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318
Rotation of a Rigid Object About a Fixed Axis
P10.82
Conservation of energy between apex and the point where
the grape leaves the surface:
mg y
mv
I
mgR
mv
mR
v
R
ff
f
f
∆=
+
−=
+
F
H
G
I
K
J
F
H
G
I
K
J
1
2
1
2
1
1
2
1
2
2
5
22
2
ω
θ
cos
af
which gives
g
v
R
f
1
7
10
2
F
H
G
I
K
J
cos
a
f
(1)
Consider the radial forces acting on the grape:
mg
n
mv
R
f
cos
2
.
At the point where the grape leaves the surface,
n
→
0.
Thus,
mg
mv
R
f
cos
=
2
or
v
R
g
f
2
=c
o
s
.
Substituting this into Equation (1) gives
gg
g
cos
cos
θθ
7
10
or
cos
=
10
17
and
=°
54 0
..
R
i
∆
y = R—R
cos
f
n
mg
sin
mg
cos
FIG. P10.82
P10.83
(a)
There are not any horizontal forces acting on the rod, so the center of mass will not move
horizontally. Rather, the center of mass drops straight downward (distance
h
/2) with the rod
rotating about the center of mass as it falls. From conservation of energy:
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .
 Fall '11
 Staff
 Physics, Conservation Of Energy, Energy

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