which leads to
2.7
0.7
2.7 .
hRrR
=−
≈
With
R
= 14.0 cm , we have
h
= (2.7)(14.0 cm) =
37.8 cm.
(b) The energy considerations shown above (now with
h
= 6
R
) can be applied to point
Q
(which, however, is only at a height of
R
) yielding the condition
gR
v
g
R
6
7
10
bg
=+
com
2
which gives us
vg
R
com
2
=
50
7. Recalling previous remarks about the radial acceleration,
Newton’s second law applied to the horizontal axis at
Q
leads to
()
2
com
50
7
v
gR
Nm
m
R
rR
r
==
−−
which (for
R
r
>>
) gives
42
2
50
50(2.80 10 kg)(9.80 m/s )
1.96 10 N.
77
mg
N
−
−
×
≈=
=
×
(b) The direction is toward the center of the loop.
8. Using the floor as the reference position for computing potential energy, mechanical
energy conservation leads to
22
release
top
top
com
11
2.
UK
U
m
g
h
m
vI
m
g
R
ω
¡
+
Substituting
Im
r
=
2
5
2
(Table 102(f)) and
=
vr
com
(Eq. 112), we obtain
2
2
com
com
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 Spring '08
 Any
 Physics, Energy, Potential Energy

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