Seely F.B. Analytical Mechanics for Engineers
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PRINCIPLE
OF WORK AND
KINETIC
ENERGY
481
to the unbalanced
(concurrent)
forces
which act on it.
Let
the
resultant of
the
forces
acting
on
P
be denoted
by
R and its
com-
ponent
tangent
to the
path
of
P
be
denoted
by
R
t
.
It will be
noted
that some
of the
particles
(cubes)
have their
velocities
changed (are
accelerated)
by
forces
exerted
only by
other
particles
of
the
body (internal
forces)
whereas
other
particles
are acted on
both
by
internal and
by
external
forces.
The work
done
by
the forces
acting
on
the
particle
as
it
moves
c
s
*
along
its
path
from
P'
to
P"
is
w= I
R
t
ds
but,
as in
Art.
189,
R
t
ds
may
be
expressed
in
terms of
the
mass, m,
and
velocity,
of the
particle by
means of the
relations,
dv
ds
R
t
=
ma
t
,
a
t
=
~Tf>
and v==
~^f
T
2
The
work
done on the
particle
=
I
Rtds
i/
S
l
r r
=
I ma
t
ds=
I
J*
l
Js
l
r
4*
=
I m-r
J
Vl
dt
=
I mvdv
M
dv
=
change
in
kinetic
energy
of
the
particle.
Therefore,
the work done
by
the
forces
acting
on
a
particle
of
a
body (whether
rigid
or
not)
during any displacement
is
equal
to
the
change
in
the
kinetic
energy of
the
particle
in
the
same
displacement.
Or, expressed
in
the form of an
equation,
w
=
Ai
II.
For a
System of
Particles.
The
principle
of
work
and
kinetic
energy
for
a
system
of
particles
(body)
may
now be
derived.
Since work
and
energy
are
scalar
quantities,
the
work
done
on a
body
is the
algebraic
sum
of the
works
done
on all
the
particles.
And,
the
change
in
the
kinetic
energy
of
the
body
is
the
algebraic
sum of
the
changes
in
the kinetic
energies
of all
the
particles.
432
WORK
AND ENERGY
But,
the
work
done
by
all the forces
acting
on all
the
particles
equals
the work
done
by
the
external forces
which
act on
the
body
plus
the work
done
by
the
internal
forces of
the
body.
Thus,
by
writing
the above
equation
for
each
particle
and
adding
both
sides
of
the
equations,
the
resulting
equation
is,
w
e -}-Wi
=
%2
That
is,
the work done
on a
system of
particles
(whether
rigid
or
not)
by
all
of
the external
and internal
forces
in
any
displacement
of
the
system
is
equal
to
the
change
in
the
kinetic
energy of
the
system
in
the
same
displacement.
III.
For a
Rigid
Body.
The
principle may
now be
expressed
for the
special
case of
a
rigid
body,
for,
as
pointed
out
in
Art.
182,
the internal forces
in
any mass-system
occur
in
pairs
of
equal
opposite
and collinear forces whether the
body
is
rigid
or
not,
but
the
work
done
by
these forces is
zero,
in
general, only
if
the
appli-
cation
points
of
each
pair
of
forces remain a
fixed
distance
apart,
which is
the case
in
a
rigid
body. Hence,
for
a
rigid
body,
w
t
=
0.
Therefore,
u;
e
=
AEfc.
That
is,
the work
done
by
the external
forces
acting
on
a
rigid
body
in
any displacement
is
equal
to the
change
in
the
kinetic
energy
of
the
body
in
the
same
displacement.
Although
an
absolutely
rigid body
does not
exist in
nature,
the
work
done
by
the external forces
in
causing
the
relative
displace-
ments
of
the
part-c.es
in
physical
bodies is
usually
negligible
in
comparison
with the
work
done
by
the
forces
in
causing
the
displacement
of
the
body
as a
whole.
194.
Application
of
Principle
to
Special
Cases
of Motion of
Rigid
Bodies.
By making
use of the
expressions
developed
in
Art. 182 and
190,
the
principle
of work and kinetic
energy may
be
expressed
for
important special
cases of
motion of
rigid
bodies as
follows :
I.
For
a
Rigid
Body Having
Uniformly
Accelerated
Rectilinear
Translation.
II.
For a
Rigid Body Rotating
about an
Axis
through
the
Mass-
center with
Constant
Angular
Acceleration.
PRINCIPLE
OF
WORK AND KINETIC
ENERGY 433
///. For
a
Rigid
Body
Having
Plane
Motion Such that the
Angular
Acceleration
is Constant and the Mass-center
has
Uniformly
Accelerated Rectilinear
Motion.
As noted
in Art. 182 the work
done
by
the forces
acting
on
any
body may
be found
by taking
the
algebraic
sum
of the works done
by
all
the forces. And in
more
general
cases of motion
than
those
stated
above,
it will
be,
as a
rule,
more
convenient to
find the work
done
in
this
way
rather than
from
the
resultant
of
the
forces.
ILLUSTRATIVE
PROBLEMS
459. An
engine capable
of
exerting
a
maximum
draw-bar
pull
of
51,000
Ib.
is
used on a certain railroad
having
small
grades
to
draw
freight
trains
having
a maximum
weight
of
2000
tons,
the
average weight
of a
freight
car
with
its
cargo
being
about 45
tons.
The train
resistance
per
ton
of
weight
varies
with the car
weight
and with the
speed.
If an
average
2000
ton
^
1
value
of
8
Ib.
per
ton
is used
and
FIG.
427.
the
engine
pulls
a
2000-ton train
while
going up
a
\
per
cent
grade,
how far will the train
travel while its
velocity
is
increasing
from
15 to
30
mi.
per
hour?
How
long
will it
take?
Solution.
The forces
acting
on the
train are shown in the
free-body
dia-
gram (Fig.
427).
The
angle
9
(exaggerated
in
the
diagram)
for
a
\
per
cent
grade
is so
small that tan
6
may
be considered to
be
equal
to
sin
6.
Hence,
1
sin
6
=
-
.
200
15,000s
=62,200X1452,
8
=
6020
ft.
Since,
_fl+^2
2
'
we
have,
-Whence,
t
=
182.5
sec.
=
3.04
min.
434
WORK
AND
ENERGY
460.
In the friction-board
type
of
drop
hammer used
in
the
production
of
drop
forgings,
the ram
(Fig.
428)
is attached to
the lower end of
a friction
board
which is lifted
by
friction
driving
rolls and these
in
turn are
driven
by spur
gears.
The rolls
are
pressed
against
the
board
by
revolving
the
eccentric
bearings
and
they may
be released
quickly
to let the
ram
drop.
The
following
data
apply
to
one
particular
hammer:
F
(
Weight
of
ram,
TT
=
1000 Ib.
Angular velocity
of
rolls,
co
=
130
r.p.m.
Diameter of
rolls,
d
=
ll
in.
Normal
pressure against
friction
board,
Eccentric
bearings
Friction board
rolls
Ram.
Coefficient of
friction,
/*
=
0.3.
Total
height
to which ram is
raised,
h
=
4.5
ft.
Find
the
distance
the
ram travels
(a)
while
being
accelerated,
that
is,
while its
velocity
is
being brought
up
to the
peripheral speed
of
the
friction
rolls;
(6)
while the ram is
coming
to
rest after the
rolls are
released;
(c)
while
the
FIG.
428.
ram ig
moving
with
constant
velocity,
after
the
acceleration
ceases and before the rolls
are released. Also
find
the
time
(seconds)
required
to
travel each of
these distances and
the
time to
complete
a
cycle
(total time).
Assume that
the
friction
rolls do not
slip
on
the
friction
board.
Although
there is some
slipping
just
as the rolls are
pressed
against
the
board,
this
assumption
will
not
introduce serious errors in
the
problem.
Solution. The
friction
force,
F',
during
the
acceleration
period
is,
F'
=^N
=
0.3
X6000
=
1800
Ib.
The
peripheral speed
of the
rolls
is,
=
130X27r
5.5
60
X
12
During
the acceleration
period,
the
forces
acting
on
the
ram are
F'
and
W,
as
shown
in
Fig.
428.
Applying
the
principle
of
work and
energy,
we
have,
1 1000
(1800
-
1000)si
=
Hence,
800s!=606.
*i=
0.758
ft.
PRINCIPLE
OF WORK
AND KINETIC
ENERGY
435
After
the rolls are
released,
the
ram
is
acted
upon
only
by
the
earth-pull
(weight)
while it
gives
up
its kinetic
energy. Thus,
W
e
1000
Therefore,
s
3
=
0.606
ft.
Hence,
S
2
=
/1-(S
1
+S
3
)
=
4.5
-(0.758
+0.606)
=
3.14 ft.
Since
the distance
traveled
in
each
period
is the
average speed
times the
time,
we
have,
Therefore,
Likewise,
Whence,
Also,
whence,
Further,
Hence,
ti
=0.243
sec.
=
duration of
acceleration
period.
0.606=^^X^3.
t
3
=
0. 194
sec.
=
duration
of
decelerating
period.
tz=
0.503 sec.
=
duration
of
constant
velocity period.
X4.5
=
0.528 sec.
=time of
falling.
=
1.468
sec.
=time of
complete
cycle.
461.
The
punching
and
shearing
machine
shown in
Fig.
429
has a
capacity
for
punching
a
2^-in.
hole
in
a
|-in.
steel
plate.
The
pinion
shaft
(and
fly-
wheel)
is driven
at
220
r.p.m.
from a counter
shaft
by
means of a
belt
drive to
the
tight pulley
on the
pinion
shaft.
Each
operation
of
punching
a
hole
(punching cycle)
causes
a fluctuaton
(decrease)
in the
speed
of the
flywheel.
The
"
coefficient
of
speed
fluctuation
"
is
0.8,
that
is,
the
speed
of
the
flywheel
decreases 20
per
cent
in
each
punching cycle.
If the
work
done in
punching
the
hole is
all
supplied
by
the
flywheel,
what
moment of inertia
should
the
flywheel
have?
What
is the
moment
of
inertia
of
the
flywheel
on the
machine
436
WORK
AND ENERGY
as
actually
designed
(as
shown in
Fig.
429),
neglecting
the
material in
the
hub
and
spokes?
If
the
difference in the belt
tensions,
T
2
T\,
is
286
Ib. and
the
pinion
shaft turns
through
an
angle
of
230
while
the hole
is
being
punched,
how much
work
is
done on the shaft
(and flywheel)
by
the
belt while the hole
is
being punched?
The
diameter
of
the
pulley
is
22
in.
Solution.
The maximum
value
of
the force
P
required
to
punch
the
hole,
Punch
attachment
r
Force exerted
by
punch
FIG.
429.
using
60,000
Ib.
per
square
inch
for the ultimate
shearing strength
of the
material
of the
plate,
is
=
7rXlX
2X60,000
=
235,500
Ib.
The work
done
in
punching
the
hole,
assuming
a
triangular
work
diagram
(see
Prob.
435),
is,
p 1 93^ ^nn
^
=
|
X
^=^^
=
58,875
in.-lb.
=
4900 ft.-lb. which is
supplied
by
the
flywheel.
For
the
flywheel
we
have
then,
Whence,
7=51.3
slug-ft.
2
PRINCIPLE
OF
WORK
AND KINETIC ENERGY
437
Hence,
the moment
of inertia of
the
flywheel
should be
51.3
slug-ft.
2
in
order
that the
speed
be decreased
not more than 20
per
cent.
Assuming
the
flywheel
to
be
made of
cast-iron
which
weighs
450 Ib.
per
cubic
foot,
the
moment of inertia of
the
flywheel
as
actually designed,
neglecting
hub
and
spokes,
is
,X1.
36X0.
375X450X6.
64
232.2
=
74.3
slug-ft.
2
.
The
work
done
on the
flywheel
while
the hole
is
being punched is,
=
286X^X230X^10
=
1050ft.-lb.
462. A solid
homogeneous
cylinder
rolls
up
an inclined
plane
without
slipping
(Fig.
430).
The
weight
of the
cylinder
is 120 Ib.
and
its diameter is
3
ft.
The
angle,
<f>,
of inclination of
the
plane
is
15.
If the
velocity
of
the
center
of
the
cylinder
is
20 ft.
per
sec.
just
as
it comes
in
contact
with
the
incline,
how
far
up
the
plane
will the
cylinder
roll?
Solution.
The forces
acting
on the
cylinder
while
rolling up
the
plane
are
shown
in
Fig.
FIG.
430.
430.
And,
the work
done
by
the
forces,
as was
shown
in Prob.
442,
is
W sin
<j)-s.
Thus,
Or,
Also
Hence,
V.
Whence,
-W
sin
W
sin
,
since
v
2
and o>
2
are
zero.
746+372
.
=
-
3T7l
=36 ft.
438
WORK AND
ENERGY
PROBLEMS
463.
A
body
weighing
80 Ib.
is
projected along
a
rough
horizontal
plane
with
a
velocity
of
8 ft.
per
sec.
It comes to rest in
a distance of
10
ft. Find
the
coefficient of kinetic
friction. Ans.
/z
=
0.099.
464.
An
automobile which
weighs
W
Ib. is
moving
at the
rate of
30
miles
per
hour when it comes
to the foot of
a
hill. Power
is then shut
off. The
slope
of the hill is
1 ft. in
50
ft. How far will
the
machine coast
up
the
hill
if
the total frictional resistance
(parallel
to the
road)
is
0.08 W"?
Ans.
s
=
301 ft.
466.
A
sphere
rotates about an axis
through
its center. Its
speed
is
increased
from 600 rad.
per
min.
to
90 rad.
per
sec. while it turns
through
10
revolutions.
If the moment
of
inertia of
the
sphere
with
respect
to the axis
of rotation
is 20
Ib.
sec.
2
ft.
(or slug-ft.
2
),
find
the moment of
the
couple
acting
on the
sphere.
466. The
winding
drum of a mine hoist is 15 ft.
in
diameter,
its radius of
gyration
is 6
ft.,
and
its
weight
is 8 tons. A
cage weighing
6
tons
is
raised
by
it.
If the
cage
is
rising
at
the
rate
of 60 ft.
per sec.,
at what level
should the
power
be
shut
off in
order that
the
cage
may
stop
at the surface without
braking?
Neglect
friction.
Ans. 103.5
ft.
467. The
hand-operated
screw
press
shown
in
Fig.
431
is
used
for
embossing,
lettering
dies,
punching
thin
plates,
etc. The diameter of
the
screw
is
2^
in. The
screw
has
triple
threads with
a
pitch
of
2|
in.
Each ball
weighs
100
Ib. and
the
diameter of
each ball
is
9 in.
If
the balls are
revolved at 60
r.p.m.,
what
is
the maximum
size
(diame-
ter)
of
hole that
can be
punched
in
a
|-in.
steel
plate,
assuming
the
shearing strength
of the steel to
be
60,000
Ib.
per
square
inch and
the
efficiency
of the
screw to
be 15
per
cent. Also
assume
the
work
dia-
punching
the hole to be
triangular.
(See
Probs.
435
and
461.)
Ans.
d
=
0.60in.
FIG. 431.
gram
for
468.
A
shearing
machine
has 3
h.p.
delivered to it
by
the belt.
Every
two
seconds
an
operation
occurs
which
requires seven-eighths
of
all the
energy
supplied
during
the
two
seconds;
the
other
one-eighth
of the
energy
is
required
to overcome
the
friction
of the
machine.
During
each
operation
the
speed
of the
flywheel
decreases
from 120
to
80
r.p.m. Assuming
that
the
work
done
in
shearing
is
done
by
the
flywheel,
what
should
be
the
CONSERVATION
OF
ENERGY
439
moment of
inertia of the
flywheel?
what
is its
radius of
gyration?
If
the
weight
of
the
flywheel
is
400
lb.~
469.
A solid
cylinder
weighing
500
Ib. is
rolled
up
an
inclined
plane
by
means of
a
descending weight
B
(Fig.
432).
The
diameter
of
the
cylinder
is
4
ft. The
pulley
over
which the
rope
runs is
assumed
to
be
frictionless
and
weightless.
Body
B
weighs
300 Ib.
If
the
cylinder
starts from
rest
at
the
bottom
of the
incline,
with what
velocity
will
its
center
reach
the
top?
470.
A
solid
disc
18
in. in
diameter
is mounted
on
a shaft 4
in.
in
diameter
(Fig.
433).
The shaft
rolls,
without
slipping,
on
two
inclined
tracks.
The
disc
and
shaft
weigh
120 Ib. The
angle
of
inclination,
<,
is
15
and the
incline
FIG. 432.
FIG.
433.
is
8 ft.
long.
If
the
disc starts from rest at the
top,
what will be the
velocity
of
its
center at the bottom
of the
incline?
Ans. v
=
3.46
ft./sec.
195.
Conservation of
Energy.
One
of the
greatest
achieve-
ments of the
nineteenth
century
was the
recognition
and
statement
of the
principle
of
the
conservation of
energy.
Like
Newton's
laws of
motion it is an
inductive
generalization
from
observation
of,
and
experience
with, physical
phenomena.
The
principle
states
that
in
any change
of
the
state
or
condition
of an
isolated
material
system
the
total
amount
of
energy
of the
system
remains
constant.
By
an
isolated
system
is meant one on
which
no
bodies
external
to
the
system
have
any
effect on the
system.
Hence,
an
isolated
system
neither
gives
nor
receives
energy. Thus,
the
distribution
of
energy
within the
isolated
system
may
be
altered
and
the
various forms of
energy
changed
into other
forms
but
the total
amount of
energy
remains
constant.
Or,
as
sometimes
stated,
energy
may
be
transformed
or
transferred but
cannot
be
created
or
destroyed.
As
noted
in
Art.
191,
the
principle
of
'conservation
of
energy
is of
particular importance
in
the
study
of
material
systems
which
possess
non-mechanical
energy, although
it is
also
of
much
value
in
the
study
of
mechanical
energy. Although
an
isolated
system
does not exist
in
nature,
certain
systems approach
440
WORK
AND ENERGY
closely
thereto,
as for
example,
the earth and a
falling
body, pro-
vided
that the action
(and
reaction)
between the
earth and
body
is
large compared
with the resistance
of the air
and
the
attractions
of other bodies
on the
falling body.
Again,
although
external
forces act
on a
system
of
bodies,
the work done
by
the
forces
may
be zero
(or negligible)
as
in the case of a
simple
swinging
pendu-
lum and
the
earth
in
which
the
pull
of the
string
on the bob is
al-
ways
normal
to
the
displacement
of
its
application point
and
the
effect
of the
air is
negligible.
196.
Relation between
Potential and
Kinetic
Energy
for Con-
servative
Systems.
For
non-rigid
mass-systems
it was
found
(Art.
193)
that,
but
w
i}
which
represents
the work done
by
the
internal
forces
in
any
displacement
of
the
system,
is
equal
(but
opposite
in
sign)
to the
change
in
potential
energy
of
the
mass-system
for
the
given
displacement,*
that
is,
w
t
=
AE
P
.
Therefore,
Now if the work
done
by
the external forces which act on
a
mass-
system
is
zero,
then,
w
e
=
and
hence,
Or,
*
This
fact
may
be
shown as
follows:
By
definition the
change
of
potential
energy
of a
body
is the work which
the
body
does
against
forces in
passing
from one
configuration
to another
configuration, providing
no
other
change
in
the
state
of the
body
(such
as a
change
in
velocity)
takes
place.
Then from the
equation,
we
obtain,
or,
W
t
=-W
e
,
since
Al?fc
is zero if no
change
of the
velocity
state occurs.
But,
by
definition
w
e
is the
potential energy
of the
system
in its initial
configuration.
As
explained
in
Art.
187,
w
e
is
a
definite
quantity
only
when
the
forces are con-
servative.