Chapter 15
459
P15.56
The kinetic energy of the ball is
Km
v
I
=+
1
2
1
2
22
Ω
,
where
Ω
is the rotation rate of the ball about its
center of mass. Since the center of the ball moves
along a circle of radius 4
R
, its displacement from
equilibrium is
sR
=
4
af
θ
and its speed is
v
ds
dt
R
d
dt
==
F
H
G
I
K
J
4
. Also, since the ball rolls without
slipping,
v
ds
dt
R
Ω
so
Ω=
=
F
H
G
I
K
J
v
R
d
dt
4
The kinetic energy is then
R
d
dt
mR
d
dt
mR
d
dt
=
F
H
G
I
K
J
+
F
H
G
I
K
J
F
H
G
I
K
J
=
F
H
G
I
K
J
1
2
4
1
2
2
5
4
112
10
2
2
2
2
2
θθ
h
5
R
R
s
FIG. P15.56
When the ball has an angular displacement
, its center is distance
hR
=−
41c
o
s
higher than
when at the equilibrium position. Thus, the potential energy is
Um
g
hm
g
R
g
−
41
c
o
s
. For small
angles,
1
2
2
−≈
cos
(see Appendix B). Hence,
g
R
g
≈
2
2
, and the total energy is
EKU
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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, Center Of Mass, Mass

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