Biomedical Engineering Reference
In-Depth Information
EXAMPLE PROBLEM 4.7
A person weighing 150 pounds has a thigh length of 17 inches. Find the moment of inertia of
this body segment with respect to its center of mass in SI units.
Solution
Thigh length in SI units is
432 m
Table 4.1 lists ratios of segment weight to body weight for different body segments. Starting
with body mass,
l thigh ¼
17 in
¼
0
:
m body ¼ð
150 lb
Þð
0
:
454 kg
=
lb
Þ¼
68
:
1kg
the thigh segment mass is
81 kg
Table 4.1 also lists body segment center of mass and radius of gyration as ratios with respect to
segment length for each body segment. Table 4.1 gives both proximal and distal segment length
ratios. Note that “proximal” for the thigh refers toward the hip and “distal” refers toward the
knee. Consequently, the proximal thigh segment length is the distance between the thigh center
of mass and the hip, and the distal thigh segment length is the distance between the thigh center
of mass and the knee. The moment of inertia of the thigh with respect to the hip is therefore
m thigh ¼ð
0
:
100
Þð
68
:
1kg
Þ¼
6
:
2
371 kg m 2
The thigh's moment of inertia with respect to the hip is related to the thigh's moment of inertia
with respect to its center of mass via the parallel axis theorem (Eq. (4.41)). Consequently, Table 4.1
data can be used to compute segment moments of inertia with respect to their centers of mass:
I thigh=hip ¼ I thigh=cm þ md
2
I thigh=hip ¼ mk
¼ð
6
:
81 kg
Þ½ð
0
:
540
Þð
0
:
432 m
Þ
¼
0
:
2
so
2
I thigh=cm ¼ I thigh=hip md
In this case, distance
d
is given by the proximal segment length data:
d ¼ð
0
:
432 m
Þð
0
:
433
Þ¼
0
:
187 m
and the final result is
371 kg m 2
2
133 kg m 2
I thigh=cm ¼
0
:
ð
6
:
81 kg
Þð
0
:
187 m
Þ
¼
0
:
EXAMPLE PROBLEM 4.8
A person weighing 160 pounds is holding a 10-lb weight in his palm, with the elbow fixed at
90 flexion (Figure 4.12 (top)). (a) What force must the biceps generate to hold the forearm in static
equilibrium? (b) What force(s) does the forearm exert on the humerus?
Continued
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