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From last class:
v
=˙
r
ˆ
r
+(
r
˙
θ
)
ˆ
θ.
(42)
By diferentiating this with respect to
t
, we can also show that
a
= (¨
r

r
˙
θ
2
)ˆ
r
r
¨
θ
+ 2 ˙
r
˙
θ
)
ˆ
θ
(43)
Circular motion
:
Consider the general case o± a particle moving around a circle with center 0 and radius
b
, not necessarily with constant speed. Since
r
=
b
, then ˙
r
=¨
r
= 0, and the velocity o± a
particle is
v
=(
b
˙
θ
)
ˆ
θ
(44)
The ±ormula ±or the acceleration becomes
a
=

b
˙
θ
2
ˆ
r
+
b
¨
θ
ˆ
θ
(45)
also remember that the circum±erential velocity
v
=
b
˙
θ
.
example:
The bob o± a pendulum moves on a vertical circle o± radius
b
, and when the
string makes an angle
θ
with the downward vertical, the velocity o± the bob is
v
2
=2
gb
cos
θ
(46)
where
g
is a positive constant. Find the acceleration o± the bob when the string is at an
angle
θ
.
solution:
From the acceleration ±ormula:
a
=

v
2
b
ˆ
r
+˙
v
ˆ
θ
=

(2
g
cos
θ
r
v
ˆ
θ
(47)
we must diferentiate to get ˙
v
:
d
dt
v
2
v
˙
v
=

(2
sin
θ
)
˙
θ
(48)
and since
v
=
b
˙
θ
,
˙
v
=

g
sin
θ
(49)
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 Spring '10
 RogerMelko

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