.
199. Work
Lost
Due to
Friction in
Bearings.
The
dissipa-
tion
of
energy
in
bearings
of
machines
of
various
kinds reduces
the
useful
energy
transmitted
or
transformed
by
the machine and
is,
therefore,
an
important
factor in
the
design
and
operation
of such
machines.
For
example,
in
textile
machinery,
which
involves
a
444
WORK
AND ENERGY
large
number of
small
parts
moving
at
relatively high
velocities, a
large
part
of
the
load
is
due to friction.
Various
types
of
bearings
are used
to meet the different
needs.
Thus,
for
shafts
subjected
to
a considerable end
thrust,
as,
for
example,
a
screw-propellor
shaft or the
shaft
of
a
centrifugal
pump,
a
collar
thrust
bearing
having
several
collars
may
be
used,
whereas
for
slow
speeds,
such
as occur
in
certain
types
of
rotary cranes,
the end thrust due to the
load and
weights
of the
moving parts
is
usually
carried
by
an
ordinary
flat
pivot
or
step
bearing.
Fig.
436
represents
an
ordinary
flat
pivot
in
which
the
work
(and
power)
lost in friction
is to
be found. Certain
assumptions
must be made. It
will be assumed that
the coefficient of
friction
for
the
rubbing
surfaces is constant over the whole
area
of contact and
that
the
normal
wear
at
any point
(wear
normal to the
rubbing
surfaces)
is
proportional
to the work
of
friction.
Thus,
according
to this as-
sumption,
the
pressure
after wear
has
occurred will not
be
uniform but will
be
a maximum at the center where
no wear
occurs and will decrease towards
the
edge,
since the relative
velocity
of the
rubbing
surfaces is
zero
at the
center
and increases
towards the
edge.
It
will,
therefore,
be
well
to cut out the material near the
center
of the
bearing
to
avoid
crushing
of
the
material.
Let
p
be
the
intensity
of
pressure
on
any elementary
area
2irpdp,
at
a
distance
p
from
the
axis of the
shaft. The
total
pressure P, then,
is
FIG.
436.
P
=
27r I
ppdp.
And from
the
assumption
made
above,
the normal
wear, n,
is
in
which
k
is
a
constant.
Therefore,
P
=
27T
\^
Jr\
k
2m
(1)
WORK
LOST DUE
TO FRICTION IN BEARING
445
The
frictional force at
any
distance
p
from the
axis
is,
And the frictional moment
T
f
for the
whole area
is,
(2)
By
eliminating
n
and
k
from
(1)
and
(2),
the
following
equation
is
obtained,
(3)
And,
if
co is
the
angular velocity
of
the shaft in
radians
per
second,
the work lost
per second, w/, is,
V=7>
=
^(r
2
+n)
....
(4)
If
P
is
expressed
in
pounds
and
ri
and
r<i
in
feet,
then w
f
is ex-
pressed
in
foot-pounds per
second and
the
horse-power
lost in
friction
is,
In order
to increase
the amount of
rubbing
surface
and
thereby
diminish
the
intensity
of
pressure,
a collar
bearing
with
two
or
more
collars
may
be
used.
Equations (3), (4),
and
(5)
also
apply
to
such
bearings.
It will be
observed
that
the
coefficient of
fric-
tion
is an
important
factor in
the
problem
of
reducing
the
dissi-
pation
of
energy
due to friction
and
this
fact
suggests
the
great
importance
of the
choice of
bearing
materials
and
of
lubricants.
Various
other
types
of
bearings may
be
treated
in
a
manner
similar
to
that
used
above.
446
WORK
AND
ENERGY
200.
A
Simple
Dynamometer. Prony
Brake.
A
simple dyna-
mometer,
commonly
called a
Prony
brake,
is
shown in
Fig.
437.
A is a
flanged pulley keyed
to a
rotating
shaft.
The
power
trans-
mitted
by
the shaft is absorbed and measured
by
the brake.
B,
B
are
bearing
blocks
against
which the
pulley develops
a fric-
tional
resistance,
the
magnitude
of which is varied
by
adjusting
the
nuts
C,
C.
D is
the
frame or
beam with its
end
E
resting
on
the
platform
of a
weighing
scale. When the
pulley
is
running,
the beam
develops
an additional
pressure
on the
platform,
due
to the
friction
of
the
pulley
on
the
bearing
blocks. The work
lost
in
friction and the
power
developed by
the
shaft
may
be
found
as
follows
:
It will be assumed that the scales are
adjusted
to
read zero
when
the
beam rests on the
platform
and
the
pulley
is not
running.
Thus,
the scale
reading
is a
measure
of the
pressure, P,
at
E due to
the friction
developed
on the
bearing
blocks when the
pulley
is
running.
Since the brake
frame is
in
equilibrium
under
the
action
of
the
forces,
N, P,
and
F
(F
denotes
the total
frictional
force
on
the
two
blocks),
the
sum
of
the moments
of
P
and
F about the
axis
of
the shaft must
equal
zero.
Or,
Fr
=
Pa.
And the
work done
by
the frictional
moment
Fr in one second
is,
Wj
=
Frw
=
Paco,
in
which
co is
expressed
in
radians
per
second.
And,
since
co
=
---
A SIMPLE
DYNAMOMETER
447
where
n is
the number of revolutions
per
minute
(r.p.m.)
of
the
shaft and
pulley,
the
expression
for w
f becomes,
27rPan
_irPan
:
~30~*
If
P
is
expressed
in
pounds
and
a in
feet,
then w
f
will be
expressed
in
foot-pounds
and the
horse-power
developed by
the frictional
moment
(and
hence
by
the
shaft)
is,
,
_
irPan
P
~
30X550
irPan
=
16,500
'
This
expression may
be
simplified
if
the
dynamometer
is
constructed
so that
a
has
a
special
value. It
should be
remembered
that the
scale
reading
should
not be
used
for
the
value of-
P
unless the scales
are
adjusted
to read
zero
when
the
pulley
is
not
running.
PROBLEMS
474. What
effort, P,
is
required
to raise a
load, Q,
of
200 Ib.
by
means of
the differential chain
hoist
shown in
Fig.
438
if
e
=
30
per
cent,
r
2
=
8
in.,
and
n=4
in.? Ans.
P
=
1671b.
475. In a
test of a
jackscrew
(see Fig.
169)
with a
screw 1.5 in. in
diameter
and
a
pitch
of
^
in.,
it was found that a
pull
of
72
Ib.
at the end
of a 15-in.
lever
was
required
to raise a load
of 2400 Ib. when no
lubricant was used
and
a
pull
of
63
Ib.
when a
hard oil lubricant was used. What is the
efficiency
of
the
jack
for each case?
Ans.
e
=
11.8%;
e
=
13.
5
per
cent.
476.
The band brake described
in Prob.
165
allows
a certain
load to
lower
at
a
constant
speed
such that
the
brake sheaves
rotate
at 120
r.p.m.,
in
a
counter-clockwise
direction. What
horse-power
is
absorbed
by
the
brake?
477.
The
dynamometer
shown in
Fig.
439 was
devised
by
Lord
Kelvin in
connection
with
the
laying
of the
Atlantic cable for
braking
the cable
drum
as
the
cable
was
laid
out.
If it is used in
the
test of
a
steam
engine,
find
the
horse-power
absorbed,
using
the
following
data :
W
=
400
Ib.
;
speed
of
engine
=
150 r.p.m.;
r
=
4
ft.;
reading
of
spring
balance
is
85 Ib.
Ans. 36
h.p.
448
WORK AND ENERGY
478.
Derive an
expression
for
the work done
by
friction in
the
spherical
pivot
shown in
Fig. 179, assuming
that normal
wear is
proportional
to
the
FIG.
439.
work
of
friction
and that the
coefficient
of
friction is the same
for all
parts
of
the
rubbing
surfaces.
Ans. Work
per
revolution
=2irnWr
a+sm
a
cos
a.
CHAPTER
XI
IMPULSE
AND MOMENTUM
201.
Preliminary.
In
Art.
176,
the
statement was made that
impulse
is a
quantity
which
involves force and
time,
and
that
momentum is
a
quantity
which involves
mass
and
velocity.
And the
fact was noted that
the use
of
these
quantities
in
the
analysis
of
the motion of
bodies
requires
no
fundamental laws
in
addition
to
Newton's laws
of
motion.
However,
methods
which
make use of
impulse
and momentum offer
advantages,
in
certain
types
of
problems,
over the
methods
of
work and
energy
(Chap.
X)
and
of
force and acceleration
(Chap.
IX).
In
determining
the effect of forces on
the motion of
bodies,
thus
far,
by
the methods of
force and acceleration and of
work
and
energy,
it has
been assumed that
the forces
have
acted on
rigid
bodies
during
a definite
(comparatively
large)
interval of
time,
and
when
the forces were not
constant
the manner
in
which
they
varied
during
the
period
was assumed to
be known. To
such con-
ditions
the methods of
impulse
and
momentum also
apply,
and
in
many
cases
offer a
simpler
method
of solution
than the methods
previously
discussed.
Forces
sometimes
act,
however,
for
a
very
short
(indefinite)
interval of
time
during
which
neither
the value
of
the force at
any
instant nor
its law of
variation
is
known.
These
forces
may,
nevertheless,
produce
very
appreciable
changes
in
the motion
(velocity)
of the
body.
Such forces
are
called
impulsive
forces.
The bodies
upon
which
impulsive
forces act
deform
under
the
excessive
pressures
produced
and
hence,
in
determining
the
motions of
bodies
under
the
influence of
impulsive
forces,
the bodies
cannot
always
be
assumed
to
be
rigid,
as
in
the
preceding chapters,
without
introducing appreciable
errors.
As
examples
of
impulsive
forces
the
following
may
be
mentioned
:
the
force exerted
on a
projectile
due to
the
explosion
of
the
pow-
der;
the action
of
one
billiard
ball
on
another;
the
force
exerted
by
the
ram
of
a
pile
driver
on
the
pile;
the
pressure
between
two
449
450
IMPULSE
AND
MOMENTUM
railway
cars when
making
a
flying
coupling;
the action of a
steam
jet
on the
blades of
a
high-speed
steam
turbine;
the
pressure
exerted
by
the
water in
a
pipe
line on a
valve which is closed sud-
denly.
The
principles
of
impulse
and
momentum are of
special
value
when
considering
the motion of bodies
under the
action of
impulsive
forces.
The
purpose
of
the
present chapter
is to
make clear
the
con-
ception
or
meaning
of
impulse
and of
momentum,
to
develop
certain
principles
which
express
relations between
these
quan-
tities,
and to
apply
these
principles
to
problems
in
kinetics.
1.
IMPULSE
202.
Impulse
and
Impact
Defined. Units.
The
impulse
of a
constant force
is
defined as the
product
of the
magnitude
of
the
force
and the time interval
during
which the force acts.
Thus,
if
Q
denotes
the
impulse
of a force
F,
the
impulse
of the force
is
defined
by
the
equation,
Q
=
F-M,
provided
that the force remains constant
during
the
time interval
At. If
the
force
varies
in
magnitude
but not
in
direction,
the
impulse
for an
indefinitely
short
period
of
time
dt,
is
F dt
and,
for
a
time
interval M
=
tz
ti,
the
impulse
is,
<?=
I Pdt.
In
order to evaluate
this
integral
by
the method
of
calculus,
F
must be
expressed
in
terms of
L
The
impulse
of a force which
acts
during
a
very
short
(indefinite)
interval
of time
(impulsive
force)
is also
given
by
the above
expression but,
as noted
in
the
preceding article,
the
impulsive
force
cannot be
expressed
in
terms
of
t,
since
its
law of variation is not known and
hence,
the
impulse
cannot be
determined
directly
but is found
in terms
of the
change
of
momentum as is
discussed
in
the
subsequent
pages.
The
impulse
of
an
impulsive
force
is sometimes
called an
impact,
that
is,
an
impact
is a sudden
impulse.
The term
impact,
however,
is also
frequently
used as
descriptive
of
the
act
of collision
of bodies.
In
some
problems
it is
convenient
to
estimate the
time
interval
of
the
impulsive
force and
to
express
the
impact
as