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10.22:
a) The acceleration down the slope is
,
sin
M
f
θ
g
a

=
the torque about the
center of the shell is
,
3
2
3
2
2
MRa
R
a
MR
R
a
I
Iα
Rf
τ
=
=
=
=
=
.
so
3
2
a
M
f
=
Solving these relations
a
for
f
and simultaneously gives
or
,
sin
3
5
θ
g
a
=
.
N
83
.
4
)
s
m
kg)(3.62
00
.
2
(
3
2
3
2
,
s
m
62
.
3
0
.
38
sin
)
s
m
80
.
9
(
5
3
sin
5
3
2
2
2
=
=
=
=
°
=
=
Ma
f
θ
g
a
The normal force is
Mg
cos
θ
, and since
,
s
n
μ
f
≤
.
313
.
0
tan
5
2
cos
sin
3
2
cos
3
2
cos
5
3
3
2
s
=
=
=
=
=
≥
θ
θ
g
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