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22
31
3
0
42
2
dE
d
d
Mv
kx
Mv a
kxv
dt
dt
dt
§·
§
·
=+
=
+
=
¨¸
¨
¸
©¹
©
¹
cm
cm
cm
cm
which leads to
=
2
3
.
a
k
M
x
cm
−
F
H
G
I
K
J
Comparing with Eq. 15-8, we see that
ω
=
23
kM
/
for this system. Since
= 2
π
/
T
, we
obtain the desired result:
TM
k
=
2
π
/.
106. (a) The potential energy at the turning point is equal (in the absence of friction) to
the total kinetic energy (translational plus rotational) as it passes through the equilibrium
position:
2
2
2
cm
cm
cm
cm
222
cm
cm
cm
11
1
1
1
1
2
2
2
2
113
244
m
v
kx
Mv
I
Mv
MR
R
Mv
Mv
Mv
=+=
which leads to
Mv
kx
m
cm
=
/
= 0.125 J. The translational kinetic energy is therefore

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