Im
r
m
L
m
L
ii
==
+
=⋅
¦
22
2
di
0.20 kg m
2
.
(b) One imagines rotating the figure (about point
A
) clockwise by 90° and noting that the
center of mass has fallen a distance equal to
L
as a result. If we let our reference position
for gravitational potential be the height of the center of mass at the instant
AB
swings
through vertical orientation, then
()
00
0
04
0
.
KU KU
m
g
hK
+=
+
¡
+
Since
h
0
=
L
= 0.50 m, we find
K
= 3.9 J. Then, using Eq. 1034, we obtain
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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

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