C a Search Many of the elements in horizontalbar arms torso

C a Search Many of the elements in horizontal-bar arms, torso (including the head), thighs, and lower legs, as shown in Figure b. Inertial parameters for a particular gymnast are as follows. exercises can be modeled by representing the gymnast by four segments consisting of Segment Mass(kg) Length (m) og (m) I (kg m2) 0.600 6.87 34.90 13.90 2.42 0.239 0.205 0.601 0.337 1.620 0.37415 0.173 0.156 Torso Thighs 0227.16 0.227 Note that rog is the distance to the center of gravity measured from the joint closest to the bar when the gymnast is hanging from the bar by his hands and the masses of the arms, thighs, and legs include both appendages. I is the moment of inertia of each segment about its center of gravity. Calculate the position of the center of gravity of the gymnast shown below. Pay close attention to the definition of reg in the table. thigh leg arm 60° MacBook Pro

Solution

According to the given problem,

Segment .. Mass(kg) .. Length(m) .. rcg(m) .. I(kg-m2)
Arms .......... 6.87 ......... 0.600 ...... 0.239 ..... 0.205
Torso ......... 34.90 ........ 0.601 ....... 0.337 .... 1.620
Thighs ....... 13.90 ........ 0.374 ........ 0.151 ... 0.173
Legs ........... 7.42............. — .......... 0.227 ... 0.156

First of all, the last column can be ignored as far as this problem goes.Second, some interpretation is required:

\"Note that in Figure(a) r_cg is the distance to the center of gravity measured from the joint closest to the bar...\" means that when looking at (a), the \"r_cg\" measurements are from the wrists, shoulders, hips and knees for the arms, torso, thighs and legs, respectively. (Normally, I think, the cg of the arms would be measured from the shoulders, not the wrists.)

Given that the measurements given in the table are from the bar, that makes diagram \"b\" fairly straightforward:

x = m*r / m

x = (6.87kg*0.239m + 34.90kg*(0.600+0.337)m+ 13.90kg*(0.600+0.601+0.151)m + 7.42kg*(0.600+0.601+0.374+0.227)m)/ (6.87+34.90+13.90+7.42)kg

where for each body segment I\'ve used its rcg plus the lengths of the preceding body segments.

x = 1.054 m [distance of cg from bar in diagram b]

Diagram (c) is basically the same process, but now there is subtraction involved. I may err, but I get

x = (6.87kg*0.239m + 34.90kg*(0.600-0.337)m + 13.90kg*(0.600-0.601-0.151)m + 7.42kg*(0.600-0.601-0.374-0.227)m)/ (6.87+34.90+13.90+7.42)kg

x = 0.0672 m [distance of cg from bar (towards head) for diagram c]

Well, that\'s the process anyway.