Seely F.B. Analytical Mechanics for Engineers
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LINEAR IMPULSE
OF A FORCE 451
the
product
of an
average
value
of the force and an assumed or
estimated
time interval
AZ
=
fe
ti
-
Thus,
Q
/>
in which
F
a
v
denotes the
average
value of the
impulsive
force which
is assumed
to act
during
the
time interval
At.
The
impulse
of
a
force
frequently
is called linear
impulse
in
contrast with
the
moment
of
the
impulse
which
is called
angular impulse. (See
Art.
204.)
Units.
The
unit of an
impulse
is a combination of
a
unit
of
force
and
of time and
hence,
is a
compound
unit.
The unit of
impulse
has no
special
name.
In
the
gravitational
or
engineer's
system
of
units,
if
the
pound
is selected as the unit of
force and the
second
for the
unit of
time,
then the
unit
of
impulse
is
the
pound-
second
(Ib.-sec.) or,
if
the
kilogram
is selected for
the
unit of
force
and the second
for the unit of
time,
the unit of
impulse
is
the kilo-
gram-second
(kg.-sec.).
203.
Components
of
Linear
Impulse.
The
impulse
of
a
force,
like
the force
itself,
is
a directed or vector
quantity,
the
sense
and action
line
of
the
impulse
being
the
same as that
of
the
force.
An
impulse
of a
force, therefore,
may
be
resolved
into
components
and
may
have a
moment with
respect
to a
point
or a line. The
component,
in
any
direction,
of
the
impulse
of
a
constant force
is
the
product
of the
component
of
the force
in
the
given
direction
and the
time
interval
At
during
which
the
force
acts.
That
is,
Q
x
=
F
x
-At;
Q
v
=
F
v
-At,
etc.
And,
if
the
force varies
in
magnitude
during
the
time
interval
At
=
t2
ti,
the
component
of
the
impulse is,
Q
x
=
|
F
x
dt]
Q
y
=
|
2
F
v
dt,
etc.
J*
Jti
Linear
Impulse of
a
Force
System.
The
linear
impulse,
in
any
direction,
of a force
system
is
the
algebraic
sum
of
the
components,
in
the
given
direction,
of
the
impulses
of
the
forces of the
system.
452
IMPULSE AND
MOMENTUM
Thus,
for a force
system
in
which
the
forces
are
constant,
the
linear
impulse,
Q
x
,
of
the
force
system
in
any
direction
x
is,
Or,
if
the forces of the
system
vary
during
the
interval, then,
C
h
&
=
S
FJt.
Jh
204. Moment of
Impulse.
Angular
Impulse.
The
moment
of the
impulse
of a constant
force
about
any point
or
axis is
the
product
of the
moment, T,
of
the force about
the
given point
or
axis,
and the
time
interval,
A,
during
which
the force
acts.
Thus, by letting
LQ
denote
the moment of
the
impulse
of a con-
stant force with
respect
to
an
axis
0,
the moment of
the
impulse
of the force is defined
by
the
equation,
And,
if
the
force varies
in
magnitude
during
the
time
interval,
At
=
fe
1\
,
the moment of
the
impulse
is
expressed
by
the
equation,
in which
TO
is
the
moment
of
the force with
respect
to
the
axis 0.
The
moment
of
the
impulse
of a force is also
called
the
angular
impulse
of
the force.
Angular
Impulse of
a Force
System.
The
angular impulse
of
a
force
system
about
any
axis is the
algebraic
sum of
the
angular
impulses
of the
forces of the
system
about
the
given
axis.
Thus,
if
the
forces
of the
system
are
constant,
the
angular
impulse, LO,
of the
system
about
the
axis
is,
And,
if the forces
vary,
the
angular impulse
of
the
force
system
about
the
axis
U,
for
the time
interval
At
=
fo
t\
,
is
expressed
by
the
equation,
MOMENTUM OF A PARTICLE
DEFINED
453
It should be noted that the linear
(and angular) impulse
of a
constant
force,
or of a variable force for which the
law
of variation
is
known,
is not as
important
a
conception
in
the
analysis
of the
motion
of
a
body
as is the
impulse
of
an
impulsive
force.
However,
as
already noted,
even for constant
forces which act
during
a
comparatively long
time
interval,
the methods
of
impulse
and
momentum
may possess advantages
over
other methods.
PROBLEMS
479.
A
body weighing
40
Ib. slides
down
an inclined
plane
in
4
sec.
The
plane
makes an
angle
of
60 with
the
horizontal.
If
the coefficient of
friction
is 0.2 find
the
component
of
the
linear
impulse
of the force
system
acting
on the
body parallel
to
the
plane.
Ans.
Q*
=
123
Ib.-sec.
480.
The
linear
impulse
of the
total
pressure
of the steam on
the
piston
of a steam
engine
is
4200
Ib.-sec.
The
diameter of
the
cylinder
is
14
in. and
the
engine
runs at
200
r.p.m.
Find the
average
(time
average) pressure
(in
Ib.
per sq.
in.)
of
the
steam
against
the
piston
during
one
stroke.
481.
A
train
having
a
weight
of 2000 tons travels
up
a
|
per
cent
grade.
The
draw-bar
pull
of the
engine
is
50,000
Ib. and
the
train resistance
is
8 Ib.
per
ton of
weight.
If it
takes
2
min.
to
travel
up
the
grade,
find
the
linear
impulse
of
the
force
system acting
on the
train
for the 2-min. interval.
Ans.
1,680,000
Ib.-sec.
482.
A
constant
frictional
moment
of 200 Ib.-ft.
is
applied
to
a
rotating
drum
by
means of
a band brake
(see
Fig.
185).
If the
moment decreases
the
angular velocity
of the drum
uniformly
from 90
r.p.m.
to
10
r.p.m.
while
the drum makes 30
revolutions,
find the
angular
impulse
of the
band
brake
on the
drum. Ans.
7200 lb.-ft.-sec.
2.
MOMENTUM
205.
Momentum of
a Particle Defined. Units.
The
momentum
of a
particle
of a
moving body,
at
any instant,
is
defined
as the
product
of the
mass of the
particle
and its
velocity
at the
instant.
Momentum,
like
velocity,
is a directed or
vector
quantity.
Furthermore,
like
force,
momentum is
represented by
a
localized
vector,
that
is,
it has a
definite
position
line.
Thus,
the
direction
of the
momentum of a
particle
is the same as that
of the
velocity
of the
particle
and
its
position
line
passes
through
the
particle.
Thus,
if
U
denotes the
momentum of a
particle
of
mass
m
and
velocity v,
the
magnitude
of
the momentum of
the
particle
is
defined
by
the
equation,
U=mv.
454
IMPULSE AND MOMENTUM
The
momentum
of
a
particle
frequently
is
called linear
momentum
in
contrast with
the
moment of
momentum
of
the
particle
which is
called
angular
momentum.
Units. The unit of momentum is a combination of a unit of
mass
and a
unit
of
time and
hence,
is a
compound
unit. The unit
of momentum
has
no
special
name.
In
the
gravitational
or
engi-
neer's
system
of
units,
the unit of mass is a derived unit as
explained
in
Art.
141;
derived from the units of
force,
length,
and time and
hence,
the
unit of
momentum
is also a derived unit.
Thus,
if
the
pound, foot,
and second are chosen
for the units of
force,
length,
and
time,
respectively,
the
unit of
momentum is
expressed
by,
1
unit of
mass
X
1
unit of
velocity
=
lib.
XI
sec
*
1
ft.
1
ft.
=
1 Ib.-sec.
X
1 sec.
It will be
observed, therefore,
that
momentum
is
expressed
in
the
same
fundamental units
as
is
impulse.
206.
Components
of Momentum.
Moment
of Momentum.
Since
the momentum
of a
particle
is a vector
quantity
it
may
be
resolved into
components and,
like
any
localized
vector,
it has a
moment
with
respect
to
any
point,
the moment
being
denned
as the
product
of
the
magni-
tude of the momentum
and the
perpendicular
distance from the
position
line
to the
point
or
moment-center.
If,
then,
U
x
and
V
y
denote the
components
of the momentum of a
particle
of mass
ra
and
of
velocity
v
FlG 440
(Fig. 440) and,
if H
denotes
the moment of momentum
of
the
particle
with
respect
to the
point 0,
the
components
of
the
momentum
of
the
particle
and
the
moment
of momentum
of
the
particle
are
expressed
by
the
equations,
and
or,
LINEAR MOMENTUM
OF
A
BODY
455
The
latter
expression
for
HQ
expresses
the
fact that
the
moment of
the
momentum
of a
particle
equals
the
algebraic
sum of
the
moments
of the
components
of the
momentum
of
the
particle.
207.
Linear Momentum
of a
Body.
The linear
momentum,
in
any
direction,
of
any
body
(mass-system)
is
the
algebraic
sum
of
the
components
in
the
given
direction of
the
momentums
of the
particles
of the
body.
In
gen-
eral the
momentums
of the
particles
of a
body
are not the
same either
in
magnitude
or
in direction.
It
will now be
shown,
however,
that the
linear
momentum
of the whole
body
may
be
found
from
the
mass of the whole
body
and
the
velocity
of the mass-center
of
the
body.
Thus,
in
Fig.
441
are shown three
particles
m', m",
and
m'" of
a
system
assumed,
for convenience
only,
to
move
in a
plane.
At
a
given
instant the
particles
and
their
mass-center are
in
the
posi-
tions
shown. The direction
of
the
momentum of each
particle
is
tangent
to
the
path
of the
particle
and
the
velocity, v,
of the
mass-
center is
tangent
to
the
path
which
the mass-center describes.
The
component,
in
the
x-direction,
of
the linear momentum
of the
mass-system is,
U
x
=
m'v
x
f
+m"v
x"+m"'v
x
'"+
.
FIG. 441.
But,
2(mv
x
) may
be shown to
be
equal
to
Vx'Zm
or
Mv
x
as follows:
The
z-coordinate, z,
of the
mass-center is
defined
by
the
principle
of
moments
(Art.
80) by
the
equation,
m
f
x'+m"x"+m'"x'"+
. . .
=x?m
=
Mx.
By
differentiating
both
sides
of
the
equation,
the
following
equa-
tion
is
obtained,
456
IMPULSE
AND
MOMENTUM
dx'
But,
-j-
is the
^-component
of the
velocity
of m'
and
hence,
dx'
Similarly,
and
dt
v
x
",
etc.,
dx
_
Thus,
the
above
equation
becomes,
mV+w'V'+m''V''+
. . .
=
Mv
x
.
!>/"
+
=
MVy.
;
and
U=Mv.
Similarly,
Whence,
That
is,
the
linear
momentum
of
any
moving mass-system
is
the
product
of
the
mass
of
the whole
system
and
the
velocity
of
the
mass-
center
of
the
system
and its direction
agrees
with
that
of
the
velocity
of
the
mass-center.
208.
Angular
Momentum of
a
Rotating
Rigid
Body.
Let
Fig.
442
represent
a
rigid
body
rotat-
ing
about a fixed axis
through
with
an
angular
velocity
co.
The
linear
velocity
of
any particle
of
the
body
at
a
distance r from
the
axis of
rota-
tion is
rco
(Art. 115)
and
the linear
momentum of the
particle
of mass
m
is mv or
rarco.
Therefore,
the moment
of
momentum
for
one
particle
about
the axis
of rotation
is,
FIG. 442.
mv r
=
rarco r
=
mr'
2
u,
and
the
algebraic
sum of
the
moments
of
the
momentums of all
the
particles
about the same axis
is,
whence,
=
coSmr
2
,
=
/oco.
ANGULAR MOMENTUM OF RIGID
BODY
457
in
which
7
is the
moment of
inertia
of
the
body
about the axis
of
rotation.
Therefore,
the
angular
momentum of
a
rotating rigid
body
about the axis of rotation is
equal
to
the
product
of
the
moment
of
inertia of the
body
with
respect
to
the axis of
rotation
and the
angular velocity
of the
body.
209.
Angular
Momentum
of a
Rigid
Body
Having
Plane
Motion. Since
a
plane
motion
of a
rigid
body
may
be
considered
as a combination of a rota-
tion
and a
translation
(Art.
134),
the
velocity
of
any
particle
P
(Fig.
443)
is the
resultant
of
a
velocity
rco
due to the rotation
of the
body
with
angular
velocity
co
about
an axis
and,
the
velocity
VQ
due to the trans-
lation
which
gives
to each
particle
the
velocity
^o
of
the
point
0.
Therefore,
the
components
of the
momen-
tum
of the
particle
P
are
mrco
and
mvo
as shown
in
Fig.
443.
And,
if
m(vo)
x
and
m(vo)
v
denote the
x- and
^/-components
of the
momentum
mvo,
then the
angular
momentum
of the
particle
with
respect
to the
axis
is,
rarco
r+m(vo)
x
-y
m(vo)
v
-
x,
and
the
angular
momentum,
H
,
for the
whole
body
with
respect
to the axis
is,
(1)
FIG. 443.
Now,
as
already
noted,
a
plane
motion of a
rigid body may
be
considered
as
a rotation about
an axis
through any
point
in the
plane
of
motion
and
a
translation,
etc.
Hence,
if the
point
is
taken as
the
mass-center
of
the
body,
then
HQ
becomes
H
and
the
above
expression
reduces
to
#=/(>,
........
(2)
458
IMPULSE AND
MOMENTUM
since
y
and
x
are then
equal
to zero and
/o
becomes
/.
Therefore,
in
general,
the
angular
momentum of a
rigid body
having plane
motion,
with
respect
to an axis
in
the
body,
is not
equal
to
the
product
of the
moment of inertia of the
body
with
respect
to
the
given
axis and
the
angular
velocity
of
the
body.
But,
for
an
axis
passing
through
the
mass-center of
the
body,
the
angular
mo-
mentum is
equal
to
the
product
of the
moment
of
inertia of
the
body
with
respect
to the axis
through
the
mass-center
and
the
angular
velocity
of the
body.
It
may
be noted
also that
if
the axis
about which
the
angular
momentum
of the
body
is taken is
the instantaneous
axis of
rota-
tion,
equation
(1)
becomes
HQ
=
!QU
since
(VQ)
X
and
(VQ)
V
are
equal
to zero.
That
is,
the
angular
momentum of a
rigid body having
plane
motion,
about
the
instantaneous axis of
rotation,
is
equal
to the
product
of the moment of
inertia of
the
body
with
respect
to the instantaneous
axis of rotation
and the
angular velocity
of
the
body.
PROBLEMS
483.
A small
body
(particle)
weighing
8
Ib. is
attached to one end
of a
string
and
is made to revolve as a conical
pendulum
(see
Fig.
324).
If the
body
revolves at a
distance of 15 in.
from the
axis
with an
angular
velocity
of
90
r.p.m.,
find
(a)
the linear momentum of the
body,
(6)
the
angular
momentum
of the
body
with
respect
to
the axis of rotation.
Ans.
(a)
E7
=
2.931b.-sec.;
(6)
tf
=
3.66 lb.-ft.-sec.
484. A
slender
rod rotates
in
a
horizontal
plane
about
a vertical axis
through
one
end of
the rod.
If the rod is 3 ft.
long,
weighs
4
Ib.
per foot,
and
rotates at
120
r.p.m.,
find its
angular
momentum
about the
axis of rotation.
Ans.
#o
=
14.05
lb.-ft.-sec.
485.
Find
the
angular momentum,
about
the
axis of
rotation,
of the
fly-
wheel
shown
on
the
punching
machine
in
Fig.
429,
when
the
flywheel
is rotat-
ing
at 225
r.p.m.
Use
the value
of
I
found
in Prob. 461.
486.
Find the
angular
momentum,
about
the
axis of
rotation,
of the two
spheres
shown
in
Fig.
431,
when
the
spheres
are
revolving
at
120
r.p.m.
Each
sphere
weighs
100
Ib. and
is
9 in.
in
diameter. Ans.
H
=
1250
lb.-ft.-sec.
487. A
solid
cylinder
weighing
128.8
Ib.
rolls,
without
slipping,
down an
inclined
plane.
The
linear
velocity
of
the
mass-center
of
the
cylinder
at a
given
instant
is 30 ft.
per
sec.
The diameter
of the
cylinder
is 18 in. Find
the
angular
momentum
of
the
cylinder
(a)
about
an
axis
through
the mass-
center
perpendicular
to
the
plane
of
motion,
(6)
about the
instantaneous
axis
of
rotation.
(Ans.) (a)
#=45
lb.-ft.-sec.;
(&)
#*
=
135
lb.-ft.-sec.
LINEAR IMPULSE AND
LINEAR MOMENTUM
459
488. The
connecting
rod
shown
in
Fig.
422 is
30 in.
long.
It
has
a
constant
cross-section
and
weighs
52
Ib. The crank
length
OB is
5 in.
and
wi
is 300
r.p.m.
The distance
from
i to
B is found
to be
34.68
in. and
r
is
found
to
be
23.75
in. Find the
angular
momentum of the
connecting
rod
(a)
about
an axis
through
the mass-center and
(6)
about
the
instantaneous axis
of
rotation,
3. PRINCIPLES
OF IMPULSE AND
MOMENTUM
210.
Preliminary.
In
order
to
determine the effect
of
a
force
system
on trr?
motion of
a
body,
that
is,
in
order to treat
the
usual
problem
in
kinetics, by
means
of
the
quantities
impulse
and
momentum,
the
relations which
exist
between
the
impulse
of
the
force
system
and
the momentum of the
body
on
which the
force
system
acts
must
be established.
These relations
are
expressed by
means
of two
principles, namely, (1)
the
principle
of
linear
impulse
and
linear momentum and
(2)
the
principle
of
angular
impulse
and
angular
momentum.
211.
Principle
of
Linear
Impulse
and
Linear Momentum.
It
was stated
in Art. 152
that
the
algebraic
sum of
the com-
ponents
in a
given
direction
of the external
forces
acting
on
any
body (whether rigid
or
not)
is
equal
to the mass of
the
body
times
the
component
of the acceleration
of
the
mass-center of
the
body
in
the
given
direction.
That
is,
if x
denotes
any
direction,
(1)
Now
if the
forces
acting
on
the
body
remain
constant,
then
a
x
will
be
constant,
that
is,
the mass-center of
the
body
will
be uni-
formly
accelerated.
Hence,
according
to
Art.
124,
in which
v
x
"
and v
x
are the
^-components
of
the
velocity
of
the
mass-center
of
the
body
at
the end
and
beginning,
respectively,
of
the
time interval
A.
Therefore,
(2)
Now
the left-hand member
of this
equation
is
the
^-component
of
the linear
impulse
of
the
external
force
system
which acts
on
the
body
(Art.
203)
and
the
right-hand
member
is the
change
in the
460
IMPULSE AND
MOMENTUM
momentum
of the
body
in
the z-direction
(Art. 207).
The
above
equation
is a
mathematical statement of
the
principle
of
linear
impulse
and
linear
momentum
for
any
body
when
acted
upon
by
constant
forces.
The
principle
applies,
however,
to the
motion
of
any
body
under
the action
of
any
force
system.
Thus,
in
general,
a
x
=
-
and
hence,
equation
(1)
may
be
written,
or,
since
M
is
constant,
Zr.-jg(Jfftr)
........
(3)
And, by
integrating
this
equation,
the
following
equation
which
expresses
the
principle
of linear
impulse
and linear
momentum
is
obtained,
r* rvx"
2/^
=
j
d(Mv
x
)
Jh
JVx
Or,
C
2
Jti
(4)
in
which,
as
before,
v
x
"
and
v
x
are the
a>components
of
the
velocity
of the mass-center
of the
body
at the end
and at the
beginning,
respectively,
of the time interval
tz
t\.
The
principle
of linear
impulse
and linear momentum
then,
as
expressed
in
equations
(2)
and
(4), may
be
stated
in
words
as
follows:
The
algebraic
sum
of
the
components,
in
any direction,
of
the
impulses
of
the external
forces
acting
on a
body
during
any
time interval is
equal
to
the
change
in the
linear momentum
of
the
body
in the same direction
during
the same
interval
of
time, or,
stated in the
form
of an
equation,
in
which
x
represents
any
direction. The
particular
forms
in
which
Q
x
and U
x
are
expressed depend upon
the
type
of
force
system,
the
type
of
motion,
and the kind of
body (see
Art.
214
for
expressions
which
apply
to
rigid
bodies
having particular
types
of
motion).
It should
be
noted
also that
equation
(3) expresses
an
impor-
tant
principle
which
may
be
stated
in
words as follows:
The
alge-
braic sum
of
the
components,
in
any
direction,
of
the external
forces