Seely F.B. Analytical Mechanics for Engineers
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SIMPLE
CIRCULAR
PENDULUM
361
the
string.
Although
these
ideal
conditions cannot be
realized
fully
in
any
physical
apparatus,
the
motion of a small
body
sus-
pended
by
a
light
thread
will be
approximately
that
of the ideal
simple
pendulum.
It
is
desired
to find
the time or
period
of
vibration
of a
simple
circular
pendulum.
Let
C
(Fig.
378)
represent
a small
body
which
is
suspended
by
a thread
from
the
point
and
is
allowed
to
swing
as a
simple pendulum
in
the
arc
BB
1
of
radius
OC
or
I.
By
resolv-
ing
forces
in
the direction of
the
tangent
to
the
path
we
have,
or
Sft-
w
W
sin
=
a*,
J/
whence,
Or
=
</
sin 0.
And
from Art.
127,
vdv=a
t
ds.
Therefore,
vdv
=
g
sin
ds.
But,
Therefore,
=
ld6.
vdv
=
lg
sin Odd.
FIG.
378.
The
integration
of this
equation,
after
expressing
v
in
terms of
t
and
6,
leads
to a
complicated
relation
between t
and
6.
However,
if
the
angle
of vibration is
small,
a
close
approximation
to the
correct
solution
may
be obtained
by
assuming
that
=
sin 6.
The last
equation
then
becomes
vdv=
IgBdB.
And
by integrating,
To
determine, Ci,
the
constant of
integration,
let
v
=
when
=
#i.
Using
these
values
in
the
above
equation,
we
have,
Hence,
or,
362
FORCE,
MASS,
AND
ACCELERATION
ds
The
time, t, may
now
be introduced
in
the
equation
since ^
-r-
And,
if
the
particle
is
moving
towards
A,
s
decreases as t
increases
ds
and hence
-j-
is
negative.
Thus,
or,
de
And, by integrating,
If time is measured
from
the instant the
body
is at
B,
then
=
Q\
when
2
=
0.
By substituting
these
values
of 6
and
t
in
the last
equation,
the value of
2
is found to be sin"
1
1 or
^-
Thus,
the
last
equation
becomes,
By
making
6
equal
to zero
in
the last
equation,
the
corresponding
value
for
t,
that
is,
the time
required
for
the
body
to move from B
to
A is
^A/-.
Since the
body gains velocity
during
the
displace-
ment
BA at
the
same rate that
it
loses
velocity
in the
displace-
ment
AB',
the
average
velocities for the two
displacements
are
equal.
Therefore,
the time
required
for a
single
oscillation
(B
to
B')
is TT
^-.
The time
or
period, P,
of a
complete
oscillation
(from
B
to
B' and back to
B)
then
is,
-2-J?
\0
This
equation
shows
that the
period
P
is
independent
of
6\,
that
is,
of
the
amplitude
of
the
oscillation.
However,
it must
be
remembered
that
the
equation
applies
to oscillations of
relatively
small
amplitudes.
Jf
the
amplitude
of
oscillation
is
not
small
enough
to
permit
of
COMPOUND
PENDULUM
363
the
assumption
that
sin
=
6,
then the time of a
complete
oscilla-
tion
is
expressed by
the
series,
in
which
k
=
sin
-.
PROBLEMS
409.
A
simple pendulum
4
ft.
long swings through
an
angle
of 60
(that
is,
^=30).
Find the
period
of
oscillation,
(1)
by
the
approximate
method and
(2)
by
the exact
method.
Ans.
(1)
P =2.21
sec.;
(2)
P
=
2.25 sec.
410. Find the
length
of the
simple
pendulum
which beats
half-seconds,
that
is,
one for which
the
period
of a
complete
oscillation
is one second.
411. What is the
length
of
a clock
pendulum
which will make one beat
per
second.
If the
clock
loses 5 sec.
per
hour,
how much should the
pendulum
be
shortened?
157.
Compound
Pendulum. A
physical
body
which
oscillates
or
swings
about
a horizontal axis under
the influence of
gravity
and the
reaction of the
supporting
axis
is
called
a
compound
or
physi-
cal
pendulum.
It is desired to find the time of
oscillation of a
compound pendulum.
Let
Fig.
379
represent
a
section
of
such
a
pendulum
which
rotates about
an
axis
through
0. From
Art.
146
we
have as one
of
the
equations
of
motion
for a
rotating body,
Or,
Whence,
Wr
sin
=
7o
a
gr
=
-
<*=
f-o
sin 0.
364
FORCE,
MASS,
AND
ACCELERATION
Now,
it was shown in
the
preceding
article
that the
tangential
acceleration
of a
simple pendulum
of
length
I
is,
a
t
=
g
sin 6.
But,
a
t
=
la
(Art.
121).
Whence,
(2)
By
comparing equations
(1)
and
(2)
it will
be
noted
that the
angular
motion of a
simple pendulum
may
be
made
exactly
the
same
as that of
a
compound pendulum (if
the
two
start from
the
same
position)
by
fixing
the
length
I of
the
simple pendulum
such
that,
A
simple pendulum
of this
length
is
called an
equivalent
simple
pendulum.
The
period, then,
of
a
complete
oscillation,
of
small
amplitude,
of a
compound pendulum is,
P f)--
I"*
**'w
158. Center
of Oscillation.
The
point, Oi,
in
the
compound
pendulum (Fig.
379)
at the distance
~
from the
center
of
rotation
is called the
center
of
oscillation. That
is,
the center of
oscillation
is that
point
at
which
the
whole
mass
of the
compound pendulum
may
be concentrated
without
changing
the
period
of
vibration.
It will be
noted that the center
of oscillation is
also
the
center of
percussion
(see
Art.
148).
Further,
the
center of oscillation
may
be made the
center of
rotation without
changing
the
period
of oscillation. That
is,
in
a
compound pendulum
the centers of
oscillations and
suspension
are
interchangeable.
This fact
may
be shown
as
follows: The
distance
G0\
from
the
center of
gravity
of the
compound pendu-
lum
to the center of oscillation
(Fig.
379) is,
But,
from
Art.
102,
TORSION PENDULUM
365
Whence,
Now
if
Oi
is made the
center
of
rotation,
then the
new
center
of
k
2
oscillation,
02,
will be
a distance
G0%
from
the
center of
gravity.
But
r
is now
equal
to
G0\
; hence,
r 'i?
Therefore
0%
coincides with
0\
and
hence the
centers of
sus-
pension
and oscillation
are
interchangeable.
PROBLEMS
412.
A
uniform slender
rod
4
ft.
long
oscillates as a
compound
pendulum
about
a horizontal axis
through
one
end of the
rod,
the
rod
being
perpendicular
to the axis,
(a)
Find
the
period
of
oscillation.
(6)
About
what other
point
could
the rod
rotate and have the same
/~\0
period
of oscillation.
Ans.
(a)
P
=
1.80
sec.; (6)
1.33
ft. from end of
rod.
413. Find the
length
of a uniform
slender bar
having
a
period
of
oscillation
of
1
sec. when allowed to
swing
as
a
compound
pendulum
about
an axis
through
one end
of
the bar.
Ans. 1
=
1.22 ft.
414. A steel
pendulum
consists
of a
circular disc 10 in.
in diameter
and
1
in.
thick,
and a
rectangular
bar
30 in.
long,
3
in.
wide,
and
1
in.
thick,
as shown
in
Fig.
380. If
the
pendulum
oscillates
about a horizontal axis
through
0,
what is
the
period
of oscillation?
159.
Torsion Pendulum.
A
torsion
pendulum
FIG. 380.
consists
of a
body
(Fig.
381)
which
is
suspended
by
a wire
or slender rod
and
allowed to
oscillate
about
the
axis
of
the
wire.
The
body
is
rigidly
attached
to
the
wire
so that
its
center
of
gravity
lies on
the axis of
the
wire
and
the wire is
rigidly
attached
at the
support.
Thus,
if
the
disc
of
Fig.
381
is
given
an
initial
angular
displacement,
61,
the
wire is twisted
and when
the
disc
is released the
wire
exerts a
turning
moment
on
the
disc,
which
starts
it to oscillate.
The
torque
or
turning
moment,
T,
exerted
by
the wire
on the
disc is
directly proportional
to the
angular
displacement,
provided
366
FORCE,
MASS,
AND
ACCELERATION
that
the elastic
limit of the material is not
exceeded. That
is,
T=
kd in
which
ft is
a
constant;
the
minus
sign
indicates
that
when
9
is
positive
T is
negative.
Since
the disc
has a
motion
of
rotation,
the
equation
of
motion
is,
Hence,
But,
a
Therefore,
dt
2
'
00
_
k
W
To
9
'
The time of an
oscillation of
the
pendu-
lum
may
now be found
by
the
integration
of
this
equation. Multiplying
each side
of
the
equation by
the factor
-^,
we
have,
_
^
a
de
dt dP
IQ
df
Whence, by
integrating,
Now,
when
(
de
\
2
k
(dt)
=
-T
Therefore,
Hence,
By
a method similar
to that
used
in
Art.
156,
the
detailed
steps
of
which
are
left to
the
student,
the time of
a
complete
oscillation
is
found
to be
in which
IQ
is
the
moment of inertia
of
the
body (disc)
with
respect
to
the
axis
of the
wire
and
B
is the
angular displacement,
corre-
DETERMINATION
OF MOMENT OF INERTIA
_367
spending
to
which
the wire
exerts the
torque
T
on the
disc.
It
will
be observed
that
the
period
of an
oscillation
is
independent
of the
initial
displacement,
that
is,
of the
amplitude
of
the vibra-
tion,
assuming
that
the elastic
limit of the wire is not
exceeded.
PROBLEMS
415. The disc
of a torsional
pendulum
is turned
through
an
angle
of
10
by
a
torque
of
1
ft.-lb. When released it
is
observed
to make five
complete
oscillations
per
second. What is the
moment of
inertia
of the
disc?
(In
using
the
formula of Art.
159,
the
angle
must be
expressed
in
radians).
Ans. 7
=
.00582slug-ft.
2
416. The
disc
of a torsional
pendulum
is
1
ft. in
diameter and
2
in. thick.
A
torque
of
1
ft.-lb. is
required
to turn the disc
through
8.
Find the
period
of
oscillation,
assuming
that the disc is made
of
cast
iron,
the
weight
of which
is
450 Ib.
per
cubic foot.
160.
Experimental
Determination
of
Moment of Inertia.
As
noted
in
Art.
106,
the calculation of the
moment
of inertia
of
many
bodies is a difficult
process
and
frequently
is
impossible.
The moment of
inertia, however,
may
be found
experimentally
by
allowing
the
body
to oscillate as a
compound
pendulum
and
observing
the
period
of
oscillation or
by
allowing
it
to oscillate
either as a
torsion
pendulum
itself or
with a
given
torsion
pendulum
to
which it
is attached.
/.
By
Means
of
a
Compound
Pendulum.
The
period
of
a
complete
oscillation of a
compound
pendulum,
from
Art.
157,
is
P
=
27TA/^.
gr
Or,
P
2
_
But,
g
Hence,
_MP
2
_
_
WP
2
r
=
~4^~
gr
~~
~b?~'
in
which
Jo
is the moment of
inertia
of
the
body
with
respect
to the
axis of
suspension
Now
W
may
be
found
by
weighing
the
body,
r
by
balancing
the
body
over
a
knife
edge,
and
P
by
observing
the
period
jDfjDScillation
.
368
FORCE,
MASS,
AND
ACCELERATION
After
7o
is
found,
the moment
of
inertia
with
respect
to a
parallel
axis
through
the mass-center
(7)
may
be
calculated from
the
equation,
II.
By
Means
of
the Torsional Pendulum.
In
order to
deter-
mine the
moment of inertia of a
body by
means
of a
torsional
pendulum,
it will
be convenient to
replace
the disc
of the
pendulum
as described
in Art. 159
by
two discs
rigidly
connected as
shown
in
Fig.
382.
If
/o
denotes
the
moment of
inertia of
the
two discs
with
respect
to the
axis of
the
wire,
the time
of a
complete
oscilla-
tion of the
pendulum
(denoted
here for
convenience
by
PO)
is, by
Art.
159,
.....
(1)
in which
C is a
constant. The
value
of
^
is a
constant of the
wire,
and
depends
on
the material of
the wire
and
its diameter.
If
this value is
known
(and
it is
readily
found),
/o
may
be
obtained
from
the
above
equation
by
observing
the
period
of
oscillation of
the
pendulum.
The
method
of
determining
the moment of
inertia of
a
body
A
is as follows: First
allow the
pendulum
to
oscillate
and observe
the
period
of
oscillation,
PO-
Next,
place
the
body
A
on the
lower
disc
(Fig.
382)
so that the center of
gravity
of
A
is
vertically
below
the
wire.
If
the moment
of
inertia
of A
with
respect
to the
axis
of the
wire
is denoted
by
I
A
and
the
period
of oscil-
lation
of the
pendulum
(and A)
is
denoted
by
P,
we
have,
by
Art.
159,
FIG.
382.
...
(2)
From
(1)
and
(2)
the
following
equation
is
obtained:
P
VT
NEED
FOR
BALANCING
363
That
is,
Hence,
It
may
be
noted that
if the discs
are
of such
form that their
moment
of inertia
may
be
calculated it
is not
necessary
to
know
the
con-
P
2
T
_
T I
L
1
A -tOl ^m
stant
;~,
of
the
wire.;
PROBLEMS
417.
The
connecting
rod
of
a steam
engine
weighs
300
Ib. and
the distance
of
the center
of
gravity
from the
crank-pin
is found
(by
balancing)
to be 50 in. When
suspended
from the
crank-pin
end and
allowed
to
vibrate
as a
compound
pendulum,
it is
found
to make
thirty
complete
oscillations
in
75 seconds.
Determine
the moment of inertia of the rod with
respect
to the
axis of the
crank-pin
and also
with
respect
to a
paral-
lel axis
through
the
center
of
gravity.
Arts.
/
=
198
slug-ft.
2
7
=
36.5
slug-ft.
2
418. The
pendulum
(Fig.
383)
of a
Charpy
impact
machine,
which
is
used to
determine the resistance
of materials
to
impact,
weighs
50.5 Ib.
and
the
distance
of the
center
of
gravity
from the
axis
of
rotation,
as
determined
by
balancing,
is found to be
27.33
in. When allowed
to
vibrate
about the axis of
rotation,
the
pendulum
is observed
to make
FIG.
383.
35
complete
oscillations
in 61 sec.
Find
the moment
of
inertia
of the
pendulum
with
respect
to
the axis
of
rotation.
BALANCING
161.
Need for
Balancing.
A
moving
part
of a
machine,
as a
rule,
has either
a
reciprocating
motion
similar to
that of
the
cross-
head
of a steam
engine
or a motion of
rotation
such
as that
of
the
crank-shaft
of an automobile
engine
or the
motor
of
a
dynamo.
In
any
case,
if the
moving
parts
are
accelerated,
forces
must be
supplied
to
produce
the accelerations. If
the
moving
parts
are
not
balanced,
the
forces which act on
the
moving
masses
are
transmitted
to them
from the
stationary
parts
of
the
machine such
as the
bearings
and
the machine
frame.
And,
in
supplying
these
accelerating
forces,
serious trouble
may
arise
such
as,
vibrations
in
automobiles,
turbines,
etc.;
defective
commutations in elec-
370
FORCE,
MASS,
AND
ACCELERATION
trical
machinery;
heavy
bearing
pressures
which
cause
undue
wear;
defective work with
grinding
discs,
high-speed
drilling
machines,
etc.;
and defective lubrication.
It is of
great impor-
tance,
therefore,
to
properly
neutralize or balance these
forces
in
various
types
of
machines.
The
moving parts
of
a machine
may
be
(1)
in
static or
standing
balance or
(2)
in
dynamic
or
running
balance.
Standing
balance
exists
if the
forces which act on the
parts,
when the
parts
are not
running,
are
in
equilibrium
regardless
of the
positions
in
which
the
parts
are
placed.
Dynamic balancing
consists in
distributing
the
moving
masses or
in
introducing
additional masses so
that
the
inertia
forces
exerted
by
the masses of the
moving system
are
in
equilibrium amongst
themselves,
thereby making
it
unnecessary
for the
stationary parts
of the machine
to
supply
any
of the
accelerating
forces.
The
complete
balancing
of a
machine,
how-
ever,
is
not
always practicable
or
possible.
The method of
balancing rotating masses,
only,
is here dis-
cussed. For methods of
balancing reciprocating
masses
and for
an
excellent discussion of
the whole
subject
of the
balancing
of
engines
see
Dalby's
"
Balancing
of
Engines."
162.
Balancing
of
Rotating
Masses.
The inertia
forces or
kinetic loads due
to
unbalanced
rotating
masses
may
be
treated,
in
general,
(1)
as
a
system
of
centrifugal
forces
or
(2)
as a
system
of
centrifugal couples
or
(3)
as a combination of the two.
(a)
FIG.
384.
A
Single Rotating
Mass.
Centrifugal
Force.
If a shaft
(Fig.
384a)
rotates at
an
angular velocity
a>
and carries a
single
mass
MI
the
center
of
gravity
of which is at the distance
r\
from
the
axis
of
rotation,
the
shaft
will be
subjected
to a
kinetic
load
(centrifugal
force)
of
magnitude
M
\riu>
2
.
This
kinetic
load
causes the shaft