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()
(
)
2
com,0
2
com,0
0
2
2
52
2
11
22
1
1
1.3 m/s
0.12 kg 1.3 m/s
9.5 10
kg m
0.12 kg 9.8 m/s
1.2 m
2
2
0.0032 m
9.4 J.
fi
i
v
KK
U
m
v
I
M
g
h
R
−
§·
=+=
+
+
¨¸
©¹
=+
×
⋅
+
=
(c) As it reaches the end of the string, its center of mass velocity is given by Eq. 211:
vv
a
t
v
gt
IM
R
com
com
com
com
com
=+=−
+
,,
.
00
0
2
1
78. (a) The acceleration is given by Eq. 1113:
a
g
R
com
com
=
+
1
0
2
where upward is the positive translational direction. Taking the coordinate origin at the
initial position, Eq. 215 leads to
yv
t
a
t
v
t
gt
R
com
com,0
com
com,0
com
=−
+
1
21
2
1
2
2
0
2
where
y
com
= – 1.2 m and
v
com,0
= – 1.3 m/s. Substituting
I
com
kg m
=⋅
0 000095
2
.,
M
=
0.12 kg,
R
0
= 0.0032 m and
g
= 9.8 m/s
2
, we use the quadratic formula and find
( )
(
)
( )
(
)
com
2
0
2
com
2
com
0
2
2
com,0
com,0
29
.8 1
.2
2
1 0.000095 0.12 0.0032
2
1
0.000095
0.12 0.0032
1
1
1.3
(1.3)
9.8
21.7 or 0.885
I
MR
gy
R
t
g
−
+
+
+−
=
−
=
B
B
where we choose
t
= 0.89 s as the answer.
(b) We note that the initial potential energy is
U
i
= Mgh
and
h
= 1.2 m (using the bottom
as the reference level for computing
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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, Acceleration

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