Biomedical Engineering Reference
In-Depth Information
into
N
slices of mass
m
i
displaced from the end by a distance
x
i
. The total mass,
M
(kg),
is the sum of the masses of the individual segments
N
M
=
m
i
(10.3)
i
=
1
with the center of mass at a distance
x
(m) from the end of the segment.
N
1
M
x
=
m
i
x
i
(10.4)
i
=
1
m
2
), about the endpoint is
The moment of inertia,
I
(kg
·
N
m
i
x
i
I
=
(10.5)
i
=
1
The moment of inertia is generally specified around the center of mass of the segment
(where it is at its lowest), and the parallel axis theorem is used to determine its value there:
Mx
2
I
0
=
I
−
(10.6)
where
I
0
is the moment of inertia around the center of mass.
Obviously this formula can also be used to determine the moment of inertia around
one of the ends of the segment if it has been provided at the center of mass.
The radius of gyration,
ρ
o
(m), of the limb segment is defined as the distance from
the center of mass for each of two point masses,
M
/
2 (kg), such that
2
0
I
0
=
M
ρ
(10.7)
Note that both the total mass and the center of mass remain unchanged in this case.
WORKED EXAMPLE
Moment of Inertia
A modern prosthetic leg has a mass of 2.5 kg, with its center of mass 200 mm from the knee
joint. The radius of gyration is 141 mm. Calculate the moment of inertia about the knee joint.
The moment of inertia around the center of mass is determined from the given mass and
the radius of gyration
2
0
I
0
=
M
ρ
10
−
3
2
=
2
.
5
×
(
141
×
)
m
2
=
0
.
0497 kg
·
The parallel axis theorem is then used to determine the moment of inertia around the knee
joint:
Mx
2
I
=
I
0
+
10
−
3
2
=
.
+
.
×
(
×
)
0
0497
2
5
200
m
2
=
.
·
0
1497 kg
