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Kinetics
Newton’s Second Law written in abstract vector form for a particle is
X
~
F
=
m~a
in which
m
is a scalar quantity called mass. Historically, mass is called
inertia
or
resistance to
motion
.
Euler later uniﬁed the theory for both solid and ﬂuid by writing “Euler’s First Law” as
X
~
F
=
d
dt
~
L
=
d
dt
(
m~v
)
in which
~
L
=
is the linear momentum of a particle or a continuous medium.
Since the above laws are written in abstract vector form, they are applicable in any coordinate
systems.
Examples
Rectlinear Motion (only one component):
X
F
=
ma
Cartesian Coordinates:
X
F
x
F
y
F
z
=
m
a
x
a
y
a
z
=
m
¨
x
¨
y
¨
z
Cylindrical Coordinates:
X
F
r
F
θ
F
z
=
m
a
r
a
θ
a
z
=
m
¨
r

r
˙
θ
2
r
¨
θ
+2˙
r
˙
θ
¨
z
Planar
nt
Coordinates:
X
F
n
F
t
F
z
=
m
a
n
a
t
a
z
=
m
v
2
/ρ
˙
v
¨
z
–3/1–
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View Full Document Example – Cylindrical Coordinates
A sphere of mass
m
is initially attached to the wall at angle
θ
by a cord. If the cord is cut, the sphere
will be allowed to swing as a pendulum. Find as a function of
θ
the ratio:
T
2
T
1
=
Tension after cut of cord
Tension before cut of cord
.
T
θ
cord
m
ˆ
e
r
ˆ
e
θ
f
r
θ
mg
T
1
FBD: before cut
T
2
FBD: after cut
ˆ
e
θ
ˆ
e
r
θ
mg
θ
Since pendulum problems are best described by polar coordinates, use
X
±
F
r
F
θ
²
=
m
±
¨
r

r
˙
θ
2
r
¨
θ
+2˙
r
˙
θ
²
Before Cut:
To obtain the sum of the forces, it is convenient to write the forces in different coordinate systems,
for example,
X
~
F
=
±
0

mg
²
xy
+
±

T
1
0
²
rθ
+
±
f
r
0
²
xy
=
±
f
r

mg
²
xy
+
±

T
1
0
²
.
To complete the sum, use the transformation matrix deﬁned as
[
Q
(
←
xy
)]
=
³
cos(
π
2

θ
)
cos(
π

θ
)
cos
θ
cos(
π
2

θ
)
´
=
³
sin
θ

cos
θ
cos
θ
sin
θ
´
.
x
y
ˆ
e
r
ˆ
e
θ
θ
θ
Therefore,
X
±
F
r
F
θ
²
=
³
sin
θ

cos
θ
cos
θ
sin
θ
´±
f
r

mg
²
xy
+
±

T
1
0
²
=
±
mg
cos
θ
+
f
r
sin
θ

T
1

mg
sin
θ
+
f
r
cos
θ
²
=
m
±
0
0
²
–3/2–
The acceleration,
~a
, is set to zero because this is a static problem. Now, by equating the orthogonal
components, the solution can be obtained as
±
f
r
=
mg
sin
θ/
cos
θ
=
mg
tan
θθ
component
T
1
=
mg
cos
θ
+
f
r
sin
θ
=
mg
cos
θ
+
mg
sin
2
cos
θr
component
Simplifying further yields
T
1
=
mg
(cos
2
θ
+ sin
2
θ
)
cos
θ
=
⇒
T
1
=
mg
cos
θ
After Cut:
At the instant the cord is cut, the acceleration is nonzero but the velocity is still zero
momentarily. For this reason, we can state that
˙
θ
=0
. Also, since the string is unstretchable,
it implies that
˙
r
=¨
r
. The equation of motion is therefore simpliﬁed to
X
±
F
r
F
θ
²
=
±
0

mg
²
xy
+
±

T
2
0
²
rθ
=
m
±
0

r
(0)
2
r
¨
θ
+ 2(0)(0)
²
=
m
±
0
r
¨
θ
²
The summation of forces, using the transformation matrix, yields
X
~
F
=
³
sin
θ

cos
θ
cos
θ
sin
θ
´±
0

mg
²
xy
+
±

T
2
0
²
=
±
mg
cos
θ

T
2

mg
sin
θ
²
.
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This note was uploaded on 04/16/2008 for the course CE 325 taught by Professor Docwong during the Spring '08 term at USC.
 Spring '08
 DocWong

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