S E C T I O N
14.5
Motion in ThreeSpace
(ET Section 13.5)
565
Substituting (1), (3) and (6) into (5) gives:
a
N
N
=
2
,
0
−
4
t
4
(
t
2
+
4
)
2
t
,
4
=
2
,
0
−
t
t
2
+
4
2
t
,
4
=
2
−
2
t
2
t
2
+
4
,
−
4
t
t
2
+
4
=
8
t
2
+
4
,
−
4
t
t
2
+
4
=
1
t
2
+
4
8
,
−
4
t
(7)
Since
a
N
=
a
×
v
v
,
a
N
is positive, hence
N
is a unit vector in the direction of
a
N
N
. Hence, by (7):
N
=
a
N
N
a
N
N
=
1
t
2
+
4
8
,
−
4
t
1
t
2
+
4
8
2
+
(
−
4
t
)
2
=
8
,
−
4
t
64
+
16
t
2
=
4 2
,
−
t
4
4
+
t
2
=
1
4
+
t
2
2
,
−
t
(8)
Combining (7) and (8) yields:
a
N
N
=
4
t
2
+
4
2
,
−
t
=
4
t
2
+
4
N
⇒
a
N
=
4
t
2
+
4
(9)
(c)
By (4) and (9) we obtain the following decomposition:
a
(
t
)
=
a
T
(
t
)
T
+
a
N
(
t
)
N
=
2
t
t
2
+
4
T
+
4
t
2
+
4
N
43.
Find the components
a
T
and
a
N
of the acceleration vector of a particle moving along a circular path of radius
R
=
100 cm with constant velocity
v
0
=
5 cm
/
s.
SOLUTION
Since the particle moves with constant speed, we have
v (
t
)
=
0, hence:
a
T
=
v (
t
)
=
0
The normal component of the acceleration is
a
N
=
κ
(
t
)v(
t
)
2
. The curvature of a circular path of radius
R
=
100 is
κ
(
t
)
=
1
R
=
1
100
, and the velocity is the constant value
v(
t
)
=
v
0
=
5. Hence,
a
N
=
1
R
v
2
0
=
25
100
=
0
.
25 cm
/
s
2
44.
At time
t
0
, a moving particle has velocity vector
v
=
2
i
and acceleration vector
a
=
3
i
+
18
k
. Determine the
curvature
κ
(
t
0
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 Fall '09
 Park
 Multivariable Calculus, Acceleration, Velocity, acceleration vector, circular path

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