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Chapter 3
Kinetics of Particles
Question 3–1
A particle of mass
m
moves in the vertical plane along a track in the form of a
circle as shown in Fig. P31. The equation for the track is
r
=
r
0
cos
θ
Knowing that gravity acts downward and assuming the initial conditions
θ(t
=
0
)
=
0and
˙
=
0
)
=
˙
θ
0
, determine (a) the di±erential equation of motion for
the particle and (b) the force exerted by the track on the particle as a function
of
θ
.
r
=
0
cos
θ
g
m
O
θ
Figure P31
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Chapter 3. Kinetics of Particles
Solution to Question 3–1
Kinematics
Let
F
be a reference frame Fxed to the track. Then, choose the following coor
dinate system Fxed in reference frame
F
:
Origin at point
O
E
x
=
along
OP
at
t
=
0
E
z
=
Out of Page
E
y
=
E
z
×
E
x
Next, let
A
be a reference frame Fxed to the direction
. Then, choose the
following coordinate system Fxed in reference frame
A
:
Origin at point
O
e
r
=
along
E
z
=
out of page
e
θ
=
E
z
×
e
r
The geometry of the bases
{
E
x
,
E
y
,
E
z
}
and
{
e
r
,
e
θ
,
e
z
}
is shown in ±ig. 31. Using
±ig. 31, we have that
E
x
=
cos
θ
e
r

sin
θ
e
θ
(3.1)
E
y
=
sin
θ
e
r
+
cos
θ
e
θ
(3.2)
e
r
e
θ
E
x
E
y
θ
θ
Figure 31
Geometry of Coordinate System for Question 3.1
Next, the position of the particle is given in terms of the basis
{
e
r
,
e
θ
,
e
z
}
as
r
=
r
e
r
=
r
0
cos
θ
e
r
(3.3)
±urthermore, since the angle
θ
is measured from the Fxed horizontal direction,
the angular velocity of
A
in
F
is given as
F
ω
A
=
˙
θ
E
z
(3.4)
63
Applying the rate of change transport theorem to
r
between reference frames
A
and
F
, we obtain the velocity of the particle in reference frame
F
as
F
v
=
F
d
r
dt
=
A
d
r
dt
+
F
ω
A
×
r
(3.5)
Now we have
A
d
r
dt
=
r
0
˙
θ
sin
θ
e
r
(3.6)
F
ω
A
×
r
=
˙
θ
E
z
×
r
0
cos
θ
e
r
=
r
0
˙
θ
cos
θ
e
θ
(3.7)
Adding the expressions in Eq. (3.6) and Eq. (3.7), we obtain the velocity in refer
ence frame
F
as
F
v
r
0
˙
θ
sin
θ
e
r
+
r
0
˙
θ
cos
θ
e
θ
(3.8)
Rewriting Eq. (3.8), we obtain
F
v
=
r
0
˙
θ(

sin
θ
e
r
+
cos
θ
e
θ
)
(3.9)
The speed in reference frame
F
is then given as
F
v
=
r
0
˙
θ
(3.10)
Dividing
F
v
by
F
v
, we obtain the tangent vector as
e
t
sin
θ
e
r
+
cos
θ
e
θ
(3.11)
Next, from the Frst of the Serret±renet formulae, we have that
F
d
e
t
dt
=
κ
F
v
e
n
(3.12)
Using the rate of change transport theorem, we have that
F
d
e
t
dt
=
A
d
e
t
dt
+
F
ω
A
×
e
t
(3.13)
where
A
d
e
t
dt
˙
θ
cos
θ
e
r

˙
θ
sin
θ
e
θ
(3.14)
F
ω
A
×
e
t
=
˙
θ
E
z
×
(

sin
θ
e
r
+
cos
θ
e
θ
)
˙
θ
cos
θ
e
r

˙
θ
sin
θ
e
θ
(3.15)
Adding the two expressions in Eq. (3.15) and Eq. (3.15), we have that
F
d
e
t
dt
2
˙
θ
cos
θ
e
r

2
˙
θ
sin
θ
e
θ
(3.16)
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Chapter 3. Kinetics of Particles
Then the principle unit normal in reference frame
F
is given as
e
n
=
F
d
e
t
/dt
±
F
d
e
t
/dt
±
=
cos
θ
e
r

sin
θ
e
θ
(3.17)
Also, the curvature is given as
κ
=
1
F
v
F
d
e
t
/dt
F
v
=
2
˙
θ
r
0
˙
θ
=
2
r
0
(3.18)
Finally, di±erentiating Eq. (3.10), we obtain
d
dt
±
F
v
²
=
r
0
¨
θ
(3.19)
The acceleration of the particle in reference frame
F
is then given as
F
a
=
d
dt
±
F
v
²
+
κ
±
F
v
²
2
e
n
=
r
0
¨
θ
e
t
+
2
r
0
r
2
0
˙
θ
2
e
n
(3.20)
which simpli²es to
F
a
=
r
0
¨
θ
e
t
+
2
r
0
˙
θ
2
e
n
(3.21)
Kinetics
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This note was uploaded on 09/30/2011 for the course EGM 3401 taught by Professor Carlcrane during the Summer '08 term at University of Florida.
 Summer '08
 CARLCRANE
 Dynamics

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