Seely F.B. Analytical Mechanics for Engineers
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COMPONENTS
OF
VELOCITY
231
circular
path
is the
product
of
its
angular
velocity
1
(with respect
to
the
center
of the
path)
and
the radius of
the
path.
If the
particle
does
not move
on a
circular
path,
the
equa-
tion
v
=
ur is also
true
if
r
is
the radius of
curvature
of
the
path
at
the
given
position
of
the
particle,
and
if
co
is
the
angular
velocity
of
the
particle
with
respect
to
the center of
curva-
ture
as the
pole.
Further,
if the
point
moves
on a curve
of
any
form
and the
center
of curvature
is not taken as the
pole
(Fig.
2736),
then
the
term
pco,
where
p
denotes the radius
vector
to
the
point,
gives
one
component only
of the
linear
velocity,
as is shown
in the
next
article.
PROBLEMS
271. A
flywheel
6
ft.
in
diameter rotates
at
120
r.p.m.
Find the linear
velocity,
in feet
per second,
of a
point
on
its circumference.
272. A rod
4 ft.
long
rotates
in
a
horizontal
plane
about
a vertical axis
through
one
end of the
rod,
so that
the
linear
velocity
of its
mid-point
is 60
ft.
per
A
K
-
50-
-
3H
sec.
Find the
angular
velocity
of the
rod,
in
r.p.m.
273. An
automobile
is
traveling
at
30
miles
per
hour on a
straight
road.
An
observer is stationed
at
(Fig.
274).
Find
the
angular
velocity
of the
automobile
with
respect
to the observer
(a)
when
the auto-
mobile is
at
A;
(6)
when at B.
116.
Components
of
Velocity.
It
is
frequently
convenient
to find
the
^
velocity
of a
moving point
by
de-
p
IG
274.
termining
its
components,
or to deal
with the
components
of
the
velocity
instead of the
total
velocity.
The
assumption
that
velocity
may
be
resolved into
components
1
When
angular velocity
is
involved
in the same
expression
with linear
velocity,
the
angular
velocity
is
expressed
in radians
per
unit
of
time. Since
the number
of radians turned
through
is
a
length along
a
circular
arc divided
by
the
length
of the
radius
and,
hence,
is
a ratio
of
length
to
length,
it
has no
dimensions. Thus
the dimensional
equation
for
and hence the
equation
is
dimensionally
correct
(Art.
19).
Likewise, angular
displacement
and
angular
acceleration
are
expressed
in
radians and radians
per
second
2
,
respectively,
under
similar
conditions.
wr
s ~
=
232 MOTION OF
A
PARTICLE
follows
from the
fact that
velocity
is
a
vector
quantity.
Its
truth,
however,
is
abundantly
confirmed
by
the
agreement
of
results,
deduced
by
the
use of
velocity components,
with
observed
facts.
Two sets
of
components
only
are
here
determined;
namely,
the
axial
components
(v
x
and
v
v
)j
parallel,
respectively,
to
the x-
and-?/
axes,
and
the
radial and
transverse
com-
ponents (VR
and
VT),
par-
allel
and
perpendicular,
respectively,
to
the
radius
vector
(Fig.
275).
Thus,
since
the
linear
velocity
of a
point
in
any
direction
is
the
rate
at
which
the
position
of
the
the
FIG 275
point
is
changing
given
direction,
the
axial
components
of
the
velocity
v
moving point
M
(Fig.
275)
are
given
by
the
expressions,
1
in
of
the
dx
v,
=
-
and
dt'
or.
if the
coordinates,
x
and
y,
of
the
particle
change
uniformly,
then,
Ax
x
2
-xi
__'.,
_Ay^y
2
-yi
At
t
2
-ti'
and
*-?-
Likewise,
at
any instant,
the transverse and
radial
components,
VT
and v
Rt
are
expressed
as follows:
_pd9
,
_c/p
1
These
equations
follow
from
the
definition of
velocity provided
that the
assumption
is made
that
the
components
of a
displacement
may
be
treated,
in all
respects,
like
the total
displacement
itself
and
that
the vector sum of
the
rates of
change
of
the
components
is the same as the rate of
change
of the
total
displacement.
The
equations may
be derived
through
an
algebraic
series
of
steps beginning
with
the
definition of a
velocity component,
v
x
=
v
cos a
and
substituting
therein
the values of v and cos
; namely,
v=-r
and
cos a
=
-^.
This
method
gives
an
apparent
definiteness
to the
conclusion
but is
as
lacking
somewhat
in its
emphasis
on the
fundamental characteristics
of
the
quantities
involved.
COMPONENTS
OF
VELOCITY
233
in which
pdd
and
dp
are
the
component
displacements
in
the
transverse
and
radial directions
during
the time
interval
dt,
as
shown
in the
preceding article,
and co is
the
angular
velocity
of
the
particle
with reference
to
the
pole
0. It
will be
noted that
when
is chosen as
the
center of curvature of the
path
the
transverse
component
of
the
velocity
becomes
the
total
velocity, tangent
to
the
path,
V
R
then
being equal
to zero.
Since
the
pole
is
arbitrarily chosen,
the
components
V
T
and
V
R
are
different for different
positions
of the
pole
or
origin,
whereas
v
x
and v
v
are
independent
of
the
origin
but
depend
on
the directions of the
coordinate axes.
Further,
if
the
path
of the
point
is
circular and the
pole
is
at the
center
or,
in
general,
if
is the center of
curvature of the
path,
at the
instant,
then
dp
and
-37-
are zero and the
expressions
for V
T
and v
are
the
same.
ILLUSTRATIVE PROBLEM
274. In the
quick-return
mechanism shown in
Fig.
276,
00i
=
18
in. and
the
crank
OA =8
in.
If the
angular
velocity
of the
crank
is
40
r.p.m.,
what
is
the
velocity
of
the block A? Find
graphic-
ally
the
component
of the
velocity
of A
perpendicular
to the rocker
arm
0\M
(the
transverse
component).
Find
also the
angu-
lar
velocity
of
0\M.
Solution.
The
block
moves on the
cir-
cular
path
of
radius
r
=
8 in. Let to
denote
the
angular
velocity
of
A
with
reference to
the
pole
and
coi
the
angular
velocity
of A
with
reference to
the
pole
0\
(that
is,
let
toi
denote the
angular
velocity
of
OiA,
or
p).
The direction of the
velocity
of
A
is
tangent
to the circle
and its
magnitude
is,
g
;^=2.79ft./sec.
60
By
resolving v,
graphically,
into its
transverse
in.=
4.ft./sec.
FIG.
276.
and
radial
components
as
shown
in
the
figure,
the
following
values
are
found.
But,
=
1.8ft./sec.
and
^=2.17
ft./sec.
By
measuring,
OiA
is found
to
be
11.7
in.
234
Therefore,
MOTION
OF
A
PARTICLE
1
1
8
1
X12
=
1
. 84
rad./sec.
=
17.6
r.p.m.
PROBLEMS
276. A
rigid body
composed
of two disks of different
diameters
(Fig.
277)
turns
about
the
center
so that a
point
on
the
circumference of each
disk
FIG. 277.
FIG. 278.
moves
with
a constant
angular velocity
of 80
r.p.m.
If
r\
=
2
ft. and
r
2
=
3
ft.,
what
is the
velocity
of
body
A
and
of
body
B? A
unwinds
from the
large
disk
as
B
winds
up
on
the small
disk.
How
far
will
A and
B travel
in 6
sec.?
276.
The link
BC
(Fig.
278)
is
2 ft.
long
and
has an
angular
velocity,
<a
if
of
50
r.p.m.
If
the
link BA
is
perpendicular
to
BC,
at
the
instant
considered,
what
is the
angular
velocity
of the
crank
A
if A is
9
in.
long?
277.
If the
angular
velocity
of the
oscillating
arm OM
(Fig.
279)
is 40
r.p.m.,
in
the
position
shown
(0
=
30),
find the trans-
verse
component
of the
velocity
of
the
block
A,
referred to
O
as
the
pole.
Also find the
total
velocity
of A.
Ans.
v
A
=9.3
ft.
/sec.
FIG. 279.
278. The
cam
shown
in
Fig.
280
revolves about
the
axis
0,
causing
the
roller A
to
change
its
^-coordinate at
the
rate
of
4 in.
per
second when
=
30.
Find
the
angular velocity
of the
bell-crank
AO\B
if
OiA
is
18
in.
Ans.
co -=4.24
r.p.m.
LINEAR
ACCELERATION
235
FIG.
280.
117. Linear Acceleration. The acceleration of a
moving
point,
at
any
instant,
is the
time
rate at which its
velocity
is
changing
at
the instant.
The
-velocity
of the
point has,
at
any instant,
a
definite
magnitude (speed)
and a definite direction. A
change
in
the
velocity
occurs if
either its
magnitude
or its direction
changes.
Hence the acceleration
of the
point
may
be the
rate of
change
of
the
velocity
due
to
a
change
in the
magnitude
of the
velocity,
or it
may
be the rate of
change
of
the
velocity
due to a
change
in the
direction of
the
velocity.
These statements are not mere arbi-
trary
statements; they
are
in
accord with our
experience,
for
experience
teaches that a
particle
tends to maintain the
direction
of its
velocity
as
persistently
as it
tends to maintain the
magnitude
of
its
velocity.
The two
properties
of
velocity
are
inherent,
independent
properties.
Hence,
Whenever
the
velocity
of a
particle
changes
in
magni-
tude
the
particle
is accelerated.
Whenever
the
velocity
of a
particle
changes
in
direction
the
particle
is accelerated.
If the
direction
and
magnitude
of
the
velocity
change
simul-
taneously,
the
acceleration of the
particle
is then
the
vector sum
of
the two
accelerations
which
arise as the
result of
the
separate
changes.
In
rectilinear
motion
the
velocity
of a
particle
can
change
in
magnitude only,
whereas
in
curvilinear motion the
velocity
of
the
236
MOTION OF A
PARTICLE
particle
may
change
in
magnitude
but must
change
in
direction.
Therefore a
particle
cannot
have a curvilinear
motion
without
being
accelerated.
PROBLEMS
279. A
quarter-mile
track
is
made
up
of
two
straight parallel
sides
con-
nected at the
ends
by
semicircles.
A
boy
runs
the
quarter
mile in 50
seconds
at uni-
form
speed.
Does the
boy
have
an
accelera-
tion
while
on the
straight
sides?
While
on
the curved ends?
Why?
280. A
particle
moves
with a
constant
speed
of
10
ft.
per
sec. on the
path
shown
in
Fig.
281. Does the
particle
have
an
acceleration when
at A?
at 5?
at C?
at
D? What is the
value of
the
accelera-
FIG.
281.
tion of
the
particle
when at
A,
and at
D?
118.
Tangential
and
Normal Acceleration.
Tangential
Accel-
eration. The
facts
stated
in the
preceding
article make it
possible
to
determine,
for
a
given
motion,
whether
or not
a
particle
is
being
accelerated. The exact
magnitude
and direction
of
the
accelera-
tion
produced
in each
of
the two
ways
mentioned
in
the
preceding
article
will
now
be found.
In
Fig.
282
let
a
point
move
so
that
its
velocity changes
in
.
Path
*
FIG.
282.
magnitude
only.
That
is,
let the
point
move
on
a
straight
line
path.
Let
vi
be
the
velocity
at one
instant and
let
V2
be
the
velocity
after the time interval At.
If
the
velocity
changes
uniformly,
the
magnitude
of the acceleration is
the
ratio
of
any
change
in the
velocity,
&v,
to the time
interval, At,
during
which
the
change
Av occurs. Since
V2
and
v\
have
the same
direction,
Av
is the
algebraic
difference of
V2
and
v\
as well
as the vector
difference.
Hence,
the
magnitude
of the
acceleration
for
uni-
formly
accelerated,
rectilinear
motion
is,
Av
V2
vi
TANGENTIAL
AND
NORMAL
ACCELERATION
237
If the
velocity
of the
moving point
in
Fig.
282 does
not
change
uniformly,
then
equation
(1)
gives only
the
average
acceleration
during
the
period
At. When
the acceleration
varies from
instant
to
instant its
value at
any
instant
is the
average
acceleration
during
a
very
small
time
interval
including
the instant.
Or, expressed
mathematically,
the
instantaneous acceleration for rectilinear
motion
is,
Av dv
a= Limit--
=
-r
......
(2)
v
being
the
velocity
of the
particle
at the
instant. In order
to
find
a from
the above
equation,
v
must
be
expressed
algebra-
ically
in
terms
of
t.
The
direction
of the
acceleration of the
particle
is
the same as
that
of the
change
of
velocity,
Ay,
but
Av is
parallel
to
v,
that
is,
along
(or tangent
to)
the
path.
If
V2
is smaller than
v\,
the
sense
of
Av is
negative,
that
is,
opposite
to that of
v,
and
hence
the
acceleration
then is
negative.
If
the
particle
moves
on a circular
path
so
that,
as
the
velocity
changes
from
vi
to
1*2,
its direction as well
as
its
mag-
nitude
changes,
then
-r.
expresses
only
that
part
of the
accelera-
U/L
tion
which
is due
to the
change
of the
magnitude
of the
velocity.
And the
direction
of
-^
at
any
instant,
as
already noted,
is
parallel
to
the
velocity,
or
tangent
to the
path
at
the
point
where the
par-
dv
tide
is
located at
the instant.
Therefore,
-r. is
called the tan-
dt
gential
acceleration and is
written,
dv
at=
di'
Hence,
-r
is
the
total
acceleration
of
a
moving
point,
only,
when
the
point
has
a
rectilinear motion.
Normal
Acceleration.
The
acceleration
produced
by
a
change
in the
direction
of
the
velocity
of a
point
will
now be
considered.
In
Fig.
283
let a
particle
move on
a
circular
path
with
constant
speed.
That
is,
let the
velocity
change
in
direction
only.
At
one
instant the
velocity
is
v\,
and
during
a time
interval At the
velocity
changes
to
^2-
The
change
in
the
velocity
during
the
time
interval,
that
is,
the vector difference
between
V2
and
v\,
is
238
MOTION OF
A
PARTICLE
Ay,
as
shown
in
Fig. 283,
since Ay
is the
velocity
that
must
be
added to
v\
to
give #2, or,
V2
is
the
resultant of
v\
and
Ay.
For
motion in
which
the
direction
of
the
velocity
does
not
change
(rectilinear
motion),
the
vector
difference
of
the
velocities of
the
point
and
the
scalar
(or
algebraic)
dif-
ference are
the
same,
but in
curvilinear
motion
(Fig. 283)
the
algebraic
difference
is
zero,
whereas
the
vector
dif-
ference,
or
change,
of
the
velocities,
is
Ay,
the
mag-
nitude
of
which
is,
Ay
sin
-
Fr
2
sin
=
-
A0
sin
,
in which
v is
the
magnitude
of
the
velocity
at
any
instant,
since
v^
=
v\
v.
Hence,
since
the
acceleration
at
any
instant
is
the
average
acceleration
during
an
indefinitely
small
time
interval
including
the
instant,
the
magnitude
of
the
instantaneous ac-
celeration due to
the
change
of
direction
of
the
velocity is,
o
2y
sin
.
a
=
Limit
At
Now,
as
A*
approaches
zero,
sin
-^
approaches
and
approaches
Zi
Z At
-T-.
Hence,
in the
limit,
dd
a=v
Tf
jn
But
-IT
is the
angular
velocity
of
the
moving point
which
corre-
TANGENTIAL AND NORMAL ACCELERATION
239
spends,
at
any instant,
to the linear
velocity
v.
Further,
v
=
cor
(Art.
115). Therefore,
The
particular
form
most
convenient for
use
depends upon
the
particular
problem
under consideration.
The
direction
of this acceleration is the
limiting
direction
of
Aw as
A
approaches
zero,
and since
A0
approaches
zero
as
A
approaches
zero,
it will
be
seen from
Fig.
283
that the
h'miting
direction of
Ay is
perpendicular
to
v\
and,
hence,
is
toward
the
center about
which the
point
is
turning,
that
is,
normal
to the
path.
This acceleration
is, therefore,
called the
normal accelera-
tion
and is
written,
v
2
a
n
=
vn
=
<a
2
r= .
If the
path
on which
the
particle
moves is not a circular
path,
the
above
expression
also holds
provided
that
r
is
the radius of curva-
ture
of the
path
at the
point
where
the
particle
is located
at the
instant,
and
that
co
is
the
angular
velocity
with
respect
to
the
cen-
ter
of curvature
as
a
pole.
Summarizing,
the
two
following
important
theorems
may
be stated:
I.
When
the
velocity,
v,
of a
particle changes
in
mag-
nitude,
an
acceleration is
produced
the
value of which is
-j-;
at
its direction at
any
instant is
parallel
to that of the
velocity,
that
is,
tangent
to the
path
at
the
point
at which
the
particle
is located at
the
instant.
Thus,
dv
II.
When the
velocity, v,
of a
particle
changes
in
direc-
tion,
an
acceleration is
produced
the
value of which is
v
2
vu
=
co
2
r
=
;
its
direction,
at
any
instant,
is
perpen-
dicular
to the
velocity,
towards the
center
of
curvature
of
the
path,
at the
point
where
the
particle
is
located at
the instant.
Thus,
v
2
a
n
=
vu
=
w
2
r
=
.
240
MOTION
OF
A
PARTICLE
The unit
of
acceleration
is
any
convenient
unit of
velocity
per
unit of
time;
such
as,
foot
per
second
per
second
(ft./sec.
2
),
mile
per
hour
per
second
(mi./hr./sec.),
etc.
ILLUSTRATIVE
PROBLEM
281.
Two
pulleys
are
connected so that
they
turn
together
about
the
center
(Fig.
284), causing
the
weight
A to
unwind and
the
weight
B
to
wind
up.
If the
angular
velocity
of the
points
M
and
P
change
uniformly
from 10
r.p.m.
to 60
r.p.m. during
a
period
of
2
sec.,
find:
(a)
the
tangential
acceleration of
each of the
two
points
at
any
instant
during
the
2
sec.;
(6)
the
acceleration of
A
and of
B',
(c)
the
total acceleration
of
M
at
the
beginning
and of
P
at
the
end of
the two-second
period.
Solution.
(a)
The
velocity
of the
point
M
changes
in
magnitude
from
10X27T.
60 :|-j
ft./sec.
to
ft./sec.
in
2
sec.
Hence
the
tangential
acceleration of M
is,
^-=^
=
-
=
1.96
ft./sec.
2
.
At Z
o
The
velocity
of
P
changes
from
=
ft./sec.
to
o
ft./sec.
hi
2
sec.
Hence,
(6)
The
magnitude
of the
velocity
of
any
point
on
the
circumference of
the
small
pulley
changes
the same amount as does the
velocity
V&\
of A.
Therefore,
the
tangential
acceleration of
M
is the
same
in
magnitude
as the
acceleration
of
A
.
Hence,
}
t
=
1.96
ft./sec.
2
aB
=
(
ap
)
t
=
2.
62
ft./sec.
2
.
(c)
The
normal
acceleration of
M
at
the
beginning
of the
2-sec.
period
is,
2
ft./sec.
2
and
the normal acceleration
of P
at the end of
2
sec.
is,
.4
ft./sec.
2
.
Therefore
the total
acceleration
of M
is,
ft./sec.