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(a) We use energy conservation in the form of Eq. 817.
KU KU
mgL
mv
mgL
11
2 2
12
2
2
01
1
2
1
+=+
+−=+−
cos
cos
θθ
bg
With
L
= 1.4 m,
θ
1
= 30°, and
2
= 20°, we have
vg
L
22
1
21
4
=−
=
cos
cos
.
b
g
ms
.
(b) The maximum speed
v
3
is at the lowest point. Our formula for
h
gives
h
3
= 0 when
3
= 0°, as expected. From
mgL
mv
33
13
2
1
2
0
+−=+
cos
we obtain
v
3
19
=
.m
s
.
(c) We look for an angle
4
such that the speed there is
vv
43
3
=
. To be as accurate as
possible, we proceed algebraically (substituting
L
3
2
1
cos
at the appropriate
place) and plug numbers in at the end. Energy conservation leads to
mgL
mv
mgL
mgL
m
v
mgL
gL
gL
gL
4 4
14
2
4
1
3
2
4
1
1
4
1
2
1
1
1
29
1
1
2
9
+−
=+−
−=+
−
−=
−
−
cos
cos
cos
cos
cos
cos
cos
where in the last step we have subtracted out
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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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