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(b) We use
τ
=
I
α
, where
τ
is the torque and
I
is the rotational inertia. The contribution of
the rod to
I
is
M
A
2
12
/
(Table 102(e)), where
M
is its mass and
A
is its length. The
contribution of each ball is
m
A
/,
2
2
b
g
where
m
is the mass of a ball. The total rotational
inertia is
I
Mm
=+
=
+
AA
22
12
2
4
640
120
12
106
2
..
kg
m
kg
m
bg
b
g
b
g
which yields
I
= 1.53 kg
⋅
m
2
. The torque, therefore, is
τ
=⋅
−
=
−
⋅
153
7 66
117
.
kg m
rad / s
N m.
ch
c
h
(c) Since the system comes to rest the mechanical energy that is converted to thermal
energy is simply the initial kinetic energy
KI
i
==
⋅
=
×
1
2
1
2
2
39
459 10
0
2
2
4
ω
kg m
rad / s
J.
2
c
h
c
h
π
(d) We apply Eq. 1013:
θω
=
+
−
0
2
2
1
2
23
9
3
2
0
1
2
766
320
tt
π
b
gb
g
c
hb
g
c
hb
g
rad / s
s
rad / s
s
2
.
which yields 3920 rad or (dividing by 2
π
) 624 rev for the value of angular displacement
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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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