
B–65142E/04
4. SELECTING A MOTORDESCRIPTIONS FOR THE α series
31
The inertia of a cylindrical object rotating about its central axis is
calculated as follows:
J
b
+
pg
b
32 980
D
b
4
L
b
J
b
: Inertia (kgf⋅cm⋅s
2
)
γ
b
: Weight of the object per unit volume (kg/cm
3
)
D
b
: Diameter of the object (cm)
L
b
: Length of the object (cm)
(kgf⋅cm⋅s
2
)
SI unit
J
b
+
pg
b
32
D
b
4
L
b
J
b
: Inertia (kgf⋅m
2
)
γ
b
: Weight of the object per unit volume (kg/cm
3
)
D
b
: Diameter of the object (m)
L
b
: Length of the object (m)
(kgf⋅m
2
)
Example)
When the shaft of a ball screw is made of steel (g = 7.8 10
3
(kg/m
3
)),
inertia J
b
of the shaft is calculated as follows:
J
b
=7.8 10
3
πB32 0.032
4
1=0.0008 [kg⋅m
2
]
(=0.0082 [kg⋅cm⋅s
2
])
J +
W
980
(
L
2p
)
2
W : Weight of the object moving along a straight line (kg)
L : Traveling distance along a straight line per revolution
of the motor (cm)
(kgf⋅cm⋅s
2
)
J + W (
L
2p
)
2
W : Weight of the object moving along a straight line (kg)
L : Traveling distance along a straight line per revolution
of the motor (cm)
(kgf⋅cm⋅s
2
)
SI unit
Example)
When W is 1000(kg) and L is 8(mm), Jw of a table and workpiece is
calculated as follows:
Jw = 1000 (0.008B2Bπ)
2
= 0.00162 (kg⋅m
2
) = 0.0165 (kgf⋅cm⋅s
2
)
D Inertia of a heavy object
moving along a straight
line (table, workpiece,
etc.)