Omron R6YXG500, Installation Manual

Page: 104 / 330

2
Installa
tion
2-46
In the same manner, the moment of inertia of a cylinder as shown in the Fig. below is given by
Moment of inertia of a cylinder
h
D
x
Center line
W
4g
=
... (6)
D
2
4
h
2
3
( + )
J=
rpD
2
h
16g
+
D
2
4
h
2
3
(
+
)
rpD
2
h x
2
4g
+
Wx
2
g
I=
rpD
2
h
16
+
D
2
4
h
2
3
( + )
rpD
2
h x
2
4
=
(kgfcmsec 2 )
m
4
D
2
4
h
2
3
( + ) + mx 2
(kgm
2
)
In the same manner, the moment of inertia of a prism as shown in the Fig. below is given by
Moment of inertia of a prism
a
c
b
x
Center line
J=
rabc(a
2
+ b
2
)
12g
W(a
2
+ b
2
)
12g
=
... (7)
+
rabcx
2
g
Wx 2
g
+
I=
rabc(a
2
+ b
2
)
12
+ rabcx
2
=
m(a
2
+b
2
)
12
m : Mass of prism (kg)
W : Weight of prism (kgf)
(kgfcmsec
2
)
+ mx
2
(kgm
2
)