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Tm
gm
v
r
Tmg
gL
L
mg
−=
¡
=+
F
H
G
I
K
J
=
2
4
5
.
With
m
= 0.092 kg, the tension is given by
T
= 4.5 N.
(c) The pendulum is now started (with zero speed) at
90
i
θ
=°
(that is,
h
i
=
L
), and we
look for an angle
such that
T
=
mg
. When the ball is moving through a point at angle
,
then Newton's second law applied to the axis along the rod yields
g
m
v
r
cos
2
which (since
r
=
L
) implies
v
2
=
gL
(1 – cos
θ
) at the position we are looking for. Energy
conservation leads to
(1
KU KU
mgL
mv
mgL
gL
gL
gL
ii
+=
+
+
−
=−
+
−
0
1
2
1
1
2
1
2
(c
o
s
)
((c
o
s
)
)
c
o
s
)
θθ
where we have divided by mass in the last step. Simplifying, we obtain
1
1
cos
71
3
−
§·
==
°
¨¸
©¹
.
(d) Since the angle found in (c) is independent of the mass, the result remains the same if
the mass of the ball is changed.
109. The connection between angle
(measured from vertical) and height
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Energy, Potential Energy

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