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()
2
2
2
3 0.040 kg m
3
2.7 kg.
2
20
.15
m
I
M
R
⋅
==
=
It also follows from the rotational inertia expression that
1
2
2
1
3
22
IM
R
ωω
=
. Furthermore,
it rolls without slipping,
v
com
=
R
ω
, and we find
K
KK
MR
mR
MR
rot
com
rot
+
=
+
1
3
1
2
1
3
.
(a) Simplifying the above ratio, we find
K
rot
/
K
= 0.4. Thus, 40% of the kinetic energy is
rotational, or
K
rot
= (0.4)(20 J) = 8.0
J.
(b) From
1
rot
3
8.0J
KM
R
(and using the above result for
M
) we find
1
015
380
27
20
.
.
.
m
J
kg
rad s
b
g
which leads to
v
com
= (0.15 m)(20 rad/s) = 3.0 m/s.
(c) We note that the inclined distance of 1.0 m corresponds to a height
h
= 1.0 sin 30° =
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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, Inertia

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