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()
2
11
1
22
2
2(0.024 m)(1.30 rad)
(6.20 kg) 9.80 m/s
(0.0910 s)
14.0 N.
R
Tmga Mg
t
θ
§·
=−
=
−
¨¸
©¹
=
(d) From the last of the above equations, we obtain the second tension:
42
2
21
2
2
(7.40 10 kg m )(314 rad/s )
14.0 N
0.024 m
4.36 N.
IR
I
TT
M
g
Rt
R
t
αθ
−
×⋅
=− =
−
−
=
−
=
69. We choose positive coordinate directions (different choices for each item) so that
each is accelerating positively, which will allow us to set
a
2
=
a
1
=
R
α
(for simplicity, we
denote this as
a
). Thus, we choose rightward positive for
m
2
=
M
(the block on the table),
downward positive for
m
1
=
M
(the block at the end of the string) and (somewhat
unconventionally) clockwise for positive sense of disk rotation. This means that we
interpret
given in the problem as a positivevalued quantity. Applying Newton’s second
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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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