This preview shows page 1. Sign up to view the full content.
Now, for this geometry,
v
=
ω
R
and
ρ
=
R
. Thus, we must solve the following equations.
N
cos
θ
−
μ
s
N
sin
θ
=
mg
N
sin
θ
+
μ
s
N
cos
θ
=
m
ω
2
R
Multiplying the first equation by
cos
θ
and the second by
sin
θ
, the sum of the resulting equations is
N
D
cos
2
θ
+sin
2
θ
i
=
mg
cos
θ
+
m
ω
2
R
sin
θ
⇒
N
=
mg
cos
θ
+
m
ω
2
R
sin
θ
Using this value of
N
in the verticalforcebalance equation, we find
D
mg
cos
θ
+
m
ω
2
R
sin
θ
i
(cos
θ
−
μ
s
sin
θ
)=
mg
or, dividing through by
m
and regrouping terms,
g
cos
θ
(cos
θ
−
μ
s
sin
θ
)+
ω
2
R
sin
θ
(cos
θ
−
μ
s
sin
θ
g
Consequently,
ω
2
R
sin
θ
(cos
θ
−
μ
s
sin
θ
g
[1
−
cos
θ
(cos
θ
−
μ
s
sin
θ
)]
=
g
(1
−
cos
2
θ
+
μ
s
sin
θ
cos
θ
)
=
g
(sin
2
θ
+
μ
s
sin
θ
cos
θ
)
=
g
sin
θ
(sin
θ
+
μ
s
cos
θ
)
Therefore, the maximum rotation rate is
ω
max
=
±
g
R
sin
θ
+
μ
s
cos
θ
cos
θ
−
μ
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 05/12/2010 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.
 Spring '06
 Shiflett

Click to edit the document details